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Homework Help: T1 topological space problem

  1. Aug 17, 2010 #1


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    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.
  2. jcsd
  3. Aug 17, 2010 #2
    This would only contradict the fact that [itex]A[/itex] is connected if both [itex]\{x_i\}[/itex] and [itex]A\setminus\{x_i\}[/itex] were open; this is clearly not true in general.

    I'd suggest trying induction. Specifically, given k distinct points [itex]x_1,\ldots,x_k[/itex] in a connected set [itex]A\subseteq X[/itex], try and find a point [itex]x_{k+1}\in A[/itex] which is distinct from all the other [itex]x_i[/itex]'s.
  4. Aug 18, 2010 #3


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    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.
  5. Aug 20, 2010 #4


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    Any further thoughts?
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