Why doesn't $\omega ^ {\omega}$ have cardinality equal to real numbers?

Question

One set of order type $2\omega$ is the set {1,2} $\times N$

One set of order type $3\omega $ is {1,2,3} $\times N$

Similarly, order type $\omega ^2$ would be the set $N \times N$

$\omega ^3$ would be $N\times N \times N$

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$\omega ^{\omega}$ would be the never-ending cartesian product $N\times N\times N\times N\times.........$

However, this infinite cartesian product simply corresponds to the set of all infinite sequences of natural numbers, aka the real numbers. Then, doesn't this mean that a set of order type $\omega ^{\omega}$ would have cardinality equal to reals?

Moreover, you could apply the diagonal argument to the set $N\times N\times N\times......$ to directly prove that it's uncountable.

Answer 1

In the case of cardinal arithmetic, where $\omega^\omega$ is the set of all functions $f\colon\omega\to\omega$, then yes, you are correct. That set has the same cardinality as the real numbers. And indeed, it corresponds to the infinite product of copies of $\omega$ itself.

However, in the case of ordinal arithmetic, where $\omega^\omega$ is the supremum of $\omega^n$ for $n<\omega$, there $\omega^\omega$ is countable. To see that, note that what it actually corresponds to is the functions in the infinitary products above which are eventually $0$. So we can identify those with the finite sequences of natural numbers, which is a countable set.

What we see here is an example of continuity of ordinal arithmetic, which is defined to be continuous: limit steps are literally the limit of their previous steps; and cardinal arithmetic, which is defined by taking cardinality of certain sets, and it is very discontinuous, especially when infinite exponentiation is involved.

Answer 2

In ordinal exponentiation, $x^\omega$ = sup($x^n$) for natural number n. As such, it is different form cardinal exponentiation.

In The Book of Numbers, by John Conway and Richard Kenneth Guy, there is a bijection from the natural numbers to a set of order type $\omega^\omega$. The bijection is specified by these two rules:

1. If the greatest prime factor of m is greater than the greatest prime factor of n, then m comes later than n.
2. If m/x comes later than n/x, then m comes later than n.

One can verify that this well-order has order type $\omega^\omega$.