I understand the Cantor's proof for why $S<2^S$, but also we know that ordinals $\omega\sim\omega^2\sim\omega^3...$. This approaches to $\omega^{\omega}$, what should be at least not less than $2^{\omega}$. What do I miss with this logic?
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This contradicts this post. – Dietrich Burde Aug 05 '20 at 10:15
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1The left one is cardinal exponentiation, while the right is ordinal exponentiation. These are not the same. – Jordan Mitchell Barrett Aug 05 '20 at 10:17
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Because ordinal exponentiation is defined differently than is cardinal exponentiation. If you were using ordinal exponentiation, you'd find that $2^\omega$ is countable. – Robert Shore Aug 05 '20 at 10:17
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If you are performing ordinal exponentiation, then $2^\omega=\omega$. This is very different from cardinal exponentiation. – Angina Seng Aug 05 '20 at 10:20
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2I maintain it is bad to write $\omega$ when you mean $\aleph_0$, especially if you also use notation $a^b$ for both exponentiation of ordinals and exponentiation of ordinals. – GEdgar Aug 05 '20 at 10:39
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You are confusing two different notions of exponentiation: the cardinal exponentiation and the ordinal exponentiation.
$2^\omega$ can mean either
- The cardinality of the power set of $\omega$ (or the set of functions $\omega\to 2$), or
- The ordinal number corresponding to the order type of finite binary sequences, ordered lexicographically.
The latter is a countably infinite ordinal (which is, in fact, equal to $\omega$, so it also happens to be cardinal). The former is an uncountable cardinal, the continuum.
Likewise, $\omega^\omega$ can mean one of the two:
- The cardinality of the set of functions $\omega\to \omega$,
- The ordinal number corresponding to the order type of finite sequences with values in $\omega$.
Here, the former is a cardinal number (in fact, equal to the continuum), while the latter is a countable ordinal (which is strictly between $\omega$ and the continuum).
Especially when there is risk of confusion, it is good practice to use $\aleph_0$ when talking about $\omega$ as a cardinal (in particular, when you mean to use cardinal exponentiation), $\omega$ when you mean to talk about it as an ordinal, and $\mathbf N$ when you mean to talk about it as a set of integers. Then it would be pretty clear that $\aleph_0=\omega=2^\omega<\omega^\omega<2^{\aleph_0}=\aleph_0^{\aleph_0}$ and $\omega\sim 2^\omega\sim \omega^\omega\not\sim 2^{\aleph_0}\sim \aleph_0^{\aleph_0}$.
tomasz
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