# Thread: Proving a set open

1. ## Proving a set open

I need help with this proof...

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

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

Nicole

2. Originally Posted by thaopanda
I need help with this proof...

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

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

Lemma 1. Let $d_1, d_2$ be two metrics for the set X. If there exists a positive number $c_1$ such that $d_{1}$(x,y) $\leq c_{1}d_{2}$(x,y) for all x,y $\in X$. Then the identity map $iX, d_2) \rightarrow (X, d_1)" alt="iX, d_2) \rightarrow (X, d_1)" /> is continuous.
Proof. Let $p \in X$ and let $\epsilon$ be a positive number. If $\delta = \frac{\epsilon}{c_1}$ and x is a member of X for which $d_2(x,p) < \delta$, then $d_1(i(x), i(p)) = d_1(x, p) \leq c_1d_2(x,p) < c_1\delta = \epsilon$. This shows that $d_1(i(x), i(p)) < \epsilon$ whenever $d_2(x,p) < \delta$. Thus $iX, d_2) \rightarrow (X, d_1)" alt="iX, d_2) \rightarrow (X, d_1)" /> is continuous.
I'll leave it to you to show that $i^{-1}X, d_1) \rightarrow (X, d_2)" alt="i^{-1}X, d_1) \rightarrow (X, d_2)" /> is continuous.
Since $iX, d_2) \rightarrow (X, d_1)" alt="iX, d_2) \rightarrow (X, d_1)" /> is continuous, then for each $d_1$-open set, $i^{-1}(O) = O$ is $d_2$-open set. This argument applies for $i^{-1}X, d_1) \rightarrow (X, d_2)" alt="i^{-1}X, d_1) \rightarrow (X, d_2)" /> as well. Thus, $d_1, d_2$ determines the same open sets in X.