# Finite set implies sequentially compact

Suppose we have a metric space ($\Bbb{N}$,d) with metric $d(m,n)=\left|\frac{1}{n} - \frac{1}{m}\right|$

I need to prove that in this metric space, a set is sequentially compact $iff$ it is finite.

I think I have an idea of the the $\Rightarrow$ part of this iff:

Assuming that a set in this metric space is not finite and showing that we cant cover S with finitely many open open balls and that the next point in a set, say, {${x_1,x_2,...x_n}$} will have to be outside $\bigcup\limits_{i=1}^{n} B^o(x_i,\epsilon)$ which is a finite subcover. $d(x_n,x_m) \geq \epsilon$ and therefore { $x_n$} will not have a convergent subsequence and so we wont have a sequentially compact set. Contradiction.

For the $\Leftarrow$ part, I am not sure how I should start.

I tried saying that if S is finite then we have at least one of the values in S appearing infinitely many times so that we could have some sequence that has a convergent subsequence. But I am not sure how I can formalise on this.

Note: I can not use the fact that a set is compact iff it is sequentially compact.

• The one catch to be careful about is that this space is not complete, since $x_n=n$ is Cauchy and not convergent. So you need to worry about sequences "converging to infinity". If you take that into account your ideas will go through.
– Ian
Jan 29, 2018 at 15:39
• What do you mean specifically about the catch? What should I note? I proved in a previous exercise that this space is not complete but how can I bring this into here as a cautionary point? Jan 29, 2018 at 15:41
• Your first part was trying to say that a sequence of distinct points has no Cauchy subsequence. But there are sequences of distinct points that have Cauchy subsequences in this space.
– Ian
Jan 29, 2018 at 15:43
• So does that make my $\Rightarrow$ argument invalid? Or should I just note that $x_n = n$ is a Cauchy sequence in this space? I am new to this, so not quite sure how the proof should go. Jan 29, 2018 at 15:46
• Your forward argument is not valid; in fact any sequence of distinct points is Cauchy in this space. The point is that also any sequence of distinct points is not convergent because they are trying to "converge to infinity".
– Ian
Jan 29, 2018 at 15:48

Finite $\implies$ sequentially compact trivially, for every sequence would have a constant, hence convergent, subsequence...

This is the point you were trying to make...

This would be true in any topological space...

• How about input on the initial $\Rightarrow$ part of the proof which I have been informed was done wrong. Jan 29, 2018 at 16:11
• Yeah, that's more interesting... let me look it over...
– user403337
Jan 29, 2018 at 16:13
• Conversely, it's easy to show the topology is discrete. Finiteness follows...
– user403337
Jan 29, 2018 at 17:06
• How would I go about showing that it is discrete? I read on some other forum that this is the case but we did not go over this in lectures. Jan 29, 2018 at 17:07
• Here's a proof: math.stackexchange.com/a/514640
– user403337
Jan 29, 2018 at 17:39