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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?

enter image description here

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 $A$.

Thanks in advance

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Ahmed Ali
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3 Answers3

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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 $\mathbb{R}$.

Siminore
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    Just after negating $\alpha - x \ge nx$, I realized how stupid I am. Thanks – Ahmed Ali May 22 '14 at 15:06
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    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, $\forall x > 0,\alpha > x \exists n, \alpha - x \leq nx \leq \alpha$. – Jakub Arnold Oct 09 '18 at 23:48
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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 $\alpha -x \lt mx$.

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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*} $$\tag*{$\square$}$$

tupben
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    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. – Lee Mosher Jul 10 '21 at 02:47