Decreasing sequence Aleph's
I need to show: there does not exist a sequence:
such that .
Supposing there exists such a sequence:
We can observe that for any . Thus we have for all . In particular, for any we have .
In a previous excercise I had to show that (without the axiom of choice): . And since is the smallest ordinal such that we obtain: .
From this I conclude we get a strictly decreasing sequence Aleph's: Since . Thus the subsequence is a sequence such that and .
Those were some of my conclusions: However I don't see where the contradiction kicks in. Can anyone offer me some suggestions? Probably there can not exist a infinite decreasing sequence of ordinals, intuitively that might sound logical but I can't proof it.
Since the set of ordinals are wellordered, our sequence: should contain a smallest ordinal. But can't we simply define that one as ?