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

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 3 '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 3 '19 at 18:22
• 2) is a good way, but why care about quotients? It's the image of the compact space $[0,1]^2$ under a continuous map, so compact. – Angina Seng Jun 3 '19 at 18:23

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.
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.