# A set of the plane recursively including “crosses”…

Definition: Call "cross with center in $(x,y) \in \mathbb R^2$" a set of $\mathbb R^2$ given by $(I_1(x)\times\{y\}) \cup (\{x\}\times I_2(y))$ where $I_1(x) \subseteq \mathbb R$ is a neighbourhood of $x$ and $I_2(y) \subseteq \mathbb R$ is a neighbourhood of $y$.

Problem: Let $A \subseteq \mathbb R^2$ be a set such that for any $z \in A$ there exists a cross with center in $z$ which is all included in $A$. Is it true that $A$ must include a nonempty open set?

(Warm up exercise: prove that $A$ can actually be not open.)

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Every set includes an open set, namely the empty set. What did you really mean to ask? A non-empty open set? – Amit Kumar Gupta Aug 10 '13 at 8:27
Also, is this homework? – Amit Kumar Gupta Aug 10 '13 at 8:28
For the warm up: Consider $A_0=\mathbb R^2\setminus \{\,(x,x)\mid x\ne0\,\}$. – Hagen von Eitzen Aug 10 '13 at 8:28
1) No this is not homework and I am not a student. This is an original problem which I invented. 2) Yes you are right, I should have said "non-empty", I'm going to correct. – Marco Disce Aug 10 '13 at 8:30

Let \begin{align}X&=\{\,(a+b\sqrt 2,a-b\sqrt 2)\mid a,b\in\mathbb Q\,\}\\&=\{\,(x,y)\in\mathbb R^2\mid x+y\in\mathbb Q\land (x-y)\sqrt 2\in\mathbb Q\,\}\end{align} and $$A=\mathbb R^2\setminus X.$$ For $(x,y)\in A$, there is at most one way to write $x=a+b\sqrt 2$ with $a,b\in \mathbb Q$, hence at most one point is missing from the line $\{x\}\times\mathbb R$. Likewise, at most one point is missing from $\mathbb R\times\{y\}$. Hence $A$ does have the cross-property.

On the other hand, $X$ is dense in $\mathbb R^2$: Given $(x,y)\in \mathbb R^2$, there are rational sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ with $a_n\to \frac{x+y}{2}$ and $b_n\to \frac{x-y}{2\sqrt 2}$. Then the sequence of points $(a_n+b_n\sqrt 2,a_n-b_n\sqrt 2)\in X$ converges to $(x,y)$. Therefore $A$ does not include any nonempty open set.

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For the most part, I think this answer is correct. However, given your A, surely the first part of your first sentence should read, "For (x,y) in A, there is no way of writing x = a + b*sqrt(2) with a,b in Q". – Adam Rubinson Aug 10 '13 at 13:06
@AdamRubinson Why do you think so? $(0,\sqrt 2)\in A$ and we clearly can write $0=0+0\sqrt 2$ and $\sqrt 2=0+\sqrt 2$. – Hagen von Eitzen Aug 10 '13 at 13:55
Surely (0,sqrt(2)) is in X. A and X are mutually exclusive. Therefore (0,sqrt(2)) cannot be in A. – Adam Rubinson Aug 10 '13 at 14:42
Unless 0 isn't a member of (your definition of) the rationals, which then means that your definition of rationals is a non-standard one. – Adam Rubinson Aug 10 '13 at 14:44
@AdamRubinson: $(0, \sqrt{2}$) is not in $X$. Indeed if it were, then one could write $0 = a + b \sqrt{2}$ and $\sqrt{2} = a - b \sqrt{2}$ for some pair $a$ and $b$ of rationals. Adding these together gives a contradiction (because $\sqrt{2}$ is irrational). – bryanj Aug 10 '13 at 15:03

Let $\mathscr{T}$ be the family of all $U\subseteq\Bbb R^2$ such for each $p\in U$, $U$ contains a cross with centre at $p$. Then $\mathscr{T}$ is a topology on $\Bbb R^2$. This topology is sometimes called the cross topology on $\Bbb R^2$ and denoted by $\Bbb R\otimes\Bbb R$. The question therefore boils down to showing that there is a set $A$ that is open in $\Bbb R\otimes\Bbb R$ but has empty Euclidean interior.

Lemma. Let $D\subseteq\Bbb R$, and let $f:D\to\Bbb R$ be injective; then $f=\{\langle x,f(x)\rangle:x\in D\}$ is a closed, discrete subset of $\Bbb R\otimes\Bbb R$. (Note that I am identifying the function $f$ with its graph.)

The proof is very straightforward, and I leave it to you.

Corollary: If $f$ is dense in the Euclidean topology on $\Bbb R^2$, then $\Bbb R^2\setminus f$ is open in $\Bbb R\otimes\Bbb R$ and has empty Euclidean interior.

To construct such an $f$, let $\mathscr{I}=\{I_n:n\in\Bbb N\}$ be an enumeration of the open intervals in $\Bbb R$ with rational endpoints, and let $\Bbb Q=\{q_n:n\in\Bbb N\}$ be an enumeration of the rationals. Let $\pi:\Bbb N\times\Bbb N\to\Bbb N$ be the pairing function, and let $\varphi=\pi^{-1}:\Bbb N\to\Bbb N\times\Bbb N$. For each $n\in\Bbb N$ let $\varphi(n)=\langle\alpha(n),\beta(n)\rangle$.

Suppose that $n\in\Bbb N$, and rational numbers $x_m=q_{k_m}$ and $y_m=q_{\ell_m}$ have been defined for all $m<n$. Let $K_n=\{k_m:m<n\}$ and $L_n=\{\ell_m:m<n\}$. (Note that the hypothesis is vacuously true for $n=0$, with $K_0=L_0=\varnothing$.) Then let $x_n=q_{k_n}$ and $y_n=q_{\ell_n}$, where $$k_n=\min\{k\in\Bbb N\setminus K_n:q_k\in I_{\alpha(n)}\}$$ and $$\ell_n=\min\{\ell\in\Bbb N\setminus L_n:q_\ell\in I_{\beta(n)}\}\;;$$ it's not hard to see that this is always possible. Let $D=\{x_n:n\in\Bbb N\}$, $E=\{y_n:n\in\Bbb N\}$, and $f=\{\langle x_n,y_n\rangle:n\in\Bbb N\}$; the construction ensures that $f$ is a bijection from $D$ onto $E$.

Now let $U$ be any non-empty Euclidean open set in $\Bbb R^2$; there are $I_k,I_\ell\in\mathscr{I}$ such that $I_k\times I_\ell\subseteq U$. Let $n=\pi(k,\ell)$; then $k=\alpha(n)$ and $\ell=\beta(n)$, so $\langle x_n,y_n\rangle\in I_{\alpha(n)}\times I_{\beta(n)}\subseteq U$, and it follows that $f$ is dense in the Euclidean topology on $\Bbb R^2$. And the function $f$ is injective, so so it follows from the corollary that $A=\Bbb R^2\setminus f$ has the desired properties.

(Since some people care, I've constructed $f$ in a way that does not require the axiom of choice; if one uses the axiom of choice, one need not deal with the pairing function.)

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