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Math Help - Non-compact metric spaces

  1. #1
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    Non-compact metric spaces

    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
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  2. #2
    Super Member girdav's Avatar
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    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 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.
    Last edited by girdav; June 28th 2011 at 11:11 AM.
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  3. #3
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    Re: Known as Bing theorem

    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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