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 .

  • $\begingroup$ $\mathbb Z$ with standard metirc is complete but not compact. $\endgroup$ – Hagen von Eitzen Sep 9 '12 at 13:14
  • $\begingroup$ @Hagen But is it complete in every topologically equivalent metric? $\endgroup$ – Ester Sep 9 '12 at 13:16
  • $\begingroup$ Oh, I see. It's topologically equivalent, not strongly equivalent. Then no: $d(n,m) = |\arctan n - \arctan m|$ is toppologically equivalent and makes $(n)$ a non-converging Cauchy sequence. $\endgroup$ – Hagen von Eitzen Sep 9 '12 at 13:24
  • $\begingroup$ Being compact is a topological property, and a metric space is compact if and only if it is complete and totally bounded. I suppose the trick is to find a way of replacing any metric with a topologically equivalent totally bounded one... $\endgroup$ – Zhen Lin Sep 9 '12 at 13:25
  • $\begingroup$ @Zhen Yes,I was trying exactly that $\endgroup$ – Ester Sep 9 '12 at 13:26

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 $|f(x_p)-f(x_q)|=0$.

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.


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