text:
\subsection{An upper bound on the chromatic threshold of $F_5$-free graphs}

In this section, we prove the upper bound in Theorem~\ref{mainF5thm}.  As in Section~\ref{secChromNearr},
we will give an upper bound on the chromatic threshold by first proving that large dimension
forces a copy of $F_5$, and then by applying Theorem~\ref{coloringthm}.
Let $(B,\gamma,F)$ be an $(r_b,r_{\gamma})$-uniform fiber bundle, and make the following definition.
A \textbf{cut} in $(B,\gamma,F)$ is a pair $(X,S)$ such that
$X \subseteq V(B)$, $S \subseteq \binom{F}{r_{\gamma}}$, and if $\gamma(x) \cap S \neq \emptyset$, then $x \in X$.  In other words,
the fibers that intersect $S$ come exclusively from $X$.
A \textbf{$k$-cut} is a cut $(X,S)$ with $\left| X \right| \leq k$.  The size of a $k$-cut is the
size of $\left| S \right|$.

We now sketch the proof of the upper bound in Theorem~\ref{mainF5thm}.  Let $G$ be
an $n$-vertex, $3$-uniform, $F_5$-free hypergraph with minimum degree at least $c\binom{n}{2}$.
Let $(G,\gamma,F)$ be the neighborhood bundle of $G$,
let $H = K_{q,q}$, and assume $\vc_H(G,\gamma,F)$ is large.  We would like to find a copy of $F_5$ in
$G$.  We first use the fact that $\vc_H(G,\gamma,F)$ is large to find a set $U$ of vertices
of $G$ such that $G[U]$ has small strong independence number.  We then argue that
because the minimum degree is large, there must be some vertices $x,y$ such that
$N(x,y) = \left\{ z : xyz \in G \right\}$ has large intersection with $U$.  Next, we show that
since $N(x,y)$ has large intersection with $U$ and $G[U]$ has small strong independence number, there must
be an edge $E$ with at least two vertices in $N(x,y) \cap U$, which gives a copy of $F_5$.

The best upper bound on the chromatic threshold will come from the lowest required minimum degree needed in the
above proof.  The minimum degree is used above to prove that there exists some $x,y$ with
$N(x,y) \cap U$ large.  If we can find a large cut $(X,S)$ in $(G,\gamma,F)$ and we make $U$ large enough,
we could remove $X$ from $U$ while still maintaining all the useful properties of $U$.
Then for all $\{x,y\} \in S$, we know that $N(x,y) \cap (U - X) = \emptyset$.
Since there are now fewer pairs $\{x,y\}$ in $\binom{F}{2}$ with $N(x,y) \cap (U - X) \neq \emptyset$,
we can require a weaker lower bound
on the minimum degree of $G$ to find $\{x,y\}$ with $N(x,y) \cap U$ large.
In other words, the larger the cut of $(G,\gamma,S)$ we can find, the better upper bound on the chromatic threshold
we can prove.  This is encoded in the following theorem, which computes the relationship
between the minimum degree and the maximum size of a $k$-cut.

\newtheorem{f5largecutbound}[thmctr]{Theorem}
\begin{f5largecutbound} \label{f5largecutbound}
Let $0 < c,s < 1$ satisfy $5c \geq 1-s$, and fix an integer $k$ and a constant $c' > c$.
Then there exists a constant $L = L(c,c',k,s)$ such that the following holds.
Let $G$ be an $n$-vertex, $F_5$-free hypergraph with $\delta(G) \geq c' \binom{n}{2}$
and let $(G,\gamma,F)$ be the neighborhood bundle of $G$.  Assume $(G,\gamma,F)$ contains a $k$-cut of size
at least $s \binom{n}{2}$.  Then $\chi(G) \leq L$.
\end{f5largecutbound}

Note that if $s = 0$ and $c = 1/5$, then $5c \geq 1-s$ and so this theorem directly proves
an upper bound of $1/5$ on the chromatic threshold of $F_5$-free hypergraphs.
The first part of the proof of Theorem~\ref{f5largecutbound} is to find a set $U$ with
small strong independence number.

\newtheorem{largedimimpliesU}[thmctr]{Lemma}
\begin{largedimimpliesU} \label{largedimimpliesU}
Let $\epsilon > 0$ be fixed.  Then there exists constants $d = d(\epsilon)$ and $q = q(\epsilon)$ such
that the following holds.  Let $G$ be an $n$-vertex, $3$-uniform hypergraph and let $(G,\gamma,F)$ be the neighborhood bundle
of $G$.  Let $H = K_{q,q}$ and assume $\vc_H(G,\gamma,F) \geq d$.
Then there exists a vertex set $U \subseteq V(G)$ such that $\left| U \right|= 5d$ and the strong
independence number of $G[U]$ is at most $(1+\epsilon)d$.
\end{largedimimpliesU}

\begin{proof}
Let $d = 100 + 100/\epsilon^2$ and $q = 3d + 2\cdot 3^d$.
Since $\vc_H(G,\gamma,F) \geq d$, there exists a matching $E_1, \ldots, E_d$ such that for each
$x_1 \in E_1, \ldots, x_d \in E_d$ the section
of $\left\{ x_1, \ldots, x_d \right\}$ contains a copy of $K_{q,q}$.  (See Figure~\ref{nearkfigure}
in Section~\ref{secChromNearr} for a picture of this structure.)
Since $q = 3d + 2\cdot 3^d$, from each of these $3^d$ copies
of $K_{q,q}$ we can pick a copy of $K_2$ such that each $K_2$ is vertex disjoint from $E_1\cup\ldots\cup E_d$
and all these $3^d$ copies of $K_2$ are vertex disjoint.
Now choose randomly $d$ copies of $K_2$, $\{y_1,z_1\},\dots,\{y_d,z_d\}$ (with repetition) from these $3^d$ vertex disjoint copies of
$K_2$, and define $Z = \left\{ y_1, \ldots, y_{d}, z_1, \ldots, z_{d} \right\}$,
$U = Z \cup E_1 \cup \ldots \cup E_d$.  Since $Z$ is disjoint from $E_1, \ldots, E_d$,
$\left| U \right| = 5d$.  We need only show that with positive probability the
strong independence number of $G[U]$ is at most $(1+\epsilon)d$.

Notice that any strong independent set in $G[U]$ contains at most $d$ vertices from
$E_1 \cup \ldots \cup E_d$ and at most $d$ vertices from $Z$.  Thus any strong independent set
in $G[U]$ with at least $(1+\epsilon)d$ vertices must have at least $\epsilon d$ vertices in
$E_1 \cup \ldots \cup E_d$ and at least $\epsilon d$ vertices in $Z$.  We need to prove
with positive probability that this does not occur.

For every $P \subseteq Z$ with $\left| P \right| = \epsilon d$ and
every $S \subseteq E_1 \cup \ldots \cup E_d$ with $\left| S \right| = \epsilon d$,
let $A_{P,S}$ be the event ``$P\cup S$ is a strong independent set in $G$''
and $X_{P,S}$ be its indicator random variable.
Let $X$ be the sum of all indicator random variables.
We claim that $X$ is small.

For any $u \in Z$, any $1 \leq i \leq d$, and any $w \in E_i$, let $\left\{ u, v \right\}$ be the
copy of $K_2$ containing
$u$.  Then, because $\left\{ u,v \right\}$ is contained in some section of $x_1 \in E_1, \ldots, x_d \in E_d$,
it is the case that $\left\{ u,v,w' \right\} \in G$ for some $w' \in E_i$.
Therefore, $Pr[\{u,v,w\} \notin G] \leq 2/3$ so that \begin{align*}
E[X] = \sum X_{P,S} \leq \binom{2d}{\epsilon d}\binom{3d}{\epsilon d} (2/3)^{\epsilon^2 d^2}
  \leq \left( \left( \frac{2ed}{\epsilon d} \right) \left( \frac{3ed}{\epsilon d} \right) \left( \frac{2}{3} \right)^{\epsilon d} \right)^{\epsilon d} <1.
\end{align*}
Also, since we have $3^d$ copies of $K_2$ and select only $d$ of them,
the probability some $K_2$ is selected twice is exponentially small.
Thus with positive probability, every event $A_{P,S}$ fails which implies the strong
independence number of $G[U]$ is at most $(1+\epsilon)d$.
\end{proof}

We can now prove Theorem~\ref{f5largecutbound}.

\begin{proof}[Proof of Theorem~\ref{f5largecutbound}]
Pick $\epsilon$ so that $c' = (1+2\epsilon)c$ and let $d = d(\epsilon)$ and $q = q(\epsilon)$ be given by
Lemma~\ref{largedimimpliesU}, and also assume that $d$ is large enough so that $5d\epsilon > k(1+2\epsilon)$.
Suppose that if $H = K_{q,q}$ then $\dim_H(G,\gamma,F) \leq d$.  Then by Theorem~\ref{coloringthm}, there exists
constants $K_1 = K_1(\epsilon,d,H)$ and $K_2 = K_2(\epsilon,d,H)$ (note that $K_1$ and $K_2$ depend only on $c,c',k,s$)
such that either $\left| F \right| < K_1$ or $\chi(G) < K_2$.  Since $\left| F \right| = \left| V(G) \right|$,
this implies that $\chi(G) < \max\{K_1, K_2\}$.

We can therefore assume that $\dim_H(G,\gamma,F) \geq d$.  By Lemma~\ref{largedimimpliesU}, there exists a set
$U \subseteq V(G)$ such that $\left| U \right| = 5d$ and the strong independence number of $G[U]$ is at most $(1+\epsilon)d$.
Let $(X,S)$ be a $k$-cut of size $s \binom{n}{2}$.
Let $G'$ be the bipartite graph with partite sets $A = U \setminus X$ and $B=\binom{V(G)}{2}\setminus S$ where $\left\{ u ,\left\{ v,w\right\}\right\}$ is an edge in $G'$ if and only if $\left\{ u , v,w\right\}$ is an edge in $G$. $\left| A \right|\ge 5d - \left| X \right|$, so $G'$ contains at least $(5d - \left| X \right|) \delta(G)$ edges. $ \left| B \right|=\binom{n}{2} - \left| S \right|$, so there is some $x \neq y$ such that $d_{G'}\left(\{x,y\}\right)$ is at least

\begin{align*}
\frac{(5d - \left| X \right|) \delta(G)}{\binom{n}{2} - \left| S \right|}
\geq \frac{(5d - k) (1+2\epsilon)c \binom{n}{2}}{(1-s) \binom{n}{2}}
\geq \frac{(5d - k) (1+2\epsilon)c}{5c}
> (1+\epsilon)d.
\end{align*}
This implies that there is some $x,y$ with $\left| N(x,y) \cap U \right| > (1+\epsilon)d$.
Since the strong independence number of $G[U]$ is at most $(1+\epsilon)d$,
there exists some edge $E$ with two vertices in $N(x,y)$.
Then $x,y$ together with $E$ form a copy of $F_5$ in $G$.  This contradiction completes the proof.
\end{proof}

\subsection{Finding a large cut in a $F_5$-free hypergraph}

In order to prove the upper bound in Theorem~\ref{mainF5thm}, we now need to show the existence of a large cut.
Note that in Theorem~\ref{f5largecutbound} the bound on the chromatic number depends
on $k$ but there are no other restrictions on $k$.  Thus to prove an upper bound on the chromatic
threshold of $F_5$-free graphs, one can pick any fixed integer $k$ and ask what is the size of the largest $k$-cut.
In the following lemma, we set $k = 5$ and prove that there exist a $5$-cut of size approximately
$4c^2 \binom{n}{2}$.  Solving $5c = 1 - s = 1-4c^2$ gives $c = (\sqrt{41}-5)/8$, the bound in Theorem~\ref{mainF5thm}.

We suspect that the bound on the chromatic threshold of $F_5$-free hypergraphs can
be improved by finding a larger cut, perhaps by increasing $k$.  In order to achieve a bound of $c = 6/49$,
we would need to find a cut of size $s\binom{n}{2}$ with $s = 1-5c=539/36 c^2 \approx 15c^2$.

\newtheorem{F5implieslargecut}[thmctr]{Lemma}
\begin{F5implieslargecut} \label{F5implieslargecut}
Let $0 < c < c'$ be fixed.  There exists a constant $n_0 = n_0(c,c')$ such that for all $n > n_0$ the following holds.
Let $G$ be an $n$-vertex, $3$-uniform, $F_5$-free hypergraph with $\delta(G) \geq c'\binom{n}{2}$. Let
$(G,\gamma,F)$ be the neighborhood bundle of $G$. Then $(G,\gamma,F)$ has a
$5$-cut of size at least $4 \binom{c(n-1)}{2}$.
\end{F5implieslargecut}

\noindent Combining Theorem~\ref{f5largecutbound} with Lemma~\ref{F5implieslargecut}, we can prove
Theorem~\ref{mainF5thm}.

\begin{proof}[Proof of Theorem~\ref{mainF5thm}]
Let $c = (\sqrt{41}-5)/8$, let $c' > c$ be fixed, and let $G$ be any $n$-vertex, $3$-uniform, $F_5$-free graph
with minimum degree at least $c' \binom{n}{2}$. Let $(G,\gamma,F)$ be the neighborhood bundle of $G$.
Let $b = (c' + c)/2$ so that $c' > b > c$.
Then by Lemma~\ref{F5implieslargecut}, either $\left| V(G) \right|$ is bounded or $(G,\gamma,F)$ contains a $5$-cut
of size at least $4\binom{b(n-1)}{2}$.  Since $b > c$, if $n$ is large enough this is at least $4c^2 \binom{n}{2}$.
Let $s = 4c^2$ and notice that $5c=1-s$, so that Theorem~\ref{f5largecutbound} implies that the
chromatic number of $G$ is bounded.
\end{proof}

\noindent The first step in the proof of Lemma~\ref{F5implieslargecut} is the following lemma.

