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I hope everything is going well for all of you. I was working through this problem and would appreciate some advice and criticisms on my proof. Anything flies, please come at me with critiques of accuracy, succinctness, and rigor. Here is the problem.

Problem

Let $K_n$, $n \in \mathbb{N}$ be a family of nested, decreasing nonempty compact sets in $\mathbb{R}$. Moreover, suppose that $\text{diam}(K_n) \to 0 \text{ as }n \to \infty$. Prove that there exists a unique point $x \in \mathbb{R}$ such that $\bigcap_{n=1}^{\infty}K_n = \{x\}$.

Some Definitions

Nested Compact Set Property - If $K_1 \supseteq K_2 \supseteq K_3 \dotsm$ is a nested set sequence of nonempty compact sets, then the intersection $\bigcap_{n=1}^{\infty} = K_n$ is not empty.

Cauchy Sequence: A sequence $(a_n)$ is called a Cauchy sequence if, for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that whenever $m,n \geq N$ it follows that $|a_n - a_m| < \epsilon$.

Theorem 2.6.2.: Every convergent sequence is a Cauchy sequence.

Compact Set: A set $K \subseteq \mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.

Closed Set: A set $F \subseteq R$ is closed if it contains its limit points.

Theorem 3.3.4. - A set $K \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

My Attempt

$\text{ }$ For all $n \in \mathbb{N}$, let $K_n \subseteq \mathbb{R}$ be a nonempty compact set such that $K_n \subseteq K_{n+1}$. We already know that $\bigcap_{n=1}^{\infty} K_n \neq \emptyset$ (Nested Compact Set Property) and $\text{diam}(K_n) \to 0$ as $n \to \infty$ and want to show that $\bigcap_{n=1}^{\infty} K_n = \{x\}$.

$\text{ }$ Given that $\text{diam}(K_n) \overset{n \to \infty}{\to} 0$, we know that $\forall \epsilon > 0$, $\exists N_{\epsilon} \in \mathbb{N}$ such that $|\text{diam}(K_n) - 0|<\epsilon$, $\forall n \geq N_{\epsilon}$. For $n, m \geq N_{\epsilon}$, we choose $a_n, a_m \in K_n$. By the definition of least upper bounds, we have that $|a_n - a_m| \leq \underset{x, y\in K_n}{\sup }|x - y|$. This means that $$|a_n - a_m| \leq \underset{x, y\in K_n}{\sup }|x - y| = \text{diam}(K_n) =|\text{diam}(K_n) - 0| < \epsilon$$ Since $\forall \epsilon > 0$, $\exists N_{\epsilon} \in \mathbb{N}$ such that $|a_n - a_m| < \epsilon$ for all $ n,m \geq N_{\epsilon}$, we have that $(a_n)_{n\in \mathbb{N}}$ is a Cauchy sequence. By Theorem 2.6.2., $(a_n) \to x$ as $n \to \infty$ for some $x \in \mathbb{R}$. By Theorem 3.3.4., we have that each $K_n$ is closed, and since closed sets contain all of their limit points, $x \in K_n$ for every $n$. Thus, $\bigcap_{n=1}^{\infty} K_n = \{x\}$.

Some Quick Questions

Is this : "For $n, m \geq N_{\epsilon}$, we choose $a_n, a_m \in K_n$." permissible and, in the case that it is, is there a more elegant way of saying it?

For this part, "...that $(a_n)_{n\in \mathbb{N}}$ is a Cauchy sequence.", which I am having a difficult time processing mentally but know is true, what would be written in the underlined area for "...that $(a_n)_{n\in \mathbb{N}}$ is a Cauchy sequence contained in _______"

