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Let $ d(x,y)= d_1 (x,y) + d_2 (x,y) $ and $ \delta (x,y) = \max(d_1 (x,y), d_2 (x,y)) $ and let $ d_1 $ and $ d_2 $ be metrics on $ M $.

Show that $ d $ and $ \delta $ are finer than $d_1$ and $d_2$. Also, show that $d_1 \sim d_2$ iff $ d, d_1 , d_2 , \delta $ are equivalent between themselves.

I showed that both $d$ and $\delta$ are metrics on $M$ and that $d$ and $\delta$ are finer than $d_1$ and $d_2$.
Not sure how to prove the iff part.

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    $\begingroup$ Hint: you can get $\sim$ with \sim. Did you pehaps mean $d_1 \sim d_1$ iff $d \sim \delta$? $\endgroup$ – mechanodroid Sep 19 '17 at 16:26
  • $\begingroup$ Thanks. Searched for the correct code but found \somethingtilde that didn't work. Kind of. I meant $d_{1} \sim d_{2}$ iff all of the four metrics are equivalent between them. So, $d \sim \delta$ and all other combinations I guess. Also I'm not sure if I proved correctly that $d$ and $\delta$ are finer than $d_{1}$ and $d_{2}$. $\endgroup$ – V. M. Sep 19 '17 at 16:37
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    $\begingroup$ Yeah, you wrote it correctly, since $d_1 \sim d_2$ iff $d \sim \delta$ iff they are all equivalent. $\endgroup$ – mechanodroid Sep 19 '17 at 16:39
  • $\begingroup$ Any idea on how to approach this question? $\endgroup$ – V. M. Sep 19 '17 at 16:42
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    $\begingroup$ Theorem: For metric spaces $(X_1,d_1), (X_2,d_2)$ and continuous $f:X_1\to X_2,$ the metric $e(u,v)=d_1(u,v)+d_2(f(u),f(v) )$ is equivalent to $d_1.$ In particular if $X_1=X_2=M$ and $d_1\sim d_2$ then $id_M$ is continuous from $(X_1,d_1)$ to $(X_2,d_2)$ so $e=d_1+d_2=d\sim d_1$. $\endgroup$ – DanielWainfleet Sep 19 '17 at 18:47
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Well, to show that $d$ and $\delta$ are finer than $d_1, d_2$ it is sufficient to notice $d_1, d_2 \le d, \delta$ since then we have:

$$B_d(x, r), B_\delta(x, r) \subseteq B_{d_1}(x, r), B_{d_2}(x, r)$$

To see this, for example take $y \in B_d(x, r)$. Then we have $d_1(x, y) \le d(x, y) < r$, thus $y \in B_{d_1}(x, r)$. The other inclusions are similar.

Now assume $d_1 \sim d_2$.

Take an arbitrary ball $B_{d_1}(x, r)$.

Since $d_1\sim d_2$, there exists $r' > 0$ such that $B_{d_1}(x, r) \subseteq B_{d_2}(x, r')$.

So, if $d_1(x, y) < r$, we have $d_2(x, y) < r'$, so $d(x, y) < r + r'$ and $\delta(x, y) < \max\{r,r'\}$.

Thus, $B_{d_1}(x, r) \subseteq B_{d}(x, r + r')$ and $B_{d_1}(x, r) \subseteq B_\delta(x, \max\{r,r'\})$.

Similarly, take the ball $B_{d_2}(x, r)$. Since $d_1\sim d_2$, there exists $r'' > 0$ such that $B_{d_2}(x, r) \subseteq B_{d_1}(x, r'')$.

So, if $d_1(x, y) < r$, we have $d_2(x, y) < r''$, so $d(x, y) < r + r''$ and $\delta(x, y) < \max\{r,r''\}$.

Thus, $B_{d_2}(x, r) \subseteq B_{d}(x, r + r'')$ and $B_{d_1}(x, r) \subseteq B_\delta(x, \max\{r,r''\})$.

Thus, the open balls of $d$ and $\delta$ nest with the open balls of $d_1$ and $d_2$ so all the metrics in question are equivalent.

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