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Thread: Proving a set open

  1. #1
    Member thaopanda's Avatar
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    Proving a set open

    I need help with this proof...

    $\displaystyle d_{1},d_{2} : X \times X \rightarrow R$ distance on X are called equivalent provided that there exists $\displaystyle C_{1},C_{2}$ > 0 such that
    $\displaystyle C_{2}d_{2}$(x,y) $\displaystyle \leq d_{1}$(x,y) $\displaystyle \leq C_{1}d_{2}$(x,y)
    $\displaystyle \forall $x,y $\displaystyle \epsilon X$

    Let $\displaystyle d_{1},d_{2}$ be equivalent distances on X and let $\displaystyle A \subseteq X$. Prove that A is open in $\displaystyle (X,d_{1})$ if and only if A is open in $\displaystyle (X,d_{2})$ (i.e. equivalent distances give the same open sets).

    please help
    Nicole
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  2. #2
    Senior Member
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    Quote Originally Posted by thaopanda View Post
    I need help with this proof...

    $\displaystyle d_{1},d_{2} : X \times X \rightarrow R$ distance on X are called equivalent provided that there exists $\displaystyle C_{1},C_{2}$ > 0 such that
    $\displaystyle C_{2}d_{2}$(x,y) $\displaystyle \leq d_{1}$(x,y) $\displaystyle \leq C_{1}d_{2}$(x,y)
    $\displaystyle \forall $x,y $\displaystyle \epsilon X$

    Let $\displaystyle d_{1},d_{2}$ be equivalent distances on X and let $\displaystyle A \subseteq X$. Prove that A is open in $\displaystyle (X,d_{1})$ if and only if A is open in $\displaystyle (X,d_{2})$ (i.e. equivalent distancesgive the same open sets).

    please help
    Nicole
    Lemma 1. Let $\displaystyle d_1, d_2$ be two metrics for the set X. If there exists a positive number $\displaystyle c_1$ such that $\displaystyle d_{1}$(x,y) $\displaystyle \leq c_{1}d_{2}$(x,y) for all x,y $\displaystyle \in X$. Then the identity map $\displaystyle iX, d_2) \rightarrow (X, d_1)$ is continuous.

    Proof. Let $\displaystyle p \in X$ and let $\displaystyle \epsilon$ be a positive number. If $\displaystyle \delta = \frac{\epsilon}{c_1}$ and x is a member of X for which $\displaystyle d_2(x,p) < \delta$, then $\displaystyle d_1(i(x), i(p)) = d_1(x, p) \leq c_1d_2(x,p) < c_1\delta = \epsilon$. This shows that $\displaystyle d_1(i(x), i(p)) < \epsilon$ whenever $\displaystyle d_2(x,p) < \delta $. Thus $\displaystyle iX, d_2) \rightarrow (X, d_1)$ is continuous.

    I'll leave it to you to show that $\displaystyle i^{-1}X, d_1) \rightarrow (X, d_2)$ is continuous.

    Since $\displaystyle iX, d_2) \rightarrow (X, d_1)$ is continuous, then for each $\displaystyle d_1$-open set, $\displaystyle i^{-1}(O) = O$ is $\displaystyle d_2$-open set. This argument applies for $\displaystyle i^{-1}X, d_1) \rightarrow (X, d_2)$ as well. Thus, $\displaystyle d_1, d_2$ determines the same open sets in X.
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