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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.

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  • $\begingroup$ Can you use 'countable dense subset' as a definition of separable? $\endgroup$
    – Ian Coley
    Feb 11 '14 at 3:12
  • $\begingroup$ Ian Coley - Yes, that is how my book defines separable. $\endgroup$ Feb 11 '14 at 3:13
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First Solution: If you know about topological spaces, observe that equivalent metrics determine the same topology. Whether a subset is countable and dense depends only on the underlying topology, not the metric, so the statement must be true.

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 $\sigma$.

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  • $\begingroup$ Pete L. Clark - Thank you for your help. $\endgroup$ Feb 11 '14 at 6:50

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