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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?

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    $\begingroup$ 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? $\endgroup$
    – Matt F.
    May 15, 2021 at 11:49
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    $\begingroup$ ... every connected subset of $[0,1]^2$ with a full projection $\endgroup$
    – Nik Weaver
    May 15, 2021 at 14:04
  • $\begingroup$ Related: en.wikipedia.org/wiki/Inductive_dimension $\endgroup$ May 15, 2021 at 14:43
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    $\begingroup$ 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. $\endgroup$ May 15, 2021 at 17:23
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    $\begingroup$ 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. $\endgroup$
    – erz
    May 16, 2021 at 18:40

2 Answers 2

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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.

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    $\begingroup$ My God, that is so much easier than I expected! $\endgroup$
    – Nik Weaver
    May 19, 2021 at 22:42
  • $\begingroup$ Excellent -- but shoudn't the credits go to @Dejan Govc ? $\endgroup$ May 20, 2021 at 6:52
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    $\begingroup$ @JochenWengenroth Perhaps I should award a new bounty for his answer to another question? $\endgroup$
    – Jim Belk
    May 20, 2021 at 9:58
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    $\begingroup$ I don't know, my comment was just a spontaneous idea. Finding a relevant source is a major point for many MO answers, so that you also deserve the credits. $\endgroup$ May 20, 2021 at 10:06
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    $\begingroup$ The answer here clearly credits Govc. I think that's sufficient. $\endgroup$
    – Nik Weaver
    May 21, 2021 at 21:26
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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 $[0,1]$.

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  • $\begingroup$ Thank you for your answer! Is the name of the book you mentioned Elements of the topology of plane sets of points? I can not find Theorem 14.3. The book has only seven chapters. $\endgroup$ May 22, 2021 at 23:38
  • $\begingroup$ @mathstackexchange31415926 yes, sorry, I didn't notice that it has a strange numeration. It's Theorem 14.3 in chapter V $\endgroup$
    – erz
    May 22, 2021 at 23:48
  • $\begingroup$ Thanks a lot! I have found the theorem! $\endgroup$ May 23, 2021 at 0:41
  • $\begingroup$ I am sorry! I have deleted the updated question and start a new question here (mathoverflow.net/questions/393520/…). Please remove your answer to the new question, so I can accept your answer! Thanks again! $\endgroup$ May 23, 2021 at 13:32
  • $\begingroup$ @mathstackexchange31415926 i don't think this is necessary $\endgroup$
    – erz
    May 23, 2021 at 19:01

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