User blog:Rgetar/Regular and singular ordinals

I was trying to extend my Veblen-like functions beyond Ωω, and there was a problem: for Veblen-like function

φαβ(X)

should be

cof(X) ≤ Ωα + 1

(cof(X), that is cofinality of X, is minimal length of increasing sequence of ordinals such as its supremum is X).

But what if cof(X) > Ωα + 1?

Then we can use

φαβ(X) = φαγ(φγβ(X))

For example,

φ26100(Ω88)

cof(Ω88) = Ω88 > Ω26 + 1

But let γ = 87, then

φ26100(Ω88) = φ2687(φ87100(Ω88))

cof(Ω88) = Ω88 = Ω87 + 1

cof(φ87100(Ω88)) = ω < Ω26 + 1

So everything is fine.

Then let β = ω

φ0ω(0)

cof(0) = 0 < Ω0 + 1 = Ω

ok.

φ0ω(1)

cof(1) = 1 < Ω

ok.

φ0ω(ω)

cof(ω) = ω < Ω

ok.

φ0ω(Ω)

cof(Ω) = Ω = Ω

ok.

φ0ω(Ω2)

cof(Ω2) = Ω2 > Ω

Not ok. Then

φ0ω(Ω2) = φ01(φ1ω(Ω2))

ok.

Similarly

φ0ω(Ω3) = φ02(φ2ω(Ω3))

φ0ω(Ω4) = φ03(φ3ω(Ω4))

φ0ω(Ω5) = φ04(φ4ω(Ω5))

...

But for φ0ω(Ωω) we need γ

φ0ω(Ωω) = φ0γ(φγω(Ωω))

such as γ + 1 = ω? However, this is impossible, since ω is not successor ordinal...

Then I thought: "Wait, cof(Ωω) is not Ωω, it is ω, since Ωω is supremum of ω, Ω1, Ω2, Ω3, Ω4, Ω5..."

Interestingly,

cof(ω) = ω

cof(Ω) = Ω

cof(Ω2) = Ω2

cof(Ω3) = Ω3

cof(Ω4) = Ω4

cof(Ω5) = Ω5

...

cof(Ωω) = ω ≠ Ωω

Apparently,

cof(Ωω + 1) = Ωω + 1

cof(Ωω + 2) = Ωω + 2

cof(Ωω + 3) = Ωω + 3

...

cof(Ωω2) = ω

(since it is supremum of Ωω, Ωω + 1, Ωω + 2, Ωω + 3, Ωω + 4, Ωω + 5...)

cof(ΩΩ) = Ω

cof(ΩΩ + 1) = ΩΩ + 1

cof(ΩΩ + ω) = ω

cof(ΩΩ 2 ) = Ω2

cof(ΩΩ 2 + 1 ) = ΩΩ 2 + 1

cof(ΩΩ 3 ) = Ω3

cof(ΩΩ ω ) = ω

cof(ΩΩ Ω ) = Ω

cof(ΩΩ Ω ω ) = ω

cof(ΩΩ Ω Ω ) = Ω

cof(ΩΩ Ω Ω Ω ...   ) = ω

(since it is supremum of ω, Ωω, ΩΩ ω, ΩΩ Ω ω , ΩΩ Ω Ω ω   , ...)

I proved that there are no ordinals of cofinality Ωω.

Proof: let cof(α) = Ωω. Then α is supremum of increasing sequence α[n], where n are all ordinals less than Ωω. Remove from this sequence all elements except α[Ωβ], where β are natural numbers. We get α[ω], α[Ω], α[Ω2], α[Ω3], α[Ω4], α[Ω5], ... α is also supremum of this new sequence of length ω, so cof(α) = ω.

And there are no ordinals of cofinality Ωω2. Proof is similar, but with α[Ωω + β] instead of α[Ωβ].

etc.

So, since there are no ordinals of cofinality Ωω, the problem with φαω(X) is solved, since cof(X) < Ωω, for example,

φ0ω(Ωω)

cof(Ωω) = ω < Ω

We even do not need γ.

And similarly for φαω2(X), etc.

So, apparently, if α is 0 or successor ordinal, then

cof(Ωα) = Ωα

and if α is limit ordinal, then possibly not.

I was wondering, if opposite true, that is if cof(Ωα) ≠ Ωα for all limit α.

I needed to check it all out. I read Wikipedia article Cofinality, and it turned out that this is really so: ordinals α such as cof(Ωα) = Ωα are called "regular", and all other ordinals are called "singular". Most ordinals are singular. Ordinals Ωα for zero or successor α are regular. Ordinals Ωα for limit α are usually singular.

But it was still unclear: are there regular Ωα with limit α? Or all Ωα with limit α are singular?

I searched "cofinality" at Googology Wiki and read article Cardinal. And the answer is "maybe". Regular Ωα with limit α are called "inaccessible cardinals". Least inaccessible cardinal is called I.