Non-compact metric spaces

**Jose27** · June 27th 2011, 10:16 PM

Following an old question (link at the bottom) that was never answered, prove that if $(X,d)$ is a non-compact metric space then there is a metric $d^*$ equivalent to d such that $(X,d^*)$ is not complete. In other words all equivalent metrics on $X$ are complete if and only if $X$ is compact.

http://www.mathhelpforum.com/math-he...ics-94835.html

**girdav** · June 28th 2011, 08:46 AM

This proof is a "translation" with my poor English of one which is given in Topologie, H. Queffelec (it seems that the proof is form A. Ancona). We assume that $d\leq 1$ (if it's not the case we will take $\frac d{1+d}$ which is uniformly equivalent to $d$ ). Since $(X,d)$ is not compact we can find a sequence $\left\{x_n\right\}$ which has no accumulation point. Let $d^*(x,y) := \sup_B|f(x)-f(y)|$ where $B =\bigcup_{n=1}^{+\infty}B_n$ and $B_n := \left\{f:X\rightarrow \mathbb{R}; |f(x)-f(y)|\leq \frac 1nd(x,y)\mbox{ and }f(x_j) =0, j>n\right\}$ .
We show that $d^*$ is equivalent to $d$ . Since $d^*\leq d$ we only have to show that $I : (X,d' )\rightarrow (X,d)$ is continuous. Let $a \in X$ . Since the sequence $\left\{x_n\right\}$ has no accumulation points for all $\varepsilon>0$ we can find $n_0$ such that for $n>n_0$ we have $d(x_n,a)\geq \varepsilon$ . Let $f(x) := \max\left(\frac{\varepsilon -d(x,a)}{n_0},0\right)$ . Since for $a,b\in \mathbb R$ we have $\max(a,0)-\max(b,0)\leq a-b$ we get $f\in B_{n_0}$ and if $d'(x,a)\leq \frac{\varepsilon}{n_0}$ then $d(x,a)\leq \varepsilon$ . This step shows that $d^*$ is a metric (the definition only gives that $d^*$ is a finite pseudometric).
Now we have to show that $\left\{x_n\right\}$ is a Cauchy sequence.
Let $\varepsilon >0$ and $N\in\mathbb{N}$ with $N\geq \frac 1{\varepsilon}$ , $p,q\geq N$ and $f\in B$ . We can find a $n$ such that $f\in B_n$ .
First case : $n\geq N$ . We have $f(x_p)-f(x_q)|\leq \frac 1nd(x_p,x_q) \leq \frac 1n\leq \frac 1N\leq \varepsilon$ .
Second case : $n<N$ . We have $f(x_p)=f(x_q) =0$ .

**Jose27** · June 29th 2011, 08:50 PM

Nice, that's a much neater proof than mine, but here it goes anyway:

Assume $(X,d)$ is not compact, then there exists a sequence $B=(x_n)_n$ with $d(x_n,x_m) \geq \varepsilon >0$ for some $\varepsilon \in \mathbb{R}$ . Define

$e_{jk}(x,y)= d(x,x_j)+|j^{-1}-k^{-1}|\varepsilon +d(y,x_k)$

$A(x,y)=\inf_{j,k} e_{jk}(x,y)$

$e(x,y)=\min \{ d(x,y), A(x,y) \}$

From the definitions it's obvious that $e$ is symmetric and that $e(x,x)=0$ . The "non-degeneracy" of the function follows by estimates analogous to the ones given below. For the triangle inequality we note that for any $x,y,z\in X$ and $j,k,l,m \geq 1$ it's true that $e_{jk}(x,y) \leq e_{jl}(x,z)+e_{mk}(z,y)$ (To see this, if $l=m$ then the triangle inequality is enough, if $l\neq m$ then the inequality reduces to $|j^{-1}-k^{-1}|\leq 1 + |j^{-1}-l^{-1}|+|m^{-1}-k^{-1}|$ and we're done).

We note that for $x\neq y$

$A(x,y) \geq d(x,B)$ whenever $x,y\notin B$

$A(x,y) \geq \min \{ \frac{ \varepsilon }{2} , \frac{\varepsilon}{j_0} \}$ whenever $x=x_{j_0} \in B, \ y\notin B$ and $d(x,y)<\frac{\varepsilon}{2}$ .

$A(x,y) \geq \varepsilon$ if $x,y \in B$

Now fix any $x\in X$ and take the ball $B_s^e(x)$ for some $s>0$ , then if $r<\min \{ s, \frac{\varepsilon}{2}, \frac{\varepsilon}{j_0}, d(x,B) \}$ it's easy to see that $B_r^d(x) \subset B_s^e(x)$ and since clearly $e\leq d$ both metrics are equivalent.

Since $e_{jk}(x_j,x_k)=|j^{-1}-k^{-1}| \geq A(x_j,x_k)$ we have that $B$ is Cauchy in $(X,e)$ and we're done.

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