# Prove that alpha = aleph_alpha where

1. Feb 5, 2015

### wj2cho

Define alpha_0 = 0, alpha_n+1 = aleph_alpha_n. Let alpha = sup{alpha_n : n is a natural number). Prove that alpha = aleph_alpha.

My attempt: As alpha <= aleph_alpha is obvious, I've been trying to prove the other direction of inequality, so that being both <= and >= implies =, but now I'm not even sure if this is the right approach. I think I cannot use (transfinite) induction because this isn't a statement about n, so I've been stuck with
sup{alpha_n : n is a natural number) >= sup{aleph_beta : beta < alpha}
where the RHS is just the definition of a cardinal aleph_gamma where gamma is a limit ordinal. Maybe I can find an injection from the RHS to the LHS but it doesn't seem to work either. Any help will be appreciated.

2. Feb 10, 2015