What "good" (illuminating) examples are there of elements in $\omega^\omega\setminus\omega^{<\omega}$?

Question

In the Von Neumann definition of ordinals every ordinal is the well-ordered set of smaller ordinals.

Let $X$ be the set of all ordinals expressible in Cantor normal form with an arbitrarily long (eventually zero) sequence of integers $S_n$ (i.e. $n\in\Bbb N_{\geq0}$) by the expression:

$\sum_n\omega^ns_n$

For example $\omega^35+\omega^2+6\in X$

I would naturally think of the set $X$ as the supremum of the above, which I think of as the ordinal $\omega^\omega$ because every ordinal below that is of the given form.

However I've seen the statement: $X=\omega^{<\omega}$ or $X=\Bbb N^{<\omega}$ which leads me to think of the above as wrong.

Presumably then, $X$ is not an ordinal and the difference between $\omega^{<\omega}$ and $\omega^{\omega}$ is akin to the difference between a sum and a product, where the former requires finitely many nonzero terms?

What "good" (illuminating) examples are there of elements in $\omega^\omega\setminus\omega^{<\omega}$ or $\omega^\omega\setminus X$? I need to see some to better grasp the dichotomy.

Answer 1

We use $\omega^{<\omega}$ to denote the finite sequences, or the eventually $0$ sequences, of natural numbers.

But the notation doesn't make a lot of sense as ordinal arithmetic. The reason is that ordinal arithmetic is continuous, so $\omega^\omega=\sup\{\omega^n\mid n<\omega\}$, which is what you'd normally expect from $\omega^{<\omega}$ as far as notation goes.

We do use $\omega^{<\omega}$ when we think about $\omega^\omega$ as the space of all sequences. This is akin to cardinal exponentiation, not ordinal exponentiation. And in that sense, an element of $\omega^\omega\setminus\omega^{<\omega}$ would be the sequence $\langle 1,2,3,4,\dots\rangle$. But again, this is not ordinal exponentiation.

Answer 2

$X$ is the ordinal $\omega^\omega$. There is no going around that. The sequence $\omega^1, \omega^2, \omega^3, \ldots$ are all contained in $X$, so $X\geq \omega^\omega$ by definition of $\omega^\omega$. But $\omega^\omega\geq X$ because $\omega^\omega\notin X$. So they are equal.

I don't even know exactly what $\omega^{<\omega}$ means. It's certainly not an ordinal with conventional ordinal exponentiation. In set (and cardinal exponentiation), it would mean (the cardinality of) the set of all finite ordered tuples of natural numbers.