# Help: Real Analysis Proof: Prove $|x| < \epsilon$ for all $\epsilon > 0$ iff $x = 0$.

I've recently started studying Real Analysis by myself(not going to any school at the moment) and need some help to review my answer to the question below.

## Question

Let $$x \in \mathbb{R}$$. Prove $$|x| < \epsilon$$ for all $$\epsilon > 0$$ iff $$x = 0$$.

Suppose for contradiction that $$x \in \mathbb{R}$$, $$|x| \geq \epsilon$$ for all $$\epsilon > 0$$.
$$|x| = |x - \epsilon + \epsilon| \leq |x - \epsilon| + |\epsilon|$$ by Triangle inequality.
Since the first term($$|x - \epsilon| \geq 0$$) and by our assumption($$\forall \epsilon > 0$$), this contradicts $$|x| \geq \epsilon$$.

## Comment

I am not sure if my answer is solid enough to lead to the contradiction. Any help or comment is appreciated.

## Source

The question is taken from the Exercise 2.12 of the pdf below.

## Edit

Thank you for all the questions!!
It seems like I've stepped towards the wrong direction from the beginning!
I will try myself to prove that there is an arbitrary $$\epsilon$$ which can take the value below $$|x|$$ unless $$|x| = 0$$ based on $$\forall \epsilon > 0$$

• You negated wrong May 5 '20 at 0:41
• Thank you for your comment. Did you mean this part? $|x| = |x - \epsilon + \epsilon|$. Can also you help me to fix it?? May 5 '20 at 0:42
• Contradiction would be $|x|\ge\epsilon$ for some $\epsilon>0$ May 5 '20 at 0:45
• @J.W.Tanner, Thank you for mentioning the critical mistake! I was checking the difference of for_all vs for_some and it seems that in this question, I shouldn't use it for the starter, doesn't it? May 5 '20 at 0:51

If $$x=0$$ then certainly, $$|x|=|0|=0<\epsilon$$ for every $$\epsilon>0$$.

Now, you want to show that if for all $$\epsilon>0$$ , $$|x|<\epsilon$$ then, $$x=0$$. Suppose $$x\neq 0$$ so $$|x|>0$$. Hence, for $$\epsilon=|x|$$, $$|x|<|x|$$. Contradiction.

• Thank you for your help! Let me take time to digest this!! May 5 '20 at 0:55

Assume that $$x \neq 0$$. Then, $$|x| = c >0$$. Set $$\epsilon for example, $$\epsilon = \frac{c}{2}$$. Then we have $$c = |x| < \epsilon = c/2$$ by assumption, meaning that $$c < c/2$$. But this of course cannot happen.

• Thank you for your help! Let me take time to digest this!! May 5 '20 at 0:55

You want to prove that

$$(\forall x\in \Bbb R)\;\;\Bigl( (\forall \epsilon>0)\; \; |x|<\epsilon\;\; \iff \;\; x=0\Bigr)$$

Proof

Let $$x\in \Bbb R$$ such that $$(\forall \epsilon>0) \;\; |x|<\epsilon$$

Assume $$x\ne 0$$.

$$x\ne 0 \implies |x|>0$$

$$\implies \;\; \exists \epsilon(=\frac{|x|}{2})>0 \;\; : \;\; |x|>\epsilon$$ which contradicts the hypothesis.

• Thank you for your help! Let me take time to digest this!! May 5 '20 at 0:55

If $$x = 0$$, the result follows. Suppose otherwise that $$|x| < \varepsilon$$ for every $$\varepsilon > 0$$ and $$x\neq 0$$. Then either $$x > 0$$ or $$x < 0$$. In the first case $$(x > 0)$$, one can choose $$\varepsilon = x/2$$, whence we get \begin{align*} |x| = x < \varepsilon = \frac{x}{2} \Longleftrightarrow x < 0 \end{align*} which leads to a contradiction. Similarly, if $$x < 0$$, we can choose $$\varepsilon = -x/2$$, whence we get that \begin{align*} |x| = -x < \varepsilon = -\frac{x}{2} \Longleftrightarrow x > 0 \end{align*} which is also a contradiction. Consequently, $$x = 0$$ and we are done.

• Thank you for your help! Let me take time to digest this!! May 5 '20 at 0:55