# Showing that for equivalent metrics, the separability of one metric space implies the separability of the other

Since $\rho$ and $\sigma$ are equivalent, there are positive numbers $c_1$ and $c_2$ such that for all $x_1, x_2 \in X$, $$c_1\cdot\sigma(x_1, x_2) \leq \rho(x_1, x_2) \leq c_2\cdot\sigma(x_1, x_2).$$

Separable means that there is a countable subset of $X$ that is dense in $X$. Also, a metric space $X$ is separable if and only if there is a countable collection $\{\mathcal{O}_n\}_{n = 1}^\infty$ of open subsets of $X$ such that any open subset of $X$ is the union of a subcollection of $\{\mathcal{O}_n\}_{n = 1}^\infty$.

I'm not sure how to relate these two characteristics of separability to the definition of equivalent metrics.

• Can you use 'countable dense subset' as a definition of separable? Feb 11 '14 at 3:12
• Ian Coley - Yes, that is how my book defines separable. Feb 11 '14 at 3:13

Second Solution: A subset $S$ of a metric space $(X,\rho)$ is dense if every $x \in X$ is the limit of a sequence $\{s_n\}$ with $s_n \in S$ for all $n \in \mathbb{Z}^+$. Suppose that $S$ is a countable dense subset with respect to the metric $\rho$, and let $x \in X$. Show that any sequence $\{s_n\}$ in $S$ which converges to $x$ with respect to $\rho$ also converges to $x$ with respect to the equivalent metric $\sigma$.