Sorry for the incomplete title; it would become too long.
The question is as follows: for the ordinal $\omega$, consider the equation card $\omega$ $\lt$ card $\xi$ $\leq$ card $2^\omega$. Denote the minimum element of the set of ordinals $\xi$ satisfying the previous equation as $\Omega$. Denote the set $\Omega'$ as the set of all ordinals less than or equal to $\Omega$.
I need to show that the set $\Omega'$ is uncountable. However, I am having trouble understanding the question and figuring out the steps I need to take to solve the problem, as it is impossible (I believe) to characterize $\xi$, and I am fairly new to set theory.
Any detailed help would be appreciated.
Thanks!
Let $\Omega$ be the minimal solution of the equation $(\star)$: card $\omega <$ card $\Omega < $ card $2^\omega$.
Recall that for any ordinal $\beta$ we have $$ \{\alpha \in \mathbb{Ord} : \alpha < \beta\}=\beta.$$
Therefore $$\Omega' = \{\alpha \in \mathbb{Ord} : \alpha < \Omega\}\cup \{\Omega\}=\Omega \cup \{\Omega\}$$ and therefore card $\Omega' = $ card $\Omega> $ card $\omega$ and thus $\Omega'$ is uncountable.