A Lower Bound

We note that since by Proposition 3.1.9 we have for any two regular norms ϕ, ψthat lh(ϕ) < lh(ψ)

implies that|ϕ|_W < |ψ|_W. Therefore (N/≡W,≤W)is stratified intoΘ-many blocks each of which

contains all norms of a given length. Furthermore we have the following result regarding regular norms of non-limit length.

Proposition 4.1.1. Letα <Θbe arbitrary. Then for any two regular normsϕ, ψwithlh(ϕ) = lh(ψ) =

α+ 1we haveϕ≡Lψ.

Proof. To see thatϕ≤L ψwe take any ∈ Rsuch thatψ(y) = αand consider the constant Lipschitz

functionf :R→ R, x 7→y. Then for allx∈ Rwe have thatϕ(x)≤ψ(f(x)), sincelh(ϕ) =α+ 1,

and soϕ≤Lψ. By symmetry of the argument we get thatϕ≡L ψ.

Thus for any ordinalαthat is not a limit ordinal, the regular norms of lengthαform a single Lipschitz degree and therefore a single Wadge degree. Thus we can expect that the regular norms of non-limit length do not contribute much to the overall value ofΣ. So we now focus on regular norms of limit length. We note that sinceΘis a cardinal there areΘ-many limit ordinals strictly belowΘ. This leads to the following definition:

Definition 4.1.2We lethλα|α <Θibe an enumeration of all limit ordinals strictly belowΘ.

For anyα <Θwe letα~ denote the regular normα~ :R→Θ, x7→α, which is of lengthα+ 1.

AssumingADandDCwe define for anyα <Θan ordinalΣαby setting

In light of Proposition 3.1.9 and Proposition 4.1.2 – assumingADandDC– we have for anyα <Θ thatΣα= ~ λα

_W. Furthermore directly from the definition of theΣαs we get that

Σ = sup{Σα|α <Θ}.

Thus if we are able to obtain lower bounds for theΣαs, we immediately get a lower bound forΣ. Our

goal throughout the rest of this subsection will be to show that for anyα <Θwe have thatΣα≥Θ·α,

which already implies thatΣ≥Θ2_.

Definition 4.1.3 For any regular normϕwe define a regular normϕ+_{by setting for allx∈}

R

ϕ+(x) :=

(

ϕ(gx+(x)) + 1, ifx(0)6= 0,

ϕ(x+_{), otherwise.}

Proposition 4.1.4. Ifϕis a regular norm such thatlh(ϕ)is a limit ordinal thenϕ <Wϕ+.

Proof. We assume towards a contradiction thatϕ+ ≤W ϕ. Let x ∈ Rbe such that the continuous

functiongxwitnesses this, i.e., for ally∈R,ϕ+(y)≤ϕ(gx(y)). Then we get

ϕ(gx(h1i_x))≥ϕ+(h1i_x) =ϕ(gx(h1i_x)) + 1> ϕ(gx(h1i_x)),

which is absurd.

Definition 4.1.5 For two regular normsϕ, ψwe say thatϕisembedded inψiff there is somex∈_Rsuch that for ally∈_R,ψ(x∗y) =ϕ(y).

Lemma 4.1.6. Ifϕis embedded inψ, thenϕ≤L ψ.

Proof. Letx∈Rbe such that for ally ∈Rwe haveψ(x∗y) =ϕ(y). Then we give a winning strategy

σ:ω<ω_{\ {∅} →ωfor PlayerIIin the gameG}≤

L(ϕ, ψ)by setting for alls∈ω

<ω_{\ {∅}} σ(s) :=    xlh(₂s), if lh(s)is even, slh(s₂)−1, if lh(s)is odd.

This strategy is indeed winning forIIsince for anyy∈Rwe haveσ(y) =x∗yand soϕ(y) =ψ(σ(y)).