\newtheorem{goodpairlemma}[thmctr]{Lemma}
\begin{goodpairlemma} \label{goodpairlemma}
In a graph $G$, we call a non-edge $uv \notin E(G)$ \textbf{good} if $N(u)\cap N(v)\ne\emptyset$.
If $G$ is a triangle-free graph with $n$ vertices and $m$ edges, then $G$ has at least $m-n/2$ good
non-edges.
\end{goodpairlemma}

\begin{proof}
We prove this by induction on $n$. It is obviously true for $n=1$ and $n=2$. Now
assume $n>2$. If some component of $G$ is not regular, then there exist vertices $u,v$ in that
component such that $u\in N(v)$ and $d(u)<d(v)$. Then $G-u$ has $n-1$ vertices and $m-d(u)$ edges.
By induction, $G-u$ has at least $m-d(u)-\frac{n-1}{2}$ good non-edges. For any vertex $w\in N(v)-u$,
$uw$ is a good non-edge, so $G$ has at least $m-d(u)-\frac{n-1}{2}+d(v)-1\ge m-n/2$ good
non-edges. If all components of $G$ are regular, then pick one component $K$. Assume $K$ is
$r$-regular, and pick a vertex $v$ in $K$and let $N_{2}(v)=\left\{u: \mbox{ there exists a $P_3$
connecting $u$ and $v$}\right\}$. If $|N_2(v)|\geq r$, then by the induction hypothesis $G-v$ has at least
$m-r-\frac{n-1}{2}$ good non-edges, and since for any vertex $u\in N_2(v)$ it is the case that $uv$ is a good non-edge,
$G$ has at least $m-r-\frac{n-1}{2}+|N_2(v)|\geq m-n/2$ good non-edges. If $|N_2(v)|< r$, then since $K$ is
triangle-free and $r$-regular, $K=K_{r,r}$ which has $r^2$ edges and $r^2-r$ good non-edges. Now $G-K$
has $n-2r$ vertices and $m-r^2$ edges, so by induction it has $m-r^2-(n-2r)/2$ good non-edges. Then $G$
has $m-r^2-(n-2r)/2+r^2-r=m-n/2$ good non-edges.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{F5implieslargecut}]
We examine the copies of $F_4$ in $G$ where $F_4$ is the
hypergraph with vertex set $\left\{ 1,2,3,4 \right\}$ and edges $\{1,2,3\}$, $\{1,2,4\}$,
and $\{2,3,4\}$.

\medskip
\noindent\textbf{Case 1} There exists a vertex $v$ of $G$ such that $v$ is not contained in
any copy of $F_4$.  Consider $L = \gamma(v)[V(G) - v]$, which is a triangle-free graph with $n-1$
vertices and at least $c \binom{n}{2}$ edges. By Lemma~\ref{goodpairlemma}, $L$ has at least
$c\binom{n}{2}-\frac{n-1}{2}$ good non-edges. Let $X=\emptyset$ and $S$ be the set of these good non-edges.
We claim that $(X,S)$ is a cut in $(G,\gamma,F)$. Suppose for contradiction that there exists some
$x\in V(G)$ and $\left\{ u,w \right\}\in S$ such that $\left\{ u,w,x \right\}\in G$. Pick a vertex
$y$ from $N_L(u)\cap N_L(w)$. Then $u,v,w,x,y$ form a copy of $F_5$ in $G$, which is a contradiction.

\medskip
\noindent\textbf{Case 2} Every vertex of $G$ is contained in some copy of $F_4$.
Pick some $U \subseteq V(G)$ such that $G[U] = F_4$, let
$U=\left\{ u_1,u_2,u_3,u_4\right\}$, and let $G'=\cup_{i=1}^{4} \gamma(u_i)$.
Consider $\gamma(u_i) \cap \gamma(u_j)$ for $i \neq j$.  If $\gamma(u_i) \cap \gamma(u_j)$
contains a matching of size two, then $G$ contains a copy of $F_5$.  Say $ab, cd \in \gamma(u_i)
\cap \gamma(u_j)$ with $a,b,c,d$ distinct.  Then since $G[U] = F_4$, there is some edge
$E = \left\{ u_i, u_j, w \right\} \in G$.  If $w \neq a$ and $w \neq b$, then $a,b,u_i,u_j,w$ form a copy
of $F_5$ and if $w = a$ or $w = b$, then $c,d,u_i,u_j,w$ form a copy of $F_5$.  Thus
$\gamma(u_i) \cap \gamma(u_j)$ is a star so has at most $n$ elements.  Since each
$\gamma(x)$ has size at least $c' \binom{n}{2}$, $G'$ has at least
$4c' \binom{n}{2} - \binom{4}{2} n > 4c \binom{n}{2}$ edges if $n$ is large enough.

Then $G'$ has $n$ vertices and at least $4c\binom{n}{2}$ edges, so there exist a vertex $v$ whose degree
in $G'$ is at least $4c(n-1)$. Let $N$ denote the neighborhood of $v$ in $G'$ and let
$N_1,\dots,N_{4}$ be a partition of $N$ such that for every $1\le i\le 4$ and every vertex
$w\in N_i, vw\in\gamma(u_i)$. Let $X=U\cup \{v\}$ and $S = \bigcup_{i=1}^{4} \binom{N_{i}}{2}$,
so that $|X|=5$ and $|S|\geq 4\binom{|N|/4}{2}= 4\binom{c(n-1)}{2}$. We claim that
$(X,S)$ is a cut in $(G,\gamma,F)$. Suppose for contradiction that there exists some $z\notin X$ such
that $\gamma(z)\cap S\neq \emptyset$. Pick $\{x,y\}\in \gamma(z)\cap S$, then $\{x,y\} \subseteq
N_i$ for some $1 \leq i \leq 4$. Now $v,u_i,x,y,z$ form a copy of $F_5$, which is a contradiction.

From these two cases we can see that $(G,\gamma,F)$ has a $5$-cut of size at least $\min\left\{c \binom{n}{2}-\frac{n-1}{2}\right.$, $\left.4\binom{c(n-1)}{2}\right\}$. $G$ is $F_5$-free, so $c\le 2/9$ and therefore $\min\left\{c \binom{n}{2}-\frac{n-1}{2},4\binom{c(n-1)}{2}\right\}=4\binom{c(n-1)}{2}$.
\end{proof}

\subsection{A construction for the lower bound} \label{f5lowersection}

To prove a lower bound on the chromatic threshold of the family of $F_5$-free hypergraphs,
we need to construct an infinite sequence of $F_5$-free hypergraphs with large chromatic number and large minimum degree.
Our construction is inspired by a construction by Hajnal~\cite{cr-erdos73} of a dense triangle-free graph with high chromatic number.
Hajnal's key idea was to use the Kneser graph to obtain large chromatic number.
The Kneser graph $\kneser(n,k)$ has vertex set $\binom{[n]}{k}$, and two vertices $F_1, F_2$ form an edge if and only if
$F_1 \cap F_2=\emptyset$.  We use an extension of Kneser graphs to hypergraphs.
Alon, Frankl, and Lov\'{a}sz~\cite{kn-alon86} considered the Kneser hypergraph $\kneser^r(n, k)$, which is the $r$-uniform hypergraph
with vertex set $\binom{[n]}{k}$, and $r$ vertices $F_1, \ldots, F_r$ form an edge
if and only if $F_i \cap F_j = \emptyset$ for $i \neq j$. They gave a lower bound
on the chromatic number of $\kneser^r(n,k)$ as follows.

\newtheorem{kneserlower}[thmctr]{Theorem}
\begin{kneserlower} \label{kneserlower}
If $n \geq (t-1)(r-1) + rk$, then $\chi(\kneser^r(n, k)) \geq t$.
\end{kneserlower}

\noindent We first show that $\kneser^r(n,k)$ is $F_5$-free for $n < 4k$.

\newtheorem{kneserf5free}[thmctr]{Lemma}
\begin{kneserf5free} \label{kneserf5free}
If $n < 4k$, then $\kneser^3(n,k)$ is $F_5$-free.
\end{kneserf5free}

\begin{proof}
Say $\left\{ a,b,c \right\}$, $\left\{ a,b,d \right\}$ and $\left\{ c,d,e \right\}$ are edges in $\kneser^3(n,k)$.  Then
by definition $a$, $b$, $c$, and $d$ are four disjoint $k$-sets in $[n]$, which is impossible because $n < 4k$.
\end{proof}

\begin{proof}[Proof of the lower bound in Theorem~\ref{mainF5thm}]
Fix $t \geq 2$ and $\epsilon > 0$.  Pick $k \geq 2t$ and $n = 3k + 2(t-1)$ and note that $n < 4k$.
By Theorem~\ref{kneserlower},
$\kneser^3(n,k)$ has chromatic number at least $t$ and by Lemma~\ref{kneserf5free} is
$F_5$-free.  For integers $u$, $v$, and $w$ where $n$ divides $u$, let $U$, $V$ and $W$ be
disjoint vertex
sets of size $u$, $v$, and $w$ respectively.  Partition $U$ into $U_1, \ldots, U_n$ such that
$\left| U_i \right| = \frac{u}{n}$ for each $i$.
Let $H$ be the hypergraph with vertex set $V(\kneser^3(n,k)) \cup U \cup V \cup W$
and the following edges.
\begin{itemize}
\item For $\left\{ S_1, S_2, S_3 \right\} \in \kneser^3(n,k)$, make
      $\left\{ S_1, S_2, S_3 \right\}$ an edge of $H$.
\item For $S \in V(\kneser^3(n,k))$, $x \in U_i$ with $1 \leq i \leq n$, and $y \in V$,
      make $\left\{ S,x,y \right\}$ an edge of $H$ if $i \in S$.
\item For $x \in U$, $y \in V$, and $z \in W$, make $\left\{ x,y,z \right\}$ an
edge of $H$.
\end{itemize}
\noindent Notice that $H$ has chromatic number at least $t$ because $\kneser^3(n,k)$ is a subhypergraph.

\medskip
\noindent\textbf{Claim 1:} $H$ contains no subgraph isomorphic to $F_5$.
\begin{proof}
Suppose $\{a, b, c\}, \{a,b,d\}$ and $\{c,d,e\}$ are the hyperedges of a copy of $F_5$ in $H$.
Notice that the hypergraph induced by $U, V, V(\kneser^3(n,k))\cup W$ is $3$-partite,
apart from those edges within $\kneser^3(n,k)$. Note that a $3$-uniform, $3$-partite hypergraph is $F_5$-free, therefore any copy of $F_5$ must contain
an edge from $\kneser^3(n,k)$.  If that edge is $\{a,b,c\}$ then $d$ must also be
contained in $V(\kneser^3(n,k))$.  But then $c$ and $d$ are both in $V(\kneser^3(n,k))$, which
means $e$ must be as well.  Because $\kneser^3(n,k)$ is $F_5$-free, this is a contradiction.
Similarly, $\{a,b,d\} \subsetneq V(\kneser^3(n,k))$.  Therefore,
$\{c,d,e\} \subseteq V(\kneser^3(n,k))$, and without loss of generality $b\in U$ and $a \in V$.
Because $\{a,b,c\}$ and $\{a,b,d\}$ are edges, $b$ must be in both $c$ and $d$, which contradicts
the fact that $\{c,d,e\}$ is an edge of $\kneser^3(n,k)$.
\end{proof}

\medskip
\noindent\textbf{Claim 2:} The minimum degree of $H$ is at least $(1-\epsilon) \frac{6}{49} \binom{\left| V(H) \right|}{2}$ if
$\left| V(H) \right|$ is large enough.

\begin{proof}
Vertices in $\kneser^3(n,k)$ have degree at least $k\frac{u}{n}v = \frac{kuv}{3k + 2(t-1)}$.
Since $t$ is fixed, we can choose $k$ large enough that
vertices $\kneser^3(n,k)$ have degree at least $(1-\epsilon/2) uv/3$.  Vertices in $A$
have degree at least $vw$, vertices in $B$ have degree at least $uw$,
and vertices in $C$ have degree at least $uv$.  Thus the minimum degree of $H$
is at least $\min\left\{ (1-\epsilon/2)\frac{uv}{3}, uw, vw \right\}$.
Choose $u$, $v$, and $w$ so that $\frac{uv}{3} = uw = vw$, we obtain that $u = v$
and $w = v/3$ and the minimum degree is at least $(1-\epsilon/2) u^2/3$.
The number of vertices is $u + v + w + \binom{n}{k} = \frac{7}{3} u + \binom{n}{k}$.  Since
$u^2/3 \approx 6/49 \binom{7u/3}{2}$, we can choose $u$ large enough so that
the minimum degree of $H$ is at least
$(1-\epsilon) \frac{6}{49} \binom{\left| V(H) \right|}{2}$.
\end{proof}

We have proved that for every fixed $t \geq 2$ and every $\epsilon > 0$, there is a constant $N_0$ such that for $N > N_0$ there exists
an $N$-vertex, $3$-uniform, $F_5$-free hypergraph with chromatic number at least $t$ and
minimum degree at least $(1-\epsilon)\frac{6}{49} \binom{\left| V(H) \right|}{2}$.  By the definition of chromatic threshold,
this implies that the chromatic threshold of the family of
$F_5$-free hypergraphs is at least $\frac{6}{49}$.
\end{proof}

\subsection{$\tkf{s}{r}$-free hypergraphs}\label{subsecTK}

For $s > r$, recall that $\tkf{s}{r}$ is the family of $r$-uniform hypergraphs such that there exists a
set $S$ of $s$ vertices where each pair of vertices from $S$ are contained together in some edge.  The set $S$
is called the set of \textbf{core vertices} of the hypergraph.
Recall also that $T_{r,s}(n)$ is the complete $n$-vertex, $r$-uniform, $s$-partite hypergraph with
part sizes as equal as possible.

