In ZFC if there is a bijection between any set and some ordinal, why use others sets in the first place? Why not assume everything is an ordinal?

298 Views Asked by At

In ZFC if there is a bijection between every set and some ordinal, why even use sets in the first place? Why not just assume everything is an ordinal? Come to think of it how do we know certain sets outside of the ordinals even exist in ZFC?

1

There are 1 best solutions below

7
On BEST ANSWER

Your last question is easy. Power sets of ordinals can be well-ordered, but they are not ordinals themselves (well, except for the cases of $\mathcal P(\varnothing)$ and $\mathcal P(\{\varnothing\})$ anyway).

But, ordinals are not enough, because we are interested in relations on sets, some structure. Just because the real numbers can be well-ordered it doesn't mean that their natural structure is somehow compatible with the natural structure of the ordinals; we need more than just that.

So that would be sets of ordinals. And indeed, a classical theorem of Balcar and Vopenka tells us that two models of ZFC with the same sets of ordinals are equal.

All in all, this is used often. For example, in many cases when one talks about iterated forcing, or some sort of structures, one starts with the implicit assumption that the underlying set which is being structured is an ordinal, or a set of ordinals.