1
$\begingroup$

I am lost in trying to solve the following:

Let $(X_1,d_1)$ and $(X_2,d_2)$ be two metric spaces. Define $$X = X_1 \times X_2 = \{ \mathbf{x}=(x_1,x_2) : x_1 \in X_1, x_2 \in X_2 \}$$

and $\forall \ \mathbf{x},\mathbf{y} \in X$

$$d_+(\mathbf{x},\mathbf{y})=d_1(x_1,y_1)+d_2(x_2,y_2) \quad d_e(\mathbf{x},\mathbf{y})= \sqrt{ (d_1(x_1,y_1)^2+d_2(x_2,y_2)^2) } \quad d_{max}(\mathbf{x},\mathbf{y})= max \{ d_1(x_1,y_1),d_2(x_2,y_2) \} $$

Show that:

If $U_1 \subset X_1$ is an open subset of $(X_1,d_1)$ and $U_2 \subset X_2$ is an open subset of $(X_2,d_2)$, then $U_1 \times U_2$ is an open subset of $(X,d_+)$. What can we say about $(X,d_e)$ and $(X,d_{max})$?

Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ have you verified the functions $d_+, d_e, d_m$ are metrics? $\endgroup$ – Alvin Lepik Aug 1 '17 at 10:41
  • 2
    $\begingroup$ Is $d_{\mathrm{max}}(\mathbf{x}, \mathbf{y})$ supposed to be defined as $\max(d_1(x_1, y_1), d_2(x_2, y_2)\}$? $\endgroup$ – Michael Lee Aug 1 '17 at 10:42
  • 1
    $\begingroup$ What definition of openness you have to work with? $\endgroup$ – Alvin Lepik Aug 1 '17 at 10:42
  • $\begingroup$ @MichaelLee yeah, my bad! Edited thanks. $\endgroup$ – Random-newbie Aug 1 '17 at 10:59
  • $\begingroup$ @AlvinLepik yeah I already checked that. As for the definition of open I am using the following: let $(X,d)$ be a metric space. A subset $U\subset X$ is an open set if, for any $x \in U$ there exists $r>0$ such that $B(x,r) \subset U$ $\endgroup$ – Random-newbie Aug 1 '17 at 10:59
3
$\begingroup$

(I). We want to show that for any $(p_1,p_2)\in U_1\times U_2$ there exists $r_3>0$ such that the open $d_+$ ball $B_{d_+}((p_1,p_2),r_3)$ is a subset of $U_1\times U_2.$

Take $r_1,r_2>0$ such that $B_{d_1}(p_1,r_1)\subset U_1$ and $B_{d_2}(p_2,r_2)\subset U_2.$ Let $r_3=\min (r_1,r_2).$ Then $$(p,p')\in B_{d_+}((p_1,p_2),r_3)\implies$$ $$\implies (\;d_1(p,p_1)<r_3\leq r_1\;\land\; d_2(p',p_2)<r_3\leq r_2\;) \implies$$ $$\implies (\; p\in B_{d_1}(p_1,r_1)\subset U_1 \;\land \;p'\in B_{d_2}(p_2,r_2)\subset U_2\;)\implies$$ $$\implies (p,p')\in U_1\times U_2.$$

Since this holds for all $(p,p')\in B_{d_+}((p_1,p_2),r_3),$ we have $B_{d_+}((p_1,p_2),r_3)\subset U_1\times U_2. $

(II). Metrics $d$ and $d'$ on a set $X$ produce the same topology iff $$(i).\quad \forall p\in X \;\forall r>0 \;\exists s>0 \;(\;B_d(p,s)\subset B_{d'}(p,r)\;),\text { and }$$ $$ (ii). \quad \forall p\in X \;\forall r>0 \;\exists s'>0\;(\;B_{d'}(p,s')\subset B_d(p,r)\;).$$

To get an idea of how to use this to show that $d_e$ and $d_{max}$ generate the same topology as $d_+,$ consider the case $X_1=X_2=\mathbb R,$ with $d_1(x,y)=d_2(x,y)=|x-y|.$ Sketch some pictures of open balls of various radii, centered at some $p\in \mathbb R^2$, with respect to these 3 metrics.

Metrics on a set $X$ that generate the same topology on $X$ are called equivalent metrics.

$\endgroup$
2
$\begingroup$

Hint:

It can be shown that the 3 mentioned metrics on $X_1\times X_2$ (even) induce the same topology.

This on base of:

  • $d_e(\mathbf{x},\mathbf{y})\leq d_+(\mathbf{x},\mathbf{y})$
  • $d_+(\mathbf{x},\mathbf{y})\leq2d_{\text{max}}(\mathbf{x},\mathbf{y})$
  • $d_{\text{max}}(\mathbf{x},\mathbf{y})\leq d_e(\mathbf{x},\mathbf{y})$
$\endgroup$

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.