# Prove two metrics are equivalent?

If $d_1$ and $d_2$ are metrics on $X$ and $Y$, respectively, and $d$ and $e$ are metrics on $X \times Y$ with $$d( (x_1, x_2) , (y_1, y_2) ) = \max \{ d_1(x_1,y_1), d_2(x_2,y_2) \}$$ and $$e ( (x_1,y_1), (x_2,y_2) ) = d_1 (x_1,x_2) + d_2 (y_1,y_2),$$ prove that these two metrics are equivalent.

I know this means that they induce the same topology, which means their open ball basis make up the same set.

I'm assuming I need to show that each open set in one contains an open set in the other, but I'm not sure how to do this.

• Can you show that $d(p_1,p_2) \leq e(p_1,p_2)$? And $e(p_1,p_2) \leq 2d(p_1,p_2)$? – Managu May 2 '16 at 18:28
• Draw a picture,representing $X$ and $Y$ by the reals, and look at $B_d(p,r)$ and $B_e(p,r)$ and $B_d(p,r/2).$ – DanielWainfleet May 2 '16 at 19:33

Observe that $$d\left(u,v\right)\leq e\left(u,v\right)\leq2d\left(u,v\right)$$

Consequently

$\left\{ v\mid e\left(u,v\right)<r\right\} \subseteq\left\{ v\mid d\left(u,v\right)<r\right\} \tag1$

$\left\{ v\mid d\left(u,v\right)<\frac{1}{2}r\right\} \subseteq\left\{ v\mid e\left(u,v\right)<r\right\} \tag2$

Let $U$ belong to the topology induced by $d$.

Then for every $u\in U$ there is a $r>0$ with $\left\{ v\mid d\left(u,v\right)<r\right\}\subseteq U$.

Then (1) implies that also $\left\{ v\mid e\left(u,v\right)<r\right\}\subseteq U$ showing that $U$ belongs to the topology induced by $e$.

The converse of this can be shown on base of (2).