Let $x \in \mathbb{R}$ and assume that for all $\epsilon > 0, |x| < \epsilon$. Prove: $x = 0$

Question

This following exercise has me kind of confused, it asks: let $x \in \mathbb{R}$ and assume that for all $\epsilon > 0, |x| < \epsilon$.

Prove that $x = 0$.

My attempt to this was to use proof by contradiction:

Proof: Let $x \in \mathbb{R}$ and assume that $x > 0.$ Then our $\epsilon=\dfrac{|x|}{2}>0.$ By assumption we have that $0\le x<\epsilon =\dfrac{ |x|}{2},$ so then $x=0$, which contradicts our $x > 0$ claim.

Will this suffice?

Well, you still need to eliminate the $x<0$ case. And it isn;t true that $0≤x<\frac {|x|}2\implies x=0$. Those inequalities are impossible. — lulu, Sep 16 '17 at 02:24
To deal with $x<0$, start instead with assuming $|x|>0$ to get the contradiction that you have. Since the contradiction says $|x|>0$ is not true, $x$ must be equal to zero. — Mathemagical, Sep 16 '17 at 02:28
Ah damn, wolfram error. Alright let me try it that way for $x<0.$ — Kono Dio Da, Sep 16 '17 at 02:30
If $x > 0$ then setting $e=x $ gives us $|x|=x <x=e$. Which is a contradiction. But we can do one better. If $x\ne 0$ then $|x|>0$. Let $e =|x|$ and we have $|x|<|x|=e $. Which is a contradiction. — fleablood, Sep 16 '17 at 02:58
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, Jun 10 '23 at 06:54
@user1110341 Same comment as 2 hours ago on that other post: I don't think it does: that post was about intuition, not formal proof. (And though related, the question is distinct.) — Anne Bauval, Jun 10 '23 at 19:48
Does this answer your question? Intuition about : $a = b \iff | a − b| < \epsilon$, for every $\epsilon > 0$ — , Oct 01 '23 at 18:10

score 7 · Accepted Answer · answered Sep 16 '17 at 02:30

It won't suffice because you have not examined small negative numbers.

I would prove it by contradiction. Assume $x \ne 0$. Then $|x| >0$ Let $\epsilon = |x|/2$. Then $|x| > \epsilon$, which contradicts the assumption that $|x| < \epsilon$ for every possible $\epsilon > 0$.

score 3 · Answer 2 · answered Aug 09 '19 at 16:31

All of the previous answers invoke contradiction, but I don't believe there's any need to. One could argue like this: By assumption, $|x|$ is smaller than every positive real number, so in particular it is different from every positive real number, so it is not positive. But, by definition, $|x|$ is non-negative. Hence, $|x|$ is zero, so $x$ itself is zero.

score 0 · Answer 3 · answered Sep 16 '17 at 03:03

0

It's way simpler than that.

$|x|\ge 0$

If $|x|>0$ then setting $\epsilon=|x|$ we get the contradictory $\epsilon =|x| >|x|$.

So $|x|=0$. So $x=0$.

answered Sep 16 '17 at 03:03

fleablood

130,341

score 0 · Answer 4 · answered Sep 16 '17 at 03:05

Almost the same proof than E.Fisher, just to use the archimedian property.

$(\mathbb R,+,\le)$ is archimedian, so for $0<|x|<\epsilon$ there exists $n\in\mathbb N$ such that $n|x|>\epsilon$.

This contradicts $|x|<\varepsilon$ for $\displaystyle \varepsilon=\frac{\epsilon}n$, thus $|x|=0\quad$ (and $x=0$ consequently)

Let $x \in \mathbb{R}$ and assume that for all $\epsilon > 0, |x| < \epsilon$. Prove: $x = 0$

4 Answers4