Show that the metric $d$ induces the euclidean topology on $\mathbb R^2$ 
Consider $\mathbb R^2$ with a metric given by $d((a_1,a_2),(b_1,b_2))=|a_1 - b_1|+ |a_2 -b_2|$.

I need to show that the metric $d$ induces the euclidean topology on $\mathbb R^2$, but I am not sure how I can show this. Can anyone please help me show this?
 A: You can show that the bases are equivalent.  To do this, you want to show that any open ball in one metric contains an open ball in the other metric and vice versa.  We can do this by showing that for any open ball in one metric centered at a specific point, it contains an open ball in the other metric centered at the same point. 
Let $(a,b)$ be a point in $\mathbb{R}^2$, let $\epsilon > 0$, and let $B_\rho((a,b), \epsilon)$ denote the open ball with radius $\epsilon$ centered on the point $(a,b)$ using the standard Euclidean metric $\rho: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow [0,\infty)$ defined by
$$\rho((x_1, y_1), (x_2,y_2)) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$
for all $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$.
Consider the open ball $B_d((a,b), \epsilon/2)$ using your given metric $d$.  If $(a',b') \in B_d((a,b), \epsilon/2)$, then since $$\rho((a,b), (a',b')) \leq d((a,b),(a',b')) < \epsilon/2 < \epsilon,$$ we have $(a',b') \in B_\rho((a,b), \epsilon)$ implying that $B_d((a,b), \epsilon/2) \subset B_\rho((a,b), \epsilon)$.  
Now consider the open ball $B_\rho((a,b), \epsilon/4)$.  If $(a'',b'') \in B_\rho((a,b), \epsilon/4)$, then since
\begin{align*}
d((a,b),(a'',b'')) &\leq \sqrt{2}\rho((a,b),(a'',b'')) < \sqrt{2}\frac{\epsilon}{4}\\ &=\frac{2^{1/2}\epsilon}{2^2} = \frac{\epsilon}{2^{3/2}} < \epsilon/2,
\end{align*}
we have $(a'',b'') \in B_d((a,b), \epsilon/2)$, implying $B_\rho((a,b), \epsilon/4) \subset B_d((a,b), \epsilon/2)$.
A: Two metrics $d,d'$ on a set $X$ induce the same topology iff for all $x\in X$ and  all  $r>0, $ we have 
(1). there exists $s>0 $ such that $B_d(x,s)\subset B_{d'}(x,r)$,.... and
(2). there exists $s'>0$ such that $B_{d'}(x,s')\subset B_d(x,r).$
On $R^2, $ with $d((a,b),(c,d))=((a-c)^2+(b-d)^2)^{1/2} $ and $d'((a,b),(c,d))=|a-c|+|b-d|, $  we may put $s=r/\sqrt 2$ in (1) and $s'=r/\sqrt 2$ in (2). It helps here to draw a diagram.
In general , when $B,B'$ are bases respectively for topologies $T, T'$ on a set $X, $ suppose that  whenever $x\in b\in B, $ there exists $b'\in B'$ with $x\in b'\subset b.$ Then for any $b\in B$ and any $x\in b, $ $$\text {let }\quad  C(x,b)=\{b'\in B': x\in b'\subset b\}.$$ $$\text {And let }\quad D(b)= \cup_{x\in b}\{C(x,b):x\in b\}.$$ We have $D(b)\subset B'\subset T'$ so $b=\cup D\in T'.$ Since this holds for all $b\in B,$ we have $B\subset T' , $ so $T\subset T'.$
If the preceding paragraph also holds with $B,B'$ interchanged and $T,T'$ interchanged, we also have $T'\subset T.$ Of course $T\subset T'\subset T\implies T=T'.$
In the case where the  topologies $T,T'$ are generated by metrics $d,d'$ we may take $B$ and $B'$ to be the families of all open $d ,  d'$ balls, respectively, which brings us to my first sentence.
