Intersection of a Chain of Subsets

Assume {Fx}xΓ is a collection of subsets (of a not-so important set!) such that every two are comparable, i.e for any x and y, either FxFy   or   FxFy  .

Considering the intersection xFx we see that many of the sets could be skipped without altering the intersection.

Question: Is it possible to attain the same intersection by taking only countably-many of the subsets?

Theorem: There is a chain of subsets of the unit interval whose intersection does not equal the intersection of any countably-many of them. They may be chosen measurable.

In order to give a proof we first show a simple lemma.

Lemma: If {Aj}j=1 is a chain then there exists a decreasing sequence out of its members

G1G2

such that jAj=iGi .

Proof: Take G1=A1 , and let j(1)=1 . We will move along the sequence Ai and pick those that are needed in the intersection, which means those that are smaller. The details are as follows:

Let j(2) be the first index bigger than j(1) such that Aj(2)Gj(1). If no such index exists then all the subsequent sets contain A1, and so we can take the constant sequence Gi=Aj(1) and it will satisfy the assertions in the claim.

Inductively, assuming that G1G2Gk, and j(1)<j(2)<<j(k) have been defined, we define j(k+1) to be the least index after j(k) such that Aj(k+1)Gk=Aj(k), and Gk+1=Aj(k+1). Again, if such an index does not exist then we could have the constant sequence Gk satisfying the assertions of the claim.

If an infinite sequence G1G2  emerges eventually, then it is the sequence claimed, because each Ai was given a chance at some point! (If, say, A2017 is none of the Gj’s, then it means that it contained one of them.)

Proof of the theorem: Take C to be the collection of all Lebesgue-measurable subsets of the unit interval with measure equal to 1 and order it with the inclusion relation “”.

We will reach a contradiction by assuming that the assertion of the theorem were false.

Take a chain F={Fx}xΓ in C. There would be a countable subcollection {Aj}j=1 giving the same intersection.

By the lemma, we can further restrict to a decreasing sequence G1G2;Gj{Fx}x and still have iGi=jAj=xFx .

It is a fact of measure theory that

μ(iGi)=limjμ(Gj)=limj1=1 .

Observe that iGi is in C, and thus a lower bound to the chain F.

What we have shown is that every chain in C has a lower bound. By Zorn’s lemma, there exists a smallest element in C. However, for any element is C removing a single point gives an even smaller set in C.

This contradiction shows that for some chains in C the intersection cannot be replaced by a countable intersection.

Note: We can work with smaller C such as the collection {[0,1]D  |  D is finite}.

 

 

About Behnam Esmayli

I started PhD in Mathematics at Pitt in Fall 2015. I have come to grow a passion for metric spaces -- a set and a distance function that satisfies the triangle inequality -- simple and beautiful! These spaces when equipped with other structures, such as a measure, becomes extremely fun to play with!
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2 Responses to Intersection of a Chain of Subsets

  1. Brandon says:

    I seem to be misunderstanding something fundamental — isn’t the intersection of any set with a subset of itself equal to that subset? And every set being pairwise related with a subset relation induces a total (quasi, since the subset operator is non-reflexive) ordering over those sets? Which means that there’s a unique minimal element, which will always be equal to the intersection of the whole chain?

    • Behnam Esmayli says:

      Consider the open interval (0,1). It is a totally ordered set. It does not have a minimum (because it does not have a minim-AL.) So, the answer to your questions is: we don’t know if an infimum exists. Note that the intersection of all the sets need not belong to the collection — if it did, it would obviously be the minimum.

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