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单词 ProofOfExtendingACapacityToACartesianProduct
释义

proof of extending a capacity to a Cartesian product


Let (X,) be a paved space such that is closed underPlanetmathPlanetmath finite unions and finite intersectionsMathworldPlanetmathPlanetmath, and (K,𝒦) be a compactPlanetmathPlanetmath paved space.Define 𝒢 to be the closureMathworldPlanetmathPlanetmath under finite unions and finite intersections of the paving ×𝒦 on X×K.For an -capacity I, define

I~:𝒫(X×K),
I~(S)=I(πX(S)),

where πX is the projection map onto X. We show that I~ is a 𝒢-capacity and that πX(S)δ whenever S𝒢δ.

Clearly, the property that I~ is an increasing set function follows from the fact that I satisfies this property. Furthermore, if SnX×K is an increasing sequence of sets with S=nSn then πX(Sn) is an increasing sequence and

I~(S)=I(πX(S))=I(nπX(Sn))=limnI(πX(Sn))=limnI~(Sn).

To prove that I~ is a 𝒢-capacity, it only remains to show that if Sn is a sequenceMathworldPlanetmathPlanetmath in 𝒢 decreasing to SX×K then I~(Sn)I~(S).Note that any S in 𝒢 can be written as S=j=1mk=1njAj,k×Kj,k for sets Aj,k and Kj,k𝒦. The projection onto X is then

πX(S)={j=1mAj,kj:kjnj,j=1mKj,kj}

which, as is closed under finite unions and finite intersections, must be in .Furthermore, for any xX,

Sx{yK:(x,y)S}=j=1m{Kj,k:knj,xAj,k}.

This shows that Sx is in the closure 𝒦* of 𝒦 under finite unions and finite intersections. Furthermore, since compact pavings are closed subsets of a compact topological spaceMathworldPlanetmath (http://planetmath.org/CompactPavingsAreClosedSubsetsOfACompactSpace), 𝒦* is itself a compact paving.

Now let Sn be a decreasing sequence of sets in 𝒢 and set S=nSn. Then πX(S)πX(Sn) for each n, giving πX(S)nπX(Sn). To prove the reverse inequality, consider xnπX(Sn). Then, (Sn)x is a nonempty set in 𝒦* for all n. By compactness, Sx=n(Sn)x must also be nonempty and therefore xπX(S). This shows that

nπX(Sn)=πX(S).

Furthermore, as we have shown that πX(Sn) and, as I is an -capacity,

I~(Sn)=I(πX(Sn))I(πX(S))=I~(S).

So I~ is a 𝒢-capacity.

We finally show that if S𝒢δ then πX(S)δ. By definition, there is a sequence Sn𝒢 such that S=nSn. Setting Sn=mnSm then, since 𝒢 is closed under finite unions and finite intersections, Sn𝒢. Furthermore, Sn decreases to S so, as shown above, πX(Sn) and

πX(S)=nπX(Sn)δ

as required.

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更新时间:2025/5/4 10:36:17