# $a\le x\le {a + \frac yn}$ for all $n \ge 1$ then $x=a$

$a\le x\le {a + \frac yn}$ for all $n \ge 1$ then $x=a$
So I start by immediately pointing out that $x-a \gt 0$ by the definition of inequality.
And we know that ${a+\frac yn}\ge x$ so:
$\frac yn \ge x-a$ then
$y \ge n(x-a)$ and
$\frac y {x-a} \ge n$

Here is where I get stuck. I'm not sure that I'm even on the right path. The only thing that I can think of is that this inequality is satisfied by $\lim_{x\to a}\frac y {x-a}\ge n$ so somehow from there x=a? I would really appreciate any hints if offered.

Also, limits haven't been defined in the book yet, so I can't use them in my proof.

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If $y$ is fixed, isn't $\lim_{n\to\infty}a+\frac{y}{n}=a$? – Clayton Mar 13 '13 at 4:23
Yes, but I can't use limits, so I need another way to prove it. – AlexHeuman Mar 13 '13 at 4:24
Note that subtracting $a$ from the inequalities, we just need to show the theorem for $w=0$. That is, if for some $y$ and all $n\in \Bbb N$ $$0\leq w\leq \frac y n$$ then $w=0$. You know that $\Bbb R$ is Archimedian, so given any positive $x$, and $y$, we can find $n$ such that $nx>|y|$. If $w$ were positive, we would find some $n$ such that $nw>y$, that is $w>y/n$, contradicting what we said in $(1)$. So $w$ is nonnegative, that is $w\leq 0$, so $w=0$. – Pedro Tamaroff Mar 13 '13 at 4:40

We know that $a \leq x$, thus if you can prove that $x \leq a$ you are done.

Assume by contradiction that $x> a$. Then,

$$x \leq a+\frac{y}{n} \Leftrightarrow \frac{x-a}{y} \leq \frac{1}{n} \Leftrightarrow n \leq \frac{y}{x-a}$$

This implies that the positive integers are bounded, contradiction....

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I think you mean $x\leq a$, not $x\geq a$, yes? – Cameron Buie Mar 13 '13 at 4:40
@CameronBuie ty, fixed – N. S. Mar 13 '13 at 4:59
No problem. +1. – Cameron Buie Mar 13 '13 at 5:00

Hint: Let $y\in[0,\infty)$ be given and assume for some $a\in\Bbb R$ we have $a\leq x\leq a+\frac{y}{n}$ for all $n\geq1$. If $x\neq a$, then $x>a$ from hypothesis, and in particular, $x-a>0$. Now, define $n$ appropriately so that $\frac{y}{n}<x-a$ and arrive at a contradiction.

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Assume that $a\le x\le {a + \frac yn}$ for all $n \ge 1$. Then $a\le x$. We claim that $a\ge x$. Suppose that $a<x$. Then $x-a>0$. Then there exists $n\in \mathbb{N}$ such that $$\frac{y}{n}<x-a.$$ This implies that $$x>a+\frac{y}{n}$$ for some $n\ge 1.$ We get a contradiction because $x\le {a + \frac yn}$ for all $n \ge 1$. The result follows.

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