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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.

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

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