1. ## Another ordinal problem

I have this kind of problem:
Show, that for every ordinal number $\displaystyle \alpha$ there exists a cardinal number $\displaystyle \kappa$ so that $\displaystyle \alpha \in \kappa$.

I have no idea how to prove that, so little help would be nice. Thanks!

2. Originally Posted by Ester
I have this kind of problem:
Show, that for every ordinal number $\displaystyle \alpha$ there exists a cardinal number $\displaystyle \kappa$ so that $\displaystyle \alpha \in \kappa$.

I have no idea how to prove that, so little help would be nice. Thanks!
What about $\displaystyle \kappa$=cardinality of powerset of $\displaystyle \alpha$? (This is the smallest ordinal equinumerous to the powerset of $\displaystyle \alpha$.)

You know that cardinality of $\displaystyle \kappa$ is greater than the cardinality of $\displaystyle \alpha$. You also should know that any two ordinals are comparable. The cardinality argument implies that $\displaystyle \kappa\le\alpha$ is not possible, hence you get $\displaystyle \alpha<\kappa$. (This is equivalent to $\displaystyle \alpha\in\kappa$.)

3. The above solution relies on the axiom of choice for the result that every every set x does have a cardinality that is indeed equinumerous with x. That is fine, but also we have a solution that does not require the axiom of choice.

By a result of Hartogs, let k be the least ordinal not dominated by a. So k is not less than nor equal to a. Then by comparability of ordinals, we have a is less than k. And k is a cardinal, for if k were equinumerous with an ordinal less than k, then k would not be the least ordinal not dominated by a.

(Granted, this requires that Hartogs's result has already been proven in the course.)