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Question: Let $S$ be a finite nonempty set of rational numbers, with minimal and maximal elements $s_1$ and $s_2$, respectively. Consider $\alpha\in\mathbb{R}$ such that $s_1\leq\alpha\leq s_2$. Prove that of all the members of $S$ there is one closest to $\alpha$.

Note: This question is due to the bolded portion of a proof from Spivak's Calculus text (see below).

I am reading through Spivak's Calculus and the question above occurred to me while reading his proof that $$ \lim_{x\to a}f(x)=0, $$ where it is assumed that $0<x<1$ and $0<a<1$ and $f$ is defined by $$ f(x)= \begin{cases} 0 & \text{if $x$ is irrational},\\[0.5em] \frac{1}{q} & \text{if $x=\frac{p}{q}$ in lowest terms}. \end{cases} $$ His proof is essentially as follows:

(Spivak proof): Consider any $\epsilon>0$. Let $n$ be a natural number so large so that $1/n\leq\epsilon$. The only numbers $x$ for which $|f(x)-0|<\epsilon$ could be false are: $$ \frac{1}{2};\frac{1}{3},\frac{2}{3}; \frac{1}{4},\frac{3}{4}; \frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\ldots;\frac{1}{n},\ldots,\frac{n-1}{n}. $$ If $a$ is rational, then $a$ might be one of these numbers. However many of these numbers there may be, there are, at any rate, only finitely many. Therefore, of all these numbers, one is closest to $a$; that is, $|p/q-a|$ is smallest for one $p/q$ among these numbers. If $a$ happens to be one of these numbers, then consider only the values $|p/q-a|$ for $p/q\neq a$. This closest distance may be chosen as the $\delta$. For if $0<|x-a|<\delta$, then $x$ is not one of $$ \frac{1}{2},\ldots,\frac{n-1}{n} $$ and therefore $|f(x)-0|<\epsilon$ is true, thus concluding the proof.

Perhaps I am just being paranoid, but doesn't the portion of his proof in bold need some justification? I will assume that a finite set has a minimum and a maximum (I read this thread), but Spivak seems to make use of the claim in my question at the top of this post. How does one prove that such a "closest element to $a$" exists? Is there some very trivial proof by contradiction? Is there some result that relies on sophisticated knowledge of bounds or topology that would leave Spivak to exclude that from a mostly basic chapter on limits (before he introduces bounds and all that)? Any sort of rigorous justification (or referral to other literature) for his proof and/or the question at the top would be most appreciated.

interrogative
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2 Answers2

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If I'm understanding you correctly, the answer to your question is in the sentence that directly follows the bolded part of your quote:

...that is, |p/q−a| is smallest for one p/q among these numbers.

The minimum element of the set $\{|\alpha - s|: s \in S\}$ is $|\alpha - s^*|$ for some $s^* \in S$. Then $s^*$ is the closest element of $S$ to $\alpha$.

user511788
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  • Oh interesting. For whatever reason I completely blanked on applying my reading of that linked thread...since $S$ is finite, then the set of distances $M={d(\alpha,s) \mid \alpha\in\mathbb R, s\in S}$ will be finite. Assuming the result proved in that linked thread, I know there will be a minimal element of $M$. The $s\in S$ that corresponds to this minimal element of $M$ is the closest rational number to $\alpha$. That's what you were emphasizing yes? – interrogative Dec 09 '17 at 07:07
  • Yes, exactly :) – user511788 Dec 09 '17 at 08:10
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What he really needs is that there is an interval $(\alpha-\epsilon,\alpha + \epsilon)$ that includes none of them aside from possibly $\alpha$ itself. As there are only finitely many, you can collect all the differences, find the minimum, divide it by $2$ and use that for $\epsilon$. One could argue the statement is wrong because there could be two at equal distance from $\alpha$ but you can just use either one of them and the proof goes through.

Ross Millikan
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