How does one show that $\prod_{n < \omega}\aleph_n \leq \aleph^{\aleph_0}_{\omega}$?
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For a more arithmetic-style way of looking at it, $\prod_{n<\omega} \aleph_n\le \prod_{n<\omega}\aleph_\omega$ by monotonicity of the product. (On as side note, we actually have equality. This is the easy direction.) – spaceisdarkgreen Apr 19 '20 at 17:51
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If we think of cardinals as sets then there is inclusion. A general element in $\prod_{n<\omega}\aleph_n$ is a function $f:\omega\to \cup_{n<\omega} \aleph_{n}$ which satisfies $f(n)\in\aleph_n$ for each $n<\omega$. But such $f$ is also a function from $\omega$ to $\aleph_{\omega}$ (because $\aleph_n\subseteq\aleph_{\omega}$ for each $n$) and hence belongs to the set of functions from $\aleph_0$ to $\aleph_{\omega}$.
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