Let $a \in \mathbb{R}$. I must prove:
$ \forall \epsilon \in \mathbb{R}^{>0}(0 \leq a<\epsilon) \to a =0$
Proof: If $a<\epsilon$ then $a< \epsilon +0$, by the property I have $a \leq 0$ and by hypothesis $a \geq 0$. Therefore $a=0$.
Is it correct?
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2$\begingroup$ Hint: If $a>0$ then $a\in \mathbb R^{>0}$. $\endgroup$– Thomas AndrewsDec 30, 2013 at 17:18
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$\begingroup$ Using the theorem from your earlier post, this seems right to me. $\endgroup$– RagnarDec 30, 2013 at 17:18
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$\begingroup$ @ThomasAndrews, thanks for hint! ;) :) $\endgroup$– mleDec 30, 2013 at 17:25
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$\begingroup$ Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? $\endgroup$– user1110341May 27 at 15:19
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4 Answers
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Yes it's correct.
There's an alternative answer: we have $$\forall \epsilon >0\quad 0\le a<\epsilon$$ so the set $\mathbb R_{>0}$ is bounded below by $a$ but $0$ is the infimum of $\mathbb R_{>0}$ so $a\le 0$ and since $a\ge0$ then $a=0$.
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To be precise, most answers seem to neglect showing that $a\ge 0$ in the first place. Assume $\forall \epsilon \in\mathbb R^{>0}(0\le a<\epsilon)$. Especially for $\epsilon = 1\in\mathbb R^{>0}$ this shows $0\le a<1$, hence $a=0$ or $a>0$. The latter case leads to the contradiction $0\le a<a$. Hence $a=0$.
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Using sequences : We have : $$\forall n \in \mathbb{N}^*, 0 \leq a < \dfrac{1}{n} \to 0$$ Then $a = 0$.
