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 \\ $ .<\/p>\n
Considering the intersection $\\cap_x F_x$ we see that many of the sets could be skipped without altering the intersection.<\/p>\n
Question: Is it possible to attain the same intersection by taking only countably-many of the subsets?<\/p>\n
Theorem:<\/strong> 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.<\/p>\n In order to give a proof we first show a simple lemma.<\/p>\n Lemma<\/strong>: If $\\{A_j\\}_{j=1}^\\infty $ is a chain then there exists a decreasing sequence out of its members<\/p>\n $$ G_1 \\supset G_2 \\supset \\cdots $$<\/p>\n such that $\\cap_j A_j = \\cap_i G_i $ .<\/p>\n 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:<\/p>\n 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.<\/p>\n Inductively, assuming that $G_1\u00a0\\supset\u00a0 G_2\u00a0\\supset\u00a0 \\cdots\u00a0\\supset\u00a0 G_k$, 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.<\/p>\n If an infinite sequence\u00a0$G_1\u00a0\\supset\u00a0 G_2\u00a0\\supset\u00a0 \\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.)<\/p>\n Proof of the theorem:<\/strong> 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 $”.<\/p>\n We will reach a contradiction by assuming that the assertion of the theorem were false.<\/p>\n Take a chain\u00a0$\\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.<\/p>\n 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\\ $.<\/p>\n It is a fact of measure theory that<\/p>\n $$\\mu (\\cap_i G_i ) = \\lim_{j \\rightarrow \\infty} \\mu (G_j) =\u00a0\\lim_{j \\rightarrow \\infty} 1 =1 \\ .$$<\/p>\n Observe that $\\cap_i G_i$ is in $\\mathfrak{C}$, and thus a lower bound to the chain $\\mathfrak{F}$.<\/p>\n 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 $\\mathfrak{C}$.<\/p>\n This contradiction shows that for some chains in $\\mathfrak{C}$ the intersection cannot be replaced by a countable intersection.<\/p>\n Note: We can work with smaller $\\mathfrak{C}$ such as the collection $\\{[0,1]\\backslash D \\ \\ | \\ \\ D \\text{\u00a0 is finite}\\}$.<\/p>\n <\/p>\n <\/p>\n