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.

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.

  • $\begingroup$ Yes, you are on the wrong track. Instead of using algebra, try geometry, namely, use Pythagorean theorem in the plane and the fact that the Euclidean metric on the plane is a metric. $\endgroup$ – Moishe Kohan Sep 14 '14 at 16:58
  • $\begingroup$ @studiosus I'm not sure I understand your hint. I would like to show it for general sets $X$ and $Y$, and not restricting myself to $\mathbb{R} \times \mathbb{R}$. Are we then still allowed to use Pythagorean theorem? $\endgroup$ – Hunter Sep 14 '14 at 17:05

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 $d_Y$.

  • 1
    $\begingroup$ What a nice proof, thanks! Small suggestion: I think in the first equation on the LHS $d_Y(x_1,x_3)$ should be $d_Y(y_1,y_3)$. $\endgroup$ – Hunter Sep 14 '14 at 17:18
  • $\begingroup$ @Hunter: you are right, updated. $\endgroup$ – Jack D'Aurizio Sep 14 '14 at 17:27
  • $\begingroup$ Sir, can we direct apply just Minkowski inequality for real numbers on $d_2((x_1,y_1),(x_2,y_2))$ ? Because $d_X(x_1,x_2), d_Y(y_1,y_2)$ are distances and hence by definition of metric they are positive real numbers. So that triangle inequality for $d_2$ is directly follows from Minkowski inequality. Sir please reply... $\endgroup$ – Akash Patalwanshi Jul 18 at 4:13

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.

  • $\begingroup$ Thanks for your answer, it is very nice to see the geometric interpretation of product spaces! $\endgroup$ – Hunter Sep 14 '14 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.