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?
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Hint: Suppose that $x > 0$, and consider $$\epsilon := \frac{x}{2}$$
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$x\leq\frac{x}{2}$??? – Trancot Aug 25 '13 at 05:34
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1@Trancot Yes. Rearranging, conclude that $2x \le x$; is this possible? – Aug 25 '13 at 05:34
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What is the setup though? I think you are hinting at a contradictory setup, but what's the setup? – Trancot Aug 25 '13 at 05:36
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"Suppose $x > 0$, intending a contradiction. Do stuff, and conclude $2x \le x$, leading to..." – Aug 25 '13 at 05:37
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No I meant to show $P$ is true show that $\neg P\implies \perp$, but I've got it now. – Trancot Aug 25 '13 at 05:41
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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.
user77404
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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 $x=0$.
Trancot
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You are kinda using the contrapositive, really: if $x>0$ we can find one $\epsilon >0$; namely $\epsilon =x/2 >0$ (or really $x/\alpha$ for any $\alpha >1$ you like) such that $$0<\frac x 2 <x $$ Thus you have proven that $x\leq \epsilon$ for each $\epsilon >0$ implies $x\leq 0$. The additional hypothesis that $x\geq 0$ implies that $x=0$. Observe how $$\neg (\forall \epsilon >0;; x\leq \epsilon)=\exists \epsilon >0(\epsilon <x)$$ – Pedro Aug 25 '13 at 05:49
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@Trancot Yes, looks good. Note that since $x>0$, we're allowed to let $\epsilon=x/2$, since $x/2$ is positive. Does it make sense why we don't need to consider the case when $x<0$? – Adriano Aug 25 '13 at 05:51