What is this called?
$$\mathbb{N}^\omega$$
Can't seem to find any mention of it when I google.
Context: comment to this answer https://math.stackexchange.com/a/1384962/243059
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It the set of all succession of natural numbers, i.e. the set of functions from $\omega$ to $\mathbb N$. – Mauro ALLEGRANZA Dec 04 '19 at 14:49
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$\omega$ is the name of the first transfinite ordinal. It corresponds to $\mathbf N$. – Bernard Dec 04 '19 at 14:51
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IMO, there is a subtle difference: $\omega$ is an ordinal: thus it has an "order". $\mathbb N$ is a set: it has no order. – Mauro ALLEGRANZA Dec 04 '19 at 14:53
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It is the set of infinite sequences of natural numbers. $\Bbb N^2$ is the set of ordered pairs of naturals, $\Bbb N^3$ is the set of triplets, and $\Bbb N^\omega$ is the set of sequences of length $\omega$
Ross Millikan
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And similarly $\mathbb{R}^\omega$ is the set of infinite real sequences? – Shuri2060 Dec 04 '19 at 15:12
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Not sure how $\mathbb{N}^\omega$ is different from $\mathbb{N}^\mathbb{N}$ – Shuri2060 Dec 04 '19 at 15:16
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Yes, $\Bbb R^\omega$ is the set of (countably) infinite real sequences. $N\Bbb ^N$ is the same as $N^\omega$. Using $\omega$ emphasizes the point that the elements of the sequence are in order because $\omega$ is an ordinal. You could think of $\Bbb {N^N}$ as generic functions from $\Bbb N$ to $\Bbb N$ without such emphasis on the fact that $\Bbb N$ is ordered. It depends on the context. – Ross Millikan Dec 04 '19 at 15:23
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In general, $A^B$ is the set of functions from $B\to A$. This makes sense with the usual set theoretic definition of natural numbers as the set of all natural numbers preceding it. So your example is the set of functions $\omega\to\mathbb N$, which are infinite sequences of natural numbers. If you consider $\omega$ and $\mathbb N$ to be the same thing, this is the same as $\mathbb N^\mathbb N$.
Matt Samuel
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