Is it consistent with ZFC that there exists an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set?


If the continuum hypothesis holds, or more generally $2^{\aleph\_{0}}<2^{\aleph\_{1}}$
, then each uncountable separable metric space contains non-Borel sets since there are only $2^{\aleph\_{0}}$ Borel sets. Therefore, in order for there to be an uncountable separable metric space $X$ where every subset of $X$ is Borel, we would need to at least have $2^{\aleph\_{0}}=2^{\aleph\_{1}}$. What other conditions are necessary in order for there to exist an uncountable separable metric space $X$ where every subset of $X$ is Borel?



Does there exist an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set?
Under Martin's Axiom plus the negation of CH, every set $X$ of reals of size $<\mathfrak c$ is a Q-set, which means that every subset of $X$ is an $F\_\sigma$-set with respect to the subspace topology that $X$ inherits from the real line. For more about such sets, see Arnie Miller's chapter, "Special subsets of the real line" in the Handbook of Set-Theoretic Topology.


