There is the question: deduce that $Y$ is a compact subset of $(X,d)$ iff the metric space $(Y,d)$ is compact. (Given that $Y$is a subset of $X$).
How to show it? (I cannot find anything to show though...)
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1This is a straightforward application of the the subspace topology. What party of the proof are your stuck on? – Sort of Damocles Apr 28 '18 at 12:58
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There is a difference between a compact subset of a metric space, and a compact metric space. This is about showing that the difference is small enough that they deserve to use the same word. – Arthur Apr 28 '18 at 13:01
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Something to help you along...is every subset of a compact metric space $(X,d)$ also compact? What could go wrong? – erfink Apr 28 '18 at 19:23
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I suppose that the context is this one: we say that $Y(\subset X)$ is compact if, for every family $(A_\lambda)_{\lambda\in\Lambda}$ such that $Y\subset\bigcup_{\lambda\in\Lambda}$, there is a finite subset $F$ of $\Lambda$ such that $Y\subset\bigcup_{\lambda\in F}$. If so, saying that $(Y,d)$ is a compact space and saying that $Y$ is a compct subset of $X$ are two different things. It's not hard to prove that they are equivalent.
José Carlos Santos
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So does it mean that I need to show sequential compactness is equivalent to compactness? – user556970 Apr 28 '18 at 13:21
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The definition that you mentioned above is the compactness of a subset, right? – user556970 Apr 28 '18 at 13:51
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But I thought that the definition above were the same as "every open cover has a finite subcover"... Could you write the methematical expressions for "every open cover has a finite subcover"? Thank you! – user556970 Apr 28 '18 at 13:56
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@user556970 For every family of open subsets of $Y$ whose union is $Y$, there is a finite subfamily whose union is still $Y$. – José Carlos Santos Apr 28 '18 at 13:59
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Does it mean that, the difference between compactness of a subset and that of a metric space: cpct of a subset only needs ⊂ but cpct of a metric space needs = ? (Is this the difference betweem two definitions?) – user556970 Apr 28 '18 at 14:05
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Be sure you are using a definition of compactness that does not depend on the ambient space, such as the traditional one that any open cover admits a finite subcover.
C Monsour
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