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

    Question
    Prove every separable metric space has a countable base

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

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

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

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

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

    Question
    Prove every separable metric space has a countable base

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

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

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

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

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