Why is $\sum_{i\in \omega} \aleph_{i}\geq \aleph_{\omega}$

Question

Why is $\sum_{i\in \omega} \aleph_{i}\geq \aleph_{\omega}$

The presentation states that $\sum_{i\in \omega} \aleph_{i}\geq \aleph_{j}$ for all $j\in \omega$.

I can see that. But my question, please, is how does that imply:

$\sum_{i\in \omega} \aleph_{i}\geq \aleph_{\omega}$

In other words, since I presume $i$ does not equal $\omega$ itself, i.e., $\omega\notin \omega$, how does the above hold?

(Ultimately they are shown to be equal. I can see how things work in the other direction.)

Thanks

Answer 1

$\aleph_\omega$ is the smallest cardinal number that exceeds all of $\aleph_j$ for finite ordinals $j.$

For any finite ordinal $j$, the sum $\sum\limits_{i\,:\,i\,<\,\omega} \aleph_i$ is at least as great as $\aleph_{j+1}$ since $\aleph_{j+1}$ is one of the terms in the sum; therefore the sum is strictly larger than $\aleph_j.$ Every cardinal number that is strictly larger than all of $\{\aleph_j : j\in\{0,1,2,3,\ldots\}\}$ is at least as large as $\aleph_\omega.$