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Consider the following family of sets $$ \begin{align*} \mathcal{F} = \{X\subseteq [0,1]\times [0,1] \mid \ &X \text{ is closed and }\\& \forall x \in \pi_0 (X) (\{y \in [0,1] \mid (x,y) \in X\} \text{ is perfect}) \text{ and}\\& \forall y \in \pi_1 (X) (\{x \in [0,1] \mid (x,y) \in X\} \text{ is perfect})\} \end{align*} $$

where $\pi_0$ and $\pi_1$ are the canonical projections. My question is the following:

  • Given any nonempty $X \in \mathcal{F}$ are there two nonempty perfect $C_0,C_1 \subseteq [0,1]$ such that $C_0 \times C_1 \subseteq X$?

Instead of $[0,1]$ I could have put any other nonempty perfect Polish space. I tried to define a counterexample via a construction à la Cantor (Cantor sets, Sierpiński carpets etc.), but without success. Any idea? Do you know if these sets have been studied somewhere?
Thanks

EDIT1: Here is a variation of the previous question:

  • Given any nonempty $X\in\mathcal{F}$ are there two $C_0,C_1\subseteq [0,1]$ such that both $C_0$ and $C_1$ have at least two elements and $C_0\times C_1 \subseteq X$?

EDIT2: I'm starting to believe that the answer to both questions may be positive and that Ramsey theory may have something to do with it, but at this moment I cannot make this feeling precise.

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    $\begingroup$ @JoelDavidHamkins yes, I edited so to disambiguate. $\endgroup$
    – Lorenzo
    Nov 16 at 14:23
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    $\begingroup$ What happens if you take the Cantor set $C$ and consider $C\times C$, but then skewed, so that the $x$th vertical slice is shifted up by $x$? That is, $\{(x,y+x)\mid x,y\in C\}$. $\endgroup$
    – Joel David Hamkins
    Nov 16 at 14:28
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    $\begingroup$ @JoelDavidHamkins in that case the origin would belong to the set, but the horizonal slice at the origin would contain only one element (which is $0$). $\endgroup$
    – Lorenzo
    Nov 16 at 14:36
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    $\begingroup$ Ah, yes. Sorry to be so useless here...It is an interesting question. (But what if we shift it mod 1, so that it wraps around? Not sure whether this still satisfies your condition on sections.) $\endgroup$
    – Joel David Hamkins
    Nov 16 at 14:40
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    $\begingroup$ Ok, I've thought about it, and the problem in this case is that the point $(0,2/3)$ would be isolated along its horizontal slice. $\endgroup$
    – Lorenzo
    Nov 19 at 8:47

1 Answer 1

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The answer to both questions is negative.

A subset $S$ of a abelian group $G$ is called a Sidon set if $a,b,c,d\in S$ and $a-b=c-d$ implies $a=b$ or $a=c.$ You can find Sidon sets of any finite cardinality, if you are allowed to pick the finite group $G.$ An example of a Sidon set of cardinality $n$ is the set of elements of $\mathbb F_3^n$ with exactly one component $1$ and all other components $0.$

Starting with $G_0$ being the trivial group, pick a sequence of finite abelian groups $G_1,G_2,\dots$ along with injective maps of sets $c_n:2\times G_{n-1}\times G_{n-1}\to G_n$ for each $n\geq 1,$ such that the image of each $c_n$ is a Sidon set. It suffices to answer the question with $T=\prod_n G_n$ in place of $[0,1].$ Define $X$ to be the set of pairs $(x,y)$ such that for each $n\geq 1$ there exists $j\in\{0,1\}$ with $x_n+y_n=c_n(j,x_{n-1},y_{n-1}).$

For any $x\in T,$ the set $\{y:(x,y)\in T\}$ is homeomorphic to $\{0,1\}^{\mathbb N}$: given a sequence $j_1,j_2,\dots\in\{0,1\}$ we can recursively construct a $y$ by the rule $y_n=c_n(j_n,x_{n-1},y_{n-1})-x_n,$ and this construction gives every $y$ with $(x,y)\in X.$ By a symmetric argument, the same is true swapping the role of $x$ and $y.$ So $X\in\mathcal F.$

