User blog comment:Alemagno12/Some more set theory questions/@comment-5529393-20180103055700

For your fixed version, there seems to be a problem: in the third rule, there can be more that one F(β) such that cof(α) = cof(F(β)), so then setting F(α) = F(β) is not well-defined; also, if the definition of α could depend on some β > α, then different choices for larger cardinals could justify different choices for α, and lead to different values of F that still satisfy the defintion. However, I will assume that we define F inductively, and always choose the smallest β < α such that cof(α) = cof(F(β)). Also, I think you mean successor cardinal in rule 2.

This function F has some interesting behavior. Of course we have that F(aleph_n) = n, F(aleph_w) = w, and then F(aleph_{w+n}) = w+n. But then F(aleph_{w2}) = w by rule three, and F(aleph_{w2+n}) = w+n again. Thus, we will have F_(aleph_{wa+n}) = w+n for any a for which cof(wa) = w. This pattern is not broken until F_(aleph_{w_1}), which will then be w2 by rule 4. Note that since cof(w2) = w != w_1, we don't get a new use out of rule 3 for cof(α) = w_1. So F(aleph_{w_1*2}) = w3, and in general F(aleph_{w_1*a}) = w(1+a), until we eventually reach F(aleph_{w_1^2}) = w_1. Now rule 3 will make any subsequent F(aleph_a) with cof(a) = w_1 equal to w_1. You can see how the pattern is going.

I believe we still get the weakly inaccessibles as the solutions to F(α) = α. Of course if α is a successor cardinal, then F(α) is a successor ordinal and not equal to α. So α must be a weak limit cardinal. Further, we have that F(α) <= cof(α). Indeed, by transfinite induction:

1. F(aleph_0) = 0 < aleph_0 = cof(aleph_0)

2. F(α) = F(β)+1 <= cof(β)+1 <= β+1 = α = cof(α)

3. Observe that each new value of F is either a previous value, or the smallest one not yet given out. Thus, if we choose β to be the smallest β such that cof(α) = cof(F(β)), F(β) will be the smallest ordinal of that cofinality, i.e. it will be a regular ordinal. So F(α) = F(β) = cof(F(β)) = cof(α).

4. We have F(α) = sup{γ < α}F(γ). But, rule 4 is only invoked if α does not have a cofinality achieved by any of the previous F(γ). Note that the smallest ordinal to achieve the cofinality of α will be the regular ordinal cof(α). So if cof(F(γ)) < cof(α) for all γ < α, then in fact F(γ) < cof(α) for all γ < α, and we will have cof(α) >= sup{γ < α}F(γ) = F(α).

Thus F(α) <= cof(α) <= α, and and therefore for F(α) to equal α, α must be regular, so α must be weakly inaccessible.

Too see that a weakly inaccessible cardinal I satisfies F(α) = α, observe that at each new regular ordinal we reach a new high value of F(α). Thus F at the regular ordinals form an increasing sequence of ordinals of length I bounded above by I, so their supremum is I. F(I) is then determined by rule 4, which gives F(I) = I.