Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $x$ – ILoveMath Aug 30 '13 at 3:07
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
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,

share|cite|improve this answer

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

share|cite|improve this answer

$\mathbb{R}^2 \setminus \mathbb{Z}^2$ 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.

share|cite|improve this answer

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?

share|cite|improve this answer
@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

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.

share|cite|improve this answer

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 $\mathbb{R}^{2}\setminus \mathbb{Z}^{2}$.

share|cite|improve this answer

Your Answer


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.