A Short Words Proof of N.I.P.

The Challenge: Write a monosyllabic version of a proof we have studied this semester.

Accepting the Challenge: My strategy was to choose a proof that was visually easy to understand and hope that I can convey the basic idea of the proof to the reader as an image. The proof I chose was the Nested Interval Property. Below is a more mathematically formal version of the proof. Even further below, you can find my simplified version.

Nested Interval Property:
Definition: For all $n\in \mathbb{N}$ , let $I_n = [a_n,b_n]=\{x\in\mathbb{R}:a_n\le x \le b_n\}$ and let $I_{n+1}\subseteq I_n$ . Then $\cap^\infty_{n=1}I_n\not=\emptyset$ .

Proof: For all $n\in \mathbb{N}$ , let $I_n = [a_n,b_n]=\{x\in\mathbb{R}:a_n\le x \le b_n\}$ and let $I_{n+1}\subseteq I_n$ . Let $A =\{a_n:n\in\mathbb{N}\}$ . Consider $b_1$ . Since $\forall$ $n\in\mathbb{N}$ , $I_n\subseteq I_1$ , it follows that $a_1\le a_n \le b_n \le b_1$ . Thus, for all $n\in \mathbb{N}$ , $a_n\le b_1$ , so $b_1$ is an upper bound for $A$ . Since A is bounded above, the axiom of completeness guarantees the existence of $x=\sup(A)$ . For all n, $b_n$ is an upper bound for A by nestedness. But $x\le b_n$ , since $x =\sup (A)$ . Therefore, $a_n \le x\le b_n$ , which implies $x\in I_n$ . Finally, $x\in \cap^\infty_{n=1} I_n$ . $\Box$

Here we go…
What It Is: Let $I_n$ be a closed span of reals such that $a_n$ is the left most part, $b_n$ is the right most part, and $a_n$ is less than $b_n$ . All the sets, $I_n$ , can be made out by their n, where n is a whole count: 1,2,3,…. As n gets big, if each $I_n$ lies in the past set, then the core of all the sets is not void.\\

Proof: First, let $I_n$ be a closed span of reals, where $a_n$ is the left most part, b is the right most part, and $a_n$ is less than $b_n$ . And, as n gets big, each $I_n$ lies in the past set. Let A be the set of left most points, $a_n$ . Let $b_1$ be the right most part of the first set, $I_1$ . Since, as n gets big, $I_n$ lies in the past set, we can say that $b_1$ is more than all $b_n$ , which we know is more than $a_n$ . Thus, $a_n$ is less than $b_1$ .

This means that $b_1$ is more than all parts in the set A and is a bound for A. Now, we use a math truth that says a set with a bound that is more than all the parts of the set will have a bound that is less than or the same as all the bounds of A. So, the set of bounds for A will have a least part. We will call this the “small bound”.

From this truth we can say that there is a small bound of A. Let x be the small bound. Then x is less than or the same as all $b_n$ . But, we know $b_n$ is more than all $a_n$ , so it too is a bound for A. So, x is less than or the same as all $b_n$ and more than or the same as all $a_n$ . Thus, $x$ must be in the set $I_n$ . Then x is in the core all the sets $I_n$ and so $I_n$ can’t be void. $\Box$

Art 1: The core of the sets $I_n$ with x in the core. The sets $I_1$ , $I_2$ , $I_3$ , $I_4$ are shown so as to give some sense of what the core of all the sets is.

Art 2: Here, the sets $I_1$ , $I_2$ , $I_3$ , $I_4$ are shown so as to give some sense of what the left most parts and right most parts are in terms of the sets.

Post Challenge: Einstein phrased it perfectly when he said, “If you can’t explain it to a six year old, you don’t understand it yourself.” I feel that this quote embodies the idea behind this challenge. By using monosyllabic words, the proof condenses down into a simple idea. Hopefully, the proof is just a bit clearer now.

References:

Stephen Abbot, Understanding Analysis 2nd Edition

Tag, By. “A Quote by Albert Einstein.” Goodreads. N.p., n.d. Web. 28 Oct. 2016.

Math 320: Real Analysis

A Short Words Proof of N.I.P.

Related

Leave a Reply Cancel reply

Sidebar

Share this:

Related

Leave a Reply Cancel reply

Sidebar