# Let D be the usual Euclidean metric on M. Are the two identity functions $I_1$ and $I_2$ continuous functions where d is the Manhattan metric on M?

Let D be the usual Euclidean metric on M. Are the two identity functions $I_1:(M,D) \to (M,d)$ and $I_2:(M,d) \to (M,D)$ continuous functions where d is the Manhattan metric on M?

The definition of continuous for a metric spaces is

Let $f:(M,d) \to (N,D)$ for two metric spaces. Then f is continuous at $x=a\in M$ if for each $\epsilon >0$ $\exists \delta_\epsilon>0$ such that when $d(x,a)<\delta_\epsilon$ we have $D(f(x),f(a)) <\epsilon$.

The Euclidean metric distance function is defined as $d(x,a)=\sqrt{(x-a)^2}=|x-a|$. The Manhanttan metric distance function is defined as $d((x_1,y_1),(x_2,y_2))=|x_1-x_2|+|y_1-y_2|$.

What I am having difficulties with is combining both of these ideas to prove the same above?

• Your Manhattan metric is not a metric. Probabaly you should flip $y_1$ & $x_2$. – user152715 Oct 5 '14 at 22:23
• You tagged your question as general topology, so I'm guessing you know the basics of that subject. Can I ask if you know how to answer the following question? "If $\tau_1$ and $\tau_2$ are two topologies on a set $X$ such that the two identity maps $(X,\tau_1) \to (X, \tau_2)$ and $(X,\tau_2) \to (X ,\tau_1)$ are continuous, what must be true about $\tau_1$ and $\tau_2$?" – Mike F Oct 5 '14 at 22:34
• @MikeF They are inverse of each other – Username Unknown Oct 6 '14 at 2:44
• More than that, the above implies that the identity map is a homeomorphism between $(X,\tau_1)$ and $(X,\tau_2)$, or in other words, that $\tau_1 = \tau_2$. So, your question is equivalent to the question: "do the Manhattan metric and the Euclidean metric generate the exact same topology?" – Mike F Oct 6 '14 at 12:49

Hint: Just show $\exists \epsilon_1,\epsilon_2,\epsilon_3$ such that $B_d(x,\epsilon_1)\subseteq B_D(x,\epsilon_2)\subseteq B_d(x,\epsilon_3)$ then you are done.
Let $(x_1,y_1)\in B_d((x,y), \epsilon)$ $\Rightarrow d((x,y),(x_1,y_1))<\epsilon\Rightarrow \sqrt {(x-x_1)^2+(y-y_1)^2}<\epsilon\Rightarrow (x-x_1)^2+(y-y_1)^2<\epsilon^2\Rightarrow(x-x_1)^2<\epsilon^2 , (y-y_1)^2<\epsilon^2\Rightarrow|(x-x_1)|<\epsilon , |(y-y_1)|<\epsilon$
Conversely, Let $(x_1,y_1)\in B_D((x,y), \epsilon)\Rightarrow |(x-x_1)|+|(y-y_1)|<\epsilon\Rightarrow|(x-x_1)|<\epsilon , |(y-y_1)|<\epsilon\Rightarrow (x-x_1)^2+(y-y_1)^2<2\epsilon^2\Rightarrow \sqrt {(x-x_1)^2+(y-y_1)^2}<\sqrt2\epsilon$. Now choosing of $\epsilon$ is in our hand. So we are done.