If $C_x(S)$ is the union of all connected subsets of $S$ which contain $x$, it is connected. I understand that, but what I don’t understand is that if $S$ is closed, then $C_x(S)$ is closed. Isn’t that like saying a subset of $S$ is closed, which is not always true?
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Let $S$ be closed in $X$. $C_x(S)$ is the component of $x$ in $S$. The components are always closed (see link, for example). If $S$ is closed, and the component of $x$ in $S$ is closed (in $S$), it is closed in $X$ as well.
Mariah
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The closure of a connected set is always connected. Hence the closure of $C_x(S)$ is another connected set containing $x$. Can you now see that this closure must be equal to $C_x(S)$? Thus $C_x(S)$ is closed in $S$. [This part is always true]. Since $S$ is given to be closed it follows that $C_x(S)$ is closed.
Kavi Rama Murthy
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