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Is saying that $x\leq\varepsilon$ $\forall\varepsilon>0$ equivalent to saying $x<0$? Why? Could anyone prove it or at least guide me to prove it?

Charlie
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2 Answers2

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Claim: $x\le \epsilon$ for all $\epsilon>0$ if and only if $x\le 0$.

Proof: "$\Rightarrow$": Suppose $x\le \epsilon$ for all $\epsilon>0$. Assume $x>0$. Then set $\epsilon=x/2>0$, therefore $x\le \epsilon=x/2$. Since $x>0$, that implies $1\le 1/2$. Contradiction. So $x\le 0$.

"$\Leftarrow$": Let $x\le 0$. Suppose $\epsilon>0$. Then $x\le 0<\epsilon$. $\square$

J.R.
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  • If I assume $x\ge0$ in the $\implies$ part, it follows a contradiction and we can deduce that $x<0$. Am I right? – Charlie Feb 08 '14 at 18:53
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    @Charlie: No. You cannot conclude by contradiction: the implication $x\le x/2\Rightarrow 1\le 1/2$ is by dividing by $x$. Therefore $x$ must not be $0$. – J.R. Feb 08 '14 at 18:55
  • How could we prove the case $x<0$? – Charlie Feb 08 '14 at 18:58
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    @Charlie: Ok, I don't really know where your misunderstanding is, so let me explain this for you again. Suppose you go out into wide world and there you meet a real number $x$. All you know about that $x$ is that is not greater than any positive number. What can you conclude? Is $x$ necessarily negative ($x<0$)? No, because the number $0$ is also not greater than any positive number. So all we can say is that $x\le 0$. All negative numbers meet the criterium of not being greater than any positive number, but $0$ does also. That is the point. – J.R. Feb 08 '14 at 19:13
  • Thank you. Everything is clear now. – Charlie Feb 09 '14 at 08:32
  • @Charlie You are welcome. – J.R. Feb 09 '14 at 11:27
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What if $x = 0$?

But certainly it rules out any $x > 0$, since then $x/2$ (say) would also be greater than $0$, but $x$ would not be less than or equal to it.

Ben Millwood
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