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We have a compact set $X$ and a closed set $E \subset X$, how to prove that $E$ is compact?

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  • $\begingroup$ Belongs means an element of, $\in$. $\endgroup$
    – Asaf Karagila
    Commented Mar 21, 2015 at 17:35
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    $\begingroup$ Please search before asking. $\endgroup$
    – user147263
    Commented Mar 21, 2015 at 17:36

2 Answers 2

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Hint: Given an open cover of $E,$ how do you find an open cover of $X,$ using the fact that $E$ is closed?

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Let $\{B_n\}$ be an open cover for $E$. Because $E$ is closed, $E^c$ is open, and $\{B_n\} \bigcup \{E^c\}$ is an open cover for $X$, which has a finite open sub-cover $\{C_n\}$ (because $X$ is compact).

If $E^c \notin \{C_n\}$ we have extracted a finite sub-cover for $E$ out of $\{B_n\}$.

If $E^c \in \{C_n\}$, then $\{C_n\} \setminus \{E^c\}$ is still open and finite cover for $E$ and consists only out of elements of $\{B_n\}$, so we again extracted a finite sub-cover for $E$ from $\{B_n\}$.

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