Non-compact metric spaces
Following an old question (link at the bottom) that was never answered, prove that if is a non-compact metric space then there is a metric equivalent to d such that is not complete. In other words all equivalent metrics on are complete if and only if is compact.
Re: Known as Bing theorem
Nice, that's a much neater proof than mine, but here it goes anyway:
Assume is not compact, then there exists a sequence with for some . Define
From the definitions it's obvious that is symmetric and that . The "non-degeneracy" of the function follows by estimates analogous to the ones given below. For the triangle inequality we note that for any and it's true that (To see this, if then the triangle inequality is enough, if then the inequality reduces to and we're done).
We note that for
whenever and .
Now fix any and take the ball for some , then if it's easy to see that and since clearly both metrics are equivalent.
Since we have that is Cauchy in and we're done.