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I'd like to find a metric in $\mathbb{R}^2$ (Denoted $d$) in which $d((3,3),(4,2))> d((3,3),(3,7))$. Is there such metric?

Adding something. I already have a pseudometric which does that (The one of $d((a,b),(c,d))=\min\{|c-a|,|d-b|\}$.

LeviathanTheEsper
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1 Answers1

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How about $$ d(\langle x,y\rangle,\langle z,w\rangle) = |10^{3-x}-10^{3-z}|+|10^{3-y}-10^{3-w}| $$

Alternatively, $$ d(\langle x,y\rangle,\langle z,w\rangle) = |x-z|+|f(y)-f(w)| $$ where $$ f(y) = \begin{cases} y & \text{for }y>2 \\ y-1000000 & \text{for }y\le 2 \end{cases} $$

hmakholm left over Monica
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  • A priori that seems to be working, (The previous one, without the $3$s too), It would be left for me to show it's a metric. Thanks. – LeviathanTheEsper Feb 22 '15 at 20:14
  • @DavidMolano: That's simple enough; it is the usual Manhattan metric pulled back through the injection $(x,y)\mapsto(10^{3-x},10^{3-y})$. So all of the conditions for being a metric follow from the well-known properties of the Manhattan metric restricted to $(\mathbb R_+)^2$. – hmakholm left over Monica Feb 22 '15 at 20:17