Why is $2^{\aleph_\omega} = \left( \sum_{\mu < \aleph_\omega} 2 ^{\mu} \right)^{\aleph_0}$?

Question

Why is $2^{\aleph_\omega} = \left( \sum_{\mu < \aleph_\omega} 2 ^{\mu} \right)^{\aleph_0}$?

The above equality is a part of a proof that I am trying to understand. I know that $$\aleph_\omega = \bigcup_{n < \omega} \aleph_n$$ I tried writing $$2^{\aleph_\omega} = 2^{\bigcup_{n < \omega} \aleph_n} = \prod_{n < \omega} 2^{\aleph_n}$$ but I don't know if I can write it like this or if this is helpful.

Please be relatively elementary in the explanation.

Answer 1

First let's argue $2^{\aleph_\omega}\geq \left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}$:

Clearly $2^{\aleph_\omega}\geq 2^\mu$ for any $\mu<\aleph_\omega$, so also $2^{\aleph_\omega}\geq \sum_{\mu<\aleph_\omega}2^\mu$. Then also $ \left(2^{\aleph_\omega}\right)^{\aleph_0}\geq \left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}$. But we have: $$ \left(2^{\aleph_\omega}\right)^{\aleph_0}= 2^{\aleph_\omega\cdot \aleph_0}=2^{\aleph_\omega} $$

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Next let's argue $2^{\aleph_\omega}\leq \left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}$:

Suppose I have a subset $X\subseteq \omega_\omega$, then I can see $X$ as a union of subsets of $\omega_n$ by considering $X_n=\{x\in X\mid x\in \omega_n\}$: we have $X=\bigcup_{n\in\omega} X_n$. Furthermore, if $X,X'$ are distinct subsets, then $X_n\neq X'_n$ for some $n\in\omega$.

So, in this way, we can send each subset $X\subseteq \omega_\omega$ injectively to an $\omega$-sequence $\langle X_n\mid n\in\omega\rangle$ where each $X_n\subseteq \omega_n$. Hence, this is an injective function from $\mathcal P(\omega_\omega)\to\left(\bigcup_{n\in\omega} \mathcal P(\omega_n)\right)^\omega$.

We're now done, since $|\mathcal P(\omega_\omega)|=2^{\aleph_\omega}$ and $$\left|\big(\bigcup_{n\in\omega} \mathcal P(\omega_n)\big)^\omega\right|=\left(\sum_{n<\omega}2^{\aleph_n}\right)^{\aleph_0}=\left(\sum_{\mu<\aleph_\omega}2^\mu\right)^{\aleph_0}.$$

Answer 2

There is in general a stronger equality which states that $2^\kappa=(2^{<\kappa})^{cf(\kappa)}$ where you have $2^{<\kappa}=\lim_{\nu<\kappa}2^\nu=\sum_{\nu<\kappa}2^\nu$

Proof:

Let $\{\kappa_{\alpha}\}_{\alpha < cf(\kappa)}$ be a set of cardinals where $\kappa_{\alpha}<\kappa$ and such that: \begin{equation*} \kappa=\sum_{\alpha < cf(\kappa)} \kappa_{\alpha} \end{equation*} Then: \begin{equation*} 2^{\kappa}=2^{\sum_{\alpha < cf(\kappa)} \kappa_{\alpha}}=\prod_{\alpha<cf(\kappa)}2^{\kappa_{\alpha}}\leq(2^{<\kappa})^{cf(\kappa)}\leq 2^{\kappa \cdot cf(\kappa)}=2^{\kappa} \end{equation*}

In your case we have $cf(\aleph_{\omega})=\aleph_0$ which you just plug in the general case.