$S$ is a compact subset of the manifold $M$ if and only if $\varphi(S)$ is compact in $R^n$, for a chart $(U,\varphi)$ such that $A \subseteq U$.

Question

Assume $S \subseteq M$ a subset of the manifold $M$, which is contained in the domain of a coordinate chart $(U,\varphi)$. I am going to prove that:

$S$ is compact in $M$ $\iff$ $\varphi(S)$ is compact in $\mathbb{R}^n$

Proof:

$\Longleftarrow$

If $\varphi(S)$ is compact in $\mathbb{R}^n$, then we suppose $\{U_\alpha\}$ an open cover of $S$, and due to $S \subseteq U$, we have $S \subseteq \bigcup_\alpha (U \cap U_\alpha)$. Hence $\varphi(S) \subseteq \bigcup_\alpha \varphi(U \cap U_\alpha)$. As $\varphi(S)$ is compact, there exist indices $\alpha_1,...,\alpha_N$ such that $\varphi(S) \subseteq \varphi(U \cap U_{\alpha_1}) \cup ... \cup \varphi(U \cap U_{\alpha_N})$, then we have $S \subseteq (U \cap U_{\alpha_1}) \cup ... \cup (U \cap U_{\alpha_N}) \subseteq U_{\alpha_1} \cup ... \cup U_{\alpha_N}$. Therefore $S$ is compact in $M$.

$\Longrightarrow$

For this direction I have no idea. Who can show me the proof?

Answer 1

I think it's just the same argument in reverse, which I sketch here:

Suppose $S$ compact in $M$. Let the sets $V_\alpha$ be an open cover of $\phi(S) \subset \phi(U)$. Let $W = \phi(U)$. $W$ is open because $\phi$ is a chart. So the sets $W \cap V_\alpha$ form an open cover of $\phi(S)$ as well. Their preimages, under $\phi$ --- call them $T_\alpha$ --- are an open cover of $S$, which is compact, so admit a finite subcover. The corresponding $V$s form a finite subcover of $\phi(S)$.