Are one-point sets always compact in any topological spaces?

Question

From the definition of compactness, I think one-point sets are always compact in any topological space. But, I am not sure about my judgement. Am I correct?

Answer 1

Every finite set is compact. This is because, one can always find a finite subcover, which can be proven inductively:

Let $X:=\{x_1, \dots x_n\}$ be a finite set. Suppose that $\{U_x\}$ covers $X$. Take any $U_{x_1}$ that covers $x_1$. Consider $\{U_x\}\setminus U_{x_1}$. Then pick one that covers $x_2$ if $U_{x_1}$ does not, etc.

Answer 2

Yes, one point set is always compact in any topological space, because it will be contained in an open set of any cover and that is the finite one.

Answer 3

Since I was asked to repost my comment as an answer:

Any open cover of any topological space is a subset of the power set of the underlying set, and power sets of finite sets are finite. So all open covers of finite spaces are already finite; in other words, finite spaces only have finitely many open sets.

Answer 4

Yes. Suppose the open sets $U_i$ with $i$ in some index set $I$ cover your one-point set $\{x\}$. Covering this set means that $x\in U_i$ for some $i\in I$. Therefore your one-point set is contained in a single open set of your open cover. This is certainly a finite subcover!

Notice that the argument did not use the fact that the sets $U_i$ are open. This is a symptom of generality: It is true for any sets $U_i$ and therefore for any topology of the ambient space at all.