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This question is a slight modified version of Compact Connected Hausdorff Space has no compact component in the complement of a point

Let $X$ be a Hausdorff Compact Connected Space. Prove that $X∖\{x\}$ has no compact component (here a component is a maximal connected subspace).

(My attempt is in the question linked above)

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  • $\begingroup$ It's true if $X$ is locally connected, at least, with an easy proof (the component would be non-trivial clopen subset of $X$). This fact should hold for all $x$, right? $\endgroup$ – Henno Brandsma Jan 17 '15 at 7:10
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In this answer the following lemma is proved:

Lemma

Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$

Also:

Let $X$ be a compact Hausdorff space, $Y$ an open subspace and $Z$ a closed subspace. Let $C$ be a connected subset of $Y \cap Z$ such that $C$ is a component of $Y$ and a component of $Z$. Then $C$ is a component of $Y\cup Z$.

which can be simplified to (as per the remark by Hamcke on that answer): if $X$ is a compact normal space and $Y$ is an open subset of $X$, then a compact connected component of $Y$ is also a connected component for $X$.

This applies directly to your question: your $X$ is compact and normal (follows from compact plus Hausdorff) and $Y = X \setminus \{x\}$ is open, and if $C$ were a compact component of $Y$ it would be one for $X$ but this cannot be, as the only connected component for $X$ is $X$ itself.

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  • $\begingroup$ Thanks for your answer. I need some time to look it trough and understand all the details. $\endgroup$ – Jonas Gomes Jan 17 '15 at 16:45
  • $\begingroup$ Actually, $C$ must be closed, not compact, in order to apply the normality. $\endgroup$ – Stefan Hamcke Jan 17 '15 at 22:22
  • $\begingroup$ @StefanHamcke compactness was in the original question, so that case is enough... And compact implies closed here anyway. $\endgroup$ – Henno Brandsma Jan 19 '15 at 21:33

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