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Sorry for the "crazy" title, but that's what it is...

Let $X$ be a topological space.

Let $E$ be a subset of $X$.

Let $F$ be a connected component of $E$.

Is the following true? Why?

For every connected component $C$ of $X \setminus F$, there exists a connected component $D$ of $X \setminus E$ such that $D \subseteq C$.

In case there is a counterexample, can we restrict the hypotheses on $X$ and $E$ to make it work? (For example, compactness of $E$ or $X = \mathbb R^2$.)

Thanks in advance for your help.

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up vote 2 down vote accepted

Revised to match the revised question:

The answer to the revised question is yes. Let $C$ be a component of $X\setminus F$. There must be a component $D$ of $X\setminus E$ such that $D\cap C\ne\varnothing$. If $D\nsubseteq C$, then $C\cup D$ is a connected subset of $X\setminus F$ properly containing the component $C$, which is impossible. Thus, $D\subseteq C$.

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I meant $D \subseteq C$ instead of $C \subseteq D$ in the question, which changes everything. Sorry for making you lose your time. – turtle Nov 14 '12 at 14:28

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