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As a corollary of König's theorem, we have $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ . On the other hand, we have $\operatorname{cf}(\aleph_\omega) = \aleph_0$.

Why the logic in the latter equation can't apply to the former one?

To be precise, why we can't have $\sup ({2^n:n<\omega}) = 2^{\aleph_0}$?

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Because it is $\aleph_0$. –  André Nicolas Nov 26 '12 at 6:21

3 Answers 3

up vote 2 down vote accepted

Should there be any reason to expect the cofinality to be a strictly increasing function? Perhaps naively, I have to admit that I don't remember my first thoughts on it right now.

Koenig's theorem assures us indeed that the continuum does not have cofinality $\omega$, from which we can deduce that $\aleph_\omega$ is never the cardinality of the continuum.

As for the second question, there is no reason to expect the continuum function to be continuous either (except the obvious similarity between the words). Indeed $2^n$ is finite, and so the supremum of all finite cardinals is $\aleph_0$ and not $2^{\aleph_0}$.

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Thank you. It's fair to say your insights from my dumb question surprises me a lot :) –  Metta World Peace Nov 26 '12 at 9:24
    
You're welcome. I have spent a fair amount of time around these questions, so I can understand where they come from. –  Asaf Karagila Nov 26 '12 at 9:40

Note that in ordinal arithmetic $2^\omega$ is the supremum of $2^n$ for all $n<\omega$, so it is $\omega$ and thus has cofinality $\aleph_0$.

$\omega$ and $\aleph_0$ are the same set but they nevertheless behave differently in practice -- because tradition is to use the notation $\omega$ for operations where a limiting process is involved and $\aleph_0$ for operations where the "countably infinite" is used all at once in a single step. Cardinal exponentatiation $2^{\aleph_0}$ is such an all-at-once process.

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As I said before, ordinals are not cardinals and cardinals are not ordinals. –  Asaf Karagila Nov 26 '12 at 9:52
    
@HenningMakholm Thank you, that's exactly where I'm confused. I just don't know how to formulate the question. –  Metta World Peace Nov 26 '12 at 10:52
    
@AsafKaragila Thanks, your link is of great help. –  Metta World Peace Nov 26 '12 at 10:55

We have $2^n < \omega$ for all $n < \omega$, so the supremum of the $2^n$ is the supremum of a set of ordinals cofinal in $\omega$, and hence is just $\omega$.

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Tons of thanks. –  Metta World Peace Nov 26 '12 at 6:36

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