proof of equivalent definitions of analytic sets for paved spaces

Let (X,β„±) be a paved space with βˆ…βˆˆβ„±, let 𝒩 be Baire spacePlanetmathPlanetmath, and let Y be any uncountable Polish spaceMathworldPlanetmath. For a subset A of X, we show that the following statements are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath.

  1. 1.

    A is β„±-analytic (

  2. 2.

    There is a closed subset S of 𝒩 and ΞΈ:β„•2β†’β„± such that

  3. 3.

    There is a closed subset S of 𝒩 and ΞΈ:β„•β†’β„± such that

  4. 4.

    A is the result of a Souslin scheme on β„±.

  5. 5.

    A is the projection of a set in (ℱ×𝒦)σ⁒δ onto X, where 𝒦 is the collectionMathworldPlanetmath of compact subsets of Y.

  6. 6.

    A is the projection of a set in (ℱ×𝒒)σ⁒δ onto X, where 𝒒 is the collection of closed subsets of Y.

(1) implies (2): As A is analytic, there exists a compact paved space (K,𝒦) and a set B∈(ℱ×𝒦)σ⁒δ such that A=Ο€X⁒(B), where Ο€X:XΓ—Kβ†’X is the projection map. Write


for An,mβˆˆβ„± and Kn,mβˆˆπ’¦. Rearranging this expression,


So, defining SβŠ†π’© by

S={sβˆˆπ’©:β‹‚n=1∞Kn,snβ‰ βˆ…}.



Setting θ⁒(n,m)=An,m gives the required expression, and it only remains to show that S is closed. So, let s1,s2,… be a sequence in S converging to a limit sβˆˆπ’©. For any kβ‰₯0 then snr=sn for all n≀k and large enough r. Hence,

β‹‚n≀kKn,sn=β‹‚n≀kKn,snrβŠ‡β‹‚n=1∞Kn,snβ‰ βˆ….

So, the collection of sets Kn,sn for n=1,2,… satisfies the finite intersection property, and compactness ( of the paving 𝒦 gives

β‹‚n=1∞Kn,snβ‰ βˆ…,

showing that s∈S and that S is indeed closed.

(2) implies (3): Supposing that A satisfies the required expression, choose any bijection Ο•:β„•β†’β„•2. Then define ΞΈ~β‰‘ΞΈβˆ˜Ο• and f:𝒩→𝒩 by f⁒(s)=t where tn=Ο•-1⁒(n,sn). As S is closed, it follows that S~=f⁒(S) will also be closed and,


as required.

(3) implies (4): Suppose that A satisfies the required expression and define a Souslin scheme (As) as follows. For any nβ‰₯1 and sβˆˆβ„•n let us set

As={θ⁒(sn),ifΒ s=t|nΒ for someΒ tβˆˆπ’©,βˆ…,otherwise.

Then, for sβˆˆπ’©,

β‹‚n=1∞As|n={β‹‚n=1∞θ⁒(sn),ifΒ s∈S,βˆ…,otherwise.

Here, if sβˆ‰S, we have used the fact that S is closed to deduce that for large n, there is no t∈S with t|n=s|n and, therefore, As|n=βˆ…. The result of the Souslin scheme (As) is then


as required.

(4) implies (5): Suppose that A is the result of a Souslin scheme (As). Let us first consider the case where Y is Cantor space, π’ž={0,1}β„•, which is a compact Polish space. Then, for any sβˆˆβ„•n, let Ks be the set of tβˆˆπ’ž such that tk=1 if k=s1+β‹―+sm for some m≀n and tk=0 for all other k<s1+β‹―+sn. These are closed and, therefore, compact sets.

Given any sequence s1βˆˆβ„•1,s2βˆˆβ„•2,… it is easily seen that β‹‚nKsn is nonempty if and only if there is an sβˆˆπ’© such that s|n=sn for all n. Define the set B in (ℱ×𝒦)σ⁒δ by


The projection of B onto X is then


which is the result A of the scheme (As) as required. The result then generalizes to any uncountable Polish space Y, as all such spaces contain Cantor space (

(5) implies (6): This is trivial, since all compact sets are closed.

(6) implies (1): This is a consequence of the result that projections of analytic sets are analytic.

Title proof of equivalent definitions of analytic sets for paved spaces
Canonical name ProofOfEquivalentDefinitionsOfAnalyticSetsForPavedSpaces
Date of creation 2013-03-22 18:48:36
Last modified on 2013-03-22 18:48:36
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Proof
Classification msc 28A05