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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?

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  • $\begingroup$ It's good. I'd have gone into more specific and explicitely state $0 < x-\epsilon < x + \epsilon$ so $0 < \frac {1}{x + \epsilon} < \frac 1{x - \epsilon}$. I'd explain clearer that any bounded set of integers is finite. Was that a theorem you have proven? If so state it, if not give a brief reason why we can assume such. $\endgroup$ – fleablood Mar 9 '18 at 16:47
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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 $S$.

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  • $\begingroup$ Can you further elaborate on the Archimedean property ? $\endgroup$ – Ling Min Hao Mar 9 '18 at 15:25
  • $\begingroup$ The archimedean properties have many forms. The basic statement is for any $b > 0$ and $M \in \mathbb R$ then exists and integer $n$ so that $nb > M$. If $b = 1$, than means for any $M$ there is an integer larger than $M$. A bit more futzing and we can prove there exists a unique integer $n$ so that $N \le M < N+1$ and thus the floor and ceiling functions are well-defined. When YAlexandrov says "Therefore only n=1,2,3,...,N" I'd be just a teensy bit clearer and say "At most, only n=1,2,3...N" can be in S. $\endgroup$ – fleablood Mar 9 '18 at 16:57
  • $\begingroup$ @fleablood, oh so he used archimedian principle to show that bounded set of integers is finite and floor and ceiling function exists right? $\endgroup$ – Ling Min Hao Mar 10 '18 at 3:46
  • $\begingroup$ Yes, it's your proof. YAlexandrov only fleshed it out. $\endgroup$ – fleablood Mar 10 '18 at 5:43
  • $\begingroup$ @fleablood,after looking at Archimedian Property, I figured another way to solve this question. Maybe you can try to help me to verify on this. By archimedian property, there is $M$ such $n$ such that $n(x+\epsilon)>1$ and thus $\frac{1}{n}<x+\epsilon$. Also, by archimedian property, there is $N$ such $n$ such that $n(\frac{1}{n})^2>x-\epsilon$ and thus $\frac{1}{n}>x-\epsilon$ From here we can generalised there are M integers such that $\frac{1}{n}<x+\epsilon$ and $N$ integers such that $\frac{1}{n}>x-\epsilon$. By taking their intersections, we have a finite set. $\endgroup$ – Ling Min Hao Mar 11 '18 at 9:24
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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 $S$?

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  • $\begingroup$ This question is introduced before limit was defined. $\endgroup$ – Ling Min Hao Mar 9 '18 at 14:57
  • $\begingroup$ Very sorry, this I did not know.You were asking for another proof. $\endgroup$ – Peter Szilas Mar 9 '18 at 14:59

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