1. ## easy question

Let $M$ a metric space and $z,x\in M$, $d$ a metric on $M$

If i have $d(z,x)<\epsilon , \forall \epsilon > 0$

is true that $z=x$ ?

if this is true, how i can prove that?

thanks

2. Suppose $z \ne x$.

Then $\exists \epsilon_1 \in \mathbb{R}^+: d (z, x) = \epsilon_1$.

Then $\exists \epsilon \in \mathbb{R}^+: \epsilon = \epsilon_1 / 2: \epsilon < d (z, x)$.

So $z \ne x \Longrightarrow \exists \epsilon \in \mathbb{R}^+: \epsilon < d (z, x)$.

The result follows by transposition.

3. I answer to my inself

Suposse that $z\neq x$

This implies that $d(z,x)>0, d(z,x)=h, h>0$

But for the arquimedean property of IR, $\exists n\in \mathbb{N}: 0<\frac{1}{n} .This implies that $\exists f:=\dfrac{1}{n} \in \mathbb{R}:d(z,x)>f$. Contradiction.

Then, $z=x$

4. Perfect.