# Is Baire Space countable?

1. May 21, 2014

### Staff: Mentor

According to Wikipedia, Baire space is defined as "the Cartesian product of countably infinite copies of the natural numbers". The page states later on that Baire space is homeomorphic to the set of irrational numbers, which seems to indicate that it is uncountable; also, its definition would seem to make it equivalent to the set of infinite sequences of natural numbers, which should also be uncountable by Cantor's diagonal argument.

However, in the sequence of transfinite ordinals, there is the ordinal $\omega^{\omega}$ (the same symbol is used for Baire space according to Wikipedia), which is clearly stated to be countable. Yet this ordinal would seem to be equivalent to Baire space. Is it? If so, is Baire space actually countable? How could it be, given the above?

2. May 21, 2014

### micromass

Hello Peter,

Baire space is uncountable as it is defined as $\mathbb{N}^\mathbb{N}$.

It is also true that $\omega=\mathbb{N}$ the ordinal exponentiation $\omega^\omega$ results in a countable ordinal. How could this be? It is in fact something that bothered me quite some time when I first read about it.

The answer is that ordinal exponentiation is not the same as cardinal exponentiation. What this means is that $\omega^\omega$ has a very specific definition that has nothing at all to do with the cardinality of the set $\omega^\omega$.

So there are two different notions of exponentiation involved here. We have the exponentation as sets, where $A$ and $B$ are sets and where $B^A$ is the set of all functions from $A$ to $B$. Or if $A=B=\mathbb{N}$, it can be seen as all sequences in $\mathbb{N}$.

But we also have a notion of exponentiation of ordinals, where $\alpha$ and $\beta$ are ordinals and where $\beta^\alpha$ is an ordinal. I will not give complete details of the construction unless you're interested. But know that the definition involved again a set of function from $\alpha$ to $\beta$, but not all functions. Rather, we just take all functions $\alpha\rightarrow \beta$ such that only finitely many elements in $\alpha$ map to a nonzero element in $\beta$. This is a severe restriction on cardinality and causes the cardinality of this set to be countable if $\alpha=\beta=\omega$. In that context, we can see $\omega^\omega$ as the set of all sequences in $\mathbb{N}$, but only those sequences which are nonzero for only finitely many terms.

Another, equivalent, definition can be given by recursion:
• $\alpha^0 = 1$
• $\alpha^{\beta + 1} = \alpha^\beta \cdot \alpha$
• If $\beta$ is a limit ordinal, then $\alpha^\beta = \textrm{sup}_{\beta_0<\beta} \alpha^{\beta_0}$

These two notions (set exponentation and ordinal exponentation) are incompatible. They do not give the same answer. Their cardinality doesn't even agree. The notation might be the same but that it. It can be confusing, but it should always be clear from context what it is (that is, if we don't mention ordinals, then it's probably not the ordinal exponentation). This is why we often write $\omega$ as $\mathbb{N}$ when we mean the ordinal. This way we can discriminate between the ordinal $\omega^\omega$ and the uncountable set $\mathbb{N}^\mathbb{N}$ (I know that $\omega^\omega$ is a notation for the Baire space, but that's quite an awful notation imo).

Finally, there's also the notion of cardinal exponentiation, where we take cardinal numbers $\kappa$ and $\lambda$ and calculate the cardinal $\kappa^\lambda$. This coincides perfectly with exponentiation of sets in the sense that if we have a set $X$ of cardinality $\kappa$ and a set $Y$ of cardinality $\lambda$, then $X^Y$ will have cardinality $\kappa^\lambda$. This notion of exponentation is again incompatible with ordinal exponentation. This is one of the reasons why we have different notations for cardinals than for ordinals, although cardinals are (or can be defined as) special ordinals. For example $\omega$ can be written as $\aleph_0$.

Last edited: May 21, 2014
3. May 21, 2014

### Staff: Mentor

Looking up the two definitions on Wikipedia, I see they are indeed different. I think I see how $\omega^{\omega}$ is countable under the ordinal exponentiation definition, but I'll have to cogitate some more to see if I have any further questions.