Product metric spaces is again a metric space

Question

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let: $$ d_2 ((x_1,y_1),(x_2,y_2)) = \left[d_X(x_1,x_2)^2 + d_Y (y_1,y_2)^2 \right]^{\frac{1}{2}} $$ for the points $(x_1,y_1)$ and $(x_2,y_2)$ in $X \times Y$. I would like to show that the triangle inequality holds, that is: $$ d_2 ((x_1,y_1),(x_3,y_3)) \leq d_2 ((x_1,y_1),(x_2,y_2)) + d_2 ((x_2,y_2),(x_3,y_3)) \tag{1} $$ but I'm really stuck.

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What I have tried so far:

Since $d_X$ and $d_Y$ are by definition metrics, we can use the triangle inequality to write: \begin{equation} \begin{aligned} d_2 ((x_1,y_1),(x_2,y_2)) & + d_2 ((x_2,y_2),(x_3,y_3)) \\& = \left[d_X(x_1,x_2)^2 + d_Y(y_1,y_2)^2 \right]^{\frac{1}{2}} + \left[d_X(x_2,x_3)^2 + d_Y(y_2,y_3)^2 \right]^{\frac{1}{2}} \\& \leq \left[[d_X(x_1,x_3)+d_X(x_3,x_2)]^2 + [d_Y(y_1,y_3)+d_Y(y_3,y_2)]^2 \right]^{\frac{1}{2}} \\& + \left[[d_X(x_2,x_1)+d_X(x_1,x_3)]^2 + [d_Y(y_2,y_1)+d_Y(y_1,y_3)]^2 \right]^{\frac{1}{2}} \end{aligned} \end{equation} where $(x_3,y_3) \in X \times Y$, and we know that: \begin{equation} \begin{aligned} d_2 ((x_1,y_1),(x_3,y_3)) = \left[ d_X(x_1,x_3)^2 + d_Y(y_1,y_3)^2 \right]^{\frac{1}{2}} \end{aligned} \end{equation} Then, substituting the above two equations into equation $(1)$ yields: \begin{equation} \begin{aligned} \Bigl[ d_X(x_1,x_3)^2 & + d_Y(y_1,y_3)^2 \Bigr]^{\frac{1}{2}} \\& \leq \left[[d_X(x_1,x_3)+d_X(x_3,x_2)]^2 + [d_Y(y_1,y_3)+d_Y(y_3,y_2)]^2 \right]^{\frac{1}{2}} \\& + \left[[d_X(x_2,x_1)+d_X(x_1,x_3)]^2 + [d_Y(y_2,y_1)+d_Y(y_1,y_3)]^2 \right]^{\frac{1}{2}} \end{aligned} \end{equation} I'm not really sure how to proceed from here onwards (or perhaps I'm already on the wrong track), and any help would be much appreciated.

Answer 1

You have to prove: $$\begin{eqnarray*}d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2 &\leq& d_X(x_1,x_2)^2+d_X(x_2,x_3)^2+d_Y(y_1,y_2)^2+d_Y(y_2,y_3)^2\\&+&2\sqrt{\left(d_X(x_1,x_2)^2+d_Y(y_1,y_2)^2\right)\left(d_X(x_2,x_3)^2+d_Y(y_2,y_3)^2\right)}\end{eqnarray*}$$ where the Cauchy-Schwarz inequality ensures that the last square root is greater or equal than: $$ d_X(x_1,x_2)\,d_X(x_2,x_3)+d_Y(y_1,y_2)\,d_Y(y_2,y_3)$$ hence it is sufficient to show that: $$d_X(x_1,x_3)^2+d_Y(y_1,y_3)^2\leq\left(d_X(x_1,x_2)+d_X(x_2,x_3)\right)^2+\left(d_Y(y_1,y_2)+d_Y(y_2,y_3)\right)^2$$ that just follows from the triangle inequality for $d_X$ and $d_Y$.

Answer 2

Here is the geometric argument I alluded to:

Suppose you have a triple of points $$ z_1=(x_1, y_1), z_2=(x_2, y_2), z_3=(x_3, y_3)\in X\times Y. $$ Then both triples of distances $$ d(x_i, x_j), 1\le i\ne j\le 3; d(y_i, y_j), 1\le i\ne j\le 3 $$ satisfy triangle inequalities. Therefore, we can realize them by triangles $\Delta(x_1', x_2', x_3'), \Delta(y_1', y_2', y_3')$ in the Euclidean plane $E^2$.

Now, define the triangle $\Delta'=\Delta(z_1', z_2', z_3')\subset E^2\times E^2=E^4$ whose vertices project to the vertices of the triangles $\Delta(x_1', x_2', x_3'), \Delta(y_1', y_2', y_3')$. By the definition of the product metric on $X\times Y$ and the Pythagorean formula, the side-lengths of the triangle $\Delta'$ are the same as $d(z_1, z_2), d(z_2, z_3), d(z_3, z_1)$. Since $E^4$ is a metric space, the distances $d(z_1, z_2), d(z_2, z_3), d(z_3, z_1)$ satisfy the triangle inequalities.