A question about connected subsets of $[0,1]^2$

Question

If $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis, must $S$ have a connected component with a full projection on the $y$-axis?

An equivalent form：

If $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis and $T⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $y$-axis, must $S\cap T\neq \emptyset$?

The motivation of this question:

The question came to me when I thought about the Brouwer fixed-point theorem:

Let $f=(f_1,f_2)$ be a continuous function mappping $[0,1]^2$ to itself. Then $$S\triangleq\{(x,y)\in[0,1]^2:f_1(x,y)=x\}$$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis and $$T\triangleq\{(x,y)\in[0,1]^2:f_2(x,y)=y\}$$ intersects every connected subset of $[0,1]^2$ with a full projection on the $y$-axis.

My further question:

If we assume that $S\subset [0,1]^2$ is a close set, what is the answer to my question, that is, if a close set $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis, must $S$ have a connected component with a full projection on the $y$-axis?

Answer 1

A counterexample to this statement was posted as a comment by Dejan Govc to the Math StackExchange question, Do partitions of a square into two sets always connect one pair of opposite edges?.

For $0 < r < \tfrac{1}{2}$, let $S_r$ be the boundary of the square $\bigl[\tfrac{1}{2}-r,\tfrac{1}{2}+r\bigr]\times \bigl[\tfrac{1}{2}-r,\tfrac{1}{2}+r\bigr]$, and let $$ S = \{(0,0),(1,0),(0,1),(1,1)\} \;\;\cup \bigcup_{r\in \mathbb{Q}\cap (0,1/2)} S_r. $$ Note that no connected component of $[0,1]^2\setminus S$ has full projection onto the $x$-axis, and therefore any connected subset of $[0,1]^2$ with full projection onto the $x$-axis must intersect $S$. However, no connected component of $S$ has full projection onto the $y$-axis.

Answer 2

This is an answer to the updated question.

Proposition: If a closed $S\subset [0,1]\times [0,1]$ intersects every connected set with a full projection onto the $x$-axis, then it has a component with a full projection onto the $y$-axis.

First, without loss of generality we may assume that $S$ does not intersect the left and the right sides of the square (otherwise, consider the same problem but for $[-1,2]\times [0,1]$. Let $\Pi=(0,1)\times (0,1)$, which is homeomorphic to the full plane.

Pick a point on the left side take a disk $D$ around this point that does not intersect $S$. Take $x$ in $D\cap \Pi$. Do the same on the right side and get $E$ and $y$. Let $S'=S\cap \Pi$, which is closed in $\Pi$.

Now $S'$ separates $x$ and $y$ within $\Pi$ (meaning any connected set in $\Pi$ that contains $x$ and $y$ has to intersect $S'$. Indeed, if a connected set $F\subset \Pi$ contains $x$ and $y$, then $F\cup D\cup E$ has a full projection onto the $x$-axis, and so has to intersect $S$, but since $D$ and $E$ do not, it follows that $F\cap S'=F\cap S\ne\varnothing$ (the first equality follows from $F\subset\Pi$).

Since $S'$ is closed in $\Pi$, which is homeomorphic to the plane, by a Theorem V.14.3 in the book Newman - Elements of topology of planar sets of points, there is a component $C$ of $S'$ that separates $x$ and $y$ in $\Pi$. Clearly, $C$ has a full projection on the $y$-axis (and so it is $(0,1)$), since otherwise we could sneak in a horizontal segment between the left and right sides, which would contradict the separation.

Since $S$ is closed, $\overline{C}$ is a connected compact subset of $S$. Hence, $C$ has a compact projection onto $y$ axis, and so this projection is $[0,1]$.