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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!

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    $\begingroup$ have you verified the functions $d_+, d_e, d_m$ are metrics? $\endgroup$
    – AlvinL
    Commented Aug 1, 2017 at 10:41
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    $\begingroup$ 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 L.
    Commented Aug 1, 2017 at 10:42
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    $\begingroup$ What definition of openness you have to work with? $\endgroup$
    – AlvinL
    Commented Aug 1, 2017 at 10:42
  • $\begingroup$ @MichaelLee yeah, my bad! Edited thanks. $\endgroup$ Commented Aug 1, 2017 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 $B(x,r) \subset U$ $\endgroup$ Commented Aug 1, 2017 at 10:59

2 Answers 2

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(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.

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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:

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