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.)
 A: 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.
A: 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.)
