## Wanted: chain of nowhere dense subsets of the real line whose union is nonmeagre, or even contains intervals

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Let $X$ be a topological space. When I call a set nowhere dense, meagre or similar without qualification, I mean that it has this property as a subset of $X$. Call a subset of $X$ weager (for weakly meagre) if it is the union of a chain (wrt containment) of nowhere dense sets. Using that finite unions of nowhere dense sets are nowhere dense, it is easy to see that meagre implies weager. Call $X$ an Astaire space (for a stronger Baire space) if every weagre subset of $X$ has empty interior. Obviously Astaire implies Baire. Two natural, but rather silly (not just because of the terminology) questions are:

Does weagre imply meagre? If not, does Baire imply Astaire?

Unsurprisingly, the 2nd (and hence also the 1st) question has a negative answer. Let $X$ be uncountable. In fact, for convenience, take $X$ to be the well-ordered set of all countable ordinals. Topologize $X$ by putting open all sets which are either empty or have countable complement. Then $X$ is a Baire space - in fact the notions countable; closed and not $X$; nowhere dense; and meagre all coincide for subsets of $X$. However, $X$ is the union of the chain of all its countable initial segments so $X$ is not an Astaire space.

The above example is somewhat unsatisfactory since the space is far from Hausdorff, but the ease with which it arose made me wonder whether my question had a positive answer even when $X = \mathbb{R}$. Adapting my example, it is at least possible to express an uncountable subset of $\mathbb{R}$ as the union of a chain of countable subsets of $\mathbb{R}$ but this is quite unhelpful because, in this context, there is no guarantee that countable implies nowhere dense, or that uncountable implies nonempty interior (or even nonmeagre for that matter). So that I don't spend too much more time today thinking about things I know nothing about and/or dreaming up silly names for concepts that probably already have much more respectable names - I pose to you the following question:

Is the real line an Astaire space? If not, are there at least weagre subsets of $\mathbb{R}$ which are not meagre?

Or, in plain English for those of you who only skimmed this nonsense:

Does there exist a chain of nowhere dense subsets of $\mathbb{R}$ whose union has nonempty interior? If not, is there such a chain whose union is not meagre?

Thank you, Michael.

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1
Andres, I do not see how this resolves my question. Although your set $T_\alpha$ is the union of a chain of sets with empty interior, this does not imply it is the union of a chain of nowhere dense sets. The rationals could very well precede $T_\alpha$ in your ordering but are certainly not nowhere dense. – Michael Jan 17 2011 at 23:20
Assuming CH, the real line is a union of a chain of finite sets, since this is true for the first uncountable ordinal. So the answer is yes in that case. – George Lowther Jan 18 2011 at 0:08
George, I think you mean to say that under CH, the real line is the union of a chain of countable sets. (Every union of a chain of finite sets can be seen to be countable.) – Joel David Hamkins Jan 18 2011 at 0:15
@Joel: yes, sorry, that was a dumb mistake. Not sure about my second comment now – George Lowther Jan 18 2011 at 0:18

## 2 Answers

Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union contains an interval.

Proof. Suppose there was such a chain $\{\ B_i \mid i\in I\ \}$, where $\langle I,\lt\rangle$ is a linear order and $i\lt j$ implies $B_i\subset B_j$. First, I claim that this chain cannot have countable cofinality, since then we could find a countable cofinal subset of $I$, and the union of the $B_j$ from this cofinal subset would also contain an interval, violating the Baire category theorem. So every countable subset of $I$ is bounded. In this case, consider the set $Q$ of rational numbers $q$ in the interval from the union $\bigcup_i B_i$. Each of them appears in some $B_{i_q}$, and the set of all $i_q$ for $q\in Q$ is a countable set and hence bounded in $I$. Thus, there is some $j\in I$ beyond all $i_q$. So $B_j$ contains all those $q$ and thus is not nowhere dense. QED

Edit. As George pointed out in the comments, essentially the same argument establishes the full property:

Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union is non-meager.

