"Every finite set is closed."

Is this true?

Question

"Every finite set is closed."

Is this true?

anonymous · Answer

Yes, because for a finite value, there is a definite number which is the last one, so you can close it.

anonymous · Answer

Yes it is

anonymous · Answer

This is what I have:

Proof: Let $A$ be a finite set. Then $A=\{x_1,x_2,...,x_n\}$ for some $n\in\mathbb{N}$. Since every point in $A$ is an isolated point. Therefore, $A'=\varnothing$, and since $\varnothing\in A$, it follows that $A$ contains all its accumulation points. Therefore, $A$ must be closed. $\blacksquare$

jamesj · Answer

You mean in R^n, under the usual topology?  Yes.

Now, in your proof, there's something odd going on here.  What is A', the compliment of A?

anonymous · Answer

It's the set of all its accumulation points.

anonymous · Answer

(or limit points)

anonymous · Answer

I have a theorem that states that if a set contains all its accumulation points, then it must be closed.

jamesj · Answer

Ok, and then you mean to write $ \emptyset \subset A $.

This is slightly obscure proof; I would use the definition of open and closed directly, as opposed to relying on such a result that comes farther down the road.

anonymous · Answer

Another hint for a direct proof: any FINITE union of closed sets is closed. (You can construct open intervals in the usual topology on R by a countably infinite intersection of closed sets - you may want to prove that...).