Equivalent metrics generate the same topology

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

Proof:

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.

Solution 1:

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".

Solution 2:

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 $\tau_2 \subseteq \tau_1$.