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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 $x = 0$.

Answer

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 $\forall \epsilon > 0$

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  • $\begingroup$ You negated wrong $\endgroup$ May 5 '20 at 0:41
  • $\begingroup$ Thank you for your comment. Did you mean this part? $|x| = |x - \epsilon + \epsilon|$. Can also you help me to fix it?? $\endgroup$
    – Rowing0914
    May 5 '20 at 0:42
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    $\begingroup$ Contradiction would be $|x|\ge\epsilon$ for some $\epsilon>0$ $\endgroup$ May 5 '20 at 0:45
  • $\begingroup$ @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? $\endgroup$
    – Rowing0914
    May 5 '20 at 0:51
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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.

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    $\begingroup$ Thank you for your help! Let me take time to digest this!! $\endgroup$
    – Rowing0914
    May 5 '20 at 0:55
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Assume that $x \neq 0$. Then, $|x| = c >0$. Set $\epsilon <c$ 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.

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    $\begingroup$ Thank you for your help! Let me take time to digest this!! $\endgroup$
    – Rowing0914
    May 5 '20 at 0:55
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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.

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    $\begingroup$ Thank you for your help! Let me take time to digest this!! $\endgroup$
    – Rowing0914
    May 5 '20 at 0:55
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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.

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    $\begingroup$ Thank you for your help! Let me take time to digest this!! $\endgroup$
    – Rowing0914
    May 5 '20 at 0:55

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