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?)

Question

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?

Answer 1

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).

Answer 2

In fact, the MO user @trutheality was right saying that the cardinal $\kappa$ is equal to the continuum. This follows from

Theorem. For any set $X\subset\mathbb R^d$ of cardinality $<\mathfrak c$ in the Euclidean space of dimension $d\ge 2$ and any closed subset $F\subset X$ there exists a closed path-connected set $P\subset \mathbb R^n$ such that $P\cap X=F$.

Proof. First we prove a partial case of this theorem for $X$ contained in the power $\mathbb I^d$ of the set $\mathbb I:=\mathbb R\setminus\mathbb Q$ of irrational numbers.

Lemma. For any subset $X\subset\mathbb I^d$ there exists a closed path-connected set $P\subset \mathbb R^d$ such that $P\cap X=F$.

Proof. Consider the open cover $$\mathcal C:=\{x+(0,1)^d:x\in\mathbb Z^d\}$$ of $\mathbb I^d$ by integer translations of the open unit cube $(0,1)^d$.

Next, for every $n\in\omega$ consider the homothetic copy $\mathcal C_n:=\{\tfrac1{2^n}C:C\in\mathcal C\}$ of the cover $\mathcal C_n$ and observe that each point $x\in \mathbb I^n$ is contained in a unique cube $C_n(x)\in\mathcal C_n$. Observe also that $(C_n(x))_{n\in\omega}$ is a decreasing neighborhood base at $x$.

Now fix any closed set $F\subset X$ and for every $x\in X\setminus F$ find the smallest number $n_x\in\omega$ such that $C_{n_x}(x)\cap F=\emptyset$. Observe that for any points $x,y\in\mathbb I^d\setminus X$ the cubes $C_{n_x}(x)$ and $C_{n_y}(y)$ are either disjoint or coincide. Indeed, if $C_{n_x}(x)$ intersects $C_{n_y}(y)$ and $n_x\le n_y$, then $C_{n_y}(y)\subset C_{n_x}(x)$ and then $C_{n_x}(y)=C_{n_y}(y)$ does not intersect $F$, which implies that $n_x=n_y$ by the minimality of $n_y$.

Then $E:=\mathbb R_d\setminus \bigcup_{x\in X}C_{n_x}(x)$ is a closed path-connected set in $\mathbb R^d$ such that $E\cap X=F$. The path-connectedness of $E$ can be proved exactly as in the answer of Sergei Ivanov (using the fact that the cover $\{C_{n_x}(x):x\in X\setminus F\}$ is disjoint and consists of open cubes with path-connected boundary in $\mathbb R^d$).$\square$

Now we consider the general case. Given any subset $X\subset \mathbb R^d$ of cardinality $<\mathfrak c$, we can find a subset $Y\subset\mathbb R$ of cardinality $<\mathfrak c$ such that $X\subset Y^n$. Choose any countable dense set $Q\subset \mathbb R$ such that $Q\cap Y=\emptyset$. By the countable dense homogeneity of $\mathbb R$, there exists a homeomorphism $h:\mathbb R\to\mathbb R$ such that $h(Q)=\mathbb Q$ and hence $h(Y)\subset h(\mathbb R\setminus Q)=\mathbb R\setminus\mathbb Q=\mathbb I$.

The homeomorphism $h$ induces a homeomorphism $g:\mathbb R^n\to\mathbb R^n$, $g:(x_i)_{i=1}^n\mapsto (h(x_i)_{i=1}^n)$, such that $g(X)\subset g(Y^n)=h(Y)^n\subset\mathbb I^n$. Given any closed set $F\subset X$, consider the its image $g(F)\subset g(X)\subset\mathbb I^d$ and using Lemma, find a closed path-connected set $E\subset\mathbb R^d$ such that $g(F)=g(X)\cap E$. Then $D:=g^{-1}(E)$ is a closed path-connected subset of $\mathbb R^d$ such that $F=X\cap D$.

Answer 3

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?