Suppose $a$ is a real number and we know that $$0 \le a \le \varepsilon$$ for every $\varepsilon \gt 0$. I need to show that $a=0$. The book I am working out of already has shown by contradiction that $0 \le a \lt \varepsilon$ for every $\varepsilon \gt 0$ implies that $a=0$. I can honestly say I am confused, since if $a=\varepsilon$ and $\varepsilon \gt 0$, then $a\neq{0}$

  • $\begingroup$ It has to hold for all $\epsilon$, not just for $\epsilon = a$. $\endgroup$ Feb 17, 2014 at 22:13
  • $\begingroup$ But isn't $a=\varepsilon$ a contradiction to what I'm trying to prove since then $a\neq{0}$? $\endgroup$
    – Faffi-doo
    Feb 17, 2014 at 22:15
  • $\begingroup$ Hint: Suppose $a > 0$. If $0 \leq a \leq \varepsilon$ holds for every $\varepsilon > 0$ then take $\varepsilon = \frac a2 > 0$. $\endgroup$
    – Sávio
    Feb 17, 2014 at 22:16
  • $\begingroup$ Why do you believe $a=\varepsilon$? Even if $a\leq \varepsilon$ everywhere, there can only be one $\varepsilon$ for which the equality occurs; all the rest must be $<$. (And in fact, you can probably see intuitively that the equality must occur at the lowest possible value.) $\endgroup$
    – tabstop
    Feb 17, 2014 at 22:17
  • $\begingroup$ It's all about the order that you pick your $a$ and $\epsilon$. First, you pick $a$. THEN for every single $\epsilon > 0$, your condition holds. You can pick $\epsilon = a$, but you could also pick it to be even smaller (assuming $a > 0$). $\endgroup$ Feb 17, 2014 at 22:17

1 Answer 1


If $\forall\varepsilon,0\le a \le \varepsilon$ then either $a>0$ or $a=0$. If $a>0$, then setting $\varepsilon<a$ we have $0\le a\le\varepsilon<a$ a contradiction. Then $a=0$.

  • $\begingroup$ This is so confusing...aren't we told that $a\le \varepsilon$? How can then you say $\varepsilon \lt a$? $\endgroup$
    – Faffi-doo
    Feb 17, 2014 at 22:21
  • $\begingroup$ Because it holds for all $\varepsilon>0$. If you say that $a>0$, then I can choose a $0<\varepsilon<a$ and reached a contradiction is like a game. $\endgroup$ Feb 17, 2014 at 22:23
  • $\begingroup$ So basically, as long as you can always find a number between $0$ and $a$, you have a contradiction to $a \gt 0$? $\endgroup$
    – Faffi-doo
    Feb 17, 2014 at 22:25
  • 1
    $\begingroup$ $a\in \mathbb{R}$, then either $a<0,a=0,a>0$ by the trichotomy. Clearly $a<0$ is not possible. Then either $a=0$ or $a>0$, but the latter yields a contradiction, because we can always pick some $0<\varepsilon<a$ because by hypothesis it holds for all numbers $>0$. Then what is the only alternative. $\endgroup$ Feb 17, 2014 at 22:28
  • 1
    $\begingroup$ It's similar in nature to the "game" a number bigger than all the numbers. If you say that is $N$ then I say $N+1$. $\endgroup$ Feb 17, 2014 at 22:34

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