The last author~\cite{rt-mubayi06-2} showed that if $s > r$ then
$\ex(n, \tkf{s}{r}) = \left| T_{r,s-1}(n) \right|$ and
$\ex(n, \linebreak[1] \tk{s}{r}) = (1 + o(1)) \left| T_{r,s}(n) \right|$.  Recently,
Pikhurko~\cite{rt-pikhurko} has shown that for large $n$ and $s > r$,
$\ex(n,\tk{s}{r}) \linebreak[1] = \left| T_{r,s-1}(n) \right|$
and that $T_{r,s-1}(n)$ is the unique extremal example.
Because $F_5$ is a member of $\tkf{4}{3}$ it follows that the chromatic threshold of
$\tkf{4}{3}$-free hypergraphs is at most $(\sqrt{41}-5)/8$.
The following simple variation on the construction from Section~\ref{f5lowersection} provides a lower
bound of $18/361$ for both $\tk{4}{3}$-free
and $\tkf{4}{3}$-free hypergraphs.

\newtheorem{tkflowerlemma}[thmctr]{Proposition}
\begin{tkflowerlemma} \label{tkflowerlemma}
The chromatic threshold of $\tkf{4}{3}$-free hypergraphs is at least $\frac{18}{361}$.
\end{tkflowerlemma}

\begin{proof}
The proof is very similar to the proof in Section~\ref{f5lowersection}, we only sketch it here.
Choose $k,n,u,v,w,U, V, W$ as in the proof of the lower bound of Theorem~\ref{mainF5thm} in Section~\ref{f5lowersection};
that is $k,n,u,v,w$ are integers and $U,V,W$ are disjoint sets of vertices of size $u,v,w$
respectively.  Divide $U$ into $U_1, \ldots, U_n$ so
that $\left| U_i \right| = u/n$ and divide $V$ into $V_1, \ldots, V_n$ such that
$\left|V_i \right| = v/n$.
Let $H$ be the hypergraph formed by taking $\kneser^3(n,k)$ and adding the complete
$3$-partite hypergraph on $U,V,W$ and the following edges.
For $S \in V(\kneser^3(n,k))$ and $x \in U_i$ and $y \in V_j$, make $\left\{ S, x, y \right\}$
an edge if $i,j \in S$.  The minimum
degree is maximized when $a = b$ and $c = a/9$, which gives minimum degree
approximately $a^2N^2/9 \approx \frac{18}{361}\cdot \binom{19a/9}{2}N^2$, where $N = u+v+w+\binom{n}{k}$
is the number of vertices in the hypergraphs.

Let $F$ be any hypergraph in $\tkf{4}{3}$ and assume that $F$
is a subhypergraph of $H$ in which $c_1, c_2, c_3, c_4$ are the four core vertices.
Because any $3$-partite hypergraph is $\tkf{4}{3}$-free,
it is easy to see that some edge of $F$ must lie in $\kneser^3(n,k)$, and so there must
be at least two core vertices in $\kneser^3(n,k)$.
If $c_1, c_2 \in \kneser^3(n,k)$ and $c_3 \in U \cup V$
then $c_3$ is in either $U_i$ or $V_i$ for some $i$.
But then $i \in c_1 \cap c_2$ (recall that vertices in $\kneser^3(n,k)$ are $k$-sets) which
contradicts the fact that $c_1$ and $c_2$ are contained together
in some edge of $\kneser^3(n,k)$.  Thus  all four core vertices must be in
$\kneser^3(n,k)$, which is not possible because $n < 4k$.
\end{proof}

This gives lower bounds on the chromatic thresholds of
$\tk{4}{3}$-free and $\tkf{4}{3}$-free hypergraphs and leads to the following questions.

\newtheorem{tkfquestion}[thmctr]{Question}
\begin{tkfquestion}
What is the chromatic threshold for $\tk{4}{3}$-free hypergraphs? It is between $18/361$
and $2/9$.  What is the chromatic
threshold for $\tkf{4}{3}$-free hypergraphs?  It has the same lower bound as for $\tk{4}{3}$-free
hypergraphs, and because $F_5 \in \tkf{4}{3}$ the upper bound is $(\sqrt{41}-5)/8$.
\end{tkfquestion}

\noindent A similar construction provides a $\tkf{s}{3}$-free hypergraph for any $s \geq 5$.
We have not optimized the values.

\newtheorem{tkflowerlemma2}[thmctr]{Lemma}
\begin{tkflowerlemma2} \label{tkflowerlemma2}
When $s \geq 5$, the chromatic threshold of $\tkf{s}{3}$-free hypergraphs is at least
$\frac{(s-2)(s-3)(s-4)^2}{(s^2 - 13)^2} = 1- \frac{13}{s} + O(\frac{1}{s^2})$.
\end{tkflowerlemma2}

\begin{proof}
Fix $t \geq 2$, $k \geq 2t$, and let $n = 3k + 2(t-1)$.  Notice that $n < 4k$.
By Theorem~\ref{kneserlower}, the chromatic number of $\kneser^3(n, k)$ is therefore
at least $t$. Fix $N \gg \binom{n}{k}$.

Partition $N$ vertices into one part of size $u$ and $s-2$ parts of size $x$, for
some $u$ that is divisible by $n$.  Include as an edge each triple that has at
most one vertex in each part.  Further partition the part of size $u$ into $n$ sets,
$U_1, \dots, U_n$, each of size $u/n$. From the remaining $s-2$ parts of size $x$,
choose two and designate them $W_1, W_2$; label the remaining $s-4$ parts
$V_1, \dots, V_{s-4}$.
Let $H$ be the $3$-uniform hypergraph
formed by taking the disjoint union of $\kneser^3(n,k)$ and the above complete $(s-1)$-partite
hypergraph, and adding the following edges.
If $S \in V(\kneser^3(n,k))$, $v \in V_i$, and $v' \in V_j$ for $i \neq j$,
add the edge $\{S, v, v'\}$.
If $S \in V(\kneser^3(n,k))$ and $u \in U_i$ and $v \in V_j$ then add the edge
$\{S, u, v\}$ if and only if $i \in S$.

\noindent Notice that $H$ has chromatic number at least $t$, and that
$V(H) = N + \binom{n}{k}$.

\medskip
\noindent\textbf{Claim 1:} $H$ contains no element of $\tkf{s}{3}$
as a subgraph.
\begin{proof} Suppose there is such a
subgraph; then at least one core vertex must be contained in $ V(\kneser^3(n,k))$, because an
$(s-1)$-partite
graph is $\tkf{3}{s}$-free.  In that case, no core vertex can be in $W_1 \cup W_2$ because
there is no edge that contains a vertex from $W_1 \cup W_2$ as well as a vertex from $ V(\kneser^3(n,k))$.
There must therefore be at least $3$ core vertices in $ V(\kneser^3(n,k))$, which means that two of
them must appear in an edge contained within $ V(\kneser^3(n,k))$.  Suppose they are $S_1, S_2$.
If another core vertex is in
$U$, say $u \in U_i$, then there must be an edge of $H$ containing $u$ and $S_1$,
and there must be an edge containing $u$ and $S_2$.  This implies that $i \in S_1 \cap S_2$,
which contradicts the fact that $S_1$ and $S_2$ appear together in an edge of $\kneser^3(n,k)$.

All core vertices must therefore be in $ V(\kneser^3(n,k)) \cup V$, which means that
there must be at least four of them
in $ V(\kneser^3(n,k))$.  Because each pair of those four core vertices must appear together
in an edge, and that edge must
be in $\kneser^3(n,k)$, those four sets must be pairwise disjoint.  This is impossible
because $n < 4k$.
\end{proof}

\noindent The minimum degree of this graph is approximately
\[
\min \left\{
\frac{1}{3}(s-4)ax + \binom{s-4}{2}x^2, \binom{s-2}{2}x^2, (s-3)ax + \binom{s-3}{2}x^2
\right\}
\]
Notice that a vertex in $W_1 \cup W_2$ has degree strictly less than a vertex in $\kneser^3(n,k)$,
and so they do not enter into the above computation.
This minimum is largest when $ u = \frac{3(2s-7)x}{s-4}$, which implies that
$x = \left(\frac{s-4}{s^2 - 13}\right) N$.  The minimum degree of $H$ is
then
\[
\frac{(s-2)(s-3)}{2}\cdot\frac{(s-4)^2}{(s^2 - 13)^2}N^2 = \left(1 - \frac{13}{x} + O\left(\frac{1}{s^2}\right) \right)\frac{N^2}{2}.
\]
\end{proof}

The construction in Lemma~\ref{tkflowerlemma2} has one part of ``type'' $U$ (which is partitioned into $n$
sets), $s-4$ parts of ``type'' $V$ (which are not partitioned, and whose vertices appear in edges
that intersect $K$), and two parts of ``type'' $W$ (which are not partitioned and have no vertices
that appear in edges intersecting $K$).
Using this strategy, one can generate similar constructions for $\tk{s}{r}$;
the above proof applies whenever there are $x$ parts of type $U$, $s - (r+1)$ parts of type
$V$, and $y$ parts of type $W$, where $x + y = r$ and $s-(r+1) + x \geq r-1$.  This last
condition is needed for the edges intersecting $K$.

\newtheorem{tkfsquestion}[thmctr]{Question}
\begin{tkfsquestion}
What is the chromatic threshold for $\tk{s}{3}$-free hypergraphs for $s > 3$? It is between
$\frac{(s-2)(s-3)(s-4)^2}{(s^2 - 13)^2} = 1 - \frac{13}{s} + O\left(\frac{1}{s^2}\right)$ and
$\left(1-\frac{1}{s-1}\right)\left(1-\frac{2}{s-1}\right) = 1 - \frac{3}{s-1} + \frac{2}{(s-1)^2}$.
The upper bound comes from  $T_{r,s-1}(n)$.
\end{tkfsquestion}

\subsection{$S(7)$-free hypergraphs}

Next, consider the Fano plane $S(7)$.
de Caen and F\"{u}redi~\cite{tdf-decaen00} showed that
$\ex(n,S(7)) = (\frac{3}{4}+o(1))\binom{n}{3}$.
The extremal hypergraph for $S(7)$, proven to be extremal by F\"{u}redi and
Simonovits~\cite{tdf-furedi05}
and also by Keevash and Sudakov~\cite{tdf-keevash05}, is the hypergraph
formed by taking two almost equal vertex sets $U$ and $V$ and taking
all edges which have at least one vertex in each of $U$ and $V$.  We can modify
the hypergraph from Section~\ref{f5lowersection}
to obtain a lower bound on the chromatic threshold of $S(7)$-free hypergraphs.

\newtheorem{fanolowerlemma}[thmctr]{Proposition}
\begin{fanolowerlemma} \label{fanolowerlemma}
The chromatic threshold of $S(7)$-free hypergraphs is at least
$9/17$.
\end{fanolowerlemma}

\begin{proof}
Fix $t\geq 2$ and $0<\epsilon \ll 1$.  Then by Lemma~\ref{kneserlower} there exists $k$ large enough
that if $n = (3 + \epsilon)k$ then $\kneser^3(n,k)$ has chromatic number at least $t$.  Fix
some such $k$, and fix $N \gg \binom{n}{k}$.

Partition $N$ vertices into two sets, $U$ and $V$, with $|U| = 9N/17$ and $|V| = 8N/17$.
Further partition $U$ into $n$ parts, $U_1,\ldots,U_n$, each of size $|U|/n$.
Include as an edge each triple that has at least one vertex in each of $U$, $V$.  Let $H$ be
the hypergraph formed by taking the disjoint union of this hypergraph and $\kneser^3(n,k)$ and
adding the following edges. For $u \in U_i$, $u' \in U_j$, and $X\in V(\kneser^3(n,k))$ include
$\{X,u,u'\}$ as an edge if $i,j \in X$ (recall that vertices in $\kneser^3(n,k)$ are subsets of $[n]$).
Let $K = V(\kneser^3(n,k)$. Notice that $H$ has chromatic number at least $t$, and that
$V(H) = N + \binom{n}{k}$.

\medskip
\noindent\textbf{Claim 1:} $H$ contains no subhypergraph isomorphic to $S(7)$.
\begin{proof}
First notice that $\kneser^3(n,k)$ is $S(7)$-free because every pair of vertices in $S(7)$
are in an edge, which would require there to be $7$ pairwise-disjoint $k$-subsets of $[n]$.
Because $n = (3 +\epsilon)k$, this would be a contradiction.
It is easy to see, by considering the partition $U,(K\cup V)$, that if $H$ contains a copy of
$S(7)$ then it must involve an edge from $H[K]$ (otherwise the extremal $S(7)$-free hypergraph also
contains a copy of $S(7)$). Call this edge $\{A, B, C\}$.

