Understanding Why If a set is open in $\mathbb{R}^n$,  Today in lectures we were doing a brief review of some metric spaces stuff and I'm not quite sure about something we did:
If we define two metrics as $d_1(x,y)= \max_{i=1,\ldots,n} |x_i-y_i|$ and the metric $d_2=\displaystyle\sum_{i=1}^n |x_i-y_i|$ show that a set is open in $(\mathbb{R}^n,d_1)$ iff it is open in $(\mathbb{R}^n,d_2)$
So this is the same in $\mathbb{R}^n$ as in $\mathbb{R}^2$ so for simplicity I'm just looking at it in $\mathbb{R}^2$ just now.
So as $\max_{i=1,2} |x_i-y_i| \leq \displaystyle\sum_{i=1}^2 |x_i-y_i| \leq \sqrt{2\times\max_{i=1,2} |x_i-y_i|^2}=\sqrt{2} \max_{i=1,2} |x_i-y_i|$
Which gives $d_1(x,y)\leq d_2(x,y)\leq\sqrt{2}d_1(x,y)$
So I understand why this shows that any open set in $(\mathbb{R}^2,d_2(x,y))$ is open in $(\mathbb{R}^2,d_1(x,y))$ as we can make a smaller open ball round points but I can't see why this shows the other direction?
Thanks for any help
 A: Suppose $d_1,d_2$ are two metrics on a set, and there is a constant $c>0$ such that for all points $x,y$ we have $d_1(x,y)\le c d_2(x,y)$.  Consider a $d_1$-open ball $B=\{z : d_1(x,z)\le r\}$.  If $z$ is in the $d_2$-open ball of radius $r/c$ about $x$, then $z$ is in $B$.  Thus $B$ is also a $d_2$-open set.  Thus every $d_1$-open ball is a $d_2$ open set, and therefore every $d_1$-open set is a $d_2$-open set.
The same proposition says that if there is also a constant $b>0$ such that for all $x,y$ we have $d_2(x,y)\le b d_1(x,y)$, then every $d_2$-open set is a $d_1$-open set.
The only thing you need to know about $c$ and $b$ is that both are positive.
A: Assume that $md_1\leq d_2 \leq Md_1$ for some $m,M>0$.Define $B_i(x,
\epsilon)=\{y\in\mathbb{R}^n:d_i(x,y)<\epsilon\}\;\;,\;i=1,2$.
Let say that $G$ is open in $(\mathbb{R}^n,d_2)$ and let $x \in G$. Exist $\epsilon>0$ 
s.t. $B_{2}(x,\epsilon))\subseteq G$.
Then $B_1(x,\frac{\epsilon}{M}))\subseteq G$ so $G$ is open in $(\mathbb{R}^n,d_1)$.
Conversely if $G$ is open in $(\mathbb{R}^n,d_1)$ and $x \in G$ exist $\epsilon>0$ 
s.t. $B_{1}(x,\epsilon))\subseteq G$.
Since $B_2(x,m\epsilon)\subseteq G \Longrightarrow G$ is open in $(\mathbb{R}^n,d_2)$.
