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?

  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


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