If A is compact and B is closed then A Intersection B is compact.
I tried to solve it using the fact that compact set is closed and bounded but the problem here they did not tell us in which topology is compact or if its compact in R.
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Yahya Ram
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See https://math.stackexchange.com/questions/406172/intersection-of-closed-and-compact-set-is-closed?rq=1. Otherwise, check this out: https://math.stackexchange.com/questions/35038/is-the-intersection-of-a-closed-set-and-a-compact-set-always-compact – morrowmh May 05 '21 at 21:20
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4Does this answer your question? Intersection of Closed and Compact Set is Closed – Conrad Crowley May 05 '21 at 21:21
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no didn't answer my question in this question they are proving that if A compact and B closed then the intersection is closed. but my question I have to prove that the intersection is compact. – Yahya Ram May 05 '21 at 21:34
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But a closed subset of a compact set is compact. – Randall May 05 '21 at 21:37
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If $A$ is compact and $B$ is closed, then because $A$ is closed, $A\cap B\subset A$ also is closed. Therefore, it suffices to show that closed subsets of compact sets are compact.
Let $X$ be a compact set, $V\subset X$ a closed subset. Any open cover $\{U_i\}_{i\in I}$ of $V$ can be extended to an open cover of $X$, e.g. by taking $X\backslash V$ and all the $U_i$. Since $X$ is compact, there is a finite subset $J\subset I$ such that $X = \bigcup_{j\in J} U_j \cup X\backslash V$. Then
$$V = \bigcup_{j\in J} V\cap U_j $$
is a finite subcover. Therefore, $V$ is also compact.
S.Farr
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