Proof. Suppose that $B_i$ for $i\in I$ is a chain of nowhere dense sets whose union $\bigcup_i B_i$ is non-meager. Again, we see that every countable subset of $I$ must be bounded, for otherwise the union would be a countable union of nowhere dense sets and hence meager. Since $\bigcup_i B_i$ is non-meager, it must be dense on an interval, and so it must have a countable subset $Q$ that is also dense on an interval. By the argument above, since $I$ has uncountable cofinality, this set $Q$ must be in some $B_j$ for large enough $j\in I$, contradicting that $B_j$ is nowhere dense. QED

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Oh, that's nice! – Andres Caicedo Jan 18 2011 at 0:18
Yes! This is great, thank you. – Michael Jan 18 2011 at 0:23
Nice. Doesn't this give a full answer to the question? That is, the union of a chain of nowhere dense sets is always meagre? I.e., suppose it isn't, so it's union is not nowhere dense, and replace $\mathbb{Q}$ in your argument by any countable dense subset of the union. – George Lowther Jan 18 2011 at 0:27
Yes, George, you are right! I will edit. – Joel David Hamkins Jan 18 2011 at 0:52
@Joel: Your nice argument is very general. What can you say if the sets are meager (rather than nowhere dense)? – Andres Caicedo Jan 18 2011 at 1:55
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(Joel's answer appeared as I was typing this.)

I think the answer is no.

Suppose to the contrary there exists a nonmeager set $A \subset \mathbb{R}$ which is the union of some chain $\{K_i\}_{i \in I}$ of nowhere dense sets. $A$ is separable, so we may enumerate a countable dense set $\{x_n\} \subset A$. Then we can find an increasing sequence $\{K_{i_n}\}$ with $x_n \in K_{i_n}$. Set $K = \bigcup_n K_{i_n}$. Since $K$ is meager $K \ne A$, so there exists $x \in A \backslash K$. Now there must be some $K_j$ with $x \in K_j$. Now for each $n$ we certainly don't have $K_j \subset K_{i_n}$, so we must have $K_{i_n} \subset K_j$ since the $K_i$ are a chain. Thus $K \subset K_j$, but then $K_j$ contains all the $x_n$ and so is not nowhere dense.

Added: This indeed shows that a second countable Baire space cannot be the union of a chain of nowhere dense subsets. Here is a stab at a counterexample in the non-second countable case.

Consider the non-separable complete metric space $\ell^\infty = \ell^\infty(\mathbb{N})$. I claim its Hamel dimension $\dim \ell^\infty$ is $\mathfrak{c}$. First, we have the natural inclusion $\ell^1 \subset \ell^\infty$; $\ell^1$ is a separable Banach space, so it is known that $\ell^1$ has Hamel dimension $\mathfrak{c}$, and thus $\dim \ell^\infty \ge \mathfrak{c}$. On the other hand, $\ell^\infty$ is the continuous dual of $\ell^1$, and thus is naturally included into the algebraic dual of $\ell^1$, which must also have Hamel dimension $\mathfrak{c}$; thus $\dim \ell^\infty \le \mathfrak{c}$. By Schroeder-Bernstein, $\dim \ell^\infty = \mathfrak{c}$.

Now suppose we believe the continuum hypothesis $\mathfrak{c} = \aleph_1$. Pick a Hamel basis $B$ for $\ell^\infty$; since it is in bijection with the least uncountable ordinal, we can well-order it in such a way that for any $x \in B$, $B_x = \{y \in B : y < x\}$ is countable. Note $B$ has no greatest element, so $\bigcup_{x \in B} B_x = B$. Let $E_x = \mathrm{span } B_x$; clearly ${E_x}$ is a chain, and $\bigcup_{x \in B} E_x = \ell^\infty$. But each $E_x$ has countable Hamel dimension and therefore is separable, so it must be nowhere dense in $\ell^\infty$.

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Nice answer too, along the same lines as Joel's, and works for any separable space with the Baire property. Interesting. – George Lowther Jan 18 2011 at 0:44
...or should that just be "any second countable space with the Baire property"? – George Lowther Jan 18 2011 at 0:47
Yeah, second countable looks like a better condition than separable, since a subset of a separable space need not in general be separable. – Nate Eldredge Jan 18 2011 at 0:59
I'll reiterate while convincing myself. If $A$ is nonmeagre then, by definition of meagre rather than the Baire Category theorem, any countable subchain is bounded. Certainly $A$ is not nowhere dense so its closure contains an interval. $\mathbb{R}$ is 2nd countable, so $A$ is 2nd countable and hence separable (I'm not sure that separability always passes to subspaces directly). Let $C \subset A$ be countable and dense in $A$. $C$ is in the union of a countable subchain, so one of our supposedly nowhere dense $K_i$ contains $C$ - contradiction since $C$ is not nowhere dense... Looks good! – Michael Jan 18 2011 at 1:00
So, I suppose one concludes that, in a 2nd countable space, a set is meagre if and only if it is the union of some (possibly uncountable) chain of nowhere dense sets? – Michael Jan 18 2011 at 1:06
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