6
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ While not exactly a duplicate, Brian M. Scott's answer to a different question provides one method to demonstrate this. $\endgroup$
    – user642796
    Commented Apr 8, 2015 at 3:31
  • $\begingroup$ @ArthurFischer Thanks for the link. I've looked at the answer but I'm not familiar with all of the terms. How does showing that X (or in this case G) is separable and metrizable and not locally compact anywhere equate to a homeomorphism? $\endgroup$
    – user2034
    Commented Apr 8, 2015 at 3:43
  • $\begingroup$ Arthur Fischer's answer to a different question actually answers your question since the set of infinite words on $\{0,1\}$ with an infinite number of $1$ is the intersection of the (open) sets of all infinite words having at least $n$ occurrences of $1$. $\endgroup$
    – J.-E. Pin
    Commented Apr 8, 2015 at 6:33

2 Answers 2

3
$\begingroup$

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.)

$\endgroup$
1
  • $\begingroup$ Thanks, that was excellent. It cleared up my confusion with Brian's answer as well as answering the question. $\endgroup$
    – user2034
    Commented Apr 8, 2015 at 12:24
0
$\begingroup$

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.

enter image description here

enter image description here

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .