so $\omega+1=\{\{0,1,2,...\},\{\omega\}\}$, it seems to be the case that $cf(\omega+1)=1$, since $1=|\{\omega\}|$ is the least cardinality of cofinal subsets of $\omega+1$, but $1$ is also the cofinality of the ordinal $2$, which is much less than $\omega+1$. How can I overcome this contradiction?
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3Rather than ${{0,1,2,...},{\omega}}$, you mean the union of this set. – Andrés E. Caicedo Aug 27 '17 at 02:19
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There's no contradiction; cofinality is not a strictly increasing function. Indeed, the cofinality of every non-zero successor ordinal (i.e., ordinals $\alpha$ such that $\alpha = \beta + 1$ for another ordinal $\beta$) is $1$.
One way to think about cofinality of an ordinal $\alpha$ is that it measures how long a sequence needs to be in order to "reach" $\alpha$, not how big $\alpha$ is.
Duncan Ramage
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But then $cf(\omega)=\omega$, and $\omega$ is less than $\omega+1$ and has a bigger cofinality? – Sid Caroline Aug 26 '17 at 21:26
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1Yes. In fact, $cf(\omega) > cf(\aleph_{\aleph_\omega} + 1)$, or indeed any other suitably monstrously huge successor ordinal. – Duncan Ramage Aug 26 '17 at 21:28
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1@SidCaroline Why would you expect $cf(\alpha)$ to be strictly increasing? – Simply Beautiful Art Aug 26 '17 at 21:29
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I guess my intuition was wrong. I thought that a bigger ordinal must have a bigger cardinality, so it must have a bigger cofinality. – Sid Caroline Aug 26 '17 at 21:34
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1Duncan, "every non-zero successor ordinal", since when is zero a successor ordinal? :D – Asaf Karagila Aug 26 '17 at 21:36
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SidCaroline Note that that isn't true even for limit ordinals. $cf(\aleph_0) < cf(\aleph_1) < ...$, but $cf(\aleph_\omega) = \omega$.
@AsafKaragila Since the first set theory textbook I used included them ;).
– Duncan Ramage Aug 26 '17 at 21:52