Assume $\{F_x\}_{x \in \Gamma}$ 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 $F_x \subset F_y \ \ $ or $\ \ F_x \supset F_y \ $ .

Considering the intersection $\cap_x F_x$ 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 $\{A_j\}_{j=1}^\infty $ is a chain then there exists a decreasing sequence out of its members

$$ G_1 \supset G_2 \supset \cdots $$

such that $\cap_j A_j = \cap_i G_i $ .

Proof: Take $G_1=A_1\ $, and let $j(1)=1\ $. We will move along the sequence $A_i$ 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 $A_{j(2)} \subsetneq G_{j(1)}$. If no such index exists then all the subsequent sets contain $A_1$, and so we can take the constant sequence $G_i=A_{j(1)}$ and it will satisfy the assertions in the claim.

Inductively, assuming that $G_1 \supset G_2 \supset \cdots \supset G_k$, and $j(1) < j(2) < \cdots < j(k)$ have been defined, we define $j(k+1)$ to be the least index after $j(k)$ such that $A_{j(k+1)} \subsetneq G_k=A_{j(k)}$, and $G_{k+1}=A_{j(k+1)}$. Again, if such an index does not exist then we could have the constant sequence $G_k$ satisfying the assertions of the claim.

If an infinite sequence $G_1 \supset G_2 \supset \cdots \ $ emerges eventually, then it is the sequence claimed, because each $A_i$ was given a chance at some point! (If, say, $A_{2017}$ is none of the $G_j$’s, then it means that it contained one of them.)

**Proof of the theorem:** Take $\mathfrak{C}$ 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 “$\subset $”.

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

Take a chain $\mathfrak{F}=\{F_x\}_{x \in \Gamma}$ in $\mathfrak{C}$. There would be a countable subcollection $\{A_j\}_{j=1}^\infty $ giving the same intersection.

By the lemma, we can further restrict to a decreasing sequence $ G_1 \supset G_2 \supset \cdots ; G_j \in \{F_x\}_x$ and still have $ \cap_i G_i = \cap_j A_j = \cap_x F_x\ $.

It is a fact of measure theory that

$$\mu (\cap_i G_i ) = \lim_{j \rightarrow \infty} \mu (G_j) = \lim_{j \rightarrow \infty} 1 =1 \ .$$

Observe that $\cap_i G_i$ is in $\mathfrak{C}$, and thus a lower bound to the chain $\mathfrak{F}$.

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

This contradiction shows that for some chains in $\mathfrak{C}$ the intersection cannot be replaced by a countable intersection.

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

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.