3
$\begingroup$

I'm having a hard time following this proof. Here $\aleph(\alpha)$ is the cardinality of $Z(\alpha)$, the set of all ordinals $\gamma$ such that $|\gamma|\leq\alpha$. Also, $\aleph_1=\aleph(\aleph_0)$ and so on.

Suppose $\kappa\lt\aleph_\alpha$. Then there exists $B\subset Z(\alpha)$ with $|B|=\kappa$. Let $\gamma$ be the ordinal of $B$. Then $\gamma\leq Z(\alpha)$. But $\gamma\neq Z(\alpha)$ since then $|\gamma|=\kappa<\aleph_\alpha=|Z(\alpha)|$. So $\gamma, so $|\gamma|\leq\alpha$.

My one question is, why does "Let $\gamma$ be the ordinal of $B$. Then $\gamma\leq Z(\alpha)$" follow? It seems like it's almost assuming the result. All I conclude from $\gamma$ being the ordinal of $B$ is that $|\gamma|=k$, since $B$ and $\gamma$ are isomorphic, right?

  • 0
    Also, for the sake of clarity, could you please edit the question to include the full definition of $\aleph$ numbers, so it won't confuse future readers?2011-05-07

1 Answers 1

5

Your cardinality is non-standard. Typically, $\aleph_\alpha$ is the $\alpha$th infinite cardinal, and it is the size of $Z(\alpha)$, the set of ordinals $\gamma$ such that $\gamma<\omega_\alpha$, where $\omega_\alpha$ is the $\alpha$th initial ordinal.

If $\alpha$ is infinite, the set of $\gamma$ such that $|\gamma|\le\alpha$ has size $|\alpha|^+$, and this cardinal is not $\aleph_\alpha$ (under the standard usage).

Is $m$ supposed to be $\alpha$? Under your definition, $Z(\alpha)$ is an ordinal itself. I suppose that saying that "$\gamma$ is the ordinal of $B$" means that $\gamma$ is the order type of $B$. Since $|B|=\kappa$, $\gamma$ is an ordinal of size $\kappa$. Since $B$ is a subset of $Z(\alpha)$, $\gamma$ is at most the order type of $Z(\alpha)$ itself, i.e., $Z(\alpha)$. This means, precisely, that $\gamma\le Z(\alpha)$. But it may be a good idea to check the definitions you are using, due to their non-standard nature.

  • 1
    Six and a half years later, I almost wrote the same comment as I did above about standard definitions and whatnot. :-)2017-11-27