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Cantor's intersection Theorem: If $(D_n)_{n=1}^\infty$ is a sequence of nonempty closed sets in a complete metric space $(X,d)$ such that $D_{n+1}\subset D_n$ for all $n\in\mathbb{N}$ and $\text{diam}(D_n)\to0$ as $n\to\infty$, then $\bigcap_{n=1}^\infty D_n$ consists of precisely one point.

Give examples with $X=\mathbb{R}$ to show that $\bigcap_{n=1}^\infty D_n$ may be empty or may contain more than one point if the hypothesis $\text{diam}(D_n)\to0$ as $n\to\infty$ is omitted.

My first idea and attempt:

Let $D_0=[\pi-1,\pi]$, so $X$ is closed and nonempty. For each $n\in\mathbb{N}$, we make the $n^{th}$ decimal place $0$, hence $D_{n+1}\subset D_n$. So $D_n$ converges to the set $[\pi-1,3]$. So it contains more than one point. And we have also omitted the hypothesis $\text{diam}(D_n)\to0$ as $n\to\infty$.

  1. Does my example answer the question?

  2. If it does, how can I write it in a more rigorous way?

Another attempt:

What if $D_n=[n,\infty)$? The set is closed right? I am not sure whether this example answer the question.

Thank you for the help!

Martin Sleziak
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user71346
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    What do you mean by $\subset$ is it "proper subset" or only "subset"? If it is only "subset" both counterexamples are fine. Else think about the first one. Also you can do the first one easier. –  Feb 08 '16 at 12:02
  • @sigmabe I think it is a proper subset, although I am not sure. Does it matter? – user71346 Feb 08 '16 at 12:24
  • For cantor's intersection theorem it doesn't matter, but if it is "proper" you must have a counterexample with "proper". Maybe the n-th decimal is already 0, so you don't get "proper" in this case. –  Feb 08 '16 at 12:45
  • You can find several older posts about hte same (or at least very similar) problem: http://math.stackexchange.com/questions/1397739/an-example-of-a-nested-decreasing-sequence-of-bounded-closed-sets-with-empty-int http://math.stackexchange.com/questions/388930/can-the-intersection-of-open-or-closed-balls-be-empty-if-their-radii-are-bounde http://math.stackexchange.com/questions/806708/examples-of-decreasing-sequences-of-closed-sets-with-constant-diameter-and-empty – Martin Sleziak Feb 08 '16 at 15:09

1 Answers1

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While your first example is (sort of) fine I'd like you to note that $D_n = D_{n+1}$ for infinitely many $n$'s, namely those for which the $n+1$st decimal place of $\pi$ is $0$. (Actually, I'm not even sure if it has been proven that the decimal representation of $\pi$ contains infinitely many $0$'s, so let's say for many $n$'s and maybe even infinitely many of them.) This can be fixed by setting the leftmost decimal place not equal to zero to zero. You may also set $D_n = [0, 1 + \frac{1}{n}]$. Clearly $D_{n+1} \subsetneq D_n$ for all $n$ and $\bigcap_{n=1}^\infty D_n = [0,1]$.

Regarding your second example: Let $D_n = [n, \infty)$. Then $\mathbb R \setminus D_n = (-\infty, n)$ is open and therefore $D_n$ is closed for all $n$. Moreover $D_{n+1} \subsetneq D_n$ for all $n$ and $\bigcap_{n=1}^\infty D_n = \emptyset$. So this is a valid example as how to violate the conclusion of Cantor's Intersection Theorem - if we drop the premise that $\operatorname{Diam}(D_n) \overset{n \to \infty} \to 0$.

Agile_Eagle
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Stefan Mesken
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