6

The nested intervals theorem says the following.

If a sequence of intervals $\langle I_n\rangle$ is decreasing, then $\bigcap_{n=1}^{\infty} I_n$ is not empty.

However, I'm trying to modify the theorem, say, if the sequence is strictly decreasing, then $\bigcap_{n=1}^{\infty} I_n$ should be a single point.

What I tried:

Let $I_n = [ a_n, b_n]$. Then $\langle a_n\rangle$ is increasing, while $\langle b_n\rangle$ is decreasing. Since $\langle a_n\rangle$ is bounded, it converges to a point, say $\alpha$. And, since $b_k$ is an upper bound $\forall k$, thus $\alpha \le b_k$.

Even if $\langle b_n\rangle$ is decreasing and bounded, I don't know how I can say $\lim_{n \to \infty} b_n = \alpha$.

Is there anyone to help me?

Nosrati
  • 29,995
Moreblue
  • 2,004
  • 1
    If all you want is to prove that the intersection may be a singleton, one example is enough. Take $a_n=-\frac1n$ and $b_n=\frac1n$, for instance. The intersection will be ${0}$ then. – José Carlos Santos Jul 27 '18 at 12:59
  • 11
    The sequence strictly decreasing is not enough here. You need the limit of the sequence $b_n-a_n$ to be zero. Then the intersection will be one point. – Mark Jul 27 '18 at 13:01
  • @José Carlos Santos I modified the expression. – Moreblue Jul 27 '18 at 13:09
  • @Mark Is there any counter-example? – Moreblue Jul 27 '18 at 13:10
  • 3
    @moreblue Consider $I_n = \left(0, 1+\frac1n\right)$. – mechanodroid Jul 27 '18 at 13:11
  • The result that the intersection is not empty is known as "Cantor's intersection theorem", and if you are adding that $\lim_{n\to\infty}(b_n-a_n)=0$ then the result is known as "cantor's lemma". A $\LaTeX$ tip, use \langle and \rangle to create $\langle$ and $\rangle$ – ℋolo Jul 27 '18 at 13:35
  • I'm pretty sure Cantor's intersection theorem is what OP meant to ask, so editing the question might be necessary here – Rushabh Mehta Jul 27 '18 at 16:21
  • I think it's worth noting that you nearly provided your own counterexample. If you're stuck on a step of a proof, see if you can come up with an example where that step fails. In your case, it would be easy to manufacture two sequences as in your problem where $\lim_{n \to \infty} b_n$ doesn't equal $\alpha$, and then you'd have your counterexample. (Of course, this is just what @JoséCarlosSantos does.) – LSpice Jul 28 '18 at 02:50
  • (Also, a minor matter of language: although what you meant is clear to you and to any professional mathematician, strictly speaking "$b_k$ is an upper bound $\forall k$" is nonsense; there is no such thing as "an upper bound" full stop, only "an upper bound of a set". Thus it is better (until you are certain that you're not mentally eliding such facts) to say something like "for all $k$, $b_k$ is an upper bound of ${a_n : n \in \mathbb Z_{> 0}}$".) – LSpice Jul 28 '18 at 02:55
  • The first theorem you state needs the intervals to be compact, otherwise $I_n=(0,1/n)$ or $I_n=[n,\infty)$ are counterexamples. – Julian Rosen Jul 28 '18 at 16:59

1 Answers1

30

The statement is false. Take $a_n=-1-\frac1n$ and $b_n=1+\frac1n$. Then the sequence $(I_n)_{n\in\mathbb N}$ is strictly decreasing, but$$\bigcap_{n\in\mathbb N}[a_n,b_n]=[-1,1].$$

José Carlos Santos
  • 427,504