What is the cardinality of $\omega^\omega$?

Question

I have started reading about Baire Space as a prerequisity for the subject of Borel hierarchies. And if I understand correctly, Baire space is $\omega^\omega$ (The space of all functions from $\omega$ to $\omega$)? I guess the Topology on this space is the product topology? also, Can I assume that the cardinality of $\omega^\omega$ is $\aleph_0$ as a countable union of countable sets $\bigcup_{k=0}^\infty \omega^k$?.

If the answer is yes, Then, what is the first ordinal that has cardinality $\aleph_1$. Also, is anyone familiar with a good source for studying Borel hierarchy (That also contain exercises if possible..)?

Thanks!! Shir

Now that I think of it, Baire space should also contain uncountable sequences.. so, what is the cardinality of $\omega^\omega$? is it $\aleph_0$ or $\aleph_1$?

Answer 1

Note that $\omega^\omega$ is not the ordinal exponentiation, but rather the set of all functions from $\omega$ to itself. It is easy, and I leave it to you, to verify that this is in fact a set of cardinality $2^{\aleph_0}$ (if you give up, it has been asked on this site infinitely many times before), so definitely not $\aleph_0$, but not necessarily $\aleph_1$ either if you don't assume more than just $\sf ZFC$.

So you shouldn't assume that this is a countable set (you can, but equally you can just assume that $0=1$ or something). The set that you suggest, which is the eventually zero sequences, is a countable set which is also dense. This set witnesses the fact that the Baire space is a separable space.

For the Borel hierarchy basic books about descriptive set theory such as Kechris' and Moschovarkis' books can be good. Jech's Set Theory also has a chapter with the basic information in the first part.

Also, the first ordinal of size $\aleph_1$ is no other than $\omega_1$. We cannot represent it in a better way using ordinal arithmetic because ordinal arithmetic preserves cardinality, so $\alpha^\beta+\gamma\cdot\delta+\eta$ is countable if all $\alpha,\beta,\gamma,\delta,\eta$ are countable. Since $\omega_1$ is not countable, it cannot be written that way.

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Related:

• Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
• Ordinal exponentiation - $2^{\omega}=\omega$
• How is $\epsilon_0$ countable?
• Is it true that $|\mathbb{R}|=2^\omega=\omega_1$?
• How to define countability of $\omega^{\omega}$ and $\omega_1$? in set theory?

Answer 2

The topology of the Baire space $\omega^\omega$ is defined by the base of open subsets $O_s:=\{f\in\omega^\omega|f\succ s\}$ for $s\in\omega^n$ for some $n\geq 0$, where $f\succ s$ shall mean, that $s$ is a finite starting sequence of $f$. This is the product topology of the discrete topologies of the $\omega$.

You should carefully distinguish the finite sequences $s\in\omega^n$ (for some $n$) from the infinite sequences $f\in\omega^\omega$. The former may be used in definitions and arguments w.r.t. Baire space but are no elements of $\omega^\omega$.