2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Can you show that $d(p_1,p_2) \leq e(p_1,p_2)$? And $e(p_1,p_2) \leq 2d(p_1,p_2)$? $\endgroup$ – Managu May 2 '16 at 18:28
  • $\begingroup$ Draw a picture,representing $X$ and $Y$ by the reals, and look at $B_d(p,r)$ and $B_e(p,r)$ and $B_d(p,r/2).$ $\endgroup$ – DanielWainfleet May 2 '16 at 19:33
3
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.