For finding the $\alpha$ I literally can't get any further than writing out the definitions.

For the second part:

Suppose $\aleph_\alpha = \aleph_\beta$ and $\beta \neq \alpha$. Then either $\alpha \in \beta$ or $\beta \in \alpha$. Assume w.l.o.g. that $\alpha \in \beta$. If $\beta$ is a limit ordinal, then $\aleph_\beta = \bigcup \{\aleph_\gamma \ | \ \gamma < \beta\}$, so maybe that is strictly greater than $\aleph_\alpha$, but I can't prove that, and I don't know what to do if $\beta$ is not a limit ordinal.

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    $\begingroup$ I'd think $\alpha$ would the ordinality of the set of cardinals $\kappa'$ such that $\aleph_0 \le \kappa' < \kappa$. $\endgroup$ – Daniel Schepler Jan 22 at 1:11
  • $\begingroup$ Do note this requires choice, and maybe heavily depend on what you already proved. $\endgroup$ – Asaf Karagila Jan 22 at 4:16

First, use transfinite induction to prove that $\alpha\le \aleph_{\alpha}$ for all ordinal alpha.

From that we get $\kappa\le\aleph_{\kappa}$, so the following is well defined: $$\alpha=\min\{\beta\in On\mid \kappa\le \aleph_{\beta}\}$$

Now try proving that this $\alpha$ is indeed the alpha you are searching for.

For the other part, define

$$\alpha=\min\{\beta\in On\mid \exists\gamma\in On(\gamma\ne \beta\land \aleph_\beta=\aleph_\gamma)\}$$ And $$\beta=\min\{\gamma\in On\mid \gamma\ne\alpha\land \aleph_\alpha=\aleph_\gamma\}$$

Now we get $\aleph_\alpha<\aleph_{\alpha+1}\le\aleph_\beta$


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