For any $x>0$, does $|a-b| < x$ imply that $|a-b|=0$?

Question

Let $a\in \mathbb{R}$. For any $x > 0$: $$ |a-b| < x $$ implies that: $$ a=b $$

Is this statement true? Why?

My attempt:

I think it is true. My reasoning is that, if $x$ is allowed to be anything larger than $0$, but not $0$, it will contain an infinitesimal that is closest possible to $0$, but not exactly $0$. I know such thing doesn't exist in the set of real numbers, but this is my intuition.

Then, if we say that there are numbers $a$ and $b$, such that their absolute difference $|a-b|$ is even smaller than $x$, then what possibilities do we have? We already said that $x$ could be the smallest positive number that is not $0$, so if $|a-b| < x$ is true, then $|a-b|$ is even smaller than the smallest positive number $x$.

If a number, e.g. $|a-b|$ (non-negative) is smaller than the smallest positive number, then it must be that $|a-b| = 0$. I cannot find any other possibility.

Answer 1

If $|a-b|> 0$, then $$|a-b|<|a-b|$$

Do you see a contradiction?

Answer 2

Just a note here: If it were $|a-b|\leq x$ instead of $<$, the conclusion still holds, simply choose (assume by the contrary) $x=\dfrac{1}{2}|a-b|$.

Answer 3

If possible let $a \neq b$ $$\Rightarrow |a-b|>0$$ $\Rightarrow$ for any $x \in (0, |a-b|) $ clearly $|a-b|>x>0$ and this is a contradiction to the given condition. $$ $$ $\therefore$ our assumption is wrong and $a=b$