Equality of Real Numbers

Question

Is the following statement provable from the axioms of $\mathbb{R}$?

If $\forall \epsilon>0$, $|r-s|\leq \epsilon$, then $r=s$.

Answer 1

yes, it should be. I would try it in the following way:

Assume that $r \neq s$ and define $\epsilon := \frac{|r-s|}{2}$ . Then we have $\epsilon >0$ and $|r-s| > \frac{|r-s|}{2}$ , by contraposition we have a proof. I think that it is well-known that the stuff that i used can be derived from any kind of axioms for the real numbers.

Answer 2

Yes. Suppose $r \neq s$ for contradiction. Then $r - s \neq 0$ and take $\epsilon = |r - s|/2 > 0$.