Strongly equivalent metrics induce equivalent topologies - trouble with radii of balls

I am trying to show that if $d_{1}$ and $d_{2}$ are strongly equivalent metrics on $X$ (I.e., that $\exists a,b>0, \forall x,y \in X$ such that $d_{1}(x,y)\leq ad_{2}(x,y)$ and $d_{2}(x,y)\leq bd_{1}(x,y)$), then they induce the same topology on $X$.

Now, I know this topic has been discussed often on MSE, but none of the proofs either 1) approach the problem the way I'm approaching it here (for one thing, I am not interested in anything about convergence of series), or 2) include the specific details I am seeking here. Read on and you'll see what I mean by this.

This is how I've begun my approach, and at the end, I'll ask a very specific question about how to get from the step where I stopped to the next step of the proof:

Suppose that $d_{1}$ and $d_{2}$ are two strongly equivalent metrics on $X$ as defined above. Suppose also that $d_{1}$ induces the topology $\mathcal{T_{1}}$ and that $d_{2}$ induces the topology $\mathcal{T_{2}}$.

WTS: $U\in \mathcal{T}_{1} \iff U\in \mathcal{T}_{2}$. I.e., that the open sets of one are precisely the open sets of the other, and vice versa.

Right now, I am stuck on the implication that $U\in \mathcal{T}_{2} \implies U\in \mathcal{T}_{1}$.

Assume $U $ is open in $\mathcal{T}_{2}$; I.e., $U \in \mathcal{T}_{2}$. Then, $\forall x_{0} \in X$, if $x_{0} \in U$, then $\exists \epsilon >0$ such that $B_{\epsilon}^{2}(x_{0})\subseteq U$ ($B_{\epsilon}^{2}(x_{0})$ is an open ball in the metric $d_{2}$). Since $B_{\epsilon}^{2}(x_{0}) = \{ x \in X \, \vert \, d_{2}(x,x_{0})<\epsilon\}$, $B_{\epsilon}^{2}(x_{0})\subseteq U \iff d_{2}(x,x_{0})<\epsilon$ for $x_{0} \in U$ implies that $x \in U$ as well.

Now, by the fact that $d_{1}$ and $d_{2}$ are equivalent metrics (and here is where I get stuck), we have that $d_{1}(x,x_{0})\leq a d_{2}(x,x_{0})<a\cdot \epsilon$.

My specific question is: suppose that open balls in $d_{1}$ are defined as $B_{\delta}^{1}(x,x_{0})= \{ x \in X \, \vert \, d_{1}(x,x_{0})<\delta\}$. How do I use the $a$ to bound the radius $\delta$ so that I can fit it inside a $d_{2}$ ball?

None of the questions and answers I have seen on MSE have gone into any detail as to how to do this. So, I would very much appreciate it if someone could walk me through how to do this.

Thank you in advance.

Solution 1:

So suppose

$$\exists a,b>0, \forall x,y \in X \text{ such that } d_{1}(x,y)\leq ad_{2}(x,y) \text{ and }d_{2}(x,y)\leq bd_{1}(x,y)$$

Then if $U$ is open under $d_1$, and $x \in U$, then $\exists r>0: B_{d_1}(x,r) \subseteq U$. Then $B_{d_2}(x, \frac{r}{a}) \subseteq B_{d_1}(x,r) \subseteq U$, because if $d_2(x,y) < \frac{r}{a}$, then $d_1(x,y) \le ad_2(x,y) < a\frac{r}{a} = r$ by the first inequality. So every $x \in U$ is a $d_2$-interior point, so $U$ is open under $d_2$.

The other inclusion is similar, interchanging $d_1$ and $d_2$ and using $\frac{r}{b}$ instead.