Let $X$ be a zero-dimensional, separable metric space. Is then $X$ a $\sigma$-space (i.e. every $G_\delta$ subset is $F_\sigma$ in $X$)?
I do not know.
Let $X$ be a zero-dimensional, separable metric space. Is then $X$ a $\sigma$-space (i.e. every $G_\delta$ subset is $F_\sigma$ in $X$)?
I do not know.
On
This is certainly not true. Recall that for a metrizable space $X$ we can define Borel pointclasses $\mathbf{\Sigma}^0_\alpha (X)$ and $\mathbf{\Pi}^0_\alpha (X)$ for $\alpha < \omega_1$ as follows:
Furthermore $\bigcup_{\alpha < \omega_1} \mathbf{\Sigma}^0_\alpha (X) = \bigcup_{\alpha < \omega_1} \mathbf{\Pi}^0_\alpha (X)$, and this is the collection of all Borel subsets of $X$.
The first few levels of this hierarchy are
The following is a theorem you can probably find in any introductory text in descriptive set theory (e.g., Theorem 22.4 in Kechris's Classical Descriptive Set Theory, and Corollary 3.6.8 in Srivastava's A Course on Borel Sets):
Theorem. If $X$ is any uncountable Polish (i.e., separable completely metrizable) space, then $\mathbf{\Sigma}^0_\alpha (X) \neq \mathbf{\Pi}^0_\alpha (X)$ for any $\alpha < \omega_1$.
In particular, the collections of Fσ- and Gδ-sets are different in these spaces, and so neither is a subfamily of the other. Thus any uncountable zero-dimensional Polish space will serve as a counterexample: e.g., the Cantor space $2^\mathbb{N}$, the Baire space $\mathbb{N}^{\mathbb{N}}$.
I think the Cantor set $C$ is a counterexample. Remove a countable dense subset from it, which leaves a dense $G_\delta$ set which is not $\sigma$-compact so not an $F_\sigma$.