Suppose that $x$ is a fixed nonnegative real number such that for all positive real numbers $\epsilon$, $0≤x≤\epsilon$. Show that $x=0$.

Question

Suppose that $x$ is a fixed nonnegative real number such that for all positive real numbers $\epsilon$, $0≤x≤\epsilon$. Show that $x=0$.

Answer 1

Hint: Use the fact that for real $x,y$ with $x < y$ there is always a real $z$ with $x < z < y$.

Answer 2

Another hint: Take a look at $\frac{x}{2}$. Now try a proof by contradiction.