(adsbygoogle = window.adsbygoogle || []).push({}); 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)[tex]\cap[/tex]B = A[tex]\cap[/tex]Cl(B) = [tex]\emptyset[/tex]). 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})[tex]\cap[/tex]A\{xi} = [tex]\emptyset[/tex]. 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}[tex]\cap[/tex]Cl(A\{xi})= [tex]\emptyset[/tex].

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

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# Homework Help: T1 topological space problem

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