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)


(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) $\tau_2 \subset \tau_1$

  • $\begingroup$ did you Typo one of the metrics in (ii)? $\endgroup$ – Riquelme Jul 11 at 17:27
  • $\begingroup$ 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. $\endgroup$ – Paulo Mourão Jul 11 at 17:28
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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?) $\endgroup$ – MrBr Jul 11 at 18:45
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    $\begingroup$ 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. $\endgroup$ – 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 $\tau_2 \subseteq \tau_1$.


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