# Thread: Separable metric spaces and bases

1. ## 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), $q\in{\mathbb{Q}}, e\in{E}$

$E$ 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

2. ## Re: Separable metric spaces and bases

Originally Posted by I-Think
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), $q\in{\mathbb{Q}}, e\in{E}$

$E$ 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 $B_\delta(x)$.