What's the difference between $ω^{<ω}$ and $ω^{ω}$?

Question

I think that the first is all finite sequences of natural numbers, while the second is all infinite sequences of natural numbers. However I'm not sure how to draw the tree for $ω^{<ω}$, nor can I visualize what is meant by a branch of $ω^{<ω}$ represents an element of $ω^{ω}$. If possible perhaps an explanation of concatenations and restrictions that usually appear with these terms. Any help is appreciated, a visual approach especially.

Answer 1

Yes. $\omega^{<\omega}$ is the set of all finite sequences of natural numbers, and $\omega^\omega$ is the set of all infinite sequences.

How to visualize it is easy. You have a root, the empty sequence, then each point has countably many immediate successors, which are indexed by the natural numbers.

Now, a branch in this tree is a maximal chain, which means it is a sequence of functions $\{f_n\mid \operatorname{dom}(f_n)=n\}$ such that for $n<m$, $f_n\subseteq f_m$. If we take the union of this branch, we have a function $f\colon\omega\to\omega$ such that $f_n=f\restriction n$.

Answer 2

You are correct, $\omega^{<\omega}$ is the set of finite sequences of naturals, and $\omega^\omega$ is the set of infinite sequences. $\omega^{<\omega}$ is a tree under the order $\sigma \leq \tau$ iff $\sigma$ is an initial segment of $\tau$. This makes the empty sequence the root of the tree. Then you have $(0),(1),(2),\ldots$ as the level one elements, then above $(n)$ you have $(n,0),(n,1),\ldots$.

The picture is $\epsilon$ at the top, with countably many children, each of these children has countably many children, and so on.

A branch of this tree is a maximal linearly ordered subset. So, you contain a sequence of every length $n$, and each of these are $\leq$ comparable. So, then we associate with this branch the sequence with $n$th element the $n$th entry of any sufficiently long sequence in the branch.