The standard statement of Cantor's intersection theorem is that, in any topological space, a nested sequence of compact, closed, nonempty sets also has a nonempty intersection.
What confuses me now is that why is "nested" needed here? For example, if I have a sequence of compact, closed sets $A_1\supset A_2\supset\dotsm$, I can simply take $B_n=A_1\cap\dotsm\cap A_n$. Since $A_n$ are assumed to be closed, a finite intersection of closed sets is closed, so $B_n$ are nested and closed. And also $\bigcap_n B_n=\bigcap_n A_n$. So if I remove "nested" from the theorem statement, it would still be true.
So my questions are:
(1) Is my reasoning above correct?
(2) Is there any "counterexample" to the Cantor intersection theorem where "nested" is violated?
Best Answer
It is not correct. If you have a sequence $(A_n)$ of (nonempty!) compact closed sets, you can define $B_n = \bigcap_{i=1]}^n A_i$. The $B_n$ are compact closed sets, the sequence $(B_n)$ is nested and $\bigcap_n A_n = \bigcap_n B_n$. But in general some $B_n$ may be empty, thus you cannot conclude that $\bigcap_n A_n \ne \emptyset$. If $(A_n)$ is nested, we have $B_n = A_n \ne \emptyset$.
As an example take the subsets $A_n = \{n\}$ of $\mathbb R$.
The assumption that you really need is known as the finite intersection property:
A family $\mathcal A = (A_i)_{i \in I}$ of sets has the finite intersection property if all finite intersections of members of in $\mathcal A$ are nonempty.
We can prove
Theorem 1. Let $\mathcal A = (A_i)_{i \in I}$ be a family of closed compact subsets of a space $X$. If $\mathcal A$ has the finite intersection property, then $\bigcap_{i \in J} A_i \ne \emptyset$.
Note that this is also true under the weaker assumption that all $A_i$ are closed and one $A_{i_0}$ is compact. Then the $A'_i = A_i \cap A_{i_0}$ form a family of compact closed subsets having the finite intersection property. Since $\bigcap_{i \in I} A_i = \bigcap_{i \in I} A'_i$, we are done.
Theorem 1. follows easily from
Theorem 2. A space $X$ is compact if and only if each family of closed subsets $\mathcal A = (A_i)_{i \in I}$ which has finite intersection property has nonempty intersection $\bigcap_{i \in I} A_i$.
The proof is easy: Observe that open covers $\mathcal U = (U_i)_{i \in I}$ of $X$ are in $1$-$1$ correspondence with families $\mathcal A = (A_i)_{i \in I}$ of closed subsets having empty intersection $\bigcap_{i \in I} A_i$ (take the complements of the members of such families). Finite subcovers of $\mathcal U$ correspond to finite subfamilies of $\mathcal A$ which have empty intersection.