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We have this theorem with this my proof that I will show to my students of an high school:

Theorem: If $x\in \mathbb{R}$ and $\forall\epsilon>0$, $|x|<\epsilon$ then $x=0$.

Proof. We supposed that $x\neq 0$. Hence $|x|>0$. If $\epsilon=|x|/2>0$, we have that by hypothesis: $$0<|x|<\epsilon=\frac{|x|}2\implies 0>|x|\left(1-\frac 12\right)=\frac{|x|}2\implies |x|<0,$$ versus the fact that $|x|>0$. So we can conclude that $x=0$.

Question: If I take numerically a big or small real $\epsilon>0$, then $|x|<\epsilon$ is not always true? How do we know that $x=0$ always? Is there a counterexample that I do not remember?

This is related, but does not duplicate, If $|x|<\epsilon, \forall \epsilon > 0$, then $x=0$.

Sebastiano
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  • It's not clear to me what you are asking, but perhaps you are interested in an alternative proof: if $x\neq0$, then there is a an $\varepsilon>0$ such that $|x|\ge\varepsilon$ (set $\varepsilon=|x|$, for instance). Now take the contrapositive of this result. – Joe Sep 18 '21 at 20:56
  • @Joe I not want a new proof. If you have $\epsilon=0.10$, is it always true that $-0.10<x<0.10, , \forall x\in \Bbb R$, but it is not necessarily true that it has to be $x=0$. – Sebastiano Sep 18 '21 at 20:58
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    It is true that there are infinitely many values of $x$ which satisfy $-0.1<x<0.1$. However, there is just one value of $x$ which satisfies $-\varepsilon<x<\varepsilon$ for every value of $\varepsilon$ (so it is not just that $-0.1<x<0.1$, but also $-0.000003454<x<0.000003454$, and so on). And that value is $x=0$. – Joe Sep 18 '21 at 21:03
  • Can I please know the reason for the negative votes? – Sebastiano Sep 18 '21 at 21:45

2 Answers2

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If you assume that $|x| < \epsilon$ for some particular number $\epsilon > 0$, then you cannot conclude that $x = 0$. But if you assume that $\forall \epsilon > 0(|x| < \epsilon)$, then you can conclude that $x = 0$. The assumption must say that the inequality holds for all positive $\epsilon$. Since this assumption starts with "$\forall \epsilon > 0$", you can assign any positive value you please to $\epsilon$. That's what makes it acceptable to make the choice $\epsilon = |x|/2$ in the proof.

Dan Velleman
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No, if you take a real number $\varepsilon>0$, the inequality $|x|<\varepsilon$ is not always verified. It fails if, for instance, $x=2\varepsilon$.

And, concerning your question “Why really as I have tried, $x=0$ always?”, it's because you have proved it.

José Carlos Santos
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  • Very kind prof. I not have taken in consideration when $x=2\varepsilon$. Excuse me very much and very thanks. +1. – Sebastiano Sep 18 '21 at 21:01