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Math Help - Compact Set (topology)

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    Question Compact Set (topology)

    Let (X, d) be a metric space.

    Show that any finite subset is compact.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    Let (X, d) be a metric space.

    Show that any finite subset is compact.
    Are you serious? Let \mathcal{V} be an open cover for some finite subset \mathcal{S} of X. There must exists some O_s\in\mathcal{V} such that s\in O_s for each s\in\mathcal{S}. Clearly then \left\{O_s\right\}_{s\in\mathcal{S}} is a subcover of \mathcal{S} and since \mathcal{S} is finite so is \left\{O_s\right\}_{s\in\mathcal{S}}.
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    Hi

    Thanks for the reply, could you explain what is this O small s?
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    These "O-sub-s" things are open sets; the union of these sets forms an open cover for your finite set.

    Recall that one of the properties of a compact set is that every open cover of that set admits a finite sub-covering.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hebby View Post
    Hi

    Thanks for the reply, could you explain what is this O small s?
    Quote Originally Posted by h2osprey View Post
    These "O-sub-s" things are open sets; the union of these sets forms an open cover for your finite set.

    Recall that one of the properties of a compact set is that every open cover of that set admits a finite sub-covering.
    While , what h2osprey said was undeniably correct, I think the finer points got lost.

    To be compact means that every open cover has a finite subcover. Thus, to show that a set is compact one way is to take an arbitrary open cover and derive a finite subcover. So, the concept is we let \mathcal{V} be some open cover of \mathcal{S}. Next, we let s\in\mathcal{S} be arbitrary, and since \mathcal{V} covers \mathcal{S} there exists some open set, which we will call O_s, in \mathcal{V} such that it contains s. So, for each s\in\mathcal{S} we take some corresponding O_s\in\mathcal{V} and construct a new class \left\{O_s\right\}_{s\in\mathcal{S}}. This is clearly a subcover, and since there is at most \text{card }\mathcal{S} elements of \left\{O_s\right\}_{s\in\mathcal{S}} it must be a finite subcover.

    The two finer points that you should definitely answer yourself is why can I choose this O_s and why is there at most \text{card }\mathcal{S} elements in \left\{O_s\right\}_{s\in\mathcal{S}}
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