There are four vertices in $S(7) \setminus \{A, B, C\}$, and at least one must be outside
$K$. No more than one can be in $V$ because there is no edge with one vertex in $K$ and
two in $V$.  No more than one can be in $U$ otherwise one of $A \cap B$, $A \cap C$, $B \cap C$
is non-empty, which contradicts the assumption that $\{A, B, C\}$ is an edge of $H[K]$.
Therefore, there must be either $5$ or $6$ vertices of $S(7)$ in $K$. Suppose $v$ is a vertex of $S(7)$
that is outside of $K$.  Then $v$ appears in three edges that overlap only at $v$, say $\{v, S_1, S_2\}$,
$\{v, S_3, S_4\}$, and $\{v, S_5, S_6\}$.  At least one of these edges must contain two vertices from $K$,
but there is no such edge in $H$.
\end{proof}

The minimum degree of $H$ is at least
\[ \min\left\{|U||V|+\binom{|U|/3}{2}, |U||V|+\binom{|U|}{2}, |U||V|+\binom{|V|}{2} \right\} = \frac{9}{34}N^2-\frac{3}{34}N. \]
\end{proof}

\newtheorem{fanoquestion}[thmctr]{Question}
\begin{fanoquestion}
What is the chromatic threshold of $S(7)$-free hypergraphs?  It is at least
$9/17$ and at most $3/4$, where the upper bound is from the extremal hypergraph
of $S(7)$.
\end{fanoquestion}

\subsection{$T_5$-free hypergraphs}

Recall that the $3$-uniform hypergraph $T_5$ has vertices $A, B, C, D, E$ and edges $\{A, B, C\}$,
$\{A, D, E\}$, $\{B, D, E\}$, and $\{C, D, E\}$.

Let $B^3(n)$ be the $3$-uniform hypergraph with the most edges among all $n$-vertex $3$-graphs
whose vertex set can be partitioned into $X_1, X_2$ such that each edge contains
exactly one vertex from $X_2$. F\"{u}redi, Pikhurko, and Simonovits~\cite{fur-pik-sim}
proved that for $n$ sufficiently large the extremal $T_5$-free hypergraph is $B^3(n)$. It follows that
the chromatic threshold for the family of $T_5$-free hypergraphs is at most $4/9$.

\newtheorem{T5lowerlemma}[thmctr]{Proposition}
\begin{T5lowerlemma} \label{T5lowerlemma}
The chromatic threshold of $T_5$-free hypergraphs is at least $16/49$.
\end{T5lowerlemma}

\begin{proof}
Fix $t\geq 2$ and $0<\epsilon \ll 1$.  Then by Lemma~\ref{kneserlower} there exists $k$ large enough
that if $n = (3/2 + \epsilon)k$ then $\kneser^3_2(n,k)$ has chromatic number at least $t$.  Fix
some such $k$, and fix $N \gg \binom{n}{k}$.

Partition $N$ vertices into two parts, $U$ and $V$, with $|U| = 4N/7$ and $|V| = 3N/7$. Further
partition $U$ into $n$ parts, $U_1,\dots,U_n$, each of size $|U|/n$.  Include as an edge any triple
with two vertices in $U$ and one in $V$.  Let $H$ be the hypergraph formed by taking the disjoint
union of this graph and $\kneser^3_2(n,k)$ and including the following edges.  If
$X \in V(\kneser^3_2(n,k))$ and $u\in U_i$ and $v \in V$ then let $\{u,v,X\}$ be an edge if
$i \in X$ (recall that vertices of $\kneser^3_2(n,k)$ are subsets of $[n]$).  Let
$K = V(\kneser^3_2(n,k))$.  Notice that $H$ has chromatic number at least $t$, and that
$V(H) = N + \binom{n}{k}$.

\medskip
\noindent\textbf{Claim 1:} $T_5$ is not a subhypergraph of $H$.
\begin{proof}
Let $H'$ be the hypergraph obtained from $H$ by deleting all edges contained in $K$, and let
$X_1 = K \cup U$ and $X_2 = V$.  It is now easy to see that $H'$ is a subhypergraph of the
extremal $T_5$-free hypergraph; if $H$ contains a copy of $T_5$ it must therefore involve an
edge from $K$.  If that edge is $\{A,D,E\}$ (see the labelling of $T_5$ above) then because
$\{B,D,E\}$ and $\{C,D,E\}$ are edges of $T_5$ it must be the case that both of $B,C$ are in $K$, but
by Lemma~\ref{newkneserT5free} $K$ does not span a copy of $T_5$.
Similarly, neither $\{B,D,E\}$ nor $\{C,D,E\}$ can be contained in $K$.

We may therefore assume that $\{A,B,C\}$ is contained in $K$.  Because $\{A,D,E\}$ is an edge, and
by Lemma~\ref{newkneserT5free}, at least one of $D,E$ is in $U$.  Suppose
that $D \in U_i$; then because $\{A, D, E\}$, $\{B, D,E\}$, and $\{C,D,E\}$ are all edges
of $T_5$ it must be the case that $i \in A \cap B \cap C$.  This contradicts the assumption that
$\{A,B,C\}$ is an edge.
\end{proof}

The minimum degree of $H$ is at least
\[ \min\left\{\frac{2|U||V|}{3}, |U||V|, \binom{|U|}{2} \right\} = \frac{8}{49}N^2-\frac{2}{7}N. \]
\end{proof}

\subsection{Co-chromatic thresholds}

There is another possibility when generalizing the definition of chromatic
threshold from graphs to hypergraphs:
we can use the co-degree instead of the degree.
Recall that if $H$ is an $r$-uniform hypergraph and
$\left\{ x_1, \ldots, x_{r-1} \right\} \subseteq V(H)$,
 then the \textbf{co-degree $d(x_1, \ldots, x_{r-1})$ of $x_1,\ldots,x_{r-1}$} is
$\left| \left\{ z : \left\{ x_1, \ldots, x_{r_1}, z \right\} \in H \right\} \right|$.
Let $F$ be a family of $r$-uniform hypergraphs.
The \textbf{co-chromatic threshold} of $F$ is the infimum
of the values $c \geq 0$ such that the subfamily of $F$ consisting of
hypergraphs $H$ with
minimum co-degree
at least $c \left| V(H) \right|$ has bounded chromatic number.
More generally, the \textbf{$k$-degree $d(x_1, \ldots, x_k)$ of $x_1, \ldots, x_k$} is
$\left| \left\{ \left\{ z_{k+1}, \ldots, z_r \right\} : \left\{ x_1, \ldots, x_k, z_{k+1}, \ldots, z_r \right\} \in H \right\} \right|$
and we can define the $k$-chromatic threshold similarly.
Given a hypergraph $H$ and subsets $U,V,W$ of $V(H)$, we say that an edge $\{u,v,w\}$
is of type $UVW$ if $u\in U, v\in V$ and $w\in W$.

The co-chromatic thresholds of $F_5$-free hypergraphs and $\tk{4}{3}$-free
hypergraphs are trivially zero because if the minimum co-degree
of $H$ is at least $10$ then $H$
contains a copy of $\tk{4}{3}$ and a copy of $F_5$.
For the Fano plane, the last author proved~\cite{mubayi-fano} that for every $\epsilon>0$ there exists $n_0$ such that any $3$-uniform hypergraph with $n>n_0$ vertices and
minimum co-degree greater than $(1/2 + \epsilon)n$  contains a copy of $S(7)$.
In 2009, Keevash~\cite{keevash-fano} improved this by proving that any $3$-uniform hypergraph with
minimum co-degree greater than $n/2$ contains a copy of $S(7)$ for $n$ sufficiently large.
Notice that the lower bound construction for the chromatic threshold described above has
non-zero minimum co-degree but the co-degree depends on the parameter $t$.  We can modify the construction
to prove a better lower bound on the co-chromatic threshold of $S(7)$-free hypergraphs.

\newtheorem{Fanonewextremal}[thmctr]{Proposition}
\begin{Fanonewextremal}\label{Fanonewextremal}
The co-chromatic threshold of $S(7)$-free hypergraphs is at least $2/5$.
\end{Fanonewextremal}
\begin{proof}
Fix $t\geq2$ and $0<\epsilon \ll 1$.  Then by Lemma~\ref{newKneser} there exists $k$ large enough that if
$n = (3/2 + \epsilon)k$ then $\kneser^3_2(n,k)$ has chromatic number at least $t$.  Fix
$N \gg \binom{n}{k}$.

Partition $N$ vertices into two parts, $U$ and $V$, of size $\frac{3N}{5}$ and $\frac{2N}{5}$
respectively. Include as an edge any triple with at least
one vertex in each part.  Further partition $U$ into $n$ sets, $U_1, \dots, U_n$, each of size
$|U|/n$.  Let $H$ be the hypergraph formed by taking the disjoint union of this hypergraph with
$\kneser^3_2(n,k)$ and including the following edges.  Include any edge of type $KUV$, where
$K = V(\kneser^3_2(n,k))$.  For any $X, Y \in K$, if $|X \cap Y| < k - 4\epsilon k$ then
include every edge of the form $\{X,Y,u\}$ where $u \in U_i$ for some $i \in X \cup Y$.
If $|X \cap Y| \geq k - 4\epsilon k$ then include every edge of
the form $\{X, Y, u\}$ where $u \in U_i$ for some $i \in X \cap Y$.
Notice that $H$ has chromatic number at least $t$ and that
$V(H) = N + \binom{n}{k}$.

\noindent\textbf{Claim 1:} The above hypergraph contains no subgraph isomorphic to $S(7)$.
\begin{proof}
First notice that the complete bipartite $3$-uniform hypergraph contains no copy of $S(7)$.  Therefore,
by considering the partition $U, V \cup K$, we can see that any copy of $S(7)$ must contain an edge
induced by $K$.  Call this edge $\{A,B,C\}$.
It also follows from Lemma~\ref{newkneserS7free} that there is no
copy of $S(7)$ completely contained in $K$.

\noindent\textbf{Claim 1a:} Any copy of $S(7)$ intersects $U$ (or $V$) in at most one vertex.
\begin{proof}
Notice that for any edge $e$ in $S(7)$, every other edge intersects $e$
in at exactly one vertex; therefore for any copy of $S(7)$ in $H$ every edge contains one of $A, B, C$.
If there were two vertices of $S(7)$ in $U$ (or in $V$) then the edge of $S(7)$ joining them would be unable
to intersect $A, B$, or $C$.
\end{proof}

\noindent\textbf{Claim 1b:} Any copy of $S(7)$ contains no vertex from $V$.
\begin{proof}
Suppose for contradiction a copy of $S(7)$ contains some vertex from $V$; then by Claim~1a it
intersects $V$ in exactly one vertex.  Every vertex of $S(7)$ is contained
in three edges, but because there is at most one vertex from $U$ involved in the copy of $S(7)$
there can be only one edge that contains the vertex from $V$.
\end{proof}

Any copy of $S(7)$ must therefore have
exactly six vertices in $K$ and exactly one vertex in $U$.  Suppose they are $A, B, C, D, E, F \in K$
and $G \in U_i$.  Suppose also that the edges of $S(7)$ induced by $K$ are
\[\{A,B,C\}, \{A, E, F\}, \{C, D, E\}, \{B, D, F\}.\]

\noindent\textbf{Claim 1c:} If $\{S_1, S_2, S_3\}$ is an edge in $K$ then
$|S_i \cap S_j| \leq k/2 + \epsilon k$ for all $i \neq j$.
\begin{proof}
This follows from the definition of the hypergraph on $K$:
\[
k = |S_1| \leq n - |S_2 \cap S_3| = (3/2 + \epsilon)k - |S_2 \cap S_3|\mbox{, so } |S_2 \cap S_3| \leq k/2 + \epsilon k,
\]
and the claim follows through symmetry.
\end{proof}

\noindent\textbf{Claim 1d:} The following intersections all have size at least
$2k - 4\epsilon k$: $A \cap D, B \cap E, C \cap F$.

\begin{proof}
We will prove that $|A \cap D| \geq 2k - 4\epsilon k$; the rest follow through symmetry.
Because $\{B,D, F\}$ is an edge, $D \subseteq (\overline{B}\cap F) \cup (B \cap \overline{F})
\cup (\overline{B} \cap \overline{F})$.
Also, because $\{A, B, C\}$ is an edge,
$|\overline{A} \cap \overline{B}| = |\overline{A}| - |\overline{A}\cap B| \leq (k/2 + \epsilon k)
- (k/2 - \epsilon k) = 2\epsilon k$.
Similarly, because $ \{A, E, F\}$ is an edge, $|\overline{A}\cap\overline{F}| \leq 2\epsilon k$. Therefore,
\[
|D \cap \overline{A}| \leq
					|\overline{A} \cap \overline{B} \cap F| + |\overline{A} \cap B \cap \overline{F}| +
					|\overline{A} \cap \overline{B} \cap \overline{F}|
					\leq |\overline{A} \cap \overline{B}| + |\overline{A} \cap \overline{F}|
					\leq 4\epsilon k,
\]
and so $|D \cap A | \geq |D| - 4\epsilon k = k - 4\epsilon k$.
\end{proof}

It follows from Claim~1d that $S(7)$ cannot be a subgraph of $H$.  Otherwise, the edges
$\{A, D, u\}, \{B, E , u\}, \{C, F, u\}$ would all appear, and by the definition of $H$,
because the intersections mentioned in Claim~1d are large, it follows that
$i \in (A \cap D) \cap (B \cap E) \cap (C \cap F)$.
In that case, however, $A \cap B \cap C$ is not empty and so $\{A, B, C\}$ is not an edge.
\end{proof}

It remains only to compute the minimum degree of $H$.
Vertices $S_1, S_2 \in K$ have co-degree at least $\frac{k-4\epsilon k}{n}|U|$ if
$|S_1 \cap S_2| \geq k-4\epsilon k$ and at least $\frac{k+4\epsilon k}{n}|U|$ otherwise.
Vertices $u_1, u_2 \in U$ have co-degree at least $|V|$ and
vertices $v_1, v_2 \in V$ have co-degree at least $|U|$.
All other pairs of vertices have co-degree at least $|U|$ or $|V|$.
The minimum co-degree is therefore at least
\[
\min\left\{\frac{k(1-4\epsilon)}{k(3/2+\epsilon)}|U|, |U|, |V| \right\}
     = \left\{\frac{2-8\epsilon}{3+2\epsilon}\cdot\frac{3}{5}N, \frac{3}{5}N, \frac{2}{5}N  \right\}.
\]
For some choice of $\epsilon$, this is approximately $\frac{2}{5}|V(H)|$.
\end{proof}

\newtheorem{cofanoquestion}[thmctr]{Question}
\begin{cofanoquestion}
What is the co-chromatic threshold of the Fano-free hypergraphs?
It is between $2/5$ and $1/2$.
\end{cofanoquestion}

In~\cite{abgkm} it was proved  that if a family $\mathcal{F}$ of graphs has positive chromatic
threshold then the chromatic threshold of $\mathcal{F}$ is in fact at least $1/3$.  We
think that a similar statement holds for hypergraphs.  For $3$-uniform hypergraphs, we believe
that the least positive chromatic threshold is achieved by the family of $\tk{4}{3}$-free
hypergraphs (see Section~\ref{subsecTK}).

