easy question

**jorgeston** · Sep 6th 2008, 01:35 PM

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

**Matt Westwood** · Sep 6th 2008, 01:44 PM

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.

**jorgeston** · Sep 6th 2008, 01:51 PM

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, $\exists n\in \mathbb{N}: 0<\frac{1}{n}<h$ .This implies that $\exists f:=\dfrac{1}{n} \in \mathbb{R}:d(z,x)>f$ . Contradiction.

Then, $z=x$

**Matt Westwood** · Sep 6th 2008, 02:04 PM

Perfect.

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