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This question came from Dugundji's $\textit{Topology}$: Given a compact, connected space $X$, let $A \subset X$ be closed. Prove that there exists a closed, connected set $B \subset X$ such that $A \subset B$ and any proper subset of $B$ is either not connected, not closed, or does not contain $A$. The text has an annoying habit of not stating whether the space is assumed to be Hausdorff.

My attempt at a proof seems to work for Hausdorff spaces, but it doesn't rely fundamentally on a lot of the hypotheses and feels rather inelegant.

If such a $B$ does not exist, then for any $B$ which is connected, closed, and contains $A$, there exists a $B' \subset B$ which also satisfies those conditions. Clearly $X$ is an example of a set which does satisfy the conditions, so at least one such set exists. Order subsets of $X$ which are closed, connected, and contain $A$ by reverse inclusion (i.e. $U > V$ if $U \subset V$ and $U \neq V$). Each chain of subsets has an upper bound: let $B_i$ satisfy the hypotheses and let $B_{n+1} \subset B_n$. Then $B_N = \cap_1^\infty B_i$ is clearly closed and contains $A$. By the result mentioned here: Arbitrary intersection of closed, connected subsets of a compact space connected? , $B_N$ is connected if $X$ is Hausdorff. Thus $B_N$ is an upper bound for the chain $\{B_i\}$. By Zorn's Lemma, this collection has a maximal element, which would have to be the set $B$ needed.

I could not find an counterexample among non-Hausdorff spaces, so I would love to see one or a proof of the general case. In addition, the hypothesis that $A$ is closed isn't used anywhere here.

I'd love to see anyone's attempts at this problem!

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Dorebell
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  • Why don't you take $\bigcap { B; ; \text{$B$ is a closed, connected subset of $X$ and $A ⊂ B$}}$? If $X$ is Hausdorff, I trust you that it's connected. It certainly is closed in $X$ and it contains anything in $A$. By definition, there can't be a smaller set with that property. – k.stm Jan 24 '13 at 09:07
  • @K.Stm.: that doesn't work. Intersection of connected subsets need not be connected. E.g. take $X = S^1$ and $A$ the union two points in $S^1$. Then two arcs containing these points as boundaries are closed and connected but their intersection is just $A$. – Marek Jan 24 '13 at 10:26
  • @Marek Oh, yeah. Right. I didn't think about that for too long. By the way, don't you need to consider uncountable chains as well? – k.stm Jan 24 '13 at 10:28
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    Dugundji defines compactness to include Hausdorffness, so the space $X$ is indeed assumed to be Hausdorff. – Brian M. Scott Jan 25 '13 at 19:01
  • @K.Stm. I believe the proof works just as well for uncountable chains. Let $B_\alpha: \alpha \in I$ be an uncountable chain (i.e. totally ordered by proper subset inclusion). The intersection is still an upper bound, and I believe the proof that it is connected doesn't use anything countability. – Dorebell Jan 25 '13 at 22:35
  • *about countability – Dorebell Jan 25 '13 at 22:42

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I think the Arturo's example from the question you linked to works as a counter-example here as well. Take as $X$ the interval with doubled-origin and as $A$ the two origins. $A$ is obviously closed. Moreover, every connected closed set containing $A$ is just a closed interval. But any such interval contains a smaller closed interval. So the Hausdorff assumption is indeed necessary.

Marek
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  • Thank you for the clarification! I assumed that space was disconnected since it is the union of two non-identical copies of the real line. I see the mistake now. – Dorebell Jan 25 '13 at 22:43