Prove that $S$ is a finite set.

Question

If $x\in\mathbb{R}$ and $0<\epsilon<x$, prove that the set $S=\{n\in\mathbb{N}:x-\epsilon<\frac{1}{n}<x+\epsilon\}$ is a finite set, that is, there are finitely many elements in $S$.

Here is my proof: To show this is a finite set, we need to show this set is bounded between two integers. \begin{eqnarray*} S&=&\{n\in\mathbb{N}:x-\epsilon<\frac{1}{n}<x+\epsilon\}\\ &=&\{n\in\mathbb{N}:\frac{1}{x+\epsilon}<n<\frac{1}{x-\epsilon}\} \end{eqnarray*} We can rewrite this because $x>\epsilon\Rightarrow x-\epsilon>0$. Then this set is bounded by $\left\lceil\frac{1}{x+\epsilon}\right\rceil\leq\frac{1}{x+\epsilon}<n<\frac{1}{x-\epsilon}\leq\left\lfloor\frac{1}{x-\epsilon}\right\rfloor$ where $\left\lceil\frac{1}{x+\epsilon}\right\rceil,\left\lfloor\frac{1}{x-\epsilon}\right\rfloor$ are nonnegative integers.

Is my proof correct?

Answer 1

It is correct.

I would have gone for making more explicit the root of the reason why it is finite and clean the proof from questions like: Why a bounded set of integers is finite? Why does the floor and ceiling exist?

From $0<\epsilon <x$ you have that $x-\epsilon >0$ and $1/(x-\epsilon)>0$. Therefore, by the Archimedean property there is $N$ such that for all $n>N$ it holds that $n>1/(x-\epsilon)$. Therefore $1/n<x-\epsilon$.

Therefore only $n=1,2,3,...,N$ could be inside $S$.

Answer 2

Consider the.sequence $y_n=1/n$, $n\in \mathbb{Z^+}$.

$y_n \gt 0$, and $\lim_{n \rightarrow \infty}y_n=0$:

For $\varepsilon \gt 0$, there is a $n_0$ such that for $n\gt n_0$:

$|1/n| \lt \varepsilon$.

This means:

We may have $y_n \ge \varepsilon$ for at most $n_0$

terms: $ y_n$, with $n=1,2,....,n_0 $.

Choose $\varepsilon = (x-\epsilon)/2$.

What does this.mean for $S$?