Consider $x,x',y,y'\in T$ with $x\neq x'$ and $y\neq y'$ and $(x,y),(x',y),(x,y'),(x',y')\in X.$ Let $n$ be large enough that $x_i\neq x'_i$ and $y_j\neq y'_j$ for some $i,j<n.$ Let $a=x_n+y_n$ and $b=x'_n+y_n$ and $c=x_n+y'_n$ and $d=x'_n+y'_n.$ These lie in a Sidon set, and $a-b=x_n-x'_n=c-d,$ so $a=b$ or $a=c.$ In the first case, using $(x,y),(x',y)\in X$ and injectivity of $c_n$ we get $(x_{n-1},y_{n-1})=(x'_{n-1},y_{n-1}),$ which implies $x_{n-1}+y_{n-1}=x'_{n-1}+y_{n-1}$ and hence $(x_{n-2},y_{n-2})=(x'_{n-2},y_{n-2})$ and so on, to give $x_i=x'_i$ for all $i<n.$ (This multiple use of injectivity isn't really necessary: we could have defined the functions $c_n$ to take all of $x_0,y_0,\dots,x_{n-1},y_{n-1}$ as arguments instead of $x_{n-1},y_{n-1}.$) The second case is similar: $y_i=y'_i$ for all $i<n.$ Either way, this contradicts the choice of $n.$

This means there is no rectangle $C_0\times C_1\subseteq X$ with $|C_0|,|C_1|\geq 2.$

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    $\begingroup$ This is very nice. Can you share how you came up with this solution? Are there similar uses of Sidon sets (I had never encountered them)? $\endgroup$
    – Iosif Pinelis
    Nov 22 at 2:53
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    $\begingroup$ (+1) Very nice; congrats!! Addressing Iosif Pinelis' comment, let me try to sketch simplifications of this proof: Take a Cantor set $C\subset(0,1)$ with $C\cup\{1\}$ linearly independent over $\mathbb Q$ (by Mycielski) and let $F=\{(x,y)\in[0,1]^2:(x+y)\bmod1\in C\}$, so if $F$ has a rectangle, then after $(x,y)\mapsto x+y$, the rectangle turns into a linearly dependent set over $\mathbb Q$, impossible. But $F\notin\mathcal F$ just barely: To fix, we swap the underlying space $[0,1]$ with an infinite profinite Polish group $G$, do the same construction in $G$, and then embed $G$ into $[0,1]$. $\endgroup$
    – Edward H
    Nov 22 at 8:44
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    $\begingroup$ Great, that's a far cleaner construction! @IosifPinelis: well, there's a much simpler argument now, but I was trying to modify the idea of taking something like $\{(x,y)\in(\mathbb Z/3)^{\mathbb N}:(\forall n)x_n+y_n\in\{0,1\}\},$ which doesn't work because you can pick perfect sets $C_0$ and $C_1$ that are non-zero on disjoint sets of indices. I don't think Sidon sets are really relevant - they are more for quantitative problems. $\endgroup$
    – Colin McQuillan
    Nov 22 at 9:26
  • $\begingroup$ @EdwardH It's not clear to me when you take the infinite profinite Polish group, why does it solve $F\not\in \mathcal{F}$? Could you please elaborate a bit more on this? Thanks to both you and Colin McQuillan for the very interesting answers! $\endgroup$
    – Lorenzo
    Nov 22 at 9:38
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    $\begingroup$ @Lorenzo Yes, sorry: My last comment was hasty and not completely thought-out, but these gaps can be closed: Take $G=\prod_{p\text{ prime}}\mathbb Z/p\mathbb Z$, and a Cantor set $C\subset G$ such that for all $c_1\ne\ldots\ne c_k\in C$ and $n_1,\ldots,n_k\in\mathbb Z\setminus\{0\}$, you have $\sum_{i=1}^nn_ic_i\ne0\in G$ by Mycielski. Let $F=\{(x,y)\in G^2:x+y\in C\}$, so that $F\subset G\times G$ has all your nice properties. Then fix a homeomorphism $f:G\to2^\mathbb N$, and $(f\times f)F\subset(2^\mathbb N)^2$ works; fix an embedding $g:G\to[0,1]$, and $(g\times g)F\subset[0,1]^2$ works. $\endgroup$
    – Edward H
    Nov 22 at 10:15

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