Example of a topological space I have realized that inserting finiteness in topological spaces can lead to some bizarre behavior. For example, it seemed natural to say that every compact subspace of a metric space is closed and bounded, which can be easily proved. But to my surprise the converse is not true. Let X be an infinite set and define d: $ X x X \rightarrow X$ by $d(x,y) = 1$ if $x=y$ and $0$ if $x \neq y$. Notice this metric induces the discrete topology on X and that every singleton set is closed. In this space, $X$ is clearly closed and bounded but not compact because the union of all singletons contains no finite subcover. I feel pretty unsettled. Is there an example of a closed and bounded subspace of some space that isn't compact, where "finiteness" isn't it in the picture? I don't think it is possible. Sorry, I know the question sounds pretty vague.
 A: Indeed, there are a lot of such examples. As you may know, in the real line (and, more generally, in finite-dimensional vector spaces), a set is compact if and only if it is closed and bounded, however this condition is not true for more general spaces. It is not hard to find counterexamples. As mathworker21 said, in any infinite-dimensional vector space (actually, this is an if and only if), the closed unit ball is not compact. But you don't need to go further in the analysis realm to find such pathologies. Note first that, in a metric space, a set is compact if and only if it it is complete and totally bounded.
Let me give you a very simple example of a pathology. Endow the rational numbers $\mathbb{Q}$ with the topology induced by the real line. It is trivially a metric space, so it might behave well, but it is not locally compact. This essentially means the following: there is no open set $V$ in $\mathbb{Q}$ with compact closure. I will show only that $A = [0,1] \cap \mathbb{Q}$ is not compact, and I hope you can carry on the details for every open set. Well, $A$ is, by definition, closed and bounded. On the other hand, $A$ is not compact, for every continuous function $f:A \to \mathbb{R}$ would be bounded, and the function $f:A \to \mathbb{R}$ defined by $f(x) = \frac{1}{x-\sqrt{2}/2}$ is an unbounded continuous function. More generally, any set of the form $C=[a,b]\cap \mathbb{Q}$ is an example of a closed and bounded set which is not compact.
