Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
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
1  
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 Aug 10 '13 at 8:30

2 Answers 2

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.)

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.