Is it possible for a separable metric and a non-separable metric to have the same Borel $\sigma$-algebra? I hope this is not too basic or obvious a question.
Let $d_1$ and $d_2$ be metrics on the same set $X$, with $d_1$ being separable and $d_2$ not being separable. Is it possible that $d_1$ and $d_2$ generate the same Borel $\sigma$-algebra? If so, does the answer change if we require $d_1$ to be Polish?
Thanks.
 A: It is possible that $(X,d_{1})$ has the same Borel sets as $(X,d_{2})$ when $d_{1}$ is separable and $d_{2}$ is not by assuming the Martin's axiom and the negation of the continuum hypothesis by the answer https://mathoverflow.net/a/155527/22277 by Andreas Blass to my question. A set $X$ is said to be a $Q$-set if every subset of $X$ is an $F_{\sigma}$-set in $X$. If Martin's Axiom holds then every subset $X$ of $\mathbb{R}$ of cardinality less than the continuum is a $Q$-set. In particular, if MA and the negation of the continuum hypothesis holds, then if $L\subseteq\mathbb{R}$ has cardinality $\aleph_{1}$, then every subset of $L$ is a Borel set. Therefore, the metric space $L$ has the same Borel sets as the Borel sets where $L$ is given some discrete metric. 
$\textbf{Added 8/17/2015}$
I claim that there is no Polish metric $\rho$ and non-separable metric $d$ on a set $X$ so that $(X,\rho)$ and $(X,d)$ have the same Borel sets. 
Suppose that $(X,d)$ is a non-separable metric space. A metric space satisfies the countable chain condition if and only if it is separable. Therefore $(X,d)$ does not satisfy the countable chain condition. Therefore, let $\mathcal{U}$ be an uncountable pairwise disjoint collection of open subsets of $X$. Let $x_{U}\in U$ for each $U\in\mathcal{U}$. Let $A=\{x_{U}|U\in\mathcal{U}\}$. If $B\subseteq A$, then $B=\overline{B}\cap\bigcup\mathcal{U}$ which is a Borel set in $X$ being the intersection of a closed set with an open set. Thus $A$ is an uncountable set where every subset of $A$ is Borel in $X$.
On the other hand, I claim that there does not exist a polish space $(X,\rho)$ and an uncountable subset $A\subseteq X$ so that every subset of $A$ is a Borel set in $X$. If $(X,\rho)$ is a Polish space and $A$ is an uncountable Borel subset of $X$, then $A$ has cardinality continuum since every Borel subset of a Polish space is either countable or has cardinality continuum. Therefore, there are at least $2^{\mathfrak{c}}$ subsets of $A$, but there are only $\mathfrak{c}$ Borel subsets of $A$. Therefore some subset of $A$ is not Borel.
