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I claim yes, and to show this, it will suffice to show that $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is open. So that for every $x \in \mathbb{R}^2 \setminus \mathbb{Z}^2$, we must find a neighborhood $N$ of $x$ such that $N \cap \mathbb{Z}^2 = \varnothing$. Let $r = \min(\|x\|, 1 - \|x\|)$ in $N = D^2(x,r)$. Suppose there exists $z \in N \cap \mathbb{Z}^2$. So $$ \|z\| \leq \|z-x\| + \|x\| < \min(\|x\|, 1 - \|x\|) + \|x\| \leq 1 - \|x\| + \|x\| = 1 $$

So, we have a contradiction, and therefore $N \cap \mathbb{Z}^2$ must be empty as desired.

Is this correct? any feedback? thanks.

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What do the double bars mean? I am guessing it is the fractional part of $x$? –  Pratyush Sarkar Aug 30 '13 at 3:06
    
means the norm of $x$ –  Euler Aug 30 '13 at 3:07
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I'm not sure I understand what you did with the definition of $r$. Can't $1 - ||x||$ be negative? –  Pratyush Sarkar Aug 30 '13 at 3:10
    
@Citizen : The first two sentences of your proof are fine. You need to construct a neighborhood $N$ like you described. What follows is probably incorrect, because your $r$ can be negative, hence your $D^2(x,r)$ can be empty. Proof by contradiction is unnecessary and confusing here. –  Stefan Smith Aug 30 '13 at 15:48
    
Another way to do it is to show that your set contains all its limit points. –  Stefan Smith Aug 30 '13 at 15:51
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6 Answers

up vote 2 down vote accepted

I think your idea with the norm is not exactly the good one. Take $(x,y) \in \mathbb R^2 \backslash \mathbb Z^2$. Then either $x$ or $y$ is not an integer. Without loss of generality, suppose $x$ is not an integer. Then there exists $z_1, z_2 \in \mathbb Z$ such that $z_1 < x < z_2$. The set $]z_1,z_2[ \, \times \, \mathbb R$ contains $(x,y)$ and is open, so it is a neighborhood of $(x,y)$. It contains no point in $\mathbb Z^2$, so is a subset of $\mathbb R^2 \backslash \mathbb Z^2$, and thus $\mathbb Z^2$ is closed.

Feedback : I think you tried to compute an open ball whose radius is smaller than something so that your ball doesn't intersect $\mathbb Z^2$, but it's just not working out (at least the way you wrote it). I took an entire open strip of the plane that doesn't cross $\mathbb Z^2$ : you could extract an open ball from it if you wanted (for instance, it could be centered at $(x,y)$ and have radius $\min \{ |x - z_1|, |x-z_2| \}$).

Hope that helps,

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$\mathbb{Z}^2$ is the set of zeros of $f(x,y)= \sin^2 x\pi + \sin^2 y\pi$, which is continuous, hence it is closed.

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Alternatively: consider a sequence $(x_n,y_n)_{n\in\mathbb{Z}_+}$in $\mathbb{Z}^2$. Suppose that this sequence converges to $(x,y)\in\mathbb{R}^2$. Can $x$ and $y$ be possibly be non-integers?

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@StefanSmith Note that the only restriction I imposed on the sequence is that it converge to $(x,y)$. Other than that, it is arbitrary! –  triple_sec Aug 30 '13 at 16:03
    
OK, I see why your argument works now. If I did it I would find it less confusing to stick with the limit point definition and show there are no limit points. –  Stefan Smith Sep 1 '13 at 15:17
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$\mathbb{R}^2 \setminus \mathbb{Z}^2$ is a union of translated strips that look like $(0, 1) \times \mathbb R$ and $\mathbb R \times (0, 1)$, each of which is open.

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Let $\{(a_i,b_i)\}\in \mathbb{Z}^2$ such that converges to element $(x,y)\in \mathbb{R}^2\setminus\mathbb{Z}^2$. That is, for all $\epsilon>0$, there exists $i_0\in\mathbb{N}$ such that for all $i\geq i_0$ we have $||(x-a_i,y-b_i)||=|x-a_i|+|y-b_i|<\epsilon/2$.

Without loss of generality, suppose that $x\in ]a_i,a_{i+1}[$ and $y\in ]b_i,b_{i+1}[$. Then

$1=|a_i-a_{i+1}|\leq |a_i-x|+|a_{i+1}-x|<\epsilon/2+\epsilon/2=\epsilon$ and that is a contradiction.

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Yes, the set $\mathbb{Z}^{2}:=\{(x,y)\in\mathbb{R}^{2} : x,y\in\mathbb{Z}\}$ is closed in $\mathbb{R}^{2}$ with its usual topology. Your idea is right, yet you may want to fine-tune your argument. Drawing a picture will help you see that the coordinates of a point in the complement of the set lie strictly between two integers.

More formally, let $[x]$ denote the integer part of $x$ (i.e. the largest integer $\leq x$). Note that for $(a,b)\in\mathbb{R}^{2}\setminus\mathbb{Z}^{2}$ either $a$ or $b$ is not an integer. Hence, at least one of $d_{1}=\text{min}\{|a-[a]|,|[a]+1-a|\}$ or $d_{2}=\text{min}\{|b-[b]|,|[b]+1-b|\}$ is strictly greater that $0$. If we let $d=\text{min}\{d_{1},d_{2}\}$, then $d>0$ and the open ball $B$ centered at $(a,b)$ having radius $d/2$ is completely contained in $\mathbb{R}^{2}\setminus \mathbb{Z}^{2}$.

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