# 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

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Mar 11, 2015

wj2cho, this may come a bit late (a month after you posted it), but if you are still interested: your definitions seems to be the cardinal equivalent to epsilon-0 ε0. (You can read about epsilon numbers at http://en.wikipedia.org/wiki/Epsilon_numbers_(mathematics).) Of course, you are referring to cardinals, but then we get into the difficulty that the alephs are not subscripted by cardinals, but rather ordinals. Therefore your definition needs to be cleaned up a little. Once it is, then you will want to look at fixed points. Google "fixed points of aleph sequence" for inspiration on how to find the fixed points of your alpha sequence.
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