A $G_δ$ subset of $2^ω$ that is homeomorphic to $ω^ω$ How do I show that there is a $G_δ$ subset of the Cantor space $2^ω$ that is homeomorphic to the Baire space $ω^ω$?
I've been given the hint to consider $G = \{x ∈ 2^ω : x\text{ is not eventually constant}\}$, but I'm not entirely sure what to do with it. Any help is appreciated.
 A: As noted in a comment to the question, a previous answer by Brian M. Scott provides one method to demonstrate that the particular subset $G$ is homeomorphic to the Baire space. In particular, he uses the following characterisation of the Baire space:

The [Baire space $\omega^\omega$] is (up to homeomorphism) the unique zero-dimensional, separable, Čech-complete metrizable space that is nowhere locally compact.

He then shows that $G$ has all of these properties to conclude that it is homeomorphic to the Baire space without explicitly constructing a homeomorphism.

If you don't have this "pile-driver" handy, you'll probably have to construct the homeomorphism by hand. 
First, to show that $G$ is Gδ, note that $G = \bigcap_n U_n$ where $$U_n := \{ \mathbf{x} \in 2^\omega : \mathbf{x}\text{ switches between }0\text{ and }1\text{ at least }n\text{ times}\}.$$ (So $\mathbf{x} \in U_n$ iff $\mathbf{x}$ has an initial segment of the form $0^{k_0} 1^{k_1} \cdots b^{k_n}$ or $1^{k_0} 0^{k_1} \cdots b^{k_n}$ where each $k_i > 0$, and $b$ is the appropriate bit.)
To construct the homeomorphism, note that given any $\mathbf{x} \in G$, we may write it as an infinite concatenation as $$\mathbf{x} = 0^{k_0} 1^{k_1} 0^{k_2} 1^{k_3} \cdots$$ where $k_0 \geq 0$, and $k_i > 0$ for all $i > 0$. Using this we define a function $\varphi : G \to \omega^\omega$ as follows: $$\varphi ( \mathbf{x} ) = ( k_0 , k_1 - 1 , k_2 - 1 , k_3 - 1 , \ldots ).$$ 
It is not too difficult to show that this function is a homeomorphism from $G$ onto the Baire space.

As a final note, it is somewhat superfluous to show that $G$ is Gδ since it is a theorem that any completely metrizable subspace of a completely metrizable space must be a  Gδ subset of that space. That is, the existence of the homeomorphism between $G$ and the Baire space shows that $G$ is a Gδ subset of the Cantor space. (Also any subspace of the Cantor space which is homeomorphic to the Baire space is a Gδ subset of the Cantor space; e.g, the subspace provided in Arthur Fischer's answer to the same question.)
A: The required claim easily follows from general facts about metrizable spaces. The Baire space $\omega^\omega$ is zero-dimensional second countable completely metrizable. Since $\omega^\omega$ is zero-dimensional second countable, it can be embedded into Cantor Cube $\{0,1\}^\omega$ (see, for instance, [Eng, 6.2.16]). Since $\omega^\omega$ is  completely metrizable, it is a $G_\delta$ sibset of $\omega$ for any its embedding (see, for instanse, [Eng, 4.3.24]).
References
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.



