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I was wondering what the relation is between a limit ordinal and the alephs. Are all limit ordinals alephs and if so can it be proven.

user340145
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One starts with the natural numbers; the cardinality of natural numbers is defined to be $\aleph_0$.The next cardinal is defined to be $\aleph_1$. Note that this is not the cardinality of the power set of $\mathbb N$ (which is also the cardinality of $\mathbb R$), this cardinality is called continuum and denoted by $\mathfrak c$. If continuum hypothesis is valid in your model then $\aleph_1=\mathfrak c$.

Basically one starts with $\aleph_0$ then the next cardinal is $\aleph_1$ and so on. However I just remarked that $\aleph$ numbers may not be constructed by taking power sets. These are called beth numbers denoted by the letter $\beth$. I.e.$|\mathbb N|=\beth_0$ and $|\mathcal P(\mathbb N)|=\beth_1$ and so on.

Having said this, I wouldn't say all limit ordinals are $\aleph$'s. Simply because not all limit ordinals are cardinals (though, of course you can associate a cardinal to them).

See here for a similar discussion: Is a limit ordinal necessarily a cardinal?

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ugur efem
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  • Regarding your third paragraph. Can it be shown that all transfinite cardinals are Alephs? – user340145 May 17 '16 at 17:39
  • yes, if you assume axiom of choice. There are non $\aleph$ cardinals without axiom of choice. See here: http://math.stackexchange.com/questions/53752/theres-non-aleph-transfinite-cardinals-without-the-axiom-of-choice – ugur efem May 17 '16 at 18:40