# 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 $\tau_2 \subset \tau_1$.

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) $$\tau_2 \subset \tau_1$$

• did you Typo one of the metrics in (ii)? – Riquelme Jul 11 at 17:27
• You should provide extra context. What have you tried? What difficulties did you have? Otherwise, it just looks like you're giving us your homework. – Paulo Mourão Jul 11 at 17:28
• Sorry, there was a typo on (ii) i corrected now, What I tried was to use the inequalities from (i) to reach $d_1(x,y) < r' \implies d_2(x,y) < r$, but it got me nowhere. – MrBr Jul 11 at 17:47

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.

• While I can see the implication (ii) $\implies$ (iii), it looks, for me, that (iii) does not implies (ii), because while the proposition (iii) give us information about open subsets, the proposition (ii) gives us information about any $x \in X$. There is something that I am missing? (Supose that I want to prove (ii) for a $x$ that it's not in any open subset, how would the proposition (iii) help me?) – MrBr Jul 11 at 18:45
• If (iii) holds and we have $x \in X$, and $r>0$ then $B:=B_{d_2}(x,r)$ is well-known to be open under $d_2$, so $B \in \tau_2$ and by (iii) $B \in \tau_1$ and as $x \in B$ and $B$ is $\tau_1$-open $r'>0$ must exist as stated and (ii) is shown. – Henno Brandsma Jul 11 at 19:12

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 $$\tau_2 \subseteq \tau_1$$.