# 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 '18 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? – ʎpoqou Jan 29 '18 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 '18 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. – ʎpoqou Jan 29 '18 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 '18 at 15:48

Finite $\implies$ sequentially compact trivially, for every sequence would have a constant, hence convergent, subsequence...
• How about input on the initial $\Rightarrow$ part of the proof which I have been informed was done wrong. – ʎpoqou Jan 29 '18 at 16:11