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Cantor's Intersection Theorem:

Let $X$ be a complete metric space and let $\{F_n\}$ be a sequence of decreasing non-empty closed subsets of $X$. If $d(F_n)\rightarrow 0$ then $F=\bigcap_{n=1}^\infty F_n$ contains exactly a unique point of $X$

Where $d(F_n)$ means the diameter of the set $F_n$ and $d(F_n)\rightarrow 0$ means that the sequence of the diameters converge to $0$.

I found that I can find an example where all the conditions except the closureness of the subsets hold but we still have the intersection a single point.

The example is, Consider $X$ to be the complex plane.For $n= 1,2,3,4,5,...$, define $F_n=\{(x,y):\sqrt{x^2 + y^2} < \frac{1}{n}\}$. So $F_n$ is disc of radius $\frac{1}{n}$.

It's clear that the sequence $\{F_n\}$ is a decreasing one. every $F_n$ is an open (but not closed) sphere of $X$ the sequence of diameters converges to $0$ but we still have:

$\bigcap_{n=1}^{\infty} F_n=\{(0,0)\}$.

So, it seems that this condition is sufficient but not necessary, Am I right?

What are the necessary conditions of this theorem?

Added:

I wonder if there is an example where only a one subset is not closed but the other conditions hold and we have an empty intersection?

FNH
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Yes, sufficient but not necessary. But to show one needs the condition consider examples like $F_n=(0,1/n)$ in the space of reals, where this time the intersection is empty (here the $F_n$ are not closed). To drop the other condition, just consider a constant sequence $F_n=[0,1]$ where this time the $F_n$ are closed but the lengths $d(F_n)$ fail to approach zero, and there are more than one point in the intersection (in fact the entire interval $[0,1]$ is the intersection).

Added: The OP has asked about what if only one of the $F_n$ is not assumed to be closed. Suppose that not necessarily closed set happens when the index $n$ is some particular natural number $n=n_1.$ All the sets $F_k$ where $k\ge n_1+1$ are then assumed to be closed. Their intersection $G$ is then nonempty by Cantor, and $G$ is a subset of $F_{n_1+1}$ because the sets are a decreasing sequence. Thus there is a point $p$ in $G$, which then lies in every $F_j$ for which $j \ge n_1+1.$ Now again from the fact that the whole sequence of sets is decreasing, $F_{n_1+1}$ is a subset of each $F_j$ for $1 \le j \le F_{n_1},$ so that the point $p$ also lies in each set $F_j$ where $1 \le j \le n_1.$ this shows that for every index $j\ge 1$ we have $p$ in $F_j,$ which is to say that $p$ is in the intersection of all the $F_j$'s.

coffeemath
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  • Cool, I've already come with examples like these. But I wonder if there is an example where only a one subset is not closed but the other conditions hold and we have an empty intersection? For some reason, I think such example doesn't exist. – FNH Oct 02 '14 at 11:26
  • If say $A_{100}$ is not closed but the rest are closed, and the sets are downwardly nested, then one can just reset the intersection to begin with $A_{101}$ and then the usual conditions are met, so nonempty intersection follows. – coffeemath Oct 02 '14 at 11:36
  • I didn't get the point, could you clarify more, please? – FNH Oct 02 '14 at 11:38
  • @MathsLover Just put something in the answer about this, in the part labeled "added". (Also changed notation to $F_n$ as in your question, rather than the $A_n$ I had used before.) – coffeemath Oct 02 '14 at 12:27
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    I think that The big intersection $A$ is contained in $B$ not the other way. $A \subset B$. but you assume that $B\subset A$ – FNH Oct 02 '14 at 12:54
  • I'll have to re-think this, will adjust the "added" part soon. – coffeemath Oct 02 '14 at 13:22
  • Let us continue this discussion in chat. – coffeemath Oct 02 '14 at 13:38
  • I think that your argument after adjusting is correct :) thank you :) – FNH Oct 02 '14 at 17:47