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  • $\begingroup$ Let me add a suggestion... you can prove the statement very quickly by using the “Nested Compact Set Property” since if there are two distinct points, the diameter of the set cannot go to $0$. $\endgroup$
    – Clayton
    Commented Apr 4, 2021 at 22:39
  • $\begingroup$ Alternative suggestion: you can try not to use the "Nested Compact Set Property” in your proof (try to show that the x you have constructed is in the intersection). $\endgroup$
    – J. Darné
    Commented Apr 4, 2021 at 22:50
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    $\begingroup$ As you say, your "for any $m, n \geq N$, we choose $a_n, a_m \in K_n$" is problematic, because you define $a_i$ several times for each integer $i$. Why don't your simply choose $a_n \in K_n$ for each $n$ ? (then you have $a_m, a_n \in K_N$ as soon as $n, m \geq N$...) $\endgroup$
    – J. Darné
    Commented Apr 4, 2021 at 22:53
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    $\begingroup$ Are you trying to prove Cantor's intersection theorem? $\endgroup$
    – rtybase
    Commented Apr 4, 2021 at 23:49
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    $\begingroup$ @rtybase Thank you for this link; I always forget to check Wikipedia, which is sad considering the idea of an accessible, expansive, free encyclopedia is a phenomenal human achievement. $\endgroup$ Commented Apr 4, 2021 at 23:57

1 Answer 1

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You are working much too hard. You already know that $\bigcap_{n\ge 1}K_n\ne\varnothing$, so let $x\in\bigcap_{n\ge 1}K_n$. Suppose $y\in\bigcap_{n\ge 1}K_n$, and $x\ne y$. Let $r=\frac12|x-y|>0$. There is an $n\in\Bbb Z^+$ such that $\operatorname{diam}(K_n)<r$, and $x,y\in K\subseteq K_n$, so $$|x-y|\le\operatorname{diam}(K_n)=r=\frac12|x-y|<|x-y|\,,$$ which is absurd. Thus, no such point $y$ exists, and $\bigcap_{n\ge 1}K_n=\{x\}$.

There are also some problems with your argument. First, you need to be more careful in choosing the points of your sequence: what you’ve written does not actually define them, since if $n\ne m$, you’re choosing $a_m$ twice, once in $K_n$ and once in $K_m$. To fix this just let $a_n\in K_n$ for each $n\in\Bbb Z^+$. Then $a_k,a_\ell\in K_n$ whenever $k,\ell\ge n$, so $|a_k-a_\ell|\le\operatorname{diam}(K_n)$ whenever $k,\ell\ge n$. For each $\epsilon>0$ there is an $n_\epsilon\in\Bbb Z^+$ such that $\operatorname{diam}(K_n)<\epsilon$ whenever $n\ge n_\epsilon$, so $|a_k-a_\ell|\le\epsilon$ whenever $k,\ell\ge n_\epsilon$, and the sequence $\langle a_n:n\in\Bbb Z^+\rangle$ is therefore Cauchy and hence convergent to some $x$. Let $n\in\Bbb Z^+$; $a_k\in K_n$ for each $k\ge n$, and $K_n$ is closed, so $x\in K_n$. Thus, $x\in\bigcap_{n\ge 1}K_n$.

Unfortunately, this actually just shows that $\bigcap_{n\ge 1}K_n$ is non-empty, and you already knew that; it does not show that the intersection is actually equal to the singleton $\{x\}$. Conceivable a different choice of the points $a_n$ could give you a sequence converging to some point different from $x$. Thus, you’re still left with the problem of proving that the intersection does not contain two distinct points, and you might as well use the argument that I gave at the top.

If you really want to get some benefit from the argument in the preceding paragraph, you could suppose that $y\in\bigcap_{n\ge 1}K_n$ and consider the sequence $\langle x_n:n\in\Bbb Z^+\rangle$, where

$$x_n=\begin{cases} a_n,&\text{if }n\text{ is odd}\\ y,&\text{if }n\text{ is even;} \end{cases}$$

clearly this sequence has a subsequence converging to $x$ and a subsequence converging to $y$. But $x_n\in K_n$ for each $n\in\Bbb Z^+$, so $\langle x_n:n\in\Bbb Z^+\rangle$ is Cauchy and therefore converges by the earlier argument. It follows that $y=x$ and hence that $\bigcap_{n\ge 1}K_n=\{x\}$.

But as I said, this is definitely working too hard!

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  • $\begingroup$ This response is quite thorough and I am very thankful for it. As with the other proof that you assisted me with, I am not disappointed in the slightest. Thank you Brian for furthering my knowledge of mathematics, I really appreciate it. I wish you the best. $\endgroup$ Commented Apr 4, 2021 at 23:55
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    $\begingroup$ @rodeo_flagellum: You’re very welcome, and the same to you. $\endgroup$ Commented Apr 4, 2021 at 23:56

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