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

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.