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 $\tau_2 \subset \tau_1$.

Question

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) $\tau_2 \subset \tau_1$

Answer 1

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.

Answer 2

Notice that $(i)$ implies for a sequence $(x_n)_n$ in $X$ if $x_n \xrightarrow{d_1} x$ then $x_n \xrightarrow{d_2} x$.

Now for every closed set $A$ in $\tau_2$ we have \begin{align} A &\subseteq \overline{A}^{\tau_1} \\ &= \{d_1-\lim_{n\to\infty} x_n : (x_n)_n \text{ is a $d_1$-convergent sequence in } A\}\\ &\subseteq \{d_2-\lim_{n\to\infty} x_n : (x_n)_n \text{ is a $d_2$-convergent sequence in } A\}\\ &= \overline{A}^{\tau_2} \\ &= A \end{align} so $A$ is closed in $\tau_1$. Taking complements gives $\tau_2 \subseteq \tau_1$.