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Let $X$ be a topological space, and $G$ an open subset. If $E$ is a connected component of $G$, then is the boundary of $E$ is contained in that of $G$?

I know that it is true if $X$ is locally connected. But I suspect the statement is generally false, could anybody give a proof if it is true, and otherwise give a counterexample. Thanks.

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Here’s a counterexample.

Take $X$ to be the Knaster-Kuratowski fan, and let $G$ be $X\setminus\{p\}$, where $p$ is the dispersion point (apex) of the fan. $G$ is totally disconnected, so if $E$ is a connected component of $G$, then $E$ is a singleton and is its own boundary, but the boundary of $G$ is $\{p\}$.

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The phenomenon is not very uncommon. It is enough if $G$ is clopen (so its boundary is empty) and $E$ is not open. Then $\partial E = \mathrm{cl}(E) \setminus \mathrm{int}(E) = E \setminus \mathrm{int}(E) \neq \emptyset$. Since any space is clopen in itself, it is enough to find an example of a space with non-open connected component - which is an elementary topological excercise.

An example: $\{0\}\cup\{\frac{1}{n}\ |\ n\in \mathbb{N} \}$, where $\{0\}$ is the a non-open component. Another example: $\mathbb Q$ - any singleton is a non-open component.

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