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This is Exercise I.11.4 of Bourbaki's General Topology.

Let $X$ be a connected space.

a) Let $A$ be a connected subset of $X$, $B$ a subset of $\complement A$ which is both open and closed in $\complement A$. Show that $A\cup B$ is connected.

b) Let $A$ be a connected subset of $X$ and $B$ a component of the set $\complement A$. Show that $\complement B$ is connected (use a)).

I have managed to show a).

My attempts to show b): Let $C$ be a nonempty clopen subset of $\complement B$. By a), $B\cup C$ is connected. Since $B$ is a component of $\complement A$, $B\cup C$ cannot be a subset of $\complement A$. So $C$ has to contain an element of $A$. Since $A$ is connected, $A\subset C$. But I'm stuck here. I can't see how this implies $C=\complement B$.

Can someone point me in the right direction?

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up vote 6 down vote accepted

Suppose $B^{c} = U \cup V$ with $U,V$ disjoint and open. Then $U$ and $V$ are also closed and if they are non-empty, they both must contain $A$ by your argument, so they can't be disjoint.

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Thank you. This argument is pretty obvious, but somehow it was totally out of my reach. – Stefan Feb 11 '11 at 10:58
@Stefan: This happens to me all the time... Cheer up, you managed the important and more difficult part :) – t.b. Feb 11 '11 at 11:05
The part b) it's very difficult! I have no idea what can I do My try: Let´s suppose that A,B form a separation of M−C i.e M−C=A∪B and also holds that A ¯ ¯ ¯ ∩B=A∩B ¯ ¯ ¯ =∅ . Clearly M=C∪A∪B is a disjoint union. And since C it´s a connected component, C it's closed. If I prove that A∪B it's closed, then I'm ready. How can I prove that no limit points of A or of B , lies in C ? Please help me! And clearly I can suppose that X lies in A without loss of generality. But also considering that I don't know how to do it. – Arkj May 2 '12 at 7:16

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