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For the following problem below, did I present a reasonable argument for why closedness of the interval cannot be dropped? Thank you

Prove if $(K_n)_{n\in N}$ is a nested sequence of nonempty closed and bounded subsets of $\mathbb{R}$ then $\cap_{n\in\mathbb{N}} K_n$ is nonempty. Show that, if the requirement of closed is omitted, then $\cap_{n\in\mathbb{N}} K_n$ can be empty.

$\textbf{Solution:}$ Suppose $K_{n+1} \subseteq K_n$ with $K_n = [a_n,b_n]$; $n\in \mathbb{N}$. So, observe that $[a_{n+1}, b_{n+1}] \subseteq [a_n,b_n]$ for all $n\ge 1$. So, $a_n \le a_{n+1} \le b_{n+1} \le b_n$ for all $n\ge 1$. So, $a_n \le a_{n+1}$ for all $n\ge 1$ and $b_{n+1} \le b_n$ for all $n\ge 1$. Therefore, $\{a_n\}$ and $\{b_n\}$ are non-decreasing and non-increasing sequence respectively. Since $K_{n+1} \le K_n$ for all $n\ge 1$, then $K_1 \ge K_2 \ge K_3 \ge K_4 \ge \dots \ge K_n \ge K_{n+1} \ge \dots.$ Thus, $a_n, b_n \in K_1$, implying $a_n, b_n \in [a_1,b_1]$ for all $n\ge 1$. Hence, the sequence $\{a_n\}$ and $\{b_n\}$ are bounded. Thus, $\{a_n\}$ is a non-decreasing sequence bounded above by $b_1$, then the sequence $\{a_n\}$ must converge to its least upper bound, call it $x.$ Therefore, $\lim_{n\to\infty} a_n = x$ where x is the least upper bound of $\{a_n\}$, implying $a_n \le x$ [*].

Furthermore, $\{b_n\}$ is a non-increasing sequence bounded below by $a_1$, then $\{b_n\}$ must converge to its greatest lower bound, call it $y$. Therefore, $\lim_{n\to\infty} b_n = y$ where $y$ is greatest lower bound of $\{b_n\}$, and $y\le b_n$ [**]$.

Moreover, observe that $a_n \le b_m$ for all $m,n \in \mathbb{N}$, implying the least upper bound $\{a_n: n\in \mathbb{N}\} \le$ greatest lower bound$\{b_m : n\in \mathbb{N}\}$, implying $x\le y$ $[***]$, from $[*]$, $[**]$, and [***], we arrive at $a_n \le x\le y\le b_n$ for all $n\ge 1$ [A]. So $[x, y] \subset [a_n , b_n]$ for all $n\ge 1$ implying $\displaystyle{[x,y] \subset \cap_{n=1}^\infty K_n} [****]$. Thus, $$\cap_{n=1}^\infty K_n \ne \emptyset.$$ Equivalently, we can say $\cap_{n=1}^\infty K_n$ is non-empty.

Now, let us consider the case in which closedness of interval is dropped. Let $K_n = (0, \frac{1}{n}); n\in \mathbb{N}$. Then, $K_{n+1} \subset K_n$, implies $(K_n)_{n\in\mathbb{N}}$ is nested sequence. Moreover, $K_n = (0, \frac{1}{n})$ is bounded for each $n$. However, $\displaystyle{\cap_{n=1}^\infty (0, \frac{1}{n}) = \emptyset}$, implying $\displaystyle{\cap_{n=1}^\infty K_n = \emptyset.}$ Hence, $\displaystyle{\cap_{n=1}^\infty K_n}$ is empty with $K_n = (0, \frac{1}{n}).$ Hence closedness of the interval cannot be dropped.

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2 Answers 2

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Your solution is wrong from the start, since it assumes that the only closed and bounded subsets of $\mathbb R$ are the intervals $[a,b]$, with $a\leqslant b$.

You can prove the statement that you want to prove in several ways. One of them is to define $a_n=\min K_n$ and then to prove that the sequence $(a_n)_{n\in\mathbb N}$ must converge and that its limit must belong to every $K_n$.

Your example is fine.

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You might want to justify why $\bigcap_n (0, \frac1n) = \emptyset$ some more:

If $x$ is in the intersection, this means that $x>0$ but then there is some $m \in \Bbb N$ such that $x < \frac{1}{m}$ (by the Archimedian property of $\Bbb R$, e.g.) and then for this $m$, $x \notin (0,\frac1m)$ and so $x \notin \bigcap_n (0, \frac1n)$, contradiction.

The example itself is fine.

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