Quick way to show $\mathbb{R}^2/ \mathbb{Z}^2$ is compact

Question

See title - I can think of 2 ways:

1) show it's homeomorphic to a torus, and apply Heine-Borel in $\mathbb{R}^3$. The problem here is showing its a homeomorphism is fiddly

2) show that $\mathbb{R}^2/ \mathbb{Z}^2 \cong [0,1]\times[0,1]/$~ where ~ is the usual equivalence relation, then apply products and quotients of compact spaces are compact.

Is there a better way?

How about showing that $\mathbb{R}^2 / \mathbb{Z}^2$ is the product of two copies of the circle $\mathbb{R} / \mathbb{Z} = S^1$, which is (more easily shown to be?) compact. — avs, Jun 03 '19 at 18:19
Show that it is the continuous image of $[0,1]\times[0,1],$ which is compact. The continuous image of a compact space is compact. — Thomas Andrews, Jun 03 '19 at 18:22

score 3 · Answer 1 · answered Jun 03 '19 at 18:23

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Let $i : \mathbb{R^2} \to \mathbb{R^2} /\mathbb{Z^2}$ be the quotient map. If $U_\alpha$ is an open cover of $\mathbb{R^2} /\mathbb{Z^2}$, then $i^{-1}(U_\alpha)$ is an open cover of $\mathbb{R^2} $, in particular, it covers the compact set $[0,1]^2$ and hence has a finite subcover.

answered Jun 03 '19 at 18:23

copper.hat

172,524

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Probably the OP wants to show that it is compact Hausdorff. – Paul Frost Jun 03 '19 at 18:25
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Any reason you suspect this, @PaulFrost ? – Thomas Andrews Jun 03 '19 at 18:28
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@ThomasAndrews See 1) in his question. But it is a guess on my part. And it is not diificult to show. – Paul Frost Jun 03 '19 at 18:35
@PaulFrost Yeah, but if OP were using the old definition of compact (necessarily Hausdorff) it is not true that the quotient of a compact Hausdorff space is not necessarily Hausdorff, while the OP says that quotients of compact spaces are compact. – Thomas Andrews Jun 03 '19 at 19:05

score 1 · Answer 2 · answered Jun 03 '19 at 18:38

If you know that $\Bbb R/\Bbb Z$ is compact (which is the same kind of problem but lets assume you know that already), and if you want to use lots of little arguments of topology you could do as follows. The map $f:\Bbb R^2\to \Bbb R/\Bbb Z\times \Bbb R/\Bbb Z$ defined by $$f:(x,y)\mapsto ([x],[y])$$ factors to a continuous bijection $\tilde{f}:\Bbb R^2/\Bbb Z^2\to \Bbb R/\Bbb Z\times \Bbb R/\Bbb Z$ wich is a homeomorphism if and only if $f$ is a quotient map. But $f$ is the product of two open maps ($\Bbb R\to\Bbb R/\Bbb Z$ is an open map because its the quotient by a subgroup in a topological group), so it is open hence quotient. You conclude using the fact that $\Bbb R/\Bbb Z\times \Bbb R/\Bbb Z$ is compact.

score 0 · Answer 3 · answered Oct 29 '23 at 20:57

Let $\Gamma\cong\mathbb{Z}^m\subset\mathbb{R}^n$ be a lattice, $m\leq n$. Consider the map \begin{align*} f:\mathbb{R}^n\to\mathbb{R}^{n-m}\times\mathbb{T}^m,(x_1,\dots,x_{n-m},y_1,\dots,y_m)\mapsto (x_1,\dots,x_{n-m},e^{2\pi iy_1},\dots,e^{2\pi iy_m}) \end{align*} where $\mathbb{T}^n=\prod_{I=1}^nS^1$. Then $f$ is surjective and the map

\begin{align*} \bar{f}: \mathbb{R}^n/\Gamma\to\mathbb{R}^{n-m}\times\mathbb{T}^m, [x]\mapsto x \end{align*} is a homeomorphism, where $x\sim y\Leftrightarrow f(x)=f(y)$. Now \begin{align*} \mathbb{R}^n/\Gamma\cong\mathbb{R}^{n-m}\times\mathbb{T}^m \end{align*} and by Tychonoff's Theorem and the fact that $S^1$ is compact, \begin{align*} \mathbb{R}^n/\Gamma \text{ is compact }\Leftrightarrow n=m. \end{align*}

Quick way to show $\mathbb{R}^2/ \mathbb{Z}^2$ is compact

3 Answers3