very basic calculus doubt Suppose $a,b \in \mathbb{R}^{\geq0}$. Let $\epsilon \in (0,1)$
$$ a \geq b \epsilon \implies a \geq b $$
My try: If $a < b $, then can find $n$ such that $b - a > \frac{1}{n} $. But I am stuck here. Maybe the result in not true??
if it is not true, why then is this book author assumes this in the green part??

 A: It is false. Let $\epsilon=1/2$, $a=1/4$, $b=1/2$. There are many other examples.
Added: The question has changed, to $a\ge b\epsilon$ for all $\epsilon$ in the interval $(0,1)$. Then it indeed follows that $a\ge b$. For suppose to the contrary that $a\lt b$. Then $a\lt \left(\frac{a+b}{2b}\right)b$. So if we take $\epsilon= \frac{a+b}{2b}$ then $a \ge b\epsilon$ fails. 
A: Your statement should read for all $\epsilon \in (0,1)$.
I think contradiction is easiest here.
First, if $b=0$, then it is clear there is nothing to prove.
Suppose $b>0$, and $b>a$.
Then $b= a + (b-a)$, and since $b>a$, we have $b-a >0$, and so $b-a > \frac{1}{2}(b-a) >0$.
Then $b > a + \frac{1}{2}(b-a)$, and so $b(1-\frac{1}{2}\frac{b-a}{b}) > a$.
Since $\frac{1}{2}\frac{b-a}{b} \in (0,1)$, we have $1-\frac{1}{2}\frac{b-a}{b} \in (0,1)$. Hence letting $\epsilon =1-\frac{1}{2}\frac{b-a}{b} \in (0,1)$, we have $b \epsilon > a$, a contradiction. 
A: If $a,b\geq 0$ and $a\geq b\epsilon$ for all $\epsilon\in(0,1)$, then $a\geq b$.
Proof:
As in copper.hat's answer, there is nothing to prove if $b=0$, so assume $b>0$.
Suppose for contradiction that $a<b$. Then $a/b<1$, so let $\epsilon$ be any number less than 1 and bigger than $a/b$. Then $b\epsilon > b(a/b)=a$, contradicting the assumption about $a$ and $b$.
A: Taking a less literal interpretation of your question than some of those in the comments, this kind of argument is quite common in theoretical analysis.  The correct statement is:
Lemma 1: If we have positive real numbers $a$, $b$ such that for all $0 < \epsilon < 1$ we have $a \geq \epsilon b$, then $a \geq b$.
This is actually equivalent to the following more common argument:
Lemma 2: If we have real numbers $a$, $b$ such that for all $\epsilon > 0$ we have $a \geq b - \epsilon$, then $a \geq b$.
Proof of Lemma 2: Suppose $a < b$; then taking $\epsilon = (b - a)/2$, we have $b - \epsilon = (a + b)/2 > 2a/2 = a$, in contradiction to the hypothesis.
To get Lemma 1, just take the logarithm:
$$a \geq \epsilon b \iff \log a \geq \log b + \log \epsilon \qquad
  a \geq b \iff \log a \geq \log b$$
Since $0 < \epsilon < 1$, we have $\log \epsilon < 0$, so Lemma 2 applies with $\epsilon$ being played by $-\log \epsilon$.
The reason this argument is so common is, roughly, that all quantities defined by limits (the central concept in analysis) can only be approached by approximation, i.e. to within a difference of $\epsilon$.  So rather than comparing two things directly we compare them to all degrees of approximation and let the error approach zero; this lemma allows the comparison to pass to the limit.
