Is zero-dimensional, separable metric space a $\sigma$-space? 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.
 A: 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 $F_\sigma$. 
A: 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:


*

*$\mathbf{\Sigma}^0_0 (X) = \{ U \subseteq X : U \text{ is open in } X \}$;

*$\mathbf{\Pi}^0_\alpha (X) = \{ X \setminus A : A \in \mathbf{\Sigma}^0_\alpha (X) \}$;

*$\mathbf{\Sigma}^0_\alpha (X) = \{ \bigcup_{n \in \mathbb{N}} A_n : \{ A_n : n \in \mathbb{N} \} \subseteq \bigcup_{\beta < \alpha} \mathbf{\Pi}^0_\beta (X) \}$ (for $0 < \alpha < \omega_1$).


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


*

*$\mathbf{\Sigma}^0_0 (X)$ the open subsets of $X$.

*$\mathbf{\Pi}^0_0 (X)$ the closed subsets of $X$.

*$\mathbf{\Sigma}^0_1 (X)$ the Fσ subsets of $X$.

*$\mathbf{\Pi}^0_1 (X)$ the Gδ subsets of $X$.


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 $\mathbb{N}^{\mathbb{N}}$.
