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For $x=(x_1, x_2, \ldots, x_n), y=(y_1, y_2, \ldots, y_n) \in \mathbb{R}^n.$ Define $d: \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ as $$d(x,y)=|x_1-y_1| + |x_2-y_2| + \cdots + |x_n-y_n|.$$ Show that $d$ is continuous by using definition in Topology, that is $d$ is continuous if for every open subset $V$ of $\mathbb{R},$ the set $d^{-1}(V)$ is open in $\mathbb{R}^n \times \mathbb{R}^n$.

I know that open sets in $\mathbb{R}$ are just intervals, but I don't know what its inverse image under $d$ looks like, so I'm stuck. Is it possible to show continuity using this definition, or should I use the $\epsilon-\delta$ definition?

Thank you!

Mashed Potato
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    Open sets in $\mathbb{R}$ are not just intervals, but arbitrary unions of intervals. Since $d^{-1}(\bigcup_iU_i)=\bigcup_id^{-1}(U_i)$, then it is enough to only check the inverse image of intervals. – logarithm May 16 '19 at 10:35
  • What are the inverse images of intervals? Say if $z \in (a,b)$ for some interval $(a,b)$, it is not clear to me how to get $d^{-1}(z).$ Please help – Mashed Potato May 16 '19 at 10:44
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    FYI, this is the taxicab metric – YuiTo Cheng May 16 '19 at 10:52
  • One simpler approach is to prove that the taxicab metric and the Euclidean metric are equivalent. Then use https://math.stackexchange.com/questions/287285/how-to-prove-the-continuity-of-the-metric-function – YuiTo Cheng May 16 '19 at 11:13
  • If $d_1,...,d_n$ are continuous functions from a domain $D$ into $\Bbb R$ then $d(v)=\sum_{i=1}d_i(v)$ is continuous. So show that each $d_i((x,y))=|x_i-y_i|$ is continuous. – DanielWainfleet May 16 '19 at 12:15
  • If you know that the open set definition is equivalent to the $\epsilon-\delta$ definition --- and the proof of equivalence is not hard at all --- then you should feel free to use whichever of the two definitions works best for the problem at hand. – Lee Mosher May 16 '19 at 17:52
  • How is the topology on $\mathbb R^n$ introduced? (Product topology obtained from the factors $\mathbb R$, Euclidean topology induced by the Euclidean norm, ...) – Paul Frost May 17 '19 at 10:44

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