Completeness and Topological Equivalence How can I show that if a metric is complete in every other metric topologically
equivalent to it , then the given metric is compact ?
Any help will be appreciated .
 A: I encountered this result in Queffélec's book's Topologie. The proof is due to A.Ancona.
It's known as Bing's theorem. We can assume WLOG that $d\leq 1$, otherwise, replace $d$ by $\frac d{1+d}$. We assume that $(X,d)$ is not compact; then we can find a sequence $\{x_n\}$ without accumulation points. We define 
$$d'(x,y):=\sup_{f\in B}|f(x)-f(y)|,$$
where $B=\bigcup_{n\geq 1}B_n$ and 
$$B_n:=\{f\colon X\to \Bbb R,|f(x)-f(y)|\leq \frac 1nd(x,y)\mbox{ and }f(x_j)=0,j>n\}.$$
Since $d'\leq d$, we have to show that $Id\colon (X,d')\to (X,d)$ is continuous. We fix $a\in X$, and by assumption on $\{x_k\}$ for all $\varepsilon>0$ we can find $n_0$ such that $d(x_k,a)>\varepsilon$ whenever $k\geq k_0$. We define 
$$f(x):=\max\left(\frac{\varepsilon -d(x,a)}{n_0},0\right).$$
By the inequality $|\max(0,s)-\max(0,t)|\leq |s-t|$, we get that $f\in B_{n_0}$. This gives equivalence between the two metrics. 
Now we check that $\{x_n\}$ still is Cauchy. Fix $\varepsilon>0$, $N\geq\frac 1{\varepsilon}$ and $p,q\geq N$. Let $f\in B$, and $n$ such that $f\in B_n$. 


*

*If $n\geq N$, then $|f(x_p)-f(x_q)|\leq \frac 1nd(x_p,x_q)\leq \frac 1n\leq \varepsilon$;

*if $n<N$ then $|f(x_p)-f(x_q)|=0$.

A: The original proof can be found in  V. Niemytzki and A. Tychonoff, Beweis des Satzes, dass ein metrisierbarer Raum dann und nur dann kompakt ist, wenn er in jeder Metrik vollständig ist, Fund. Math. 12 (1928), 118-120.
