Let $X$ be a set. Two metrics $d_1, d_2: X\times X\to\mathbb{R}$ are equivalent, if constants $\alpha,\beta > 0$ exist such that for all $x,y\in X$ holds:

$\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$

Show that equivalent metrics generate the same topology


Let $d_1, d_2: X\times X\to\mathbb{R}$ be metrics and $\tau_1, \tau_2$ the induced topolgies.

We have to show, that $\tau_1=\tau_2$.

  1. $\tau_1\subseteq\tau_2$.

Let $U\in\tau_1$ open. Then exists for every $x\in U$ a $\epsilon >0$ such that $B_{d_1}(x,\epsilon)\subseteq U$.

It is $B_{d_1}(x,\epsilon)=\{y\in X|d_1(x,y)<\epsilon\}$.

Since $d_1$ and $d_2$ are equivalent, there are constants $\alpha,\beta >0$ such that $\alpha d_1(x,y)\leq d_2(x,y)\leq \beta d_1(x,y)$. Take $\epsilon':=\beta^{-1}\epsilon >0$.

We get $d_2(x,y)\leq \beta d_1(x,y)<\beta\cdot \beta^{-1}\cdot\epsilon=\epsilon$.

Hence $U\in\tau_2\checkmark$.

The other inclusion $\tau_1\supseteq\tau_2$ works analogously.

Is this proof correct? Thanks in advance.

  • $\begingroup$ I'd say so. You basically need to show that open balls in one topology correspond to open balls in the other. If you can always fit an open ball in one topology into the other, you're done. $\endgroup$ – Adrian Keister Jul 15 '18 at 2:01

Why does $d_2(x,y) < \varepsilon$ imply $U \in \tau_2$? This does not follow, you're leaving out parts of the argument, and going the wrong way with inequalities:

So at the start you know that we have $\alpha, \beta>0$ such that for all $x,y$ we have $$\alpha d_1(x,y) \le d_2(x,y) \le \beta d_1(x,y)$$

It's OK to start with $U \in \tau_1$. We want to show $U \in \tau_2$, so let $x \in U$. As $U \in \tau_1$ there exists an $\varepsilon > 0$ such that $B_{d_1}(x,\varepsilon) \subseteq U$. Then I claim that

$$B_{d_2}(x,\alpha \varepsilon) \subseteq B_{d_1}(x,\varepsilon)$$

Suppose $y \in B_{d_2}(x,\alpha\varepsilon)$, then $d_2(x,y) < \alpha\varepsilon$ and so $d_1(x,y) \le \frac{1}{\alpha} d_2(x,y) < \frac{1}{\alpha}\alpha\varepsilon = \varepsilon$ and so $y \in B_{d_1}(x,\varepsilon)$. Then as $B_{d_1}(x,\varepsilon) \subseteq U$ we see that also $B_{d_2}(x,\varepsilon\alpha) \subseteq U$ and thus $x$ is an interior point of $U$ wrt $d_2$. As $x \in U$ was arbitrary, $U \in \tau_2$ and so $\tau_1 \subseteq \tau_2$. As an exercise, I'd suggest to do the reverse inclusion in the same detail as I did above, not be lazy and say "it's analogous".

  • $\begingroup$ How do you get the inequality $d_1(x,y)\leq \alpha d_2(x,y)<\alpha\cdot\frac{\epsilon}{\alpha}$? Should you know take $B_{d_2}(x,\alpha\epsilon)$? Since you get $d_1(x,y)\leq \alpha^{-1}d_2(x,y)<\alpha^{-1}\alpha\epsilon=\epsilon$. If you divide $\alpha$ you get $d_1(x,y)\leq\alpha^{-1}d_2(x,y)$. Thats why I am confused right now. $\endgroup$ – Cornman Jul 16 '18 at 5:42
  • $\begingroup$ @Cornman Yes, I was mistaken; I edited. It should be OK now. $\endgroup$ – Henno Brandsma Jul 16 '18 at 15:31

For completeness I also wish to add the other inclusion on this same page.

Just as above we have the same $\alpha, \beta>0$ such that for all $x,y \in X$ we have $$\alpha d_1(x,y) \le d_2(x,y) \le \beta d_1(x,y)$$

Let $U \in \tau_2$. We will show $U \in \tau_1$. We take $x \in U$. As $U \in \tau_2$ there exists an $\varepsilon > 0$ such that $B_{d_2}(x,\varepsilon) \subseteq U$. Then I claim that: $$B_{d_1}(x,\varepsilon / \beta) \subseteq B_{d_2}(x,\varepsilon)$$

Suppose $y \in B_{d_1}(x,\varepsilon/\beta)$, then $d_1(x,y) < \varepsilon/\beta$ and so by the equivalence of metrics we have $d_2(x,y) \le \beta d_1(x,y) < \beta \cdot\varepsilon/\beta = \varepsilon$ and so $y \in B_{d_2}(x,\varepsilon)$. We see that $B_{d_1}(x,\varepsilon / \beta) \subseteq B_{d_2}(x,\varepsilon) \subseteq U$.

We deduce that $x$ is an interior point of $U$ with respect to $d_1$. As $x \in U$ was arbitrary, $U \in \tau_1$ and so $\tau_2 \subseteq \tau_1$.


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