How do you show that $\mathbb{Z}^2$ is a closed set in $\mathbb{R}^2$?

Question

In $(\mathbb{R}, \tau_{st})$, we can write $\mathbb{R} \setminus \mathbb{Z} = \bigcup_{n \in \mathbb{Z}} (n,n+1)$, and hence $\mathbb{R} \setminus \mathbb{Z}$ is an open set. Thus the complement, $\mathbb{Z}$, is closed.

In $(\mathbb{R}^2, \tau_{st})$ I am not sure how to write an expression for $\mathbb{R}^2 \setminus \mathbb{Z}^2$ in terms of open balls. So the first part of my question is whether we can write this in a 'neat' expression like above?

I can try to write $\mathbb{R}^2 \setminus \mathbb{Z}^2 = \mathbb{R}^2 \setminus \bigcup_{n,m \in \mathbb{Z}} \big\{(n,m)\big\}$. So knowing that $\big\{(n,n)\big\}$ is a closed set, we have the complement of the union of closed sets. But this is an infinite union, so I can't actually conclude that $\mathbb{Z}^2 = \bigcup_{n,m \in \mathbb{Z}} \big\{(n,m)\big\}$ is actually closed in this manner.

I know that we can say $\mathbb{Z}^2$ is closed by arguing its set of limit points is empty, but is there any other way I can show this? In particular, is there a way to show (analogous to the $\mathbb{R}^1$ case above) that $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is closed, without relying on an argument by limit points? I.e. Can I show the set as an explicit construction of union of open balls?

Answer 1

$\mathbb{R}^2 \setminus \mathbb{Z}^2$ is the union of the open balls centered on points both having irrational coordinates and radius small enough to not intersect $\mathbb{Z}^2$.

That makes a lot of balls... but it works!

Answer 2

Set $\mathbb{Z}$ is closed in $\mathbb{R}$ (see: this post). Next, we know if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A×B$ is closed in $X × Y$.

Answer 3

Here are already enough nice answers just using elementary topology. Let me propose something slightly different. We can define $$ f: \mathbb{R}^2 \rightarrow \mathbb{R}, \ f(x,y) = \sin(\pi x)^2 + \sin(\pi y)^2. $$ Then one notes that $f$ is continuous and that $\mathbb{Z}^2=f^{-1}(\{0\})$ and therefore, $\mathbb{Z}^2$ is closed in $\mathbb{R}^2$ in the euclidean topology.

Answer 4

You could show this similarly to what you did rather easily. Define $C_m=(m,m+1)\times \mathbb{R}$ and $R_m=\mathbb{R} \times(m,m+1)$ for all $m\in\mathbb{Z}$, which are open strips in the plane. Define $$U=\bigcup_{m\in\mathbb{Z}}\big(C_m \cup R_m \big)$$ and notice that $\mathbb{Z}^2=\mathbb{R}^2\setminus U$.

Answer 5

Alternatively, a converging sequence in $\Bbb{Z}^2$ is eventually constant so its limit remains in $\Bbb{Z}^2$. Hence, $\Bbb{Z}^2$ is closed in $\Bbb{R}^2$.

Answer 6

I shall prove that $\mathbb{Z}^n$ is a closed set of $\mathbb{R}^n$, for each positive integer $n$. The idea is to show that $U_n:=\mathbb{R}^n\setminus\mathbb{Z}^n$ is an open set.

Let $B^n(x,r)\subseteq\mathbb{R}^n$ denote the open ball centered at $x\in\mathbb{R}^n$ with radius $r>0$. Observe that $$U_n=\bigcup_{x\in\left(\mathbb{Z}+\frac12\right)^n}\,B^n\left(x,\frac{\sqrt{n}}{2}\right)$$ is a union of open sets. Here, $\mathbb{Z}+\dfrac12$ is the set of half-integers (i.e., numbers of the form $k+\dfrac12$, where $k\in\mathbb{Z}$).

Note that, when $n=1$, we get $$B^1\left(k+\frac12,\frac{\sqrt{1}}{2}\right)=(k,k+1)$$ for each $k\in\mathbb{Z}$. Therefore, $$U_1=\bigcup_{k\in\mathbb{Z}}\,(k,k+1)\,,$$ which recovers the same proof that you have in the first paragraph of your question.

Answer 7

The following has kind of the same flavor as a few of the above, but is, I think, distinct enough to merit posting. The approach is to show that $\mathbb{R}^n \setminus \mathbb{Z}^n$ is open for for any $n \in \mathbb{N}$.

To show that $X = \mathbb{R}^n \setminus \mathbb{Z}^n$ is open, it is sufficient to show that if $x \in X$, then there is some $r > 0$ such that $B(x,r)$ (the open ball with radius $r$ centered at $x$) is contained in $X$. To that end, fix some $x \in X$ and take $$ r = \min_{y\in \mathbb{Z}^n} d(x,y),$$ where $d(x,y)$ denotes the (Euclidean) distance from $x$ to $y$.

That is the entirety of the argument. One might ask whether or not such a minimal $r$ exists and is positive. In this case, it must, as any ball of finite radius can contain at most a finite number of elements of $\mathbb{Z}^n$, and the distance from $x$ to any one of those elements must be positive. A finite set of positive numbers must have a minimal element, hence $r$ is well-defined.

Answer 8

For $x \in \Bbb R$ define

$$h(x)=\begin{cases}1&\text{when }x=0\\ x&\text{otherwise} \end{cases}$$

Define

$\quad \delta(x,y)= h(x-n) \times (n+1-x) \times h(y-m) \times (m+1-y) \text{ where } $ $\quad \quad (\exists \text{ unique } n,m \in \Bbb Z) \; [n \le x \lt n+1] \land [m \le y \lt m+1 ]$

Exercise 1: Show that $\delta(x,y)$ is well-defined on $\Bbb R^2$, taking on strictly positive values on $\displaystyle \Bbb R^2 \setminus \Bbb Z^2$.

For each $(x,y) \in \Bbb R^2 \setminus \Bbb Z^2$ let $B^{\,\delta(x,y)}_{(x,y)}$ be the open ball about $(x,y)$ of radius $\delta(x,y)$.

Exercise 2: Show that

$\quad \Bbb R^2 \setminus \Bbb Z^2 = \displaystyle \bigcup_{(x,y) \in \Bbb R^2 \setminus \Bbb Z^2} B^{\,\delta(x,y)}_{(x,y)}$