$x=0$: A Condition Proof I'm asked to prove that if $x\geq 0$ and $x\leq \epsilon$ for all $\epsilon >0$, then $x=0$, but I'm not sure where to go. I have that the logically equivalent statement is
$$x\neq 0\implies\exists\epsilon\leq0~\text{s.t.}~x<0\lor x>\epsilon,$$
but what does this tell me?
 A: Hint: Suppose that $x > 0$, and consider $$\epsilon := \frac{x}{2}$$
A: There are a few techniques of proof. One of them is called proof by contradiction. Where you assume the negation, and show it is impossible (arrive at a contradiction). So in this case suppose $x \ne 0$, by assumption this means $x > 0$ now from the hint your given by T.Bongers, you should be able to arrive at a contradiction.
Keep in mind your two assumptions are $x \ge 0$ and  $x \le \epsilon. $ If $\epsilon = \frac{x}{2}$ then by assumption $x \le \frac{x}{2}$ but there is only one $x$ that satisfies this inequality, if you think about what that x must be, you will see the contradiction. 
A: This is what I've got from what you've all said:
Assume that $x>0$, and let $\epsilon=\frac{x}{2}$, but by assumption $x\leq \epsilon$, namely $x\leq \frac{x}{2}$, but since $x>0$, then division by $x$ on both sides produces a contradiciton, namely that $1\leq \frac{1}{2}$, and so it must be true that if $x\geq 0$ and $x\leq \epsilon$ for all $\epsilon>0$, then $x=0$.
