Borel Liftings of Borel Sets: Some Decidable and Undecidable by Gabriel Debs

By Gabriel Debs

One of many goals of this paintings is to enquire a few common homes of Borel units that are undecidable in $ZFC$. The authors' start line is the next effortless, even though non-trivial end result: contemplate $X \subset 2omega\times2omega$, set $Y=\pi(X)$, the place $\pi$ denotes the canonical projection of $2omega\times2omega$ onto the 1st issue, and feel that $(\star)$ : ""Any compact subset of $Y$ is the projection of a few compact subset of $X$"". If in addition $X$ is $\mathbf{\Pi zero 2$ then $(\star\star)$: ""The limit of $\pi$ to a few really closed subset of $X$ is ideal onto $Y$"" it follows that during the current case $Y$ is usually $\mathbf{\Pi zero 2$. realize that the opposite implication $(\star\star)\Rightarrow(\star)$ holds trivially for any $X$ and $Y$. however the implication $(\star)\Rightarrow (\star\star)$ for an arbitrary Borel set $X \subset 2omega\times2omega$ is reminiscent of the assertion ""$\forall \alpha\in \omegaomega, \,\aleph 1$ is inaccessible in $L(\alpha)$"". extra exactly the authors end up that the validity of $(\star)\Rightarrow(\star\star)$ for all $X \in \varSigma0 {1 \xi 1$, is comparable to ""$\aleph \xi \aleph 1$"". despite the fact that we will express independently, that once $X$ is Borel you can, in $ZFC$, derive from $(\star)$ the weaker end that $Y$ is usually Borel and of a similar Baire type as $X$. This final end result solves an previous challenge approximately compact protecting mappings. in reality those effects are heavily regarding the next normal boundedness precept Lift$(X, Y)$: ""If any compact subset of $Y$ admits a continual lifting in $X$, then $Y$ admits a continuing lifting in $X$"", the place by way of a lifting of $Z\subset \pi(X)$ in $X$ we suggest a mapping on $Z$ whose graph is contained in $X$. the most results of this paintings will supply the precise set theoretical power of this precept looking on the descriptive complexity of $X$ and $Y$. The authors additionally turn out an identical end result for a version of Lift$(X, Y)$ within which ""continuous liftings"" are changed via ""Borel liftings"", and which solutions a question of H. Friedman. between different functions the authors receive an entire option to an issue which works again to Lusin in regards to the life of $\mathbf{\Pi 1 1$ units with all parts in a few given category $\mathbf{\Gamma$ of Borel units, enhancing previous effects via J. Stern and R. Sami. The facts of the most outcome will depend upon a nontrivial illustration of Borel units (in $ZFC$) of a brand new sort, related to a large number of ""abstract algebra"". This illustration was once at the beginning constructed for the needs of this facts, yet has numerous different functions.

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Additional info for Borel Liftings of Borel Sets: Some Decidable and Undecidable Statements

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F : h −→ h f, f −1 holomorph . : offene Teilmenge von C), dann heißen X und Y biholomorph ¨ aquivalent, falls ein f : X −→ Y existiert, sodass f bijektiv ist und f, f −1 holomorph sind. 9 hist biholomorph ¨ aquivalent zu D := z ∈ C |z| < 1 . Beweis: Betrachte die Abbildung f : h −→ D, f (z) := z−i z+i . |f (z)| < 1 ⇐⇒ |z + i|2 > |z − i|2 ⇐⇒ (y + i)2 > (y − i)2 ⇐⇒ 2y > −2y ⇐⇒ y > 0 Bemerkung: 1. Die Umkehrabbildung zu f ist f −1 = f f −1 (w) = 1 w+1 i w−i 2. Es gibt zu f : h Aut(D). ): fA (z) = = az + b = cz + d (az + b)(c¯ z + d) |cz + d|2 (ad − bc) z |cz + d|2 Also fA ∈ Aut(h) ⇐⇒ A ∈ GL(2, R)+ .

1 i (Eigenwertgleichung). Die andere Richtung folgt durch Nachrechnen! 33 3 Die Riemannschen Fl¨achen C, C und h Also zu zeigen: Aut(D)0 = fN N = α 0 0 α = g ∃ s ∈ S 1 : g(z) = sz (Dabei ist S 1 := z ∈ C |z| < 1 ). Das Unterstrichene folgt aber aus dem folgenden Lemma. 2 Ist h ∈ Aut(D), dann |h(z)| ≤ |z|, dto. h. 11 (Lemma von Schwarz) Sei f : D −→ C holomorph, f (0) = 0, |f (0)| < 1 f¨ ur |z| < 1, dann gilt: 1. |f (z)| ≤ |z| ∀ |z| ≤ 1 2. Ist |f (z0 )| = |z0 | f¨ ur ein z0 = 0, dann ist f (z) = λz mit geeignetem λ ∈ C, |λ| = 1 1 Beweis: Setze g(z) := f (z) ur z da f (z) = 0, folgt g holomorph in D.

Damit ist ℘(z + γ) = ℘(z) + c(γ) (∗) (wobei c(γ) eine Konstante ist, die von γ abh¨ angt). Zeige c(γ) = 0: Offensichtlich ist ℘(−z) = ℘(z) (℘ ist eine gerade Funktion). 1 Ell(Γ) := f ∈ M er(C) ∀ γ ∈ Γ : f (z + γ) = f (z) heißt Ko ¨rper der elliptischen Funktionenelliptische Funktionen zu Γ. Konvention: Ist z0 Pol von f , dann setze f (z0 ) = ∞ ∈ C. 1 Ell(Γ) ist ein K¨orper. 1 Sei f ∈ Ell(Γ). F¨ ur z0 ∈ C, γ ∈ Γ gilt: ordz0 (f ) = ordz0 +γ (f ) Beweis: Es gibt eine offene Umgebung U von z0 und eine auf U holomorphe Funktion g mit g(z0 ) = 0, so dass f (z) = (z − z0 )n g(z) f¨ ur z ∈ U, z = z0 und n = ordz0 f (∗) (∗) Dann gilt aber f¨ ur z ∈ γ + U (γ + U ist offene Umgebung von z0 + γ): f (z) = f (z − γ) = n (z −(z0 +γ)) g(z −γ).