Now we can establish a lower bound for the number of regular norms of a given limit length. Lemma 4.1.7. Ifλ <Θis a limit ordinal andα < Θarbitrary, then there is a strictly≤W-increasing sequencehϕν |ν < αisuch that for allν < αwe havelh(ϕν) =λ.

Proof. Sinceα <Θwe fix a surjections:R→α. Furthermore we fix a regular normϕwithlh(ϕ) =λ.

Now we define a strictly≤W-increasing sequencehϕν |ν < αiof norms of lengthλrecursively onν as

follows. We setϕ0:=ϕ. Then we assume thatϕξis already defined forξ < νand define a regular norm

ϕ∗νby setting for allx, y∈R

ϕ∗_ν(x∗y) :=

(

ϕs(x)(y), ifs(x)< ν, ϕ(y), ifϕ(y).

Since by induction hypothesis we havelh(ϕξ) =λthis definition ensures that alsolh(ϕ∗ν) =λ. Further-

more for allξ < νwe have by Lemma 4.1.6 thatϕξ ≤W ϕ∗ν. Now we letϕν := (ϕ∗ν)+. Then we still

havelh(ϕν) =λand furthermore by Lemma 4.1.4 we get thatϕξ <Wϕν for allξ < ν, concluding the

Theorem 4.1.8. We assumeADandDC. Letα <Θ. Then we haveΣα≥Θ·α. Forα < ωwe even

have thatΣα≥Θ·(α+ 1).

Proof. We assume towards a contradiction that this claim is not true. Then letαbe the least ordinal such that there is a regular normϕwithlh(ϕ) = λα+ 1and such that|ϕ|_W <Θ·(α+ 1), ifα < ω, and

|ϕ|_W<Θ·αotherwise. We fix such a regular normϕand distinguish the following cases.

Case 1is thatα= 0. Thenlh(ϕ) =ω+ 1and|ϕ|_W<Θ. But then by Lemma 4.1.7 there is a strictly ≤W-increasing sequencehψν |ν <|ϕ|W+ 1iof regular norms withlh(ψν) =ωfor allν <|ϕ|W+ 1. But then for anyν <|ϕ|_W+ 1we can inductively establish that we have|ψν| ≥ν. So in particular for

ψ:=ψ|ϕ|_Wwe have|ψ|W ≥ |ϕ|W, whilelh(ψ)<lh(ϕ), which is absurd.

Case 2is thatαis a successor ordinal, sayγ+ 1. Without loss of generality we assume thatα > ω. The argument forα < ωis very similar. Letψbe a norm of lengthλγ+ 1. Then by minimality ofαwe

have that|ψ|_W≥Θ·(γ+ 1). Since, however,αwas a counterexample, there is an ordinalζ <Θsuch that|ϕ|_W = Θ·(γ+ 1) +ζ. But then again by Lemma 4.1.7 we get a strictly≤W-increasing sequence hψν |ν < ζ+ 1iof regular norms withlh(ψν) =λαfor allν < ζ+ 1. Then since|ψ0|W≥Θ·γwe get that for allν < ζ+ 1,|ψν|W ≥Θ·γ+ν and so in particular|ψζ|W ≥Θ·γ+ζ =|ϕ|W, while

lh(ψ)<lh(ϕ), which is absurd.

Case 3is thatαis a limit ordinal. ThenΘ·α=S

γ<αΘ·γ. So if|ϕ|W<Θ·αthere isγ < αsuch that|ϕ|_W ≤Θ·γ. But by minimality ofγwe can take a regular normψwithlh(ψ) =λγ+ 1and get

that|ψ|_W≥Θ·γ≥ |ϕ|_Wand soϕ≤Wψ, whilelh(ψ)<lh(ϕ), which is absurd. This now immediately gives us the desired lower bound forΣ.

Theorem 4.1.9. We assumeADandDC. Then we have thatΘ2≤Σ. Proof. Using Theorem 4.1.8 we can calculate that

Σ = sup{Σα|α <Θ} ≥sup{Θ·α|α <Θ}= Θ2.