A set is open in $(X_1 \times X_2, d)$ if and only if it is open in $(X_1 \times X_2, p)$.

Let $$(X_1, d_1), (X_2, d_2)$$ be metric spaces and define $$(X_1 \times X_2, d)$$, $$(X_1 \times X_2, p)$$ where $$d[(x_1, x_2),(y_1, y_2)] = d_1(x_1, y_1) + d_2(x_2,y_2)$$ and $$p[(x_1, x_2),(y_1, y_2)] = \max\{d_1(x_1,y_1), d_2(x_2,y_2)\}$$

My attempt:

Notice that $$p[(x_1, x_2),(y_1, y_2)] \leq d[(x_1, x_2),(y_1, y_2)]$$, that is because the metric $$d$$ is equal to the metric $$p$$ + $$\min\{d_1(x_1,y_1), d_2(x_2,y_2)\}$$.

Let $$G$$ be an open subset of $$X_1 \times X_2$$ with the metric $$d$$, so, for any element $$t = (t_1, t_2)$$ of $$G$$ there exists an $$r > 0$$ such that $$B(t; r) \subseteq G$$, that is $$\{(z_1, z_2) \in X_1 \times X_2: d[(t_1, t_2),(z_1, z_2)] < r\}$$, since $$p[(t_1, t_2),(z_1, z_2)] \leq d[(t_1, t_2),(z_1, z_2)]$$, we have $$p[(t_1, t_2),(z_1, z_2)] < r$$, therefore for any element in $$G$$, you can choose the same $$r$$ that guarantees that the open ball with metric $$d$$ is contained in $$G$$. That same $$r$$ is also going to guarantee that the open ball with metric p is contained in $$G$$.

Is that correct? I haven't been able to prove it on the other direction, I tried this.

let $$d_1(x_1, y_1) > d_2(x_2, y_2)$$, and consider an open subset with metric $$p$$, so $$d_1(x_1, y_1) < r$$ I tried adding $$d_2(x_2, y_2)$$ to the inequality and I got $$r + d_2(x_2, y_2)$$, but this doesn't guarantee that the open ball with metric $$d$$ is going to be contained in the open subset. That's where I'm stuck. Any ideas?

• In $\Bbb{F}^2$, the norms $\| (\xi,\eta)\|_1:=|\xi|+|\eta|$ and $\|(\xi,\eta)\|_\infty :=\max\{|\xi|,|\eta|\}$ are equivalent. That is, there are constants $A, B>0$ such that $A\| \cdot \|_1 \leq \| \cdot \|_\infty \leq B \| \cdot \|_1$. Looks like you’ve already found $B$. Find $A$ and you’ll be done. – Alonso Delfín Aug 25 at 21:58

For non-negative $$a$$ and $$b$$ the following holds: $$\max(a,b)\le a+b\le 2\max(a,b)$$. From this you can show that each $$d$$-ball is contained in a $$p$$-ball and each $$p$$-ball is contained in a $$d$$ ball.