Thread: easy question

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

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


is true that $\displaystyle z=x$ ?

if this is true, how i can prove that?

thanks

Suppose $\displaystyle z \ne x$.

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

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

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

The result follows by transposition.

I answer to my inself

Suposse that $\displaystyle z\neq x$

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

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

Then, $\displaystyle z=x$

Perfect.