Equality and inequality with infinitesimal number Is the following statement true?
Suppose $a$ and $b$ are real numbers.
If $0\leq a-b < \epsilon$ for every $\epsilon>0$, then $a=b$.
What if I replace '$<\epsilon$' by '$\leq \epsilon$' and make the following statement?
Suppose $a$ and $b$ are real numbers.
If $0\leq a-b \leq \epsilon$ for every $\epsilon>0$, then $a=b$.
 A: The following two statements are equivalent:


*

*"$0 \leq a-b < \epsilon$ for every $\epsilon > 0$" implies "$a = b$"

*"$0 \leq a-b \leq \epsilon$ for every $\epsilon > 0$" implies "$a = b$"


Indeed, suppose the first.
Then to prove the second, suppose $0 \leq a-b \leq \delta$ for every $\delta$. Fix arbitrary $\epsilon > 0$.
Let $\delta = \frac{\epsilon}{2}$; so $0 \leq a-b \leq \frac{\epsilon}{2} < \epsilon$.
So by the first property, $a=b$.
The converse is trivial.
A: 
Proposition
Suppose that $a$ and $b$ are real numbers that for every $\varepsilon>0$ satisfy $0\leq a-b<\varepsilon.$ Then $a = b$.

Proof. Assume for contradiction that $a\neq b$. By the original inequality we have that 
$$b\leq a < b+\varepsilon. $$
Since $a\neq b$ and $b\leq a$ we have that $b<a$. This means there exists a strictly positive number $\delta>0$ such that $b+\delta = a$, and hence $a = b+\delta < b + \varepsilon$, which implies that $0<\delta < \varepsilon.$ The inequality holds for any $\varepsilon>0$ so choose $\varepsilon = \delta/2>0$ from which the inequality is $0<\delta <\delta/2$ which is a contradiction.
We conclude that $a = b$. $_\square$
The proof can easily be adjusted to the second claim.
