If $|x|<\epsilon, \forall \epsilon > 0$, then $x=0$

Question

How do I start a proof like this? I am new to epsilon-delta proofs - I think this is a basic proof, I just want to see how someone would structure it properly.

Answer 1

Suppose $x\neq 0$. Then $|x|>0$. Take $\epsilon = |x|/2$ and we have $|x|<\epsilon = |x|/2$, a contradiction.

Answer 2

First of all let's say you are wrong, $x\ne 0$, now if $x\ne 0$ then $|x|>0$. Because you said for all $\epsilon$ any expression of $epsilon$ that is greater than $1$ should be valid. So let's try $\epsilon\equiv \frac{|x|}{n}, n\in\Bbb R^+, n>1$, because both $n$ and $|x|$ are positive $\epsilon$ is also positive. Now it is given that $|x|<\epsilon$ so $$|x|<\epsilon\implies |x|<\frac{|x|}{n}\implies 1<\frac{1}{n}$$because that $n>1$ we have contradiction. So $x=0$

Answer 3

By hypothesis, \begin{equation} |x| < \frac{1}{n},~\forall n \in \mathbb{N} \tag{1}\label{eq}. \end{equation}

Notice that \eqref{eq} holds if and only if \begin{equation} -\frac{1}{n} < x < \frac{1}{n},~\forall n \in \mathbb{N}. \end{equation}

By the squeeze theorem for sequences, we must have $x=0$.

Answer 4

I like the answer that Hugo C Botós provided, nevertheless I can provide same solution but clearest.

In the language of mathematical logic, our proposition is, $$\forall \epsilon > 0~(|x|<\epsilon \Longrightarrow x=0)$$

Proof.

Let $\epsilon \in \mathbb{R}$ such that $\epsilon > 0$. Suppose, to arrive at a contradiction, that $x\not=0$. Thus $|x|\geqslant 0$ and $x\not=0$, $|x|>0$. Given the class of the set of positive numbers, thus $|x|>0$ and $\frac{1}{2}>0$, $\frac{|x|}{2}>0$.

Since the quantifier is $\forall \epsilon >0$, in particular, it is for $\epsilon = \frac{|x|}{2}>0$. Clearly $\epsilon = \frac{|x|}{2}<|x|$ and, by hypothesis, $|x|<\epsilon$. Using the transitivity of the relation $<$, thus $\epsilon = \frac{|x|}{2}<|x|$ and $|x|<\epsilon$, $\epsilon < |x| < \epsilon$, which is a contradiction because $\epsilon \not< \epsilon$.

The contradiction comes from supposing that $x\not=0$. Therefore, $x=0$.

$\therefore \forall \epsilon > 0~(|x|<\epsilon \Longrightarrow x=0)$.