Proof : $\forall \epsilon \in \mathbb{R}^{>0}(0 \leq a<\epsilon) \to a =0$

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?

• Hint: If $a>0$ then $a\in \mathbb R^{>0}$. Dec 30, 2013 at 17:18
• Using the theorem from your earlier post, this seems right to me. Dec 30, 2013 at 17:18
• @ThomasAndrews, thanks for hint! ;) :)
– mle
Dec 30, 2013 at 17:25
• Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon&gt;0$ then $a\leq b$? May 27 at 15:19

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 $a=0$.

• thanks for alternative answer! :) ;)
– mle
Dec 30, 2013 at 17:27
• Thanks for saying those kind words. :) Dec 30, 2013 at 17:54

If $a>0$, then we have $0\le a<a$, a contradiction.

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 $a=0$.

Using sequences : We have : $$\forall n \in \mathbb{N}^*, 0 \leq a < \dfrac{1}{n} \to 0$$ Then $$a = 0$$.