Assume {Fx}x∈Γ 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 Fx⊂Fy or Fx⊃Fy .
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
G1⊃G2⊃⋯
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 G1⊃G2⊃⋯⊃Gk, 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 G1⊃G2⊃⋯ 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}x∈Γ 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 G1⊃G2⊃⋯;Gj∈{Fx}x and still have ∩iGi=∩jAj=∩xFx .
It is a fact of measure theory that
μ(∩iGi)=limj→∞μ(Gj)=limj→∞1=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}.
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?
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.