# 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 .

• $\mathbb Z$ with standard metirc is complete but not compact. – Hagen von Eitzen Sep 9 '12 at 13:14
• @Hagen But is it complete in every topologically equivalent metric? – Ester Sep 9 '12 at 13:16
• 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. – Hagen von Eitzen Sep 9 '12 at 13:24
• 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... – Zhen Lin Sep 9 '12 at 13:25
• @Zhen Yes,I was trying exactly that – Ester Sep 9 '12 at 13:26

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