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

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?

• 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, 2021 at 11:49
• ... every connected subset of $[0,1]^2$ with a full projection May 15, 2021 at 14:04
• May 15, 2021 at 14:43
• 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. May 15, 2021 at 17:23
• 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, 2021 at 18:40

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.

• My God, that is so much easier than I expected! May 19, 2021 at 22:42
• Excellent -- but shoudn't the credits go to @Dejan Govc ? May 20, 2021 at 6:52
• @JochenWengenroth Perhaps I should award a new bounty for his answer to another question? May 20, 2021 at 9:58
• 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. May 20, 2021 at 10:06
• The answer here clearly credits Govc. I think that's sufficient. May 21, 2021 at 21:26

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 $$[0,1]$$.

• 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. May 22, 2021 at 23:38
• @mathstackexchange31415926 yes, sorry, I didn't notice that it has a strange numeration. It's Theorem 14.3 in chapter V
– erz
May 22, 2021 at 23:48
• Thanks a lot! I have found the theorem! May 23, 2021 at 0:41
• 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! May 23, 2021 at 13:32
• @mathstackexchange31415926 i don't think this is necessary
– erz
May 23, 2021 at 19:01