By Gabriel Debs

One of many goals of this paintings is to enquire a few ordinary houses of Borel units that are undecidable in $ZFC$. The authors' place to begin is the subsequent straight forward, although non-trivial consequence: think about $X \subset 2omega\times2omega$, set $Y=\pi(X)$, the place $\pi$ denotes the canonical projection of $2omega\times2omega$ onto the 1st issue, and believe 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 particularly closed subset of $X$ is ideal onto $Y$"" it follows that during the current case $Y$ can be $\mathbf{\Pi zero 2$. 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 resembling 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 $, is reminiscent of ""$\aleph \xi \aleph 1$"". notwithstanding we will exhibit independently, that after $X$ is Borel you possibly can, in $ZFC$, derive from $(\star)$ the weaker end that $Y$ is usually Borel and of an analogous Baire type as $X$. This final end result solves an previous challenge approximately compact protecting mappings. actually those effects are heavily concerning the subsequent 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 means 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 provide the precise set theoretical power of this precept reckoning on the descriptive complexity of $X$ and $Y$. The authors additionally turn out the same consequence for a version of Lift$(X, Y)$ during which ""continuous liftings"" are changed by way of ""Borel liftings"", and which solutions a query of H. Friedman. between different functions the authors receive an entire way to an issue which matches again to Lusin about the lifestyles of $\mathbf{\Pi 1 1$ units with all parts in a few given category $\mathbf{\Gamma $ of Borel units, enhancing past 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 kind, related to a large number of ""abstract algebra"". This illustration was once at the start constructed for the needs of this facts, yet has a number of different purposes.

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**Extra resources for Borel Liftings of Borel Sets: Some Decidable and Undecidable Statements**

**Sample text**

If p is a periodic point of Ot with period T and {p} is a closed set, then OT (p) = p. Proof. Let A =IT > 0 1 0T(p) = p}. By the definition of period, T = inf A. So, T E cl(A). Because {p} is closed, (T, p) E cl(A) x {p} = cl(A x {p}). 2. Periodic Points 29 Since 0 is continuous and {p} is closed, OT (p) = O(T, p) E cl(q(A x {p})) = cl({p}) = {p}. Therefore, OT (p) = p. 5. Let qt be a flow on a topological space X. If p E X and O' (p) = p for some real number T, then OnT (p) = p for every integer n.

So, each orbit lies within a level set of F. To determine the Poincare recurrent set of Ot we analyze the level sets of F. Case I. F(x, y, z) = 0. In this case the level set of F is the z-axis. Along the z-axis, x = y = 0, and the system of differential equations reduces to the constant differential equation z=1. Thus, the z-axis is an orbit of Ot, and no point on the z-axis is Poincare recurrent. Case II. F(x, y, z) = 2 for r > 0. r 2. Recurrent Points 54 Parametrize the level set F(x, y, z) = 1/r2 by x(a, 8) _ r2 (r -} - 1 cos a) cos 0, = (r + r2 - 1 cos a) sin 9, r2 - 1 sin a, z(a, 9) = y(a, 9) where a e S' and 9 E S1.

The fixed set of a flow on a Hausdorff topological space is closed. Proof. Let Ot be a flow on a Hausdorff topological space X. We shall prove that X \ Fix(gt) is open. Let x E X \ Fix(gt). Thus, there exists a real number T such that 0'(x) L x. Since X is Hausdorff, there exist disjoint open subsets U and V of X such that 0'(x) E U and x E V. Let A = 0--r(U) n V. Since 0' is continuous, A is an open set. We shall show that A C X \ Fix(ot). If y c A, then 0'(y) E U and y E V. Because U and V are disjoint, qT(y) y.