Show that if two metrics induce the same topology, one metric space is compact iff the other one is.

Question

Question: Given two metric spaces $(X, d)$ and $(X,e)$, where $\tau(X,d)$ = $\tau(X,e)$, show that $(X,e)$ is compact if and only if $(X,d)$ is compact.

I'm aiming to prove this using the definition of compactness whereby a metric space is compact exactly when each sequence contains a convergent subsequence. I can't see how the existence of a convergent subsequence (of a sequence) in $(X,d)$ could guarantee one in $(X,e)$. Any idea what comes next?

Answer 1

If two metrics $d$ and $e$ induce the same topology, then a sequence $(x_n)_{n\in\mathbb N}$ converges to $x$ in $(X,d)$ if and only if it converges to $x$ in $(X,e)$. Therefore, a sequence has a convergent subsequence in $(X,d)$ if and only if it has a convergent subsequence in $(X,e)$.

Answer 2

$x_n \to x$ in a topological space iff for all open subsets $O$ that contain $x$, $O$ contains all but finitely many $x_n$.

So convergence only depends on the topology not on the metric, and so if $d$ and $e$ are equivalent metrics, they have the same open sets and thus the same sequences converge to the same points in either metric. So if $(X,d)$ is sequentially compact, so is $(X,e)$ and vice versa. That's really all there is to it.

Answer 3

A sequence $(x_n)_{n=1}^{\infty}$ is eventual in $B_d(x,\delta)$ in $(X,d)$ iff the sequence $(x_n)_{n=1}^{\infty}$ is eventual in $B_e(x,\delta)$ in $(X,e)$ as the two metrics $d$ & $e$ generate the same topology. That is $(x_n)_{n=1}^{\infty}$ is convergent in $(X,d)$ iff it is convergent in $(X,e)$. Hence if one of $(X,d)$ & $(X,e)$ is compact, then the other is also so.