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Math Help - Open sets

  1. #16
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    In general, if you are given a subset of a topological space, which is itself referred to as a topological space, then it is assumed to have the subspace topology. Any subset of a topological space inherits the ambiant topology in a natural way. Let \iota : S^2 \to \mathbb{R}^3 be the inclusion map. The topology on S^2 is the topology generated by the open sets of the form \iota^{-1}(U), where U is open in \mathbb{R}^3.

    Now it is easy to solve the problem. Show that any open cover of a set in S^2 lifts to an open cover of the same set, considered as a subset of \mathbb{R}^3. Conversely, any open cover of some subset of S^2 (considered as a subset of \mathbb{R}^3) can be considered as an open cover of the same set as a set of S^2.

    By the way, the notation S^n for the n-sphere is universal!
    Lifting maps are way, way overkill for a simple problem like this, why introduce it?

    The more simplistic idea is as follows...

    Let X be a topological space and Y a subspace. Then, Z\subseteq Y\subseteq X is a compact subspace of Y if and only if it's a compact subspace of X

    To see this first assume that Z is a compact subspace of Y and let \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} be an open cover Z with the subspace topology inherited from X. Then, U_\alpha=V_\alpha\cap Z where V_\alpha is open in X. But, note that V_\alpha\cap Z=V_\alpha\cap Z\cap Y=\underbrace{\left(V_\alpha\cap Y\right)}_{\text{ope}\text{n in }Y}}\cap Z and so \left\{V_\alpha\cap Y\cap Z\right\}_{\alpha\in\mathcal{A}} has a finite subcover (since it's an open cover of Z as a subspace of Y) from where the conclusion follows.

    The other direction is similar.
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  2. #17
    MHF Contributor Bruno J.'s Avatar
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    I'm not sure what you're saying. I'm not speaking of "lifting" in the sense of lifting a path from a manifold to some covering! I'm just saying that every open set o of S^2 (as a topological space) can be obtained as the intersection of some open set o' of \mathbb{R}^3 with S_2 (as a subset).

    What you wrote is exactly what I was suggesting.
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  3. #18
    MHF Contributor Bruno J.'s Avatar
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    I'd also like to point out that the map U_\alpha \mapsto V_\alpha which I described does constitute a lift in the sense of category theory.
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  4. #19
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    Ok, i think i solved this, more or less, thanx all for the help and ideas

    And let me say again, that it was a mistype in the header, it shouldnt be "open set" just "any set".

    S^2 is compact if it is closed and bounded, and as S^2 contains all its 'peripheral points' (that is probably not the right word in english, but i hope it speaks for itself). Also there exist a 'ball'
    containing S^2, so S^2 is compact, therefore K is compact in S^2 if and only if K is closed.. Since K ⊆ S^ 2 ⊆ R^3, then K will be compact in R^3 <=> K is closed.
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  5. #20
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    Great.. I completely missed Drexel28's answer, which was more or less what i was after... Anyhoo i think its more or less solved now, thanx all
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