## If Q is a subset of the plane of size less than continuum, then does every closed F in Q extend to a closed connected G in the plane with the same trace on Q? (Or is this independent of ZFC?)

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This question arises in connection with this MO question and especially with Sergei Ivanov's wonderful answer, which showed that for any countable set $Q\subset\mathbb{R}^2$ and every closed set $F\subset Q$, there is a closed connected $G\subset\mathbb{R}^2$ with $G\cap Q=F$. (In fact, he makes $G$ path-connected.)

My question is about the extent to which this phenomenon might generalize to higher cardinals, when the Continuum Hypothesis fails. For example, if the continuum $2^\omega$ is very large, then can we hope to handle uncountable sets $Q$ in the way Sergei handled the countable sets, provided that they have size less than the continuum? Or perhaps the best possible is always just the countable sets? Or is this independent of ZFC?

It seems sensible to introduce what seems to be a new cardinal characteristic here. Specifically, let $\kappa$ be the size of the smallest counterexample, that is, the smallest cardinal size of a set $Q\subset\mathbb{R}^2$ having a closed subset $F\subset Q$ for which there is no closed connected $G\subset\mathbb{R}^2$ with $G\cap Q=F$.

Sergei proved that this cardinal $\kappa$ is uncountable, and obviously $\kappa$ is at most the continuum (it is easy to make counterexamples of size continuum), and so $$\omega_1\leq\kappa\leq 2^\omega.$$ So the question is, what can we say about $\kappa$ in ZFC?

If the Continuum Hypothesis holds, of course, then the two endpoints above are identical and so $\kappa=2^\omega$. But is it consistent with ZFC that $$\omega_1\leq \kappa\lt 2^\omega?$$ Perhaps one can achieve particular values of $\kappa$ by forcing? Is the cardinal $\kappa$ related to other well-known cardinal characteristics? Perhaps the value of $\kappa$ is pushed up to the continuum $2^\omega$ by some of the standard forcing axioms?

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For clarity, when I refer to a closed set $F\subset Q$, what I mean is that $F$ is closed in the subspace $Q$; equivalently, $\bar F\cap Q=F$. So the issue  is whether every $\bar F$ extends to a closed  connected $G$ with the same trace on $Q$. – Joel David Hamkins Nov 17 2010 at 1:15

## 4 Answers

It seems to me that Sergei Ivanov's proof can be generalized to show that Martin's Axiom for countable posets implies that $\kappa$ is the continuum. By the characterization of $cov(\mathcal{M})$ (the smallest number of meager sets required to cover the real line) as the smallest cardinal for which MA(countable) fails, it follows that $cov(\mathcal{M})\leq\kappa$.

Given closed $F\subseteq Q$ both of size less than continuum we define a countable poset $\mathbb{P}$ as follows. First fix a countable collection $\mathcal{U}$ of open balls so that for any rational $q\in\mathbb{Q}^2$ and any rational $\epsilon_1<\epsilon_2$ there is $U\in\mathcal{U}$ centered at $q$ with some radius $\epsilon$ such that $\epsilon_1<\epsilon<\epsilon_2$ and the boundary of $U$ is disjoint from $Q$. (We can do this because $Q$ has size less than continuum and there are continuum many choices for $\epsilon$).

Now let $\mathbb{P}$ be the collection of finite disjoint unions of members of $\mathcal{U}$ which are disjoint from $F$. Then $\mathbb{P}$ is countable. For each $x\in X$, the set $D_x$ of $p$ with $x$ in $p$ is dense; we prove this as follows. Let $p\in\mathbb{P}$ and $x\in Q\setminus F$ be given. Take $\delta$ so that the $\delta$-ball $O$ around $x$ is disjoint from $F$ and $p$ (possible because $x$ doesn't lie on the boundary of any of the balls comprising $p$). Pick a rational $q$ within $\delta/3$ of $x$. Then there is a member $U$ of $\mathcal{U}$ insides $O$ and containing $x$. So $q=p\cup U$ belongs to $D_x$.