\newtheorem{thresholdconj}[thmctr]{Conjecture}
\begin{thresholdconj}
If a family $\mathcal{F}$ of $3$-uniform hypergraphs has positive chromatic threshold
then the chromatic threshold of $\mathcal{F}$ is at least $18/361$.
\end{thresholdconj}

\subsection{An insightful attempt at proving Theorem~\ref{coloringthm}} \label{vcmotivation}

The proof of Theorem~\ref{coloringthm}, which appears in Sections~\ref{vcdimfullproof}
and \ref{vcseclargeoverlap}, is complex, but
it started as a simple idea built on three key ideas.
In this section we attempt to motivate the essential ingredients behind the proof.

Consider the following proof strategy for Theorem~\ref{coloringthm}: assume the chromatic number
of $B$ is large and give an algorithm that produces $d$ edges which witness that $\vc_H(B,\gamma,F) \geq d$.
The first key idea is to use the greedy algorithm, which selects the edges one by one while maintaining that
all sections are large enough to force a copy of $H$.  Initially, we can pick any edge $E_1$, since
any section of $x \in E_1$ is the set $\gamma(x)$, and by assumption $\gamma(x)$ is large enough to force
a copy of $H$.  So where could the greedy algorithm get stuck?  Assume the greedy algorithm
selected $E_1, \ldots, E_i$ but cannot continue.  That is, for every other edge $E$, there
exists some section $S$ of $x_1 \in E_1, \ldots, x_i \in E_i$ and there exists some $x \in E$ such that
$S \cap \gamma(x)$ is too small to force a copy of $H$.

The second key idea in the proof is to assume that every edge of $B$ has small overlap;
that is, we assume that $(B, \gamma, F)$ satisfies the condition that for every $x \neq y \in E \in B$,
the number of edges in the hypergraph $\gamma(x) \cap \gamma(y)$ is small.
With this assumption, the $r^i$ sections $S_1, \ldots, S_{r^i}$ of
$x_1 \in E_1, \ldots, x_i \in E_i$ are almost disjoint.
Let $S_{r^i+1} = \binom{F}{r-1} \setminus S_1 \setminus \ldots \setminus S_{r^i}$ so that
$S_1, \ldots, S_{r^i+1}$ almost form a partition.
Recall that the greedy algorithm could not continue because for every edge
$E$ disjoint from $E_1, \ldots, E_i$, there existed some $S_j$ and some $x\in E$ with
$S_j \cap \gamma(x)$ small.  Because $S_1, \ldots, S_{r^i+1}$ is almost a partition, this implies
that there is some other $S_{j'}$ with $S_{j'} \cap \gamma(x)$ large.
In other words, the greedy algorithm cannot continue if there exists sets $S_1, \ldots, S_{t+1}$ such that
for every edge $E$, there exists some $S_j$ and some $x \in E$ such that $S_j \cap \gamma(x)$ is large.

The third key idea is to use this in a density increment argument, similar to that used in the
proof of the Regularity Lemma.
If we have a partition $P$ of $(B,\gamma,F)$ (a partition will be formally defined later), we apply
the greedy algorithm from the last two paragraphs in each part of $P$ to refine the partition;
the sets $S_1, \ldots, S_t$ are used to refine each part.
If we define the density of a partition correctly, we can show that each time when we apply the greedy algorithm
the density will increase by a constant amount and add only a constant number of new parts in the partition.

The last part of the proof is to reduce the full theorem to the case where $(B,\gamma,F)$ satisfies
the condition that for every $x \neq y \in E \in B$, $\gamma(x) \cap \gamma(y)$ is small.

\subsection{A conditional proof of Theorem~\ref{coloringthm}} \label{vcdimfullproof}

\theoremstyle{definition}
\newtheorem*{condition1}{Condition 1}
\newtheorem*{condition2}{Condition 2}
\theoremstyle{plain}

\newcommand{\lambda}{\lambda}
\newcommand{\eta}{\eta}
\newcommand{\cGreedy}[1]{\psi_{#1}}
\newcommand{\alpha}{\alpha}
\newcommand{\beta}{\beta}
\newcommand{L_1}{L_1}
\newcommand{L_2}{L_2}

\begin{condition1}
Throughout this section, fix $r_b,r_{\gamma} \geq 1$, $d \in \mathbb{Z}^{+}$, $0 < \epsilon < \frac{1}{4} r_b^{-d}$,
and $H$ an $r_{\gamma}$-uniform hypergraph with zero Tur\'{a}n density.
Also, let $(B,\gamma,F)$ be an $(r_b,r_{\gamma})$-uniform fiber bundle for which $\vc_H(B,\gamma,F) < d$
and if $b \in V(B)$, then $\left| \gamma(b) \right| \geq \epsilon \binom{\left| F \right|}{r_{\gamma}}$.
\end{condition1}

\noindent In this section, we will prove the following theorem.

\newtheorem{conditionalcolorthm}[thmctr]{Theorem}
\begin{conditionalcolorthm} \label{conditionalcolorthm}
Let $(B,\gamma,F)$ be a fiber bundle satisfying Condition~1.  Then there exists constants
$0 < \lambda = \lambda(r_b,r_{\gamma},d,\epsilon,H) < \epsilon$, $L_1 = L_1(r_b,r_{\gamma},d,\epsilon,H)$, and
$L_2 = L_2(r_b,r_{\gamma},d,\epsilon,H)$ such that the following holds.
Assume $\left| F \right| \geq L_1$ and for every $x \neq y \in E \in B$,
$\left| \gamma(x) \cap \gamma(y) \right| \leq \lambda \binom{\left| F \right|}{r_{\gamma}}$.
If $r_b = 1$, then $\left| B \right| \leq L_2$ and if $r_b \geq 2$ then
$\chi(B) \leq L_2$.
\end{conditionalcolorthm}

The only differences between this theorem and Theorem~\ref{coloringthm} are that we allow $r_b = 1$ and also assume
a restriction on $\gamma(x) \cap \gamma(y)$ for $x \neq y \in E \in B$.  We will
remove the need for this assumption in the next section.

\noindent Define the following constants.
\begin{align*}
\alpha = \frac{1}{1000} \left(\frac{\epsilon}{4}\right)^{d+1},   \qquad \eta = \frac{1}{4} \epsilon^2 \alpha,
  \qquad \beta = \alpha^{1/\eta}, \qquad L_2 = \left\lceil r_b d(r_b^d+2)^{1/\eta} \right\rceil.
\end{align*}
Next, pick $L_1$ large enough so that if $\left| F \right| \geq L_1$ and $S \subseteq \binom{F}{r_{\gamma}}$ with
$\left| S \right| \geq 4^{-d} \epsilon^{d+1} \beta \binom{\left| F \right|}{r_{\gamma}}$, then $S$ contains a copy of $H$.
Lastly, pick $\lambda = \frac{\epsilon^{d+1} \beta}{d r_b^2 4^d}$.

\begin{condition2}
If $(B,\gamma,F)$ is a fiber bundle, then for every edge $E$ in $B$ and every $x\neq y \in E$, the section of $x,y$
has at most $\lambda \binom{\left| F \right|}{r_{\gamma}}$ edges.
\end{condition2}

If $(B,\gamma,F)$ is a fiber bundle, a \textbf{partition} $P$ of $(B,\gamma,F)$ is a family
$P = \{ (X_1,S_1), \linebreak[1] \ldots, (X_p,S_p) \}$ such that $X_1, \ldots, X_p$ is a partition of $V(B)$
and $S_1, \ldots, S_p$ is a partition of $\binom{F}{r_{\gamma}}$, where we allow $X_i = \emptyset$ or $S_i = \emptyset$.
A partition $Q$ is a refinement of a partition $P$ if
for each $(X,S) \in P$, there exist $(Y_1,T_1), \ldots, (Y_q,T_q) \in Q$ such that $X = \cup Y_i$ and $S = \cup T_i$.
For $X \subseteq V(B)$ and $S \subseteq 2^F$, the \textbf{density of $(X,S)$} is
\begin{align*}
d(X,S) = \begin{cases}
1 & S = \emptyset \text{ or } X = \emptyset, \\
\min \left\{ \frac{\left| \gamma(x) \cap S \right|}{\left| S \right|} : x \in X \right\} & \text{otherwise},
\end{cases}
\end{align*}
and define
\begin{align*}
d(P) = \min\left\{  d(X,S) : (X,S) \in P \right\}.
\end{align*}
A partition $P$ is a \textbf{partial coloring} if for every $(X,\emptyset) \in P$,
we have that $B[X]$ independent.
The \textbf{rank} of a partition $P$ is
the minimum of $\left| S \right|$ over all $(X,S) \in P$ with $S \neq \emptyset$.

Recall from the sketch in Section~\ref{vcmotivation} that the general structure of our proof is to show
how to refine a partial coloring $P$ to a partial coloring $Q$ where $d(Q) \geq \eta + d(P)$.  This will
imply that we can only refine the partition a constant number of times, at which point we will have a coloring
with a bounded number of parts.
The key lemma to facilitate this refinement is the following.

\newtheorem{partitionlem}[thmctr]{Lemma}
\begin{partitionlem} \label{partitionlem}
Let $(B,\gamma,F)$ be a fiber bundle satisfying Conditions~1 and 2.
Let $X \subseteq V(B)$ and $S \subseteq \binom{F}{r_{\gamma}}$ with $X \neq \emptyset$, $d(X,S) \geq \epsilon$,
$\left| F \right| \geq L_1$, and
$\left| S \right| \geq \beta \binom{\left|  F  \right|}{r_{\gamma}}$.
Then there exists a partition $Y_1, \ldots, Y_n,Z$ of $X$
and a partition $T_1, \ldots, T_n$ of $S$ such that $n \leq r_b^d+1$ and
\begin{itemize}
\item $\left| T_i \right| \geq \alpha \left| S \right|$,
\item $d(Y_i,T_i) \geq \min \left\{ 1, \eta + d(X,S)\right\}$,
\item $B[Z]$ is independent.
\end{itemize}
\end{partitionlem}

\noindent This lemma has an easy corollary.

\newtheorem{refinementlem}[thmctr]{Corollary}
\begin{refinementlem} \label{refinementlem}
Let $(B,\gamma,F)$ be a fiber bundle satisfying Conditions~1 and 2 and $\left| F \right| \geq L_1$.
Let $P$ be a partial coloring of $(B,\gamma,F)$ where
$P$ has rank at least $\alpha^{k} \binom{\left|F\right|}{r_{\gamma}}$ with $k \leq \frac{1}{\eta}$.
Then there exists a refinement $Q$ of $P$ such that
\begin{itemize}
\item $\left| Q \right| \leq (r_b^d+2) \left| P \right|$,
\item $Q$ is also a partial coloring,
\item the rank of $Q$ is at least $\alpha^{k+1} \binom{\left|F\right|}{r_{\gamma}}$,
\item $d(Q) \geq \min\left\{1, \eta + d(P)\right\}$.
\end{itemize}
\end{refinementlem}

\begin{proof}
For each pair $(X,S) \in P$ with $X \neq \emptyset$ and $S \neq \emptyset$, apply
Lemma~\ref{partitionlem}.  Since $k \leq \frac{1}{\eta}$,
$\left| S \right| \geq \alpha^{k} \binom{\left| F \right|}{r_{\gamma}}
                \geq \alpha^{1/\eta} \binom{\left| F \right|}{r_{\gamma}}
                \geq \beta \binom{\left| F \right|}{r_{\gamma}}$.
Lemma~\ref{partitionlem} produces
$Y_1, \ldots, Y_n,Z$ and $T_1, \ldots, T_n$ with $n \leq r_b^d+1$.  We replace the pair $(X,S)$ with the pairs
$(Y_1,T_1), \ldots, (Y_n,T_n), (Z,\emptyset)$.  The resulting partition satisfies all the required properties.
\end{proof}

We can now easily prove Theorem~\ref{conditionalcolorthm}.

