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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?

mathoverflow12345
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    Shorter version: If $S \subset [0,1]^2$ intersects every 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? –  May 15 '21 at 11:49
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    ... every connected subset of $[0,1]^2$ with a full projection – Nik Weaver May 15 '21 at 14:04
  • Related: https://en.wikipedia.org/wiki/Inductive_dimension – Gerald Edgar May 15 '21 at 14:43
  • @mathstackexchange31415926 I made that comment above. I'd suggest editing your question to put it in Matt F.'s form (corrected) --- otherwise people have to go to the comments to see the "easy" version. – Nik Weaver May 15 '21 at 15:24
  • @Matt F Thanks for your shorter form. – mathoverflow12345 May 15 '21 at 16:34
  • @Nik Weaver Thanks for your suggestion. – mathoverflow12345 May 15 '21 at 16:34
  • You're welcome, good question BTW. – Nik Weaver May 15 '21 at 16:54
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    I parsed the grammar wrong at first. I think the hypothesis is "For every connected subset $C$ which has a full projection (i.e. $\pi_x(C) = [0,1]$), we have $S \cap C \ne \emptyset$". Right? I first thought it was saying that the intersection should have full projection. – Nate Eldredge May 15 '21 at 17:23
  • @Nate Eldredge Yes, your later thoughts are right. – mathoverflow12345 May 15 '21 at 17:55
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    Another equivalent statement: if $S$ separates the left and right sides of the square, then it connects the top and the bottom sides. It seems that if $S$ was closed, then there would be a component of $S$ that separated the sides. It then is easy to see that this component intersects both top and the bottom. – erz May 16 '21 at 18:40
  • FWIW I thought the original form of the question was less ambiguous, this new form I read as requiring $\pi_1(S \cap A) = [0,1]$ for every connected $A$ and had to pull up the edit history. – Ville Salo May 17 '21 at 05:18
  • About your additional question: usually, it is better to start a new question. – Anton Petrunin May 22 '21 at 23:18
  • @Anton Petrunin: Thank you for your reminding. I have deleted the new question and start a new question. – mathoverflow12345 May 23 '21 at 13:13
  • Too late --- you get an answer already. (I wrote it for the future.) – Anton Petrunin May 24 '21 at 04:03

2 Answers2

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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.

Timothy Chow
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Jim Belk
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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 $[0,1]$.

erz
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