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I have Cantor intersection theorem:
Let $X$ be a complete metric space, and let $\{F_n\}$ be a decreasing sequence of non-empty closed subsets of X such that $d(F_n)$ converges to $0$.
Then $F=\bigcap_{n=1}^\infty F_n$ contains exactly one point.
What affects whether the condition $d(F_n)$ converges to zero is removed?
Give an example to show that the set $F$ in the intersection of Cantor theorem may be empty if this happens?

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This sounds like homework. Any ideas? – Alex Youcis Apr 22 '12 at 2:39
What is $d(F_n)$? Diameter, right? – user21436 Apr 22 '12 at 2:45
The book suggests that the reader ask the question but more that I think I find the example – Chalie Her Apr 22 '12 at 2:46
yes, but I can not find the example – Chalie Her Apr 22 '12 at 2:48
See also this question: An empty intersection of decreasing sequence of closed sets – Martin Sleziak Apr 22 '12 at 4:41
up vote 3 down vote accepted

Note: The following examples show that the conditions $\displaystyle \lim_{x\to\infty} d(F_{n})=0$ and that $F_{n}$ are closed sets both are necessary for the validity of the theorem.

Example: Let $X$ be the real line $R$ and let $F_{n}=[n,\infty).$ Now we know that $X$ is complete, $F_{1}\supset F_{2}\supset F_{3}....$ and $F_{n}$ are closed sets. But $\displaystyle \bigcap_{n=1}^{\infty} F_{n}=\phi.$Note that $\displaystyle \lim_{n\to\infty}d(F_{n})\neq 0.$

Example: Let $X$ be the real line $R$ and let $\displaystyle F_{n}= ( 0, \frac{1}{n}].$ Now we know that $X$ is complete, $F_{1}\supset F_{2}\supset F_{3}....$ and $\displaystyle \lim_{n\to\infty}d(F_{n})=0.$ But $\displaystyle \bigcap_{n=1}^{\infty} F_{n}=\phi.$ Note that the $F_{n}'s$ are not closed.

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The easiest (and I think necessary in some sense) type of example where d(Fn) does not converge to 0 and the intersection is void is when d(Fn) = infinity. For instance R and the intervals [n,infinity) are all closed, yet their intersection is empty. Besides if you didn't have d(Fn) to 0 you wouldn't get a unique point (like say d(Fn)--->c < infinity)

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