\begin{proof}[Proof of Theorem~\ref{conditionalcolorthm}]
By assumption, $(B,\gamma,F)$ satisfies Conditions~1 and 2.
Start with the partition $P = \left\{ (V(B), \binom{F}{r_{\gamma}}) \right\}$ and
apply Corollary~\ref{refinementlem} repeatedly until the partition satisfies $d(P) = 1$.  Since the
value of $d(P)$ increases by $\eta$ at each step, the partition is refined at most $1/\eta$ times,
and so
the resulting partition $P$ has at most $(r_b^d+2)^{1/\eta}$ parts.
Consider a part $(X,S) \in P$.
If $S = \emptyset$, then since $P$ is a partial coloring $B[X]$ must be independent, so $\chi(B[X]) = 1$.
If $S \neq \emptyset$,
we know that $\left| S \right| \geq \beta \binom{\left| F \right|}{r_{\gamma}}$, which by the choice of $\beta$ and $L_1$
forces a copy of $H$ in $S$.  Since $d(X,S) = 1$ we must have $S \subseteq \gamma(x)$ for every $x \in X$,
so that a matching of size $d$ in $B[X]$ witnesses that $\vc_H(B,\gamma,F) \geq d$.  Therefore, the maximum size
of a matching in $B[X]$ is $d-1$. If $r_b = 1$, then $B[X]$ has at most $d-1$ edges, so the total number of
edges of $B$ is $d(r_b^d+2)^{1/\eta}$.  If $r_b \geq 2$, then since the size of a maximal
matching in $B[X]$ is $d-1$, it is the case that $\chi(B[X]) \leq r_b(d-1)+1$.
This implies that the chromatic number of $B$ is at most $r_b d(r_b^d+2)^{1/\eta}$.
\end{proof}

\newcommand{\delta}{\delta}

All that remains is to prove Lemma~\ref{partitionlem}.  Before proving this lemma, we make
some definitions.
If $E_1, \ldots, E_t \in B$ and $S \subseteq \binom{F}{r_{\gamma}}$, then
the \textbf{minimum section density of $E_1, \ldots, E_t$ with respect to $S$} is
\begin{align*}
\delta(E_1, \ldots, E_t,S) = \min\left\{
    \frac{\left| \gamma(x_1) \cap \ldots \cap \gamma(x_t) \cap S \right|}{\left| S \right|}
    : x_1 \in E_1, \ldots, x_t \in E_t \right\}.
\end{align*}
Notice that if $E_1, \ldots, E_d$ are disjoint, $\delta(E_1, \ldots, E_d,S) > 0$,
$S$ is a constant fraction of $\binom{F}{r_{\gamma}}$, and $F$ is large,
then $E_1, \ldots, E_d$ witness that $\vc_H(B,\gamma,F) \geq d$.
Define constants $\cGreedy{1}, \ldots, \cGreedy{d}$ with $\cGreedy{1} = 1$ by
$\cGreedy{m+1} = \frac{1}{4} \epsilon \cGreedy{m}$ for $1 \leq m \leq d-1$.

For the proof of Lemma~\ref{partitionlem}, recall the outline from Section~\ref{vcmotivation}.
We will greedily select edges $E_1, \ldots, E_i$ as long as we can maintain that (using our new notation)
$\delta(E_1, \ldots, E_i,S)$ is large enough to force a copy of $H$.  Since $\vc_H(B,\gamma,F) < d$,
the greedy algorithm must terminate before choosing $d$ edges.  Once the greedy
algorithm terminates, we will let the sets $T_1, \ldots, T_{n-1}$
(the sets that we must find to prove Lemma~\ref{partitionlem})
be all sections of $x_1 \in E_1, \ldots, x_i \in E_i$.  If we let $Y_i$ be the set of vertices $y$
such that $\left| \gamma(y) \cap T_i \right| / \left| T_i \right|$ is at least
$d(X,S)+\eta$, we will satisfy almost all of the requirements in Lemma~\ref{partitionlem}.  We need only
prove that if $Z$ are the vertices not in any $Y_i$ then $B[Z]$ is independent.  We do this by showing that
if $B[Z]$ contained an edge $E$ we could have continued the greedy algorithm by selecting $E$.
In order for this to be true, we need the greedy algorithm to require a weaker lower
bound on $\delta(E_1, \ldots, E_{i+1}, S)$ than the algorithm required on
$\delta(E_1, \ldots, E_i,S)$.  The constants $\cGreedy{i}$ are used to define the greedy
algorithm: the greedy algorithm will select edges while maintaining
$\delta(E_1, \ldots, E_i, S) \geq \epsilon \cGreedy{i}$.  Since $\cGreedy{i+1}/\cGreedy{i} = \frac{1}{4} \epsilon$,
we lose a fraction of $\epsilon$ in each step.  The careful choice of $L_1$ guarantees that even after
losing a fraction of $\epsilon$ for the $d$ steps per refinement over a maximum of $1/\eta$ refinements
we still have enough edges to force a copy of $H$.

\begin{proof}[Proof of Lemma~\ref{partitionlem}]
Start by greedily selecting disjoint edges $E_1, \ldots, E_i$ of $B[X]$ such that
$\delta(E_1, \ldots, \linebreak[1] E_i,S) \geq \epsilon \cGreedy{i}$.
Since for every $x \in V(B)$
\begin{align*}
\frac{\left| \gamma(x) \cap S \right|}{\left| S \right|} \geq d(X,S) \geq \epsilon \cGreedy{1},
\end{align*}
the greedy algorithm can start with any edge $E_1$ in $B[X]$.
Assume the greedy algorithm has selected
$E_1, \ldots, E_m$ with $\delta(E_1, \ldots, E_m,S) \geq \epsilon \cGreedy{m}$ but for every other
edge $E$ in $B[X]$
disjoint from $E_1, \ldots, E_m$, we have $\delta(E_1, \ldots, E_m, E,S) < \epsilon \cGreedy{m+1}$.

First, we prove that $\vc_H(B,\gamma,F) \geq m$.
Since $d(E_1, \ldots, E_m,S) \geq \epsilon \cGreedy{m} \geq \epsilon \cGreedy{d}$, we have
that every section of $x_1 \in E_1, \ldots, x_m \in E_m$ has size at least
$\epsilon \cGreedy{d} \left| S \right| \geq \epsilon \cGreedy{d} \beta \binom{\left| F \right|}{r_{\gamma}}$.
By the choice of $L_1$, the section of $x_1, \ldots, x_m$ contains a copy of $H$, and so
$m \leq d$.  We make the following definitions.

\begin{itemize}
\item Let $R_1, \ldots, R_t$ be all $r_b^m$ sections of $v_1 \in E_1, \ldots, v_m \in E_m$ intersected with $S$.
    If some $I \subseteq F$ appears in more than one $R_i$, remove it from all but the least indexed $R_i$.
\item For $1 \leq i \leq t$, start with $T_i = R_i$ and remove elements from $T_i$ (recall that elements of $T_i$ are subsets
    of $F$) until
    $\left| T_i \right|$ is smaller than $2 \epsilon \left| S \right|$.  (If $R_i$ is already smaller than $2 \epsilon \left| S \right|$,
     nothing needs to be removed.)
\item Let $T_{t+1} = S \setminus T_1 \setminus \ldots \setminus T_t$.
\item For $1 \leq i \leq t+1$, define
 \begin{align*}
  Y_i = \left\{ x \in X : \frac{\left| \gamma(x) \cap T_i \right|}{\left| T_i \right|}
                          \geq \min\left\{1, \eta + d(X,S) \right\} \right\}.
 \end{align*}
 If some $x$ appears in more than one $Y_i$, remove it from all but the least indexed $Y_i$.
\item Let $Z = X \setminus Y_1 \setminus \ldots \setminus Y_{t+1}$.
\end{itemize}
 By the definition of $Y_i$, $d(Y_i,T_i) \geq \min\{1, \eta + d(X,S)\}$.
Therefore, to finish the proof we need to check that $\left| T_i \right| \geq \alpha \left| S \right|$ and $B[Z]$ is independent.

\medskip
\noindent \textbf{Claim:} $\left| T_i \right| \geq \alpha \left| S \right|$ for all $1 \leq i \leq t+1$.

\begin{proof}
Since $\delta(E_1, \ldots, E_m,S) \geq \epsilon \cGreedy{m}$,
before removing anything from $R_i$, each $R_i$ has size at least $\epsilon \cGreedy{m} \left| S \right|$.
Consider some $I \in R_i \cap R_j$ for some $j \neq i$.  Since $j \neq i$, there must be some $E_k$ such that
$R_i$ selects $\gamma(x)$ and $R_j$ selects $\gamma(y)$ for $x \neq y \in E_k$.
Thus $I \in \gamma(x) \cap \gamma(y)$, which has size at most $\lambda \binom{\left| F \right|}{r_{\gamma}}$ by
Condition~2.  In other words, every element removed from $R_i$ is contained in $\gamma(x) \cap \gamma(y)$
for some $x \neq y \in E_k$.  There are at most $mr_b(r_b-1)$ choices of $x \neq y \in E_k$, so
the maximum number of elements removed from $R_i$ is $mr_b^2\lambda \binom{\left| F \right|}{r_{\gamma}}$.
Since $m \leq d$ and $\left| S \right| \geq \beta \binom{\left| F \right|}{r_{\gamma}}$,
we remove at most
$d r_b^2 \lambda \beta^{-1} \left| S \right|$ elements from $R_i$.  By the choice of constants,
$d r_b^2 \lambda \beta^{-1} \leq \frac{1}{2} \epsilon \cGreedy{d} \leq \frac{1}{2} \epsilon \cGreedy{m}$, so
\begin{align} \label{eqsizeRi}
\left| R_i \right| \geq \frac{1}{2} \epsilon \cGreedy{m} \left| S \right|.
\end{align}
Since $\cGreedy{m} \leq 1$ and we remove elements from $R_i$ to form $T_i$ only if $|R_i| \geq 2\epsilon \left| S \right|$, equation \eqref{eqsizeRi} implies
\begin{align} \label{eqsizeTi}
\left| T_i \right| \geq \frac{1}{2} \epsilon \cGreedy{m} \left| S \right|,
\end{align}
which is at least $\alpha \left| S \right|$ by the choice of $\alpha$.
Now consider the size of $T_{t+1}$.  Since each $T_i$ with $i \leq t$ has size at most $2 \epsilon \left| S \right|$
and we assumed that $\epsilon < \frac{1}{4} t^{-1}$ in Condition~1, the set $T_{t+1}$ has at least
$\frac{1}{2} \left| S \right| \geq \alpha \left| S \right|$ elements.
\end{proof}

\medskip
\noindent \textbf{Claim:} $B[Z]$ is independent.

\begin{proof}
Assume $E$ is an edge in $B[Z]$.  We would like to show that
there exists some $x \in E$ and some $T_j$ such that
\begin{align} \label{eqdensityTj}
\frac{\left| \gamma(x) \cap T_j \right|}{\left| T_j \right|} \geq \min\left\{1, \eta + d(X,S) \right\},
\end{align}
since this would show that $x \in Y_j$, contradicting that $x \in Z$.
Assume $E$ intersects some $E_i$ for some $1 \leq i \leq m$, with $x \in E \cap E_i$.
Since $x \in E_i$ there are many sections $R_j$ that select $x$, since the sections $R_j$ were formed by choosing one vertex
from each of $E_1, \ldots, E_d$.  Fix some such section $R_j$ that selects $x$, in which case $R_j \subseteq \gamma(x)$.
Then $T_j \subseteq R_j \subseteq \gamma(x)$ and
$\left| \gamma(x) \cap T_j \right|/\left| T_j \right| = 1$ so \eqref{eqdensityTj} is satisfied.

Now assume $E$ is disjoint from $E_1, \ldots, E_m$.  Since the greedy algorithm could not continue,
$\delta(E_1, \ldots, E_m, E, S) < \epsilon \cGreedy{m+1}$, which implies that there exists some
$v_1 \in E_1, \ldots, v_m \in E_m, x \in E$ such that
\begin{align*}
\left| \gamma(v_1) \cap \ldots \cap \gamma(v_m) \cap \gamma(x) \cap S \right| <
\epsilon \cGreedy{m+1} \left| S \right|.
\end{align*}
By the definition of $T_i$, there exists some $T_i$ such that $T_i \subseteq \gamma(v_1) \cap \ldots \cap \gamma(v_m) \cap S$.
Therefore,
\begin{align*}
\left| \gamma(x) \cap T_i \right| < \epsilon \cGreedy{m+1} \left| S \right| \leq \frac{2\cGreedy{m+1}}{\cGreedy{m}} \left| T_i \right|,
\end{align*}
where the last inequality uses \eqref{eqsizeTi}.
Assume that for every $j \neq i$, \eqref{eqdensityTj} fails.  Then
\begin{align*}
\left| \gamma(x) \cap S \right|
&= \left| \gamma(x) \cap T_i \right| + \sum_{j \neq i} \left| \gamma(x) \cap T_j \right|
\leq \frac{2\cGreedy{m+1}}{\cGreedy{m}} \left| T_i \right| +
   \sum_{j \neq i} (\eta + d(X,S)) \left| T_j \right|.
\end{align*}
Dividing through by $\left| \gamma(x) \cap S \right|$ we obtain
\begin{align*}
1 \leq \frac{2\cGreedy{m+1}}{\cGreedy{m}} \frac{\left| T_i \right|}{\left| S \right|} \frac{\left| S \right|}{\left| \gamma(x) \cap S \right|} +
   (\eta + d(X,S)) \left( 1 - \frac{\left| T_i \right|}{ \left| S \right|} \right) \frac{\left| S \right|}{\left| \gamma(x) \cap S \right|}.
\end{align*}
Because $\left| S \right|/\left| \gamma(x) \cap S \right| \leq \frac{1}{d(X,S)} \leq \frac{1}{\epsilon}$,
\begin{align} \label{vccolorbadeq}
1 \leq \frac{2\cGreedy{m+1}}{\cGreedy{m}\epsilon} \frac{\left| T_i \right|}{\left| S \right|} + \left(\frac{\eta}{\epsilon} + 1\right) \left( 1 - \frac{\left| T_i \right|}{\left| S \right|} \right).
\end{align}
Let $w = \left| T_i \right|/\left| S \right|$. The right hand side of the
above inequality is a weighted average of $\frac{2\cGreedy{m+1}}{\cGreedy{m}\epsilon}$ and
$(1+\frac{\eta}{\epsilon})$:
\begin{align*}
\frac{2\cGreedy{m+1}}{\cGreedy{m} \epsilon} w + \left(1 + \frac{\eta}{\epsilon} \right) (1 - w).
\end{align*}
Since $2\cGreedy{m+1}/(\cGreedy{m} \epsilon) = \frac{1}{2} < 1 + \frac{\eta}{\epsilon}$,
this will be maximized when $w$ is as small as possible.
By \eqref{eqsizeTi}, $w \geq \alpha$, so
\begin{align*}
\frac{2\cGreedy{m+1}}{\cGreedy{m} \epsilon} \alpha + \left(1 + \frac{\eta}{\epsilon} \right) (1 - \alpha)
&< \frac{1}{2} \alpha + 1 + \frac{\eta}{\epsilon} - \alpha
\leq 1 + \frac{\eta}{\epsilon} - \frac{1}{2} \alpha < 1.
\end{align*}
This implies that for any $w \geq \alpha$, the inequality in \eqref{vccolorbadeq} is false.
This contradiction shows that there must be some $j \neq i$ such that
$\left| \gamma(x) \cap T_j \right| / \left| T_j \right|$ is at least $\eta + d(X,S)$, which contradicts
that $E$ is contained in $B[Z]$.
\end{proof}
Thus $B[Z]$ is independent and the proof is complete.
\end{proof}

