Equality and inequality with infinitesimal number

Question

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 $a=b$.

Answer 1

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.

Answer 2

The following two statements are equivalent:

1. "$0 \leq a-b < \epsilon$ for every $\epsilon > 0$" implies "$a = b$"
2. "$0 \leq a-b \leq \epsilon$ 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.