Consider a metric space $(\Bbb N, d)$ where $d(m,n) = \frac{\vert m-n \vert} {1+\vert m-n \vert}$.
Need to prove that any infinite subset $X \subset \Bbb Z$ is not compact and not sequentially compact.
(Have previously proven that every set in $(\Bbb N, d)$ defined above is open and closed, and that $(\Bbb N, d)$ is complete.)
I find the question a bit odd, as there are 2 spaces involved: $\Bbb N$ and $\Bbb Z$. So I just take it to mean "not compact and not sequentially compact" in $(\Bbb Z, d)$ rather than $(\Bbb N, d)$
So I suspect the trick for this question is that $\Bbb Z$ is topologically equivalent to $\Bbb N$ ("homeomorphism"?) and I need to consider infinite subsets $X' \subset \Bbb N$ and use my previous results about $(\Bbb N, d)$
Also, to answer this question, I suppose I just need to prove either X is not compact or not sequentially compact, since the 2 are equivalent for any metric space. I also think that (correct me if I'm wrong) in any metric space, a subset is compact iff it is closed and bounded. But I don't really know how to proceed. Any help would be much appreciated.
I strongly suspect that $\Bbb Z$ is simply a typo for $\Bbb N$, and that you’re actually to show that no infinite subset of $\Bbb N$ is compact or sequentially compact.
It is not in general true that in metric spaces the compact sets are precisely the closed and bounded sets. For example, the set $\Bbb Q\cap[0,1]$ is closed and bounded in the metric space $\Bbb Q$ with the usual Euclidean metric, but it is not compact. You should start with an arbitrary infinite $X\subseteq\Bbb N$ and show directly that it is neither compact nor sequentially compact.
To show that $X$ is not compact, find an open cover of $X$ that has no finite subcover. HINT: You’ve already shown that $\{x\}$ is open for each $x\in\Bbb N$.
To show that $X$ is not sequentially compact, find an infinite sequence in $X$ that has no convergent subsequence. HINT: Prove that any infinite sequence with no repeated terms has this property.