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Consider a subset of rational numbers $S = \{\frac{a}{b}; a \in \mathbb{N}, b \in \mathbb{N}, a < b\}$. I want to prove that $\sup S = 1$. By the definition of supremum, for $\epsilon > 0$, it suffices to show that there exists $\frac{a}{b} \in S$ such that $1 - \epsilon < \frac{a}{b}$.

I tried to prove it using archmedian property ($\forall x \in \mathbb{R}, \exists n \in \mathbb{N}$ such that $n \geq x$) by setting $a = 1$ or $b$ being a multiple of $a$ for the purpose of deriving the value of remaining variable by fixing one variable. However, none of them worked. I feel like I ran out of trick. How should one prove it?

Tianlalu
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Ted
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4 Answers4

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First of all it is easy to prove that $1$ is an upper bound. Now to show it is the supremum take any $\epsilon>0$. I assume you know there is a number $n\in\mathbb{N}$ such that $\frac{1}{n}<\epsilon$. So take such $n$ and assume it is at least $2$, otherwise just take a bigger $n$. And now note that $\frac{n-1}{n}\in S$, and also:

$\frac{n-1}{n}=1-\frac{1}{n}>1-\epsilon$

Mark
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Let a=b-1. Then $a/b =1-1/b$. Choose $b > 1/\epsilon$.

marty cohen
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Hint $$\sup_{n \in \mathbb N}({ \frac{n}{n+1}}) \leq \sup(\frac{a}{b} \cdots) \leq 1$$

Bargabbiati
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Let $s=sup(S)$

Suppose $s<1$

Between two real numbers $x<y$ there's always $q\in\mathbb{Q}$ (i.e. $q \in (x, y)$)

Let $q=\frac{q_1}{q_2}$ be a rational number in $(max(s, 0), 1)$

By definition $s<q$

then choose $a=q_1$ and $b=q_2$ $\Rightarrow s<q=\frac{a}{b}$

But $q\in S$.

So $s \geq 1$

But $S$ is bounded by $1$, then $s = 1$

texdditor
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