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Math Help - Separable metric spaces and bases

  1. #1
    Senior Member I-Think's Avatar
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    Separable metric spaces and bases

    Need help verifying/completing proof

    Here's a definition before the question

    Separable: A metric space is separable if it has a countable dense subset

    Base: A collection [V_\alpha] of open sets of metric space X is called a base for X if the following is true: For every x\in{X} and every open set G\subset{X} such that x\in{G}, we have x\in{V_{\alpha}}\subset{G}

    Question
    Prove every separable metric space has a countable base

    Proof
    Let E be a countable dense subset of metric space X and let [V_{\alpha}] be a base for X

    Let G be a neighborhood of e: d(x,e)<q, q\in{\mathbb{Q}}, e\in{E}

    E dense, so either \forall{x\in{E}} (in which case X countable, so proof follows) or
    \forall{x\in{E}} are limit points of X

    So x\in{[V_\alpha]}\subset{G}, for some \alpha and \forall{x}

    There are a countable number of sets G, so the base is countable

    QED
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Separable metric spaces and bases

    Quote Originally Posted by I-Think View Post
    Need help verifying/completing proof

    Here's a definition before the question

    Separable: A metric space is separable if it has a countable dense subset

    Base: A collection [V_\alpha] of open sets of metric space X is called a base for X if the following is true: For every x\in{X} and every open set G\subset{X} such that x\in{G}, we have x\in{V_{\alpha}}\subset{G}

    Question
    Prove every separable metric space has a countable base

    Proof
    Let E be a countable dense subset of metric space X and let [V_{\alpha}] be a base for X

    Let G be a neighborhood of e: d(x,e)<q, q\in{\mathbb{Q}}, e\in{E}

    E dense, so either \forall{x\in{E}} (in which case X countable, so proof follows) or
    \forall{x\in{E}} are limit points of X

    So x\in{[V_\alpha]}\subset{G}, for some \alpha and \forall{x}

    There are a countable number of sets G, so the base is countable

    QED
    I'm having a hard time following what precisely you mean, but I think you have the right idea. Namely, let (M,d) be separable metric space with countable dense subset \mathfrak{D} and define \mathfrak{B}=\left\{B_{q}(d):q\in\mathbb{Q}^+\text  { and }d\in\mathfrak{D}\right\} then \mathfrak{B} is a (countable) basis (really you just need \left\{B_{a_n}(d):n\in\mathbb{N}\text{ and }d\in\mathfrak{D}\right\} for some positive sequence \{a_n\} with \displaystyle \inf_{n\in\mathbb{N}}|a_n|=0). What I don't see is where you actually prove it's a basis. The idea would go something like let x\in M be arbitrary and U an arbitrary neighborhood. Choose \delta such that B_{\delta}(x)\subseteq U. Since \mathfrak{D} is dense you know there is some d\in B_{\delta}(x) different from x. You then finagle with neighborhod radii so that you can find a ball around d which contains x and stays within B_\delta(x).
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