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Theorem. If there are strictly positive constants $c_1$ and $c_2$ such that

$$c_1 d_1(x,y) \leq d_2 (x,y) \leq c_2 d_1 (x,y)$$

for all $x,y \in X$, then $d_1$ and $d_2$ are topologically equivalent metrics on $X$.

Proof. (for one direction) Let $U$ be open in $(X, d_1)$. Then for $a \in X$ there > exists $\epsilon > 0$ such that

$$\{x \in X; d_1(x,a) < \epsilon\} \subset U.$$

But then

$$\{x \in X; d_2(x,a) < c_1\epsilon\} \subset \{x \in X; d_1(x,a) < \epsilon\} \subset U$$

and $U$ is open in $(X, d_2)$.

The line I don't understand is the one that says $\{x \in X; d_2(x,a) < c_1\epsilon\} \subset \{x \in X; d_1(x,a) < \epsilon\}$. Why is the former a subset of the latter? I don't see the justification of this.

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    $\begingroup$ Hint: the inequality $c_1d_1(x,y)\le d_2(x,y)$ can be written $d_1(x,y)\le\frac{1}{c_1}d_2(x,y)$, and then apply $d_2(x,y)<c_1\epsilon$. $\endgroup$ – Jesus RS Mar 7 '15 at 16:54
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We know that $c_{1}d_{1}(x,y) \leq d_{2}(x,y)$ for all $x,y \in X$.

For membership in $\{x \in X: d_{2}(x,a) < c_{1} \epsilon \}$, we have that $d_{2}(x,a) < c_{1} \epsilon$.

Therefore, we conclude that $c_{1}d_{1}(x,a) \leq d_{2}(x,a) < c_{1} \epsilon$, thus $c_{1}d_{1}(x,a) < c_{1} \epsilon$, and thus we can say $d_{1}(x,a) < \epsilon$ since $c_{1}$ is positive.

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  • $\begingroup$ I can figure out why $d_1(x,a) < \epsilon$, but I don't see how $d_2(x,a) < c_1\epsilon$ is $< d_1(x,a) < \epsilon$, since it's supposed to be a subset $\endgroup$ – mr eyeglasses Mar 7 '15 at 17:04
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    $\begingroup$ If you can conclude that $d_{1}(x,a) < \epsilon$ from $d_{2}(x,a) < c_{1} \epsilon$, then you are done. You are showing that the restriction on the first set guarantees the restriction on the second, so the former must be a subset. $\endgroup$ – Jonathan Hebert Mar 7 '15 at 17:12

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