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

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 $\aleph_1$?

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.

Related:

• "if you give up, it has been asked on this site infinitely many times before" How can it be infinitely many times? Overall there are only finitely many questions on this forum. I know this question is very old, but I always thought mathematicians wish to be precise about things. May 2, 2019 at 11:30
• I guess you also thought that mathematicians are devoid of sarcasm. May 2, 2019 at 11:31

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 $\omega^\omega$.