\subsection{Fiber bundles with large overlap} \label{vcseclargeoverlap}

In the previous section, we proved Theorem~\ref{coloringthm} restricted to fiber bundles that satisfy
Condition~2.  To prove Theorem~\ref{coloringthm}, we will divide the edges of $B$ into two pieces.  Let $B'$ be
the subset of edges of $B$ that satisfy Condition~2; that is for every $x \neq y \in E \in B'$,
$\gamma(x) \cap \gamma(y)$ has density at most $\lambda$.  We apply
Theorem~\ref{conditionalcolorthm} to $(B',\gamma,F)$ to bound the chromatic number of $B'$.  For the remaining
edges, we will merge $x$ and $y$ into a new vertex $z$ if
$\gamma(x) \cap \gamma(y)$ has density at least $\lambda$
(we define $\gamma(z)$ to be $\gamma(x) \cap \gamma(y)$).
Let $(M,\psi,F)$ be the fiber bundle after merging all such vertices.
Since all edges of $B - B'$ have some such pair $x,y$, all edges of $M$ will have size
at most $r_b - 1$.  Then we apply induction on $r_b$ to bound the chromatic number
of $M$.  To be able to apply induction,
we need to verify that $\vc_H(M,\psi,F) \leq \vc_H(B,\gamma,F)$ and that there is a lower bound on the density
of $\gamma(m)$ for $m \in V(M)$.  The definition of $\gamma(z) = \gamma(x) \cap \gamma(y)$ satisfies
both of these requirements.  First, $\vc_H(M,\psi,F) \leq \vc_H(B,\gamma,F)$ because
any copy of $H$ in $\gamma(z)$ will be in both $\gamma(x)$ and $\gamma(y)$. Also, there is a lower bound
on the density of $\gamma(z)$ because we only merge if $\gamma(x) \cap \gamma(y)$ has density
at least $\lambda$.  The magic in this proof is that Condition~2, the extra requirement needed
in the previous section, fits exactly with the requirements to be able to apply induction
after merging.

For technical reasons, our induction statement needs to be slightly stronger than Theorem~\ref{coloringthm};
we no longer assume $B$ is a uniform hypergraph.
Instead, we allow the edges of $B$ to have size between one and $r_b$.  This is because
after merging, all we know is that the edges have size between one and $r_b - 1$.
This is also why we need to allow 1-uniform hypergraphs in Theorem~\ref{conditionalcolorthm}.

\newtheorem{coloringwithone}[thmctr]{Theorem}
\begin{coloringwithone} \label{coloringwithone}
Let $r_b, r_{\gamma} \geq 1$, $d \in \mathbb{Z}^{+}$, $0 < \epsilon < 1$, and $H$ be an
$r_{\gamma}$-uniform hypergraph with zero Tur\'{a}n density.  Then there exists constants $K_1 = K_1(r_b,r_{\gamma},d,\epsilon,H)$
and $K_2 = K_2(r_b,r_{\gamma},d,\epsilon,H)$ such that the following holds.
Let $(B,\gamma,F)$ be any fiber bundle where the edges of $B$ have size between $1$ and $r_b$,
$\vc_H(B,\gamma,F) < d$, and for all $b \in V(B)$, $\gamma(b)$ is an $r_{\gamma}$-uniform hypergraph and
$\left| \gamma(b) \right|  \geq \epsilon \binom{\left| F \right|}{r_{\gamma}}$.
Let $A$ be the set of edges of size $1$ in $B$.
If $\left| F \right| \geq K_1$, then $\left| A \right| \leq K_2$ and $\chi(B - A) \leq K_2$.
\end{coloringwithone}

\begin{proof}
The proof is by induction on $r_b$.  When $r_b = 1$, we can directly apply
Theorem~\ref{conditionalcolorthm}, since Condition~2 is trivially satisfied.  So assume $r_b \geq 2$,
and let $\lambda = \lambda(r_b,r_{\gamma},d,\epsilon,H)$ be the constant from Theorem~\ref{conditionalcolorthm}.
Define a subhypergraph $B'$ of $B$ as follows:
\begin{align*}
B' = \left\{ E \in B : \left| E \right| = r_b \text{ and } \forall x\neq y\in E
              \quad \left| \gamma(x) \cap \gamma(y) \right| \leq \lambda \binom{\left| F \right|}{r_{\gamma}} \right\}.
\end{align*}
Then $(B',\gamma,F)$ is an $(r_b,r_{\gamma})$-uniform fiber bundle to which we can apply Theorem~\ref{conditionalcolorthm}
to bound the chromatic number of $B'$.  To complete the proof, we will bound the chromatic number
of $B - B'$.

Initially, let $B_0 = B - B'$ and $\gamma_0 = \gamma$.  At stage $i$, assume we have defined $(B_i,\gamma_i,F)$.
Let $E$ be some edge of $B_i$ with $\left| E \right| = r_b$,
where there exists $x \neq y \in E$ be such that
$\left| \gamma_i(x) \cap \gamma_i(y) \right| \geq \lambda \binom{\left| F \right|}{r_{\gamma}}$.
We form $(B_{i+1}, \gamma_{i+1}, F)$ by merging the vertices $x$ and $y$.  More precisely,
let $z$ be a new vertex and define
\begin{align*}
V(B_{i+1}) &= V(B_i) - x - y + z, \\
E(B_{i+1}) &= \left\{ E \in B_i : E \cap \left\{ x,y \right\} = \emptyset \right\} \cup \left\{ E - x - y + z : E \in B_i, E \cap \left\{ x, y \right\} \neq \emptyset \right\}, \\
\gamma_{i+1}(w) &=
\begin{cases}
\gamma_i(w) & w \neq z, \\
\gamma_i(x) \cap \gamma_i(y) & w = z.
\end{cases}
\end{align*}
We do this until every edge of $B_i$ has size at most $r_b-1$; say this occurs at step $s$.

Through this modification, the dimension cannot increase.  Consider step $i$,
when we merge the vertices $x$ and $y$ in $(B_i, \gamma_i, F)$ to form $(B_{i+1}, \gamma_{i+1}, F)$.
Let $S$ be any section in $(B_i, \gamma_i, F)$ that selects $x$ or $y$, and let $S'$ be a section
in $(B_{i+1}, \gamma_{i+1}, F)$ that is identical to $S$ except that it selects $z$ instead of $x$ or $y$.
If $S'$ contains a copy of $H$, then this copy of $H$ is in $\gamma(z)$, which implies it is in both $\gamma(x)$
and $\gamma(y)$.  Therefore, $H$ is in $S$.

For each vertex $z$ we add, we have $\left| \gamma(z) \right| \geq \lambda \binom{\left| F \right|}{r_{\gamma}}$.
Therefore, we can apply induction on $r_b$ to $(B_s, \gamma_s,F)$ using $\epsilon = \lambda$ to
bound the chromatic number of $B_s$.
We now consider ``un-merging'' the vertices of $B_s$ to obtain a coloring of $B_0$.
Consider un-merging $z \in V(B_i)$ to obtain $x,y \in V(G_{i-1})$.
If $z$ was contained in an edge of size one, we color $x$ and $y$ using two new colors.
Otherwise, we color $x$ and $y$ the same color as $z$.
After the un-merging, we will have a proper coloring of $B_0 = B - B'$.
By induction, there is a bounded number of edges of size one, so we will use a bounded number of new colors.

The colorings on $B_0$ and $B'$ can be combined to obtain a coloring
of $B$ using a bounded number of colors.  Let $K_1$ (the minimum size of $F$) be the maximum
between the required size from Theorem~\ref{conditionalcolorthm} and the size required by induction.
Lastly, any edge of size one in $B$ will also appear in $B_s$, and by induction $B_s$ has a bounded
number of edges of size one.
\end{proof}

\section{Introduction}

\input{intro.tex}

\section{Results}

\input{results.tex}

\section{Fiber Bundles and Fiber Bundle Dimension} \label{secFiberDef}

\input{fiberdef.tex}

\section{Chromatic threshold for near $r$-partite hypergraphs} \label{secChromNearr}

\input{nearkchrom.tex}

\section{Coloring hypergraphs with bounded dimension} \label{secpartitionproof}

\input{partitionproof.tex}

\section{Extremal results for critical hypergraphs}\label{extremal}

\input{extremalcycle.tex}

\section{Chromatic threshold of $F_5$-free hypergraphs} \label{secF5}

\input{f5bounds.tex}

\section{Generalized Kneser hypergraphs} \label{secnewKneser}

\input{newkneser.tex}

\section{Open Problems and Partial Results} \label{secopenproblems}

\input{openprob.tex}

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


text:
CinemaCon SIGGRAPH SMPTE AFCI Global Directory
January 31, 2014 9:00am PT by Carolyn Giardina
Film vs. Digital? 3 Things to Know Before Saturday's ASC Awards
Roger Deakins on the set of "Prisoners."
Cinematographers chime in on filmmaking technology and breaking the rules. "It's about the final look," ASC president Richard Crudo says.
This story first appeared in the Feb. 7 issue of The Hollywood Reporter magazine.
The focus will be on how movies look when the American Society of Cinematographers hands out its 28th annual ASC Awards for Outstanding Achievement in features and television Feb. 1. Here's what you need to know:
PHOTOS: 5 Top Cinematographers Reveal Secrets of Their Craft
There is no dominant technology
The nominees are Sean Bobbitt for 12 Years a Slave, Barry Ackroyd for Captain Phillips, Philippe Le Sourd for The Grandmaster, Emmanuel Lubezki for Gravity, Bruno Delbonnel for Inside Llewyn Davis, Phedon Papamichael for Nebraska and Roger Deakins for Prisoners.
That list is a testament to how filmmaking is changing. Grandmaster, 12 Years and Davis were shot on film; Nebraska and Prisoners used Arri's Alexa digital cinematography camera; Phillips combined multiple formats; and Gravity used the Alexa for live action while also employing virtual cinematography techniques. Observing that "it's a great year for cinematography," ASC president Richard Crudo emphasized that the tools are irrelevant when it comes to the nominated images. He says, "It's about the final look."
PHOTOS: Oscar Nominees
Take lessons from a rule breaker, cinematographer Dean Cundey
The director of photography, who has filmed lots of VFX-heavy movies such as Jurassic Park, earned an Oscar nom for Who Framed Roger Rabbit and also shot Apollo 13 and the Back to the Future trilogy. Recalling when he teamed with director Robert Zemeckis on Rabbit, Cundey, who will receive the ASC's Lifetime Achievement Award, says, "We went to Disney, which had combined live action and animation [in movies like Mary Poppins], and they gave us guidelines -- use wide shots, make sure the lighting is very even, don't have the characters interact too much with the rest of the world. After leaving that meeting, Bob said, 'There's the rules we're going to break.' "
Oscar doesn't take its cues from ASC
Five of the ASC nominees also have been nominated for an Oscar: Le Sourd, Lubezki, Delbonnel, Papamichael and Deakins. But in the past 10 years, the ASC winner has matched the Oscar winner only four times. In 2013, Deakins won his third ASC award for Skyfall, but the Oscar went to Life of Pi. In fact, Deakins, revered by fellow cameramen, has received 11 Oscar noms but has yet to win.
The ASC Awards take place Feb. 1 in the Ray Dolby Ballroom at Hollywood & Highland.
Carolyn Giardina
THRnews@thr.com CGinLA

