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Let $\Omega$ be the first ordinal with cardinality $2^\mathfrak c$. Take now the set of all ordinals $\alpha < \Omega$ which are limit ordinals. Is the cardinality of this set countable or is it $\mathfrak c$?

Thank you!

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    $\begingroup$ Do you think the first ordinal of cardinality $\mathfrak{c}$ is a limit ordinal or a successor ordial? Is it smaller than $2^\mathfrak{c}$? $\endgroup$ – Michael Greinecker Jan 24 '14 at 12:40
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Define an equivalence relation on the ordinals, $\alpha\sim\beta$ if and only if the interval between $\alpha$ and $\beta$ is finite.

The definition is equivalent to saying that there is a limit ordinal $\delta$ and $n,k<\omega$ such that $\alpha=\delta+n$ and $\beta=\delta+k$. If you don't see it immediately, try proving that.

Now it is easy to prove that every equivalence class is countable. In fact order isomorphic to $\omega$. Limiting ourselves to $\Omega$, we have to have exactly $2^\frak c$ equivalence classes.

Finally, note that picking the unique limit ordinal from each equivalence class is exactly a system of representatives. Therefore there are $2^\frak c$ limit ordinals below $\Omega$, and their order type is exactly $\Omega$.

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  • $\begingroup$ So, using the same considerations, can I say that, If $\Omega$ be the first ordinal with cardinality $\mathfrak c$. Then, the set of all ordinals $\alpha < \Omega$ which are limit ordinals Is of cardinality $\mathfrak c$? $\endgroup$ – topsi Jan 24 '14 at 14:04
  • $\begingroup$ Shir, that is correct. The only ordinal where this is not true is the first ordinal which is countably infinite. $\endgroup$ – Asaf Karagila Jan 24 '14 at 14:10
  • $\begingroup$ ok, got it. thank you! $\endgroup$ – topsi Jan 24 '14 at 14:12
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$\Omega$ will be a cardinal. It is either a successor cardinal or a limit cardinal. In either case, a simple induction shows that for every $\alpha<\Omega$ the $\alpha$th limit ordinal is also less than $\Omega$. Thus, the cardinality of the limits less than $\Omega$ is $\Omega$.

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Call the number you're looking for $\lambda$.

Each limit ordinal in $\Omega$ is followed by exactly $\omega$ successor ordinals, so as a matter of cardinal arithmetic we must have $\Omega = \omega\times\lambda$. But, at least assuming the axiom of choice we have $\omega\times\lambda = \max(\omega,\lambda) = \lambda$, so $\lambda=\Omega=2^{\mathfrak c}$.

Did you mean to ask about initial ordinals rather than limit ordinals?

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  • $\begingroup$ I just noticed that both our answers have the same core idea in them. But what does the continuum hypothesis has to do with anything? $\endgroup$ – Asaf Karagila Jan 24 '14 at 13:36
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    $\begingroup$ @Asaf: It doesn't -- it was an (extremely failed) attempt to type "axiom of choice" :-) $\endgroup$ – Henning Makholm Jan 24 '14 at 13:48
  • $\begingroup$ I figured you'd say that... :-) Just to be nitpicky, if we already assume it is an ordinal, the axiom of choice is not needed for the calculation. $\endgroup$ – Asaf Karagila Jan 24 '14 at 14:02

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