show that torus is compact I am having difficuties in showing a torus is compact. Initially I wanted to use Heine-Borel theorem, but after that I realise we are not working in $\mathbb{R}^n$ space. So a simple way to show torus is compact is by definition. But after defining an open cover for torus, I don't know how to proceed. Can anyone guide me?
 A: If you are thinking of the torus as $S^1 \times S^1$:


*

*Products of compact sets are compact (you only need the finite case)

*$S^1$ compact (as it is a closed and bounded subset of $R^2$).


If you are thinking of the torus as $R^2 / Z^2$:


*

*This quotient map makes the same identifications as the exponential map $(e^{2 \pi i x}, e^{2 \pi i y})$, so since both are quotients $R^2 / Z^2$ is homeomorphic to $S^1 \times S^1$. (You will have to prove that if $q_1: X \to Y_1$ and $q_2 : X \to Y_2$ are quotient maps, and $q_1(x) = q_1(x')$ iff $q_2(x) = q_2(x')$, then $Y_1$ and $Y_2$ are (naturally) homeomorphic.)

A: If $X$ is compact then the quotient space $X/\sim $ is compact.  Proof.  The quotient map $p : X \to X /\sim, \ x \mapsto [x]_{\sim}$, is surjective and continuous by definition.
As you've heard the product of compact spaces is compact.
A: One possibility is to show that $T=[0,1]^2/\sim$ is homeomorphic to one of the other definitions of the torus, which are obviously compact. One such possible equivalent definition is $S^1\times S^1$.
Alternatively, let $\phi:[0,1]^2\rightarrow T$ be the quotient map. Take an open cover $\{U_\alpha\}$ of $T$. Then $\{\phi^{-1}(U_\alpha)\}$ is an open cover of $[0,1]^2$, which is compact. Thus there exist a finite subcover $\{\phi^{-1}(U_k)\}_{k=1}^n$ of $[0,1]$. Then $\{U_k\}_{k=1}^n$ is a finite subcover of $\{U_\alpha\}$ for $T$.
Notice that this proof works for the general statement: the quotient of a compact space is compact.
