I need to decide whether this proposition is true or not :"A countable set is always compact".
My first thought is NO.
As a simple proof I would present a counter example: since $\mathbb{N}$ is countable if the claim was true it would imply that it's compact. Compactness (by the of Heine–Borel theorem) requires Closeness (that in this case is satisfied) and Boundness (which fails to hold). For this reason the claim is not true.
Now, I know that in certain spaces the Heine-Borel property does not apply, and that's why I would like to know if in this very generic case (the exercise does not provide any other information) this "proof" is a valid one.
If not, I would like to know what extra conditions should I impose to make it true.
Thanks in advance.
Application of Heine–Borel theorem.
general-topologyreal-analysis
Best Answer
The open-ness of this question really comes down to what set, and what topology on that set the asker has in mind. If both of those pieces of data are left open to interpretation on the part of the answerer, then your answer is perfectly fine. What your answer shows is that countable sets are not always compact, in the most general sense.
The Heine-Borel theorem says that a subset of $\mathbf R$, equipped with its usual topology, is compact if and only if it is closed and bounded. The collection $\mathbf N$ of natural numbers, thought of as a subset of $\mathbf R$, is countable, closed, and unbounded, so it is not compact by Heine-Borel.
If we wanted to modify the question to only consider certain types of topological spaces, that is, sets equipped with certain kinds of topologies, then maybe we can get different answers. For instance, any countable subset (in fact any subset whatsoever) of a set equipped with the indiscrete topology is compact.