We have $ f :R^2\to R$ be defined by $f(x,y)=x+y$. Let $R^2$ have the taxicab metric and let $R $ have the usual metric. Show that $f $ is continuous.

My try:To prove this I have to show that inverse image of open set of $R$ is open in $R^2$.But the problem is that I don't know about taxicab metric.Thank you.

  • $\begingroup$ The basic open sets look like squares $\endgroup$ – MatheMagic Feb 12 '17 at 16:58

For any vector $v\in\Bbb R^2$, $\|v\|_2\le\|v\|_1\le2\|v\|_2$, where $\|\cdot\|_1$ is the taxicab norm and $\|\cdot\|_2$ is the Euclidean norm.

This implies that $\Bbb R^2$ with the taxicab metric has the same topology as with the Eucliedan metric. Then $f$ is continuous.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.