# If there exists a $c>0$ that for any $x,y \in X$ we have $c \cdot d_1(x,y) \geq d_2(x,y)$ then $\tau_2 \subset \tau_1$.

by MrBr   Last Updated July 11, 2019 19:20 PM

Prove that if $$d_1$$ and $$d_2$$ are metrics over $$X$$, $$\tau_1$$ and $$\tau_2$$ are the family of open subsets of their respective metric spaces then:

(i) $$\implies$$ (ii) $$\iff$$ (iii)

Where:

(i) There exists a $$c>0$$ that for any $$x,y \in X$$ we have $$c \cdot d_1(x,y) \geq d_2(x,y)$$

(ii) For any $$x \in X$$ and $$r > 0$$, there exists a $$r'>0$$ where $$B_{d_1}(x,r') \subset B_{d_2}(x,r)$$

(iii) $$\tau_2 \subset \tau_1$$

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If (i) holds, then for given $$x \in X$$ and $$r>0$$, define $$r'=\frac{r}{c}$$.

Then if $$y \in B_{d_1}(x,r')$$ then $$d_1(x,y) < r'$$. Also by (i): $$d_2(x,y) \le c\cdot d_1(x,y) < c \cdot r' = c \cdot \frac{r}{c}=r$$

so that $$y \in B_{d_2}(x,r)$$ , showing the inclusion. and so $$\text{(i)} \implies \text{(ii)}$$ holds.

The equivalence of (ii) and (iii) is quite obvious from the definitions. What does $$O \in \tau_2$$ mean? Also recall that open balls are open sets in their induced topologies.

Henno Brandsma
July 11, 2019 17:41 PM