How to find the cardinality of $A = \{ (a \rightarrow 2^{\omega}) : a < 2^{\omega} \}$ under CH?

Question

Assume the continuum Hypothesis. You may use the fact that if $A =_{c} A', B =_{c} B'$ then $(A \rightarrow B) =_{c} (A' \rightarrow B')$. Also, assume that $|\bigcup A| \leq \text{sup} \{|A|, \text{sup} \{ |a| : a\in A \} \}$. Then determine the cardinality of $|\bigcup A|$ where $$A = \{ (a \rightarrow 2^{\omega}) : a < 2^{\omega} \}$$ and a in the definition of $A$ ranges over all ordinals strictly less than $2^{\omega}$. It should be stated as an arithmetic expression in terms of cardinals among $0, 1, 2, ... , \omega $.

Anyone knows how to find this cardinality?

Thank you in advance.

Answer 1

HINT: You know that $|(a\to 2^\omega)|=|(\omega\to 2^\omega)|$ for each $a$ satisfying $\omega\le a<\omega_1$ (and hence for each $a$ satisfying $\omega\le a<2^\omega$, since we’re assuming $\mathsf{CH}$). Can you see why

$$|(\omega\to 2^\omega)|=|((\omega\times\omega)\to 2)|=|(\omega\to 2)|=2^\omega\;?$$

Answer 2

Assuming CH $2^{\omega}=\omega_1$ so if $a<\omega_1$ $a$ is countable, thus $|(a\to 2^{\omega})| =2^{\omega}$.Thus $|A|=2^{\omega}$.