TITLE: 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 [11 upvotes]: 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?

REPLY [3 votes]: 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 $F=X\cap D$.