1
$\begingroup$

This is a problem from Munkres' Topology.

Let $X$ be a metric space.

(a) Suppose that for some $\epsilon \gt 0$, every $\epsilon$-ball in $X$ has compact closure. Show that $X$ is complete.

(b) Suppose that for each $x \in X$, there is an $\epsilon \gt 0$ such that the ball $B(x, \epsilon)$ has compact closure. Show by means of an example that $X$ need not be complete.

I'm currently stuck on (a), I'm trying to show that given a Cauchy sequence ($x_n$), some tail of the sequence belongs entirely on a ball $B(x,\epsilon)$ and thus since a compact set is sequentially compact in metric space, the sequence has a convergent subsequence and so the Caucy sequence converges to the same limit. However, I'm having trouble constructing this process. Also, for (b), I cannot think of any such example. Any solution, hint, or suggestions would be appreciated.

$\endgroup$
  • $\begingroup$ Take a close look at the definition of a Cauchy sequence again. $\endgroup$ – Daniel Fischer Dec 13 '14 at 14:19
  • $\begingroup$ Ohh... I have been mixing up (a) with (b)...Should've been so simple. $\endgroup$ – nomadicmathematician Dec 13 '14 at 14:26
3
$\begingroup$

Apply the definition of a Cauchy sequence to $\frac{\epsilon}{2}$. So there exists $N \in \mathbb{N}$, such that for all $n,m \ge N$, $d(x_n, x_m) < {\epsilon \over 2}$. What ball of radius $\epsilon$ does the tail lie in?

The last part follows because of the following fact:if a subsequence of a Cauchy sequence converges to $p$, the whole sequence converges to $p$.

As to an example: consider $\{\frac{1}{n}: n \in \mathbb{N}\}$.

$\endgroup$
1
$\begingroup$

For part a) fix the relevant $\epsilon$ and use the definition of Cauchy. You should easily see that the some tail is contained in some epsilon ball.

For part b) consider $\{1/n:n\in \mathbb{N}\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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