# T1 topological space problem

1. Aug 17, 2010

1. The problem statement, all variables and given/known data

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.

3. 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)$$\cap$$B = A$$\cap$$Cl(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})$$\cap$$A\{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}$$\cap$$Cl(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.

2. Aug 17, 2010

### foxjwill

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.

3. Aug 18, 2010

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.

4. Aug 20, 2010