Please check my proof for : Prove $|x| < \epsilon$ for all $\epsilon > 0$ iff $x = 0$.

Question

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

Question from Exercise 2.12 of this PDF.

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.

Edit

Thank you for all the questions! I 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 $\forall \epsilon > 0$.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.