# Closed property of nonempty finite set

I came across this text in Rudin's book where it has been mentioned that a non-empty finite set is closed.

But a closed set is a set which contains all of it's limit points in the set itself but none of the elements of a non-empty finite set can possibly have a limit point because a neighborhood of a limit point has infinite points . So how come a non-empty finite set be a closed set ??

• A singleton is obviously closed (in the context of Rudin), and then you have that a finite set is the union of finitely many closed sets, and thus closed. – TSF Oct 25 '17 at 11:29
• In the world of general topology, your statement is false, as you only specify a finite set but no topology. However, a non-empty finite subset of, say, $\mathbb{R}$ or $\mathbb{R}^n$ (with the standard topology), is always closed. – MartianInvader Oct 25 '17 at 18:22
• @MartianInvader Well I am considering it in terms of real analysis but thanks for mentioning the fact - I will remember it when I will read topology – Aashish Loknath Panigrahi Oct 25 '17 at 18:56
• By the way, in the more general setting, a topological space in which all finite sets are closed is called a T₁ space. Spaces used in analysis are always T₁. By the way, you do not need the word non-empty, of course, since the empty subset is closed as well. – Jeppe Stig Nielsen Oct 25 '17 at 19:38
• It is also easy enough to directly check that the complement is open, which is another way of seeing why this is true. – John Coleman Oct 25 '17 at 22:18

"$A$ Contains all of its limit points" is short for "for all $x$ such that $x$ is a limit point of $A$ we have that $x \in A$".

This statement is voidly true if there are no limit points (as in this case).

Equivalently put, the condition can be written as $A' \subseteq A$ and the empty set is a subset of all sets.

Disregarding possible fine print of definitions the resolution is quite simple:

If there are no limit points, then of course all limit points are trivially contained in the set.

Alternatively a set $A$ is closed iff $A=\bar A= A \cup \{\text{set of all limit points}\}.$

In your case, $\{\text{set of all limit points}\}=\emptyset.$ Thus $A=\bar A$.