Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that a totally bounded complete metric space $X$ is compact.

I can use the fact that sequentially compact $\Leftrightarrow$ compact.

Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$ $\forall\epsilon>0$, $X$ can be covered by a finite number of balls of radius $\epsilon$. I'm trying to show that all sequences in $X$ have a subsequence that converges to an element in $X$. I don't see how to go from convergent Cauchy sequences and totally bounded to subsequence convergent $in$ $X$.

share|improve this question
    
Write out the definition of Cauchy sequence, and write out the totally bounded condition. Observe that they both start with $\forall \epsilon > 0$. Apply the pigeonhole principle... –  Zhen Lin Feb 15 '12 at 7:33

1 Answer 1

up vote 7 down vote accepted

You need to show that if $X$ is totally bounded, every sequence in $X$ has a Cauchy subsequence. Let $\sigma=\langle x_n:n\in\mathbb{N}\rangle$ be a sequence in $X$. For each $n\in\mathbb{N}$ let $D_n$ be a finite subset of $X$ such that the open balls of radius $2^{-n}$ centred at the points of $D_n$ cover $X$. $D_0$ is finite, so there is a point $y_0\in D_0$ such that infinitely many terms of $\sigma$ are in $B(y_0,1)$. Let $$A_0=\{n\in\mathbb{N}:x_n\in B(y_0,1)\}\;,$$ so that $A_0$ is infinite. Now $D_1$ is finite, so there is a $y_1\in D_1$ such that $$A_1=\{n\in A_0:x_n\in B(y_1,2^{-1})\}$$ is infinite. Repeat: if $A_k$ is an infinite subset of $\mathbb{N}$, there must be a $y_{k+1}\in D_{k+1}$ such that $$A_{k+1}=\{n\in A_k:x_n\in B(y_{k+1},2^{-(k+1)})\}$$ is infinite, and the process can continue.

Now choose a strictly increasing sequence $\langle n_k:k\in\mathbb{N}\rangle$ of natural numbers in such a way that $n_k\in A_k$ for every $k\in\mathbb{N}$. Can you show that $\langle x_{n_k}:k\in\mathbb{N}\rangle$ is Cauchy?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.