# Why is there a unique ordinal $\alpha$ for every infinite cardinal $\kappa$ such that $\kappa = \aleph_\alpha$?

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.

• I'd think $\alpha$ would the ordinality of the set of cardinals $\kappa'$ such that $\aleph_0 \le \kappa' < \kappa$. – Daniel Schepler Jan 22 at 1:11
• Do note this requires choice, and maybe heavily depend on what you already proved. – 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.
$$\alpha=\min\{\beta\in On\mid \exists\gamma\in On(\gamma\ne \beta\land \aleph_\beta=\aleph_\gamma)\}$$ And $$\beta=\min\{\gamma\in On\mid \gamma\ne\alpha\land \aleph_\alpha=\aleph_\gamma\}$$
Now we get $$\aleph_\alpha<\aleph_{\alpha+1}\le\aleph_\beta$$