## anonymous 4 years ago "Every finite set is closed." Is this true?

1. anonymous

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

2. anonymous

Yes it is

3. anonymous

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. $$\blacksquare$$

4. JamesJ

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?

5. anonymous

It's the set of all its accumulation points.

6. anonymous

(or limit points)

7. anonymous

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

8. JamesJ

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.

9. anonymous

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...).