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Thread: easy question

  1. #1
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    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
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  2. #2
    MHF Contributor Matt Westwood's Avatar
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    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.
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  3. #3
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    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, $\displaystyle z=x$
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  4. #4
    MHF Contributor Matt Westwood's Avatar
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    Perfect.
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