redpajama_set_name: RedPajamaCommonCrawl


text:
I'm Ready for Divorce!
Translator: Latte Editor: Sasha
"What do you mean, Marie?" Anna asked.
"Cough, nevermind," Marie said as she shook her head.
"Anyway, it looks like madame's condition has improved…"
Marie recalled the name of Elody's illness that she heard at a meeting with the other servants a while ago.
The Siren's Tears.
"I heard it from Sir Heinz! How the hell does that guy know about madame's disease?"
Heinz was the leader of a faction of the knights that used to be mercenaries. Everybody knew that they held some sort of prejudice against the madame.
Because of that, most of the servants dismissed Heinz's words as nothing but a mere lie. They couldn't trust someone who disliked the madame. And Marie also held the same opinion.
'I should talk about this with the lord…'
Marie decided to continue watching the situation more for the time being.
'Ugh, I don't think I'll have time to sleep again…' Caville thought as he walked up the stairs to his bedroom after having breakfast.
Yesterday as well, Caville was forced to stay up all night due to the never-ending amount of paperwork. Thus, he was only able to sleep a little at dawn.
Caville sighed as he watched Elody sleeping quietly. Though he was terribly sleep-deprived, he was still grateful to be given a beautiful sight to see every morning when he woke up. He loved seeing his wife's peaceful expression as she slept and would do anything to keep seeing it again. Even if it meant he had to give up sleeping his entire life.
He was so happy he could die.
Of course, he was still greedy for more. But he decided that it would be best to appreciate the fact that he didn't have to sleep separately from his wife anymore.
'You used to hug me to sleep whenever winter came…'
Caville believed that over time, Elody would be able to love him the same way he loved her.
At first, Caville thought that even if his wife couldn't reciprocate his feelings, he would still be fine with it. As long as she didn't hate him, he would always be satisfied with what he had.
However, he found that he started becoming more and more greedy.
Caville looked at the window and saw that it was already dawn.
Caville, who wasn't able to sleep at all last night, watched the sun as it slowly rose from the window. Then, he slowly reached out to Elody.
He held Elody's tiny pinky and glanced at it for a moment.
The feeling of him holding her finger at that time was still vivid. He had only touched her little finger, but Caville was happy enough to be given the chance to touch her again.
Elody was surprised to see Heinz visiting her in the lab so early in the morning.
'Why does he keep visiting me?'
Elody was still worried about his forehead as the scars hadn't fully healed yet.
'Has he not used the medicine yet? It should've been healed by now unless… he's doing this on purpose. But why? To show off to the others?'
"May I help you, Sir Heinz?"
"I heard that madame needed someone to get rid of the rocks in the forest. So I called my subordinates here to help!"
"…Oh, that's alright. You don't really have to…. The apprentices or squires would suffice."
"No! We'll do it instead!"
Elody glanced at Dame Solar who stood next to her with a troubled expression. Solar shrugged her shoulder and replied by whispering into Elody's ear.
"Just let them pamper you. They're really strong."
However, Elody was uncomfortable with the mercenary-knights.
She remembered them glaring at her wherever she went, and their eyes were full of hostility… In addition, they seemed to have a close relationship with Princess Larissa.
Today, the knights were… not as hostile to her as she thought they would be. Rather, they seemed to look a bit… dispirited.
"Alright. Then, can you start by moving that boulder over there?"
When Elody spoke reluctantly, Heinz replied by telling her not to worry. Then, he dragged his men to the forest.
"Madame, why is Sir Heinz doing that for you?" Solar asked, confused.
She was also curious about it herself. What in the world made him act like this?
"How suspicious… He used to despise you, but now he acts as if he's your number one supporter. I wonder what made him change his mind… You must be wary of him, madame," Dame Solar warned.
Elody nodded her head in response. He was definitely suspicious to say the least.
The boulders were too heavy to be carried alone, so several knights had to work together to destroy the pieces first.
Elody felt bad for the knights since it looked very tiresome.
They already hated her, but the fact that they had to do this made her feel even worse about herself. What if they hated her even more after this?
Well, it didn't really matter if they actually did, but she still couldn't help feel but uncomfortable at the thought.
As Elody watched the knights, she suddenly thought of something.
"Madame, what's wrong?"
Then, Elody had her servants bring a large bowl and packs of ice from the castle.
"I'm going to make a fruit punch."
"Oh, fruit punch?"
"Yes. I think it would be great if we're all able to drink it together."
Elody started making fruit punch with the peaches she had from the greenhouse and the raspberries she harvested from the forest.
Anna lifted her thumb up after she tasted it.
"Delicious! Shall I give these to the knights?"
"Yes. Everyone has had a rough time carrying those rocks. This is my way of thanking them for their hard work."
Anna thought that it was a waste to give them to the knights, but in the end, she thought that they deserved it since they had been suffering for quite a while.
Seeing the maids distribute the fruit punch to the knights made Elody smile. She was happy to see such a lively atmosphere in the mansion.
Caville called her from behind.
"Caville, what's going on?"
"I was supposed to catch you a fish from the river."
"Do you really want to? You must be busy with work…"
"I've finished them all."
The knights and servants glanced at the ducal couple having a friendly conversation.
Elody personally guided Caville across the forest. And the person following behind them was none other than Sir Brien.
"Sir Vedos, are you alright?"
Elody asked after seeing his terrible complexion.
Truth to be told, none of them looked fine. Caville had dark eye-bags underneath his eyes while Sir Brien looked as if he had been crying all night.
"Don't cry, you'll make other people feel bad…"
Elody looked at the river. The water seemed to be denser than yesterday. Just the mere sight of it made her feel refreshed.
"See you in a minute, my wife," Caville said, kissing the back of her Elody's hand.
Elody was startled and immediately ran back to the forest.
"I'm gonna die from excessive cuteness," Caville murmured while staring at Elody's back, while Brien creased his eyebrows in return.
"Well done, everyone."
Elody wanted to express her gratitude to the knights who had worked hard.
"Thanks to you guys, we were able to finish our work quickly as well. Thank you."
"You don't need to thank us. It was nothing," Heinz replied, blushing and grinning at the same time.
The men who stood behind him glanced at each other. They thought it was strange that the duchess had personally thanked them herself.
"By the way, madame. The fruit punch you gave to us was really delicious."
"I'm glad to hear it. Do you want more?"
"Can we?"
Elody only said that for courtesy's sake, but Heinz's eyes were shining so brightly it was hard to refuse.
Thus, Elody nodded her head and instructed Marie.
"Bring more peaches from the greenhouse to make the fruit punch."
"Yes, madame."
Marie glanced at Heinz before heading to the greenhouse with her maids.
Now it was time to start working.
Elody rolled up her sleeves and began harvesting the berries and mushrooms with the maids.
'Did the trees grow again?'
The tree where she picked the mushrooms yesterday was now growing with new mushrooms.
Just like the peach tree in the greenhouse.
"The duchess, she's so kind."
"Yeah, she's not like the rumors."
"She seems to have a good relationship with the leader…."
Heinz's men were sitting under the shade of a tree while enjoying their time off.
They chatted while drinking the fruit punch that the maids gave them.
It was really refreshing.
Drinking the sweet fruit punch while the cool breeze of fresh air swept away their sweat.
Ah yes, this was life.
The sun was shining, the flowers were blooming, the birds were singing, and the atmosphere was lively.
They thought that the duchess was a cruel and cold-hearted person. But she gave them fruit-punch and even thanked them in person.p
They thought she would ignore them and looked down on them for being former mercenaries, but the duchess had no such expression on her face.
Today, they had finally felt like they had actually settled in the estate. That feeling made them very excited.
There was only one thing they had always wanted. It was to be recognized and accepted.
They were those who returned with the young lord who was discriminated against on the battlefield. They had helped him fight in war, so they wanted to be recognized.
However, the knights who remained in the manor disregarded their hard work and secretly discriminated against them for being a mercenary.
They thought that the duchess would be the same.
However, Elody actually acknowledged their hardships and treated them fairly from the very beginning.
"This sounds like a nice place to live."
The knights ended their conversation with a smile on their faces. They thought that it was a good idea to have come here to the duchy with the leader.
Elody set aside her harvest and headed for the river, carrying the fruit punch she had left for Caville and Brien.
"Madame! I'll hold it for you. Why are you carrying these things by yourself?" Heinz said as he took the tray from her hands.
Elody purposely brought it herself to let the maids take a break…
Heinz smiled and looked at Elody.
"Why don't you use the medicine I gave you for the wound on your forehead?"
"This is my proof of loyalty to you, madame," He answered proudly.
Did he pick up the wrong medicine somewhere and ate it?
'I heard that when a person suddenly changes, it's likely that they are dying… '
"I'm sorry, but you should give that loyalty to Caville. He deserves it more than me."
Elody again took the tray from Heinz and headed to Caville.
But after a few steps, Elody stopped her tracks.
"My wife!"
Caville smiled when seeing his loved one in front of him.
The bucket by the river was already full of big fishes.
'Why is he naked?'
Elody bit her lips as she looked at Caville's naked torso.
It was the second time she saw Caville's grown-up body.
But last time, she was too concerned about Caville's wounds, so she felt nothing when she saw him shirtless.
Elody, feeling her cheeks burning, turned around and handed the tray over to Heinz.
"Madame?"
She immediately ran away and headed for the forest again. Her cheeks, ears, and neck felt hot, it was as if she was burning from fire.
It was the moment when Ifrit's advice to show his abs finally worked.
However, Caville only blinked and looked at Elody's back without knowing what had just happened.
"Why is she running away from me…?"
Likewise, Heinz, who did not know a thing about love, was confused by Elody's bizarre actions.
Nonetheless, he disregarded it as an impulse and approached the duke who stood in the river.
"Leader, this fruit punch is made by the madame. It tastes amazing. The vice-commander should try it too!"
Caville ran out of the water with a gloomy expression.
"Gosh, it's hot."
Heinz took out a handkerchief from his bosom and wiped the sweat from his forehead.
And at that moment…
When Caville's low voice sounded a bit grumpy, Heinz raised his head with a dumbfounded look on his face.
"…L-leader?"
When Heinz saw Caville's terrifying expression, he got chills all over his body as if a bucket of ice had just fallen on top of his head.
Caville parted his mouth and spoke, "Why do you have that handkerchief?"
[T/N]: It was at this moment Heinz knew, he f***** up. LOL
[ED/N]: Caville is ready to not sleep for the rest of his life for Elody. This much should be expected. XD
Dandere General and his Lord
I Am Your Sister
Your Meaning
Age Progression, Alternate World, Antihero Protagonist, Child Protagonist, Female Protagonist, Handsome Male Lead, Romance, Royalty, Transmigration, Villainess Noble Girls


I'm neither a millionaire nor an expat but live a quite average life with my family, job, and cleaning my kitchen … what means I really have to work for my flight tickets! But I'm always on the go to leave once getting the time and opportunity.
I've traveled independently all along, mostly backpacking. When heading for my next destination, I barely have a clue about how and where to spend the next days or weeks. Because I pretty much love the unforeseen, the unexpected. I like to drift, to let things happen.
Where exactly I'm driftin' to and what happens (to me) there – that's what is told on this website.
There was a time of travel reviews, compiled as word files and spread via email. With only few pics in order to keep the attachment thus loading times low … ;) What has begun just as a sign from abroad for family and friends spread itself out after all. It was astonishing so see who else knew about where we've been and what we were doing, begging to be added to the next tour mailing list directly.
For you my lovelies, and all those we've met (and will meet) on tour, and of course for all others who are interested in – for you all is this website, mostly the blog posts. Here you find im- and expressions related to my travels as ever.
All those who prefer to travel autonomously and off the beaten paths know how inspiring trip-related tips from the online community are. Apart from the fact that it is hard sometimes to get necessary details, most of us (me as well) more and more stand back from carrying hardcopy travel guides – what makes the www the exclusive and preferred source of information.
Also for you, dear globetrotters, is this website – as on my tour details page I publish massed hard facts and figures related to my/our journeys. Depending on the tour type it may involve route plannings, prices, tips for accommodation and other specific information. And I guess this particular information is what distinguishes sauerkraut adventures from other travel blogs.
But as a main intention sauerkraut adventures is to be my tribute to the beauty and diversity of this planet. To all its wonderful people that have crossed my paths and guided and inspired me, touched me deeply occasionally. This all and you all are my motivation to keep on traveling.
Enjoy reading, get inspired! There is a whole world out there, so stop not traveling!


Your private tour begins with pickup from your hotel. Your professional guide will build excitement in your group as he/she describes the adventure you are embarking on. Our guides are passionate about and knowledgeable of the Florida Keys and you will love the fun facts you learn on your way to Key Largo. When we arrive in Key Largo, a tropical forest sets the mood for our destination and the incredible snorkeling found within its boundaries. The shallow waters of John Pennekamp State Park, also known as the National Marine Sanctuary, have an abundance of soft and hard corals and is loaded with tropical fish. There are more than a dozen different reefs to choose from and the reefs only average 6 -12 feet deep. Everyone in your group will be able to enjoy the underwater beauty found in the Park.
After the 2.5 hour snorkeling adventure in Pennekamp State Park, we take you to one of the most fun places to eat in the Florida Keys….Alabama Jacks. Alabama Jacks is the perfect embodiment of a Florida Keys Restaurant and bar, and their food is so good that the Food Network has featured it multiple times on its tv shows. Eating and drinking at Alabama Jacks is the perfect way to relax and wind down after a day in the water. You will love it!


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Product No. 9308A, AquaGuard (Model 7500).
Intertap Quarter Turn Drinking Water Tap with Ceramic Disc Valve for easy, smooth lever operation and greater reliability (as pictured above).
This high quality filter system is a 1 micron block carbon filter to reduce sediment, chlorine, Volatile Organic Compounds (VOC's) like pesticides and insecticides, heavy metals such as lead and cysts (microscopic single cell parasites) including Giardia and Cryptosporidium.
1850 Litre Capacity per cartridge.
Delicious tasting water: Produces clear water ideal for preparing food; perfect for cooking pasta, washing and preparing fruits and vegetables.
Independently certified filter: Unlike many other Carbon filters, ours have been independently tested and certified to NSF International Standard 42 to deliver continuous water quality performance throughout the lifetime of the cartridge.
Simple and easy to install: Fits easily and neatly under the kitchen sink.
Efficient and economical to run: Does not require electricity.
Peace of mind: Each filter is fitted with a MacGuard™ which prevents the user from drinking unfiltered water. This unique and patented devise measures the amount of water passing through each filter and stops the flow of water once the filter cartridge is exhausted.
Easy filter metering system: Fitted with a PureMometer® which shows you how much filter life remains so you know when it's time to obtain a replacement cartridge.
12 month warranty: On parts excluding filter cartridge and labour.