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I have been asked to show whether a number of sets are metric spaces (according to the definition in Baby Rudin), but I couldn't prove the following question:

$R^2$ with the metric $d((a, b), (c, d)) := (|a - c|^{3/2} + |b-d|^{3/2})^{2/3}$

Note that all I am concerned with is the Triangle Inequality property. The others are fairly intuitive.

Darío G
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user3495690
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  • @EricTowers I made an edit. I managed to solve ii. – user3495690 Nov 01 '16 at 18:08
  • Then the question still stands. What have you tried? How does your solution to ii not inform your solution to i? – Eric Towers Nov 01 '16 at 18:09
  • @EricTowers I can't see how my proof to i informs my solution to i.. I've tried to solve i a number of times (probably hundreds of different ways) but haven't arrived at any solutions, yet. – user3495690 Nov 01 '16 at 18:16
  • You might want to take a look at this: http://math.stackexchange.com/questions/447747/how-to-prove-triangle-inequality-for-p-norm – Darío G Nov 01 '16 at 18:25
  • @Wore Thanks for the suggestion, I've already looked at that question; however, the original ii was: $R^2$ with the metric $d((a, b), (c, d)) := (|a - c|^{1/2} + |b-d|^{1/2})^{2}$. Which is kind of the "reverse" of the p-norm (unless the p-norm also works with fractions, which I've never seen before..) – user3495690 Nov 01 '16 at 18:36
  • Please note that when I say "reverse" I'm referring to the exponents – user3495690 Nov 01 '16 at 18:37
  • The usual $p$-norm is defined as $|(x_1,\ldots,x_n)|p=\left(\sum{k=1}^n |x_i|^p\right)^{1/p}$, for $p\geq 1$. In your question, $p=3/2>1$. – Darío G Nov 01 '16 at 18:44
  • @Wore Okay, I've been trying to relate that proof to my solution all day, but I still can't manage.. I'm having trouble relating it to $R^2$, in that I have to show: $(|a-c|^{3/2} + |b-d|^{3/2})^{2/3} <= (|a-e|^{3/2} + |b-f|^{3/2})^{2/3} + (|e-c|^{3/2} + |f-d|^{3/2})^{2/3}$ – user3495690 Nov 02 '16 at 19:07
  • That is, $d((a, b), (c,d)) <= d((a, b), (e, f)) + d((e, f), (c, d))$ – user3495690 Nov 02 '16 at 19:10

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