Proof of this property: $x=0$ if and only if $|x|0$ I'm trying to prove a property that we've seen in our analysis lecture, but I have a little doubt about my way of doing it. The property is the following:
$$x=0 \Leftrightarrow |x| < z \text{ for all } z>0$$
...and we work with real numbers only.
I tried to prove it that way:
1) First step: $x=0 \Rightarrow |x|<z \text{ for all } z>0$
We know by a property of absolute values that $x=0 \Leftrightarrow |x|=0$, so $x=0$ means that $|x|=0$, which is smaller than $z$ for all $z>0$. Thus, $x=0 \Rightarrow |x|<z \text{ for all } z>0$.
2) Second step: $|x|<z \text{ for all }z>0 \Rightarrow x=0$
Let's suppose, on the contrary, that $|x|<z \text{ for all }z>0 \Rightarrow x \ne 0$, i.e. $x<0$ or $x>0$.
If $x<0$, then $|x|=-x < z$. This must be true for all $z \in \mathbb{R}$, but if $x=-2$, for example, then there exists $z$ (for example: $z=1$) such that the inequality above doesn't hold (we would have 2 < 1). So it's a contradiction.
If $x>0$, then $|x|=x < z$. Again, if $x=2$ and $z=1$, we have a contradiction.
As $x$ cannot be smaller or higher than $0$ without leading to a contradiction, the only possibility is that $x=0$.
I'm not sure about the second part of my proof, as it doesn't look as formal as usual... we usually only use axioms and properties derived from those axioms, so I'm not very happy with using counterexamples here. Does anyone know if there is another and nicer way to prove it?
 A: Your proof is flawed because you cannot prove a general fact from listing examples. Also the negation of "$|x| < y$ for all positive $y$" is "there exists some positive $y$ such that $|x| \geq y$", whereas you have its negation written as "$|x| < z$ for all $z \implies x \neq 0$, i.e. $x < 0$ or $x > 0$". Here is is how you could go about doing this:
Suppose first that $x = 0$. Then $|x| = 0$, which is smaller than any positive real number.
You attempted to prove the next part by contradiction, which could actually work, but the cleanest way is by contrapositive. Here is how that might work:
Suppose that $x \neq 0$. Then $|x| \neq 0$ and $|x| \geq 0$, so $|x| > 0$. In particular there is a positive number which is less than or equal to $|x|$, $|x|$ itself.
By contrapositive, if $x \in \mathbb{R}$ is such that $|x| < y$ for every positive $y \in \mathbb{R}$, then $x = 0$.
A: For the second part, we have $|x| < z$ for all $z>0$ and suppose for the sake of contradiction that $|x| \neq 0$. Since $|x| \geq 0$, this means, in this case, $|x| > 0$. But then, $|x| + |x| > |x|$. That is, $|x| > \frac{|x|}{2} >0 $, contradicting that $|x| < z$ for all $z > 0$. Therefore, our assumption that $|x| \neq 0$ isn't possible and it must be that $|x| = 0$.
