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Let $x$ be a real number. If $|x|< \varepsilon$ for any $\varepsilon>0$, then $x=0$. That was a lemma in my lecture today. My Dr. proceeded as follows:

  • Suppose that $x≠0$
  • Choose $\varepsilon =|x|/2>0$
  • We have $|x|<|x|/2$ (according to our assumption)
  • Which is a contradiction
  • Thus $x=0$

But I have a little question here:

  • If I took $\varepsilon=1>0$

  • I have $|x|<1$

  • Then $-1<x<1$

  • Thus we can't assure the value $x=0$, it has more values, since $x$ is a real number in the interval $]-1,1[$

    Is my Dr.'s solution accurate or am I missing something?

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Abed Shaar
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    Any means every in that context. – Clement C. Sep 28 '17 at 17:10
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    If you believe that there can be other $x$ values, you can say it. For example, does $x= \frac 13$ work? No, because it would fail for $\varepsilon = \frac 14$. In fact, any $x \neq 0$ fails. – ThePortakal Sep 28 '17 at 17:12
  • @ThePortakal Do you mean that if I chose a value for x≠0 I am still able to find a value for ε such that |x|>ε because ε can be any value? – Abed Shaar Sep 28 '17 at 17:26
  • I think I'm getting it right here. I have a problem with Quantifiers. Is there a way to facilitate their concepts? – Abed Shaar Sep 28 '17 at 17:27
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    Consider using LaTeX symbols instead of cut-and-pasting. The command \varepsilon between dollar signs will give you $\varepsilon$, e.g. 0 < \varepsilon \ll 1 will give $0 < \varepsilon \ll 1$. If you prefer $\epsilon$, then just use \epsilon instead. – Fly by Night Sep 28 '17 at 17:28
  • Thank you @FlybyNight – Abed Shaar Sep 28 '17 at 17:32

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