How do I show that if I have a discrete metric space $X$ which is separable then it is countable?
As $X$ is separable, it contains a countable, dense subset $S$.
The metric on $X$ is defined by
$d(x,y)=1$ if $x\neq y$
$d(x,y)=0$ if $x= y$
[Math] Discrete metric separable is countable
functional-analysisgeneral-topologymetric-spaces
Related Question
- [Math] If $X$ is a separable metric space and $M \subset X$ is a metric subspace then $M$ is separable.
- [Math] Subset of separable metric space can have at most a countable amount of isolated points
- [Math] Prove that $\langle\mathbb{R}$, discrete metric$\rangle$ is not separable
- Subset of a basis of a separable metric space
Best Answer
A set $D\subseteq X$ is dense in $X$ if and only if $D$ intersects every non-empty open set in $X$. If $X$ is discrete, $\{x\}$ is open for each $x\in X$, and $D\cap\{x\}\ne\varnothing$ if and only if $x\in D$. Thus, the only dense subset of a discrete space is the whole space, and in order for the space to have a countable dense subset, it must itself be countable.