Let $X$ be a Hausdorff Compact Connected Space. Prove that $X\setminus\{x\}$ can't be expressed by the disjoint union of two connected sets with one them being compact.(lets assume the empty set is connected)

So, I tried a demonstration by contradiction:

If $X\setminus\{x\}$ is connected, then if $X\setminus\{x\}$ is compact, it's a closed subspace of $X$ (because $X$ is Hausdorff) and therefore it's a clopen on $X$. So, $X\setminus\{x\}$ can't be disconnected. Assume that there are disjoint connected sets $U$ and $V$ of $X\setminus\{x\}$ such that $$X\setminus\{x\} = U \cup V$$ And $U$ is a compact set. $U$ can't be open on $X\setminus \{x\}$ (or otherwise it'd be a clopen set of $X$). But as $U \cup V =X= \cup_{i\in I} A_i$ with $A_i$ open connected subsets of $X\setminus \{x\}$. Because $U$ is connected, $U \subset A_{i_1}$ and the same goes to $V \subset A_{i_2}$. $X \subset U \cup V \subset A_{i_1} \cup A_{i_2}$. Because $X$ is not connected, $i_1 \neq i_2$. Also, because $A_{i_2} \cap U = \emptyset$ and then $A_{i_2} = V$ and therefore $U = A_{i_1}$ and $U$ is open, a contradiction.

As it is stated, the proposition is false (check Brian's answer). I've corrected it in another question: Complement of a point of a Compact Connected Hausdorff Space has no compact maximal connected subspace

  • $\begingroup$ I suspect that you are missing the condition that the subsets in question are connected components of $X -\{x\}$. $\endgroup$ – Moishe Kohan Jan 15 '15 at 23:48

Revised and corrected to match the edited question.

The statement is false. Let $X=[0,2]$ with the usual topology, and let $x=0$. Then


where $(0,1)$ and $[1,2]$ are disjoint and connected, and $[1,2]$ is compact.

  • $\begingroup$ Why would I use Hasdorfness to show that $K$ is open in $X$? To show that $X\setminus \{x\}$ is open? $\endgroup$ – Jonas Gomes Jan 15 '15 at 22:44
  • $\begingroup$ @Jonas: Yes. (Of course you’re really just using the $T_1$ property here.) $\endgroup$ – Brian M. Scott Jan 15 '15 at 22:49
  • $\begingroup$ Thanks! But I'm sorry I made a terrible mistake in the question. I'll edit it. My components are not necessarily open. $\endgroup$ – Jonas Gomes Jan 15 '15 at 22:52
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    $\begingroup$ @Stefan: Depends on who’s using it and on the context. Here I would not understand it that way. If it is intended that way, the stated result is true and fairly trivial: both sets must be open in $X$ as well as in $X\setminus\{x\}$, so $K$ is clopen in $X$, which is impossible. $\endgroup$ – Brian M. Scott Jan 15 '15 at 23:50
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    $\begingroup$ Oh, right, I did not see that the result is trivial in that case. So I think you are right and it indeed means that $X$ is just the disjoint union of these sets. $\endgroup$ – Stefan Hamcke Jan 15 '15 at 23:57

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