Suppose $x$ is a non-negative real number such that for all $\epsilon >0$ we have $x < \epsilon$. Then $x=0$

Question

Is the following statement true?

Suppose $x$ is a nonnegative real number such that for all $\epsilon >0$ we have $x < \epsilon$. Then $x=0$.

Intuitively speaking yes since $0$ is an infimum of the set $\{\epsilon \space|\epsilon >0\}$. I was trying to come up with an axiomatic reason for this but all I could find was algebraic properties of $0$ on $\mathbb{R}$ which didn't seem too relevant. But can this result be derived from the order axioms of $\mathbb{R}$ if true?

Answer 1

If $x>0$, take $\varepsilon=x$. Then $x<\varepsilon=x$, which is impossible.