I have this kind of problem:
Show, that for every ordinal number there exists a cardinal number so that .
I have no idea how to prove that, so little help would be nice. Thanks!
You know that cardinality of is greater than the cardinality of . You also should know that any two ordinals are comparable. The cardinality argument implies that is not possible, hence you get . (This is equivalent to .)
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.)