# Non-compact metric spaces

• Jun 27th 2011, 10:16 PM
Jose27
Non-compact metric spaces
Following an old question (link at the bottom) that was never answered, prove that if $\displaystyle (X,d)$ is a non-compact metric space then there is a metric $\displaystyle d^*$ equivalent to d such that $\displaystyle (X,d^*)$ is not complete. In other words all equivalent metrics on $\displaystyle X$ are complete if and only if $\displaystyle X$ is compact.

http://www.mathhelpforum.com/math-he...ics-94835.html
• Jun 28th 2011, 08:46 AM
girdav
Known as Bing theorem
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 $\displaystyle d\leq 1$ (if it's not the case we will take $\displaystyle \frac d{1+d}$ which is uniformly equivalent to $\displaystyle d$). Since $\displaystyle (X,d)$ is not compact we can find a sequence $\displaystyle \left\{x_n\right\}$ which has no accumulation point. Let $\displaystyle d^*(x,y) := \sup_B|f(x)-f(y)|$ where $\displaystyle B =\bigcup_{n=1}^{+\infty}B_n$ and $\displaystyle B_n := \left\{f:X\rightarrow \mathbb{R}; |f(x)-f(y)|\leq \frac 1nd(x,y)\mbox{ and }f(x_j) =0, j>n\right\}$.
We show that $\displaystyle d^*$ is equivalent to $\displaystyle d$. Since $\displaystyle d^*\leq d$ we only have to show that $\displaystyle I : (X,d' )\rightarrow (X,d)$ is continuous. Let $\displaystyle a \in X$. Since the sequence $\displaystyle \left\{x_n\right\}$ has no accumulation points for all $\displaystyle \varepsilon>0$ we can find $\displaystyle n_0$ such that for $\displaystyle n>n_0$ we have $\displaystyle d(x_n,a)\geq \varepsilon$. Let $\displaystyle f(x) := \max\left(\frac{\varepsilon -d(x,a)}{n_0},0\right)$. Since for $\displaystyle a,b\in \mathbb R$ we have $\displaystyle \max(a,0)-\max(b,0)\leq a-b$ we get $\displaystyle f\in B_{n_0}$ and if $\displaystyle d'(x,a)\leq \frac{\varepsilon}{n_0}$ then $\displaystyle d(x,a)\leq \varepsilon$. This step shows that $\displaystyle d^*$ is a metric (the definition only gives that $\displaystyle d^*$ is a finite pseudometric).
Now we have to show that $\displaystyle \left\{x_n\right\}$ is a Cauchy sequence.
Let $\displaystyle \varepsilon >0$ and $\displaystyle N\in\mathbb{N}$ with $\displaystyle N\geq \frac 1{\varepsilon}$, $\displaystyle p,q\geq N$ and $\displaystyle f\in B$. We can find a $\displaystyle n$ such that $\displaystyle f\in B_n$.
First case : $\displaystyle n\geq N$. We have $\displaystyle f(x_p)-f(x_q)|\leq \frac 1nd(x_p,x_q) \leq \frac 1n\leq \frac 1N\leq \varepsilon$.
Second case : $\displaystyle n<N$. We have $\displaystyle f(x_p)=f(x_q) =0$.
• Jun 29th 2011, 08:50 PM
Jose27
Re: Known as Bing theorem
Nice, that's a much neater proof than mine, but here it goes anyway:

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

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

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

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

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

We note that for $\displaystyle x\neq y$

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

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

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

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

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