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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 $\gamma<Z(\alpha)$, 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?

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As Andres said, this is a very nonstandard notation, as it would imply $\aleph_0=1$ and $\aleph_\omega=\omega_1$ and so on. Strange indeed. –  Asaf Karagila May 7 '11 at 7:15
    
Maybe I misstranscribed the notation. The actual notes say $\aleph(m)=|Z(m)|$, but then define $\aleph_1=\aleph(\aleph_0)=|Z(\aleph_0)|$, and so forth. So maybe it's more in line with what is standard then? –  Dani Hobbes May 7 '11 at 7:32
    
This is a lot more in line, in fact this can be viewed as the standard definitions, once you declare $\aleph_0=\omega$ and $\aleph_{\alpha+1} = \aleph(\aleph_\alpha)$, although it does not deal with limit cases. –  Asaf Karagila May 7 '11 at 7:34
    
While we're at it, do you have any recommendations on books that a have a nice treatment of alephs? I think these notes are a little messy. –  Dani Hobbes May 7 '11 at 7:36
    
That depends on your background and what you aim to study. If you want general set theory stuff, Jech's Set Theory (3rd Millennium edition) is very good for the basic stuff and a great reference for the more advanced things. –  Asaf Karagila May 7 '11 at 7:41

1 Answer 1

up vote 5 down vote accepted

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.

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Thanks Andres, yeah, $m$ was supposed to be $\alpha$, I wasn't thinking when I copied down the proof. I see that $Z(\alpha)$ is an ordinal, but in your last paragraph, why would $\gamma\in B$ if $\gamma$ is the ordinal of $B$? –  Dani Hobbes May 7 '11 at 3:52
    
Oh, I had misread the key line in your question. :-) I fixed it. –  Andres Caicedo May 7 '11 at 5:21
    
Thanks! ${}{}{}{}$ –  Dani Hobbes May 7 '11 at 7:10
    
Andres: Note the discussion on the main question, it appears that the definition is in fact the standard one. –  Asaf Karagila May 7 '11 at 7:42
    
Asaf: Now with Hartogs' $\aleph(\alpha)$ rather than than Cantor's $\aleph_\alpha$, sure. –  Andres Caicedo May 7 '11 at 14:16

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