Let $(X,d)$ be a complete metric space and $\{A_n\}$ be the decreasing sequence of non empty closed subsets of $X$. Then Cantor's intersection theorem tells that if $d(A_n) \to 0$ then the set $A = \cap_{i} A_i$ consists a single point.
If any of the above conditions are dropped then $A$ would not be a singleton set. Suppose if we drop the condition of decreasing sequence then $A$ may not be bounded or if we drop closedness of $A_i$ then $A$ may be empty.
Now if we drop the condition $d(A_i) \to 0$ then I intuitively feel that $A$ should be non empty even though I'm not able to prove that or to find any counter examples. Does there exist a sequence like the diameter of them does not go to 0 but intersection is empty?
Best Answer
Let $A_n$ be the closed interval $[n,\infty)$. They are decreasing but have empty intersection.