A rather non-$F_\sigma$ Borel set I asked this question at MSE a week ago, but received no answer, so I cross-post it here.
I obtained a negative answer to this MSE question provided each metric space $X$ such that $|X|=\frak c$ and density $d(X)<\frak c$, contains a Borel set $B$ such that $|B\setminus C|=\frak c$ for each $F_\sigma$-subset $C$ of $X$ with $C\subset B$. My question is whether the latter claim holds. I guess this is  known (and true), but it is hard to find a reference. Thanks.
My try. I guess using Theorem 22.4 from [Kech] I can show the claim when $X$ is Polish. To prove the claim for a separable $X$, by Proposition 12.1 from [Kech], it suffices to prove it for subspaces of an arbitrary fixed Polish space.
References
[Kech] A. Kechris, Classical Descriptive Set Theory, Springer, 1995.
 A: Here's an argument that the statement is false if the Continuum Hypothesis fails and the covering number for the null ideal is the same as the continuum. Wellorder the Borel sets of reals as $\langle B_{\alpha} : \alpha < \mathfrak{c} \rangle$. Choose for each $\alpha < \mathfrak{c}$ an $F_{\sigma}$ set $C_{\alpha} \subseteq B_{\alpha}$ such that $B_{\alpha} \setminus C_{\alpha}$ is null and a real $x_{\alpha}$ not in $B_{\beta} \setminus C_{\beta}$ for any $\beta < \alpha$. Let $X = \{ x_{\alpha} : \alpha < \mathfrak{c}\}$. Then any Borel subset of $X$ is $B_{\alpha} \cap X$ for some $\alpha$. Furthermore, $C_{\alpha} \cap X$ is an $F_{\sigma}$ subset of $B_{\alpha} \cap X$, and $(B_{\alpha} \cap X) \setminus (C_{\alpha} \cap X)$ is contained in $\{ x_{\beta} : \beta \leq \alpha\}$ which has cardinality less than $\mathfrak{c}$.
As for the consistency of the statement that CH is false and $\mathrm{cov}(\mathcal{N}) = \mathfrak{c}$, this is a standard consequence of MA + not-CH (see Theorem 26.39 of the 2003 edition of Jech's Set Theory). The consistency of MA and not-CH is Theorem 16.13 of Jech.
I have to revise my earlier claim that the statement in question follows from CH. This appears to be true for spaces $X$ having a Borel subset which is not $F_{\sigma}$, by the idea in my original comment : if $B$ has an $F_{\sigma}$ subset $C$ such that $B \setminus C$ is countable, then $B$ is $F_{\sigma}$. On the other hand, one can run the proof in the first paragraph of this answer under CH to produce a set of reals of cardinality $\mathfrak{c}$ such that every Borel set is $F_{\sigma}$. So the statement would fail for such a space.
