I am reading this Bolzano-Weierstrass property saying that a metric space X is sequentially compact if every sequence in X has a convergent subsequence. I also read a theorem saying that sequentially compact is equivalent to compact.
So I am thinking of this sequence$a_{2k-1}=1, a_{2k}=n$. this sequence obviously has a convergent subsequence, but it is not bounded above. How to prove that an unbounded space is compact? given that compact preserves boundedness in real metric.
An unbounded metric space is not compact, as given any $x \in X$, the open balls $B(x, n)$ for $n \in \mathbb{N}$ is an open cover that does not have a finite subcover.
The sequence above does indeed have a convergent subsequence, but to apply the property that sequentially compact is equivalent to compact for metric spaces, you need all sequences to have a convergent subsequence, not just that particular one.