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Consider $X \times Y$ for metric spaces $X$ and $Y.$

I'm trying to show that $d_E(p,q) = \sqrt{d_X(p,q)^2 + d_Y(p,q)^2}$ fulfills the triangle inequality, but I can't workout the algebra so that things fall into place and it is clear that $$d_E(p_1,p_3) \leq d_E(p_1,p_2) + d_E(p_2,p_3).$$

I avoided writing out everything that I have written on paper because it is messy, but here is a little bit:

Consider $$d_E(p,p'') = \sqrt{d_X(p,p'')^2 + d_Y(p,p'')^2}$$ and $$d_E(p,p') = \sqrt{d_X(p,p')^2 + d_Y(p,p')^2}$$ and $$d_E(p',p'') = \sqrt{d_X(p',p'')^2 + d_Y(p',p'')^2}.$$ It must be shown that $$d_E(p,p'') \leq d_E(p,p') + d_E(p',p''),$$ or that $$d_E(p,p'')^2 \leq (d_E(p,p') + d_E(p',p''))^2.$$

I then proceeded to expand it out via the square, and it became messy and not, in the end, clear that the left was less than or equal to the right.

Rafael Vergnaud
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  • What you have doesn't make sense to start with. Is $p\in X$? Is $p\in Y$? Is $p\in X\times Y$? If $X$ and $Y$ are different metric spaces how can $p$ and $q$ be in both to measure their distance? – Matt Oct 09 '18 at 19:37
  • Sorry. Let me make an edit. – Rafael Vergnaud Oct 09 '18 at 19:38
  • I see your point. I suppose I was considering as a model for visualization R^2. Then, would not $d_X(p,p)$ just be the distance between the points considering only the x-axis? – Rafael Vergnaud Oct 09 '18 at 19:42
  • I will copy verbatim from the book: Define a metric on the Cartesian product $M = X \times Y$ of two metric spaces $X$ and $Y.$ Consider the metric $d_E(p,p') = \sqrt{d_X(x,x')^2+d_Y(y,y')^2},$ where $p = (x,y)$ and $p' = (x',y')$ belong to $M.$ I realize that what I wrote is ambiguous, but what i mean by $d_X(p,p')$ is $d_X(p_X,p'_X).$ – Rafael Vergnaud Oct 09 '18 at 19:51
  • That is, the distance in $X$ of the $X$ components of $p$ and $p'.$ – Rafael Vergnaud Oct 09 '18 at 19:52
  • https://math.stackexchange.com/questions/1052511/prove-that-the-product-space-is-a-metric-space – Matt Oct 09 '18 at 20:11
  • In essence it's the Cauchy-Schwartz inequality. – Henno Brandsma Oct 09 '18 at 20:12

1 Answers1

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Note that $d_X(x_1,x_3)≤d_X(x_1,x_2)+d_X(x_2,x_3)$ and $d_Y(y_1,y_3)≤d_Y(y_1,y_2)+d_Y(y_2,y_3)$. Let us put $r_1=d_X(x_1,x_2),s_1=d_X(x_2,x_3),r_2=d_Y(y_1,y_2),s_2=d_Y(y_2,y_3)$.

Note that, $[(r_1+s_1)^2+(r_2+s_2)^2]^\frac{1}{2}≤(r_1^2+r_2^2)^\frac{1}{2}+(s_1^2+s_2^2)^\frac{1}{2}$(expand both sides to reduce shorter form and then apply cauchy-schwarz). Hence $d_E((x_1,y_1),(x_3,y_3))≤[(d_X(x_1,x_2))^2+(d_Y(y_1,y_2))^2]^\frac{1}{2} +[(d_X(x_2,x_3))^2+(d_Y(y_2,y_3))^2]^\frac{1}{2}=d_E((x_1,y_1),(x_2,y_2))+d_E((x_2,y_2),(x_3,y_3))$.

Sumanta
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