How can one prove that every connected subset of a T1 space is infinite?

[radou]

Homework Statement

Let X be a non empty T1 space (i.e. such one that for every two distinct points each one of them has a neighborhood which doesn't contain the other one). One needs to show that every connected subset of X, containing more than one element, is infinite.

The Attempt at a Solution

Let A be a connected subset of X containing more than one element. Assume A is finite, with cardinality k, so A = {x1, ..., xk}. Take any element from A, let's say xi. I claim that then the sets {xi} and A\{xi} are two non-empty separate sets whose union is A, which is a contradiction with the fact that A is connected. (Two sets A and B are separate if Cl(A)\capB = A\capCl(B) = \emptyset). The proof of this claim is the part I'm not quite sure about: Since X is T1, {xi} is a closed set, so Cl({xi})\capA\{xi} = \emptyset. Now, to fulfill the other requirement for separation, I need to show that xi is not in the closure of A\{xi}. If it would be, then every neighborhood of xi would intersect A\{xi}. But then X\(A\{xi}) is a neighborhood of xi which doesn't intersect A\{xi}, so {xi}\capCl(A\{xi})= \emptyset.

Perhaps there's a more easy way to prove it, but I didn't manage to do so, if this works at all.

 

[foxjwill]

I claim that then the sets {xi} and A\{xi} are two non-empty separate sets whose union is A, which is a contradiction with the fact that A is connected.

This would only contradict the fact that A is connected if both \{x_i\} and A\setminus\{x_i\} were open; this is clearly not true in general.

I'd suggest trying induction. Specifically, given k distinct points x_1,\ldots,x_k in a connected set A\subseteq X, try and find a point x_{k+1}\in A which is distinct from all the other x_i's.

 

[radou]

This would only contradict the fact that A is connected if both \{x_i\} and A\setminus\{x_i\} were open; this is clearly not true in general.

My proof is based on a theorem which states that the following statements are equivalent:

(i) X is connected
(ii) if X is the union of two separate sets A and B, then either A or B is empty.

In our case, I have shown that A is the union of two separate non-empty sets, so it can't be connected.

Perhaps I'm really missing something here.

 

[radou]

Any further thoughts?