Proof for: If $x>-a$ for $a>0$, then $x\geq0$

Question

I came across a claim that

If $$ x>-a \text{ }\forall \text{ } a>0 $$ then $$ x\geq0 $$ This claim is mentioned here (the answer).

When graphed, $x$ does occupy places between $-a$ and $0$ and therefore I don't understand how $x\geq0$ except when $a$ is close enough to zero such that the only 'next' number on the number line is zero so that $x\geq0$. How could I formally prove this is true (if it is true) or otherwise?

$$x>-a\qquad\forall a>0$$ means that $x$ is strictly greater than any strictly negative number (alternatively, $x\ge\sup{-a:a\in\mathbb R, a>0}=0$.) — Maximilian Janisch, Jul 29 '21 at 14:58
Does this answer your question? Proof of $\forall\epsilon>0:a<b+\epsilon\implies a\le b$ — Martin R, Jul 29 '21 at 16:09

Antonio · Accepted Answer · 2021-07-29T15:06:14.677

4

If $\;x\;$ were negative, by letting $\;a=-x>0\;,\;$ we would get that $\;x>-a=x\;,\;$ but it is a contradiction, so $\;x\;$ cannot be negative, hence $\;x\geqslant0\;.$

edited Jul 29 '21 at 15:06

answered Jul 29 '21 at 14:59

Antonio

414

score 0 · Answer 2 · answered Jul 29 '21 at 14:59

0

By contradiction if $x \lt 0$, then

$$0 \gt - (-x /2) \gt x$$ with $-(x/2) \gt 0$. A contradiction with the hypothesis $x \gt -a$ for all $a \gt 0$.

answered Jul 29 '21 at 14:59

mathcounterexamples.net

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Proof for: If $x>-a$ for $a>0$, then $x\geq0$

2 Answers2