My question comes from $\aleph_{\omega}$ being singular because it has a cofinal subset $\{\aleph_{i}: i\in\omega\}$.
My question, please, is what else is there in $\aleph_{\omega}$?
I can see that $\{\aleph_{i}: i\in\omega\}$ has only $\omega$ elements. But $\aleph_{\omega}$ is the $\text{sup}\{\aleph_{i}: i\in\omega\}$ which (I think) is the union of those sets.
So, although it is evidently incorrect, why is it wrong to think $\{\aleph_{i}: i\in\omega\}$ is actually all of $\aleph_{\omega}$ rather than a subset?
Thanks
You are correct that $$ \aleph_\omega = \sup \{ \aleph_n \mid n < \omega \} = \bigcup \{ \aleph_n \mid n < \omega \}. $$
So for any $\alpha \in \aleph_\omega$ there is some $n < \omega$ such that $\alpha \in \aleph_n$. This does however not mean that all elements of $\aleph_\omega$ are of the form $\aleph_n$.
Take for example $\aleph_0 + 1$ (ordinal addition). We have $\aleph_0 + 1 \in \aleph_1 \subseteq \aleph_{\omega}$ but clearly $\aleph_0 + 1$, as a successor ordinal, is not itself of the form $\aleph_n$.
So, yes, it is wrong to think of $\aleph_\omega$ as $\{ \aleph_n \mid n < \omega \}$. There are many (in fact $\aleph_\omega$ many) elements of $\aleph_\omega$ that are not themselves in $\{ \aleph_n \mid n < \omega \}$. Here are some additional examples:
every finite ordinal $n$,
more generally, every successor ordinal $\alpha < \aleph_{\omega}$,
$\aleph_n + \lambda$ for any $\lambda < \aleph_{\omega}$ of cofinality $\le \aleph_n$,
more generally $\aleph_n + \lambda$ for every $\lambda \in \aleph_\omega \setminus \{ \aleph_m \mid n < m < \omega \}$,
every singular ordinal $\lambda < \aleph_\omega$,
Take any strictly increasing sequence $(\xi_i \mid i < \omega)$ of ordinals $\xi_i \in \aleph_\omega$ such that $\xi := \sup_{i < \omega} \xi_i \not \in \{ \aleph_0, \aleph_\omega\}$. Then $\xi \not \in \{ \aleph_n \mid n < \omega \}$ as $\xi$ has cofinality $\omega$ whereas the $\aleph_n$ (except for $n = 0$) are regular of higher cofinality.
...