# Sequences of nested intervals determine at most one point

(under construction ... this turned out to be difficult ... maybe better avoided ...)

(part of the [coronvirus-in-R course](https://github.com/alexhkurz/coronavirus-in-R/blob/master/README.md)) ... ([previous](https://hackmd.io/TDuL28cASH6xfAo8psNU_g?both))

**Proposition:** For a sequence of nested intervals (in either $\mathbb Q$ or $\mathbb R$) there can be at most one element that is in all intervals.

*Proof:* Let $([x_n,y_n])_{n\in\mathbb N}$ be a sequence of nested intervals. We want to show that if $x,y$ in all intervals, then $x=y$.

We show this by contradiction. That is, we assume that there are $x<y$ such that $x_n\le x$ and $y\le y_n$ for all $n\in\mathbb N$.

Let $d = y-x$.

It follows from the definition of a sequence of nested intervals that there are $x_n,y_n$ such that $$y_n-x_n\le \epsilon$$

for all $\epsilon >0$, hence also for $\epsilon = \frac 1 2 d$.

(Why is $\frac 1 2 d>0$?)

But together with $x_n\le x$ and $y\le y_n$ and $d = y-x$ it now follows $0<d\le \frac 1 2 d$, which is impossible.

We have shown that it is not the case that all intervals contain two distinct points $x,y$.

By the logical principle of double negation, we know that "not not A" holds if and only if "A" holds.

Therefore, there can be at most one point that is contained in all intervals. QED

**Remark:** In the history of mathematics there has been a fierce debate on whether double negation should be accepted as valid inference principle. Various schools of mathematics known as intuitionism or constructivism reject double negation. But most mathematicians do accept double negation.

**Remark:** Proofs that use double negation are often called proofs by contradiction. But not all proofs by contradiction use double negation. In fact, all proofs of a negative statement proceed by contradiction. Those who object to proofs by contradiction as not being constructive only object to the proof of a postitive statement by using contradiction.