Help with proving triangle law to prove these distance functions are metric spaces

Question

Let $(X, d_x)$ and $(Y, d_y)$ be metric spaces. I was given distance functions:

$d_1((x_1,y_1),(x_2,y_2)) = d_x(x_1, x_2)+d_y(y_1,y_2)$ $d_\infty ((x_1,y_2), (x_2, y_2) = max \{d_x(x_1, x_2), d_y((y_1, y_2)\}$.

I'm trying to prove that these are metric spaces by proving they meet the three axioms (positivity, symmetry and triangle law). The first two conditions are quite trivial. I'm having difficulty proving the third one, where $d(x,y) \leq d(z,y) + d(x,z)$. Help would be very much appreciated.

I don't understand your first definition. Should that be a $+$ instead of a comma? — Alexis Olson, Oct 06 '16 at 03:13
For the second one, this has been posted before: http://math.stackexchange.com/questions/1020808/maximum-of-two-metrics-is-a-metric — Alexis Olson, Oct 06 '16 at 03:15
@AlexisOlson Right you are. Thanks for pointing that out! And also giving the link to the second one. — Nikitau, Oct 06 '16 at 03:20

score 1 · Accepted Answer · answered Oct 06 '16 at 03:32

It's just a matter of writing out all the notation, starting from the complicated part and manipulating it into the simpler part. The proof itself is pretty straightforward. Given any $x_1, x_2, x_3 \in X$ and any $y_1, y_2, y_3 \in Y$, notice that: \begin{align*} ~~&d_1((x_1, y_1), (x_3, y_3)) + d_1((x_3, y_3), (x_2, y_2)) \\ &= [d_x(x_1, x_3) + d_y(y_1, y_3)] + [d_x(x_3, x_2) + d_y(y_3, y_2)] \\ &= [d_x(x_1, x_3) + d_x(x_3, x_2)] + [d_y(y_1, y_3) + d_y(y_3, y_2)] \\ &\geq d_x(x_1, x_2) + d_y(y_1, y_2) \\ &= d_1((x_1, y_1), (x_2, y_2)) \end{align*} as desired. $~~\blacksquare$

Ah, thank you so much! For some reason I was "stuck" trying to justify line 3 and 4 ... and then I realized I was being very silly. Thank you! — Nikitau, Oct 06 '16 at 03:54

Help with proving triangle law to prove these distance functions are metric spaces

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