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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

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Jonas Gomes
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1 Answers1

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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

$$X\setminus\{x\}=(0,2]=(0,1)\cup[1,2]\;,$$

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

Brian M. Scott
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  • Why would I use Hasdorfness to show that $K$ is open in $X$? To show that $X\setminus {x}$ is open? – Jonas Gomes Jan 15 '15 at 22:44
  • @Jonas: Yes. (Of course you’re really just using the $T_1$ property here.) – Brian M. Scott Jan 15 '15 at 22:49
  • Thanks! But I'm sorry I made a terrible mistake in the question. I'll edit it. My components are not necessarily open. – Jonas Gomes Jan 15 '15 at 22:52
  • @Jonas: The statement in its present form is false: see my revised answer. – Brian M. Scott Jan 15 '15 at 23:31
  • Thanks a lot, I lost a lot of time trying to prove a false statement with an obvious counter-example! – Jonas Gomes Jan 15 '15 at 23:36
  • @JonasGomes: Doesn't disjoint union of $A$ and $B$ mean that the space has the topology with respect to these subspaces, i.e. a set is open if its intersection with each of the two sets is open in that set? – Stefan Hamcke Jan 15 '15 at 23:41
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    @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. – Brian M. Scott Jan 15 '15 at 23:50
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    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. – Stefan Hamcke Jan 15 '15 at 23:57
  • The components are supposed to me maximal connexed sets, I'll post another question with this modification, thanks BrianMScott and @StefanHamcke – Jonas Gomes Jan 16 '15 at 19:05
  • I've posted the new question here: http://math.stackexchange.com/questions/1107062/complement-of-a-point-of-a-compact-connected-hausdorff-space-has-no-compact-maxi – Jonas Gomes Jan 16 '15 at 19:12