What is the cardinality of all limit ordinals $\alpha$ s.t. $\alpha < 2^\mathfrak c$ 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!
 A: $\Omega$ 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 $\Omega$.
A: 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?
A: 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 $\Omega$.
