# Proof of archimedean property

I am trying to self-study Baby Rudin (and it's proving quite challenging to me)

Could someone clarify where the underlined part comes from?

Text:

(a) If $$x \in R, y \in R,$$ and $$x > 0$$, then there is a positive integer $$n$$ such that $$nx > y$$.

Proof (a) Let $$A$$ be the set of all $$nx$$, where $$n$$ runs through the positive integers. If (a) were false, then $$y$$ would be an upper bound of $$A$$. But then $$A$$ has a least upper bound in $$\mathbb{R}$$. Put $$\alpha = \sup A$$. Since $$x > 0$$, $$\alpha - x < \alpha$$, and $$\alpha - x$$ is not an upper bound of $$A$$. $$\underline{\text{Hence \alpha - x < mx for some positive integer m}}$$. But then $$\alpha < (m+1)x \in A$$, which is impossible, since $$\alpha$$ is an upper bound of $$A$$.

Since $\alpha -x$ is not an upper bound of $A$, there must be an element in $A$, call it $mx$, bigger than $\alpha-x$, namely $\alpha -x < mx$. It is the logical negation of the property of being an upper bound for a subset of $\mathbb{R}$.

• Just after negating $\alpha - x \ge nx$, I realized how stupid I am. Thanks May 22, 2014 at 15:06
• i don't think that's why. The reason why there is mx is because the distance between $\alpha - x$ and $\alpha$ is exactly $x$, so no matter where $\alpha$ is, the interval is large enough to have one multiple of $x$ land in it, kinda like a pidgeonhole principle. You could probably prove it as a separate lemma, $\forall x > 0,\alpha > x \exists n, \alpha - x \leq nx \leq \alpha$. Oct 9, 2018 at 23:48

We know $\alpha$ - $x$ is not an upper bound of $A$. i.e. there exists some element of A greater than $\alpha - x$. So let this element, greater than $\alpha - x$ be written as $mx$ for some m, an element of the postive integers. So then $\alpha -x \lt mx$.

The following seems far simpler to me, but there must be something wrong if Rudin doesn't include it. Maybe the $$\lceil$$ceiling$$\rceil$$ function is not defined yet? Thoughts?

Statement:

If $$x,y \in \mathbb{R}$$ and $$x >0$$, then there is a positive integer $$n$$ such that $$nx>y$$.

Proof:

Given $$x,y>0$$, let $$n=\lceil{\frac{y+1}{x}}\rceil$$. Then,

\begin{align*} nx&= \lceil {\frac{y+1}{x}} \rceil \cdot x \\ &\ge y+1\\ &>y. \end{align*} $$\tag*{\square}$$

• Indeed, the definition of the ceiling function is $\lceil x \rceil =$ the least integer $n$ such that $n \ge x$, so if you haven't yet proved the existence of one such integer, then that definition is not applicable. Jul 10, 2021 at 2:47