Let $X$ be a compact connected hausdorff space. Let $A\subseteq X$ such that $A$ is closed and $A\neq\emptyset,X$. Let $C$ be a (connected) component of $A$. Is it true that $C\cap\partial A\neq\emptyset$?
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Do you mean by component a subset? – aGer Jun 25 '15 at 21:40
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1@aGer: No, the OP means a connected component. – Brian M. Scott Jun 25 '15 at 21:41
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I edited my answer. Sorry about the previous mistake. – shalop Jun 27 '15 at 04:42
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Lemma 1: Connected components of any topological space are always closed sets in that space.
Proof: let $D$ be a connected component of some arbitrary topological space $Y$. Since closures of connected sets are connected, it follows that $\overline{D}$ is a connected set intersecting $D$. By definition of connected component, any connected set intersecting $D$ is contained in $D$, so it follows that $\overline{D} \subseteq D$. Thus $\overline{D}=D$, so $D$ is closed). $\Box$
Lemma 2: Let $X$ be a compact Hausdorff space, and let $U \subset X$ be open. If $C$ is a compact connected component of $U$, then $C$ is a connected component of $X$.
Proof: (This is the hardest part.) Apply the second lemma in this answer, with $Y=U$ and $Z=C\cup(X \backslash U)$. $\Box$
Returning to the original question, we see by Lemma $1$ that $C$ is a closed (hence compact) subset of $X$.
Assume for contradiction that $C \cap \partial A = \emptyset$. Since $C \subset A$, this means that $C$ and $\overline{X \backslash A}$ are disjoint closed subsets of the normal space $X$. Thus there exist disjoint open sets $U,V$ such that $C \subset U$ and $\overline{X \backslash A} \subset V$. Since $C$ is a connected component of $A$ and since $U \subset A$, it follows that $C$ is a connected component of $U$. By Lemma $2$, we see that $C$ is also a connected component of $X$. By connectedness of $X$ this implies that $C=X$ or $C=\emptyset$, contradiction.
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To prove that second lemma, he uses the first lemma, but I'm having trouble seeing why $C\cap B=\emptyset$ in the if part of that first lemma. In that lemma, the first one, $K$ is a compact subset of $X$, which is Hausdorff, and $B$ is $\partial K$ in $X$. In my question $X$ is Hausdorff, and compact and connected. If my $X$ is Hausdorff too, $A$ is compact too (because $A$ is closed in a compact) and my $C$ is a component of $A$, why there, in the if part of the first lemma, $C\cap\partial A=\emptyset$, but you prove that $C\cap\partial A\neq\emptyset$. What is going on?! – Quique Ruiz Jun 27 '15 at 22:15
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@QuiqueRuiz: The reason that $C \cap B = \emptyset$ is that $C$ has an open neighborhood (namely, the interior of $K$) which does not intersect $X \backslash K$. – shalop Jun 28 '15 at 03:42
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