Now if $G$ is a filter intersecting each $D_x$, then $\cup G$ is a collection of pairwise disjoint balls disjoint from $F$ and containing every member of $Q\setminus F$. Let $C$ be the complement of the union. Then $C$ is as desired; it is path connected by exactly Sergei Ivanov's argument (some people seemed concerned about the radii of the balls but it doesn't appear to matter).

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Fantastic! I was playing around with much larger partial orders having the same goal, but you seem to have found the right way to do it. Since all nontrivial countable forcing is isomorphic to adding a Cohen real, this is a very weak forcing axiom. So this shows that it is consistent with ZFC that the continuum is large and Sergei's fact still holds for all sets of size less than the continuum. Now we need the converse result, where $\kappa$ is small and the continuum is large... – Joel David Hamkins Nov 17 2010 at 10:20

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Congratulations! I was trying to do this with balls centered at the points of $Q-F$, with radii in appropriate countable sets. But not only isn't that poset countable, it doesn't even seem to be ccc, so not even the full MA would complete the proof. – Andreas Blass Nov 17 2010 at 14:41

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Andreas, I was doing the same thing with the same result---it is definitely not ccc if Q-F is large. I think Justin's answer is really great, and deserves many up-votes. – Joel David Hamkins Nov 17 2010 at 15:05

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I suspect that Sergei Ivanov's proof can be extended to the case when $\mathbb R^2 \backslash Q$ is connected. I also suspect that the only case when $\mathbb R^2 \backslash Q$ is disconnected is precisely when $Q$ is a continuum.

(Consider this "answer" a comment, but with the details filled in, it would imply that $\kappa = 2^\omega$.)

Edit: I'm not so sure about the second point as the first: What is the cardinality of the smallest set $Q$ such that $\mathbb R^2 \backslash Q$ is disconnected?

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It seems that Sergei's argument used the countablility of $Q$, so I'm not clear on your proposal. (For your question in the edit, any set of size less than continuum has path-connected complement, since there is a continuous foliation of disjoint paths from $a$ to $b$, and one of them must be OK.) – Joel David Hamkins Nov 16 2010 at 22:48

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Your second point is correct; removing fewer than continuum many points will leave the rest pathwise connected (by polygonal paths consisting of just two line segments). I'm not at all convinced by your first point, because Sergei Ivanov's proof depended on being able to choose disks (around the omitted points) that stay away from previously chosen disks, of which there were only finitely many. It will be much harder when there are infinitely many. Note also that, when there are uncountably many of these disks, uncountably many of them will have radii bounded away from zero. – Andreas Blass Nov 16 2010 at 22:51
The way I see the proof is that the countability and the covering of $Q \backslash F$ with disks was merely a means to avoid the "bad" points in $Q$ when path-connecting the points in $F$. Perhaps this works: For any $Q$ that has cardinality less than a continuum $\mathbb R^2 \backslash (Q \backslash F)$ is path connected. Let $G$ be a path in this set that passes through every point in $F$. Conveniently, paths are closed, so $F$ is closed. – trutheality Nov 16 2010 at 23:12
* I meant $G$ is closed. – trutheality Nov 16 2010 at 23:13
@trutheality: If $Q=F=\mathbb{Q}^2$ then how do you construct a closed path which passes through every point of $\mathbb{Q}^2$? – Guillaume Brunerie Nov 16 2010 at 23:49
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I think you do not use the fact, that there is no point on the boundary of any of the balls. You only use the fact that $\kappa$ is less than continuum in the definition of MA. (actually the generic covering will cover everything from $G$ \ $F$ and will be disjoint from $F$)

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The no-points-on-the-boundary property is used in showing that the complement is path-connected, since you draw the straight line, and then whenever it is obstructed by a circle, you follow the boundary of the circle around the obstruction; and it is also used to know that the circles do not overlap, since otherwise you'd want to place another circle covering the boundary point. – Joel David Hamkins Nov 17 2010 at 15:08

Yes, partly you are right, I still think it is not a problem if there is an $F$ point on the boundary, since not $G$ \ $Q$ is connected, just $G$. (or can we make $G$ \ $Q$ connected? probably.) But it is a problem if there are some $Q$ \ $F$ points on the boundary.

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