Notes - Metric Spaces MT23, Product spaces
Flashcards
Suppose we have two metric spaces $(X, d _ X)$ and $(Y, d _ Y)$. How is the product space $(X \times Y, d)$ defined?
\[d((x_1, y_1), (x_2, y_2)) = \sqrt{d_X(x_1, x_2)^2 + d_Y(y_1, y_2)^2}\]
Proofs
Quickly prove that if we have two metric spaces $(X, d _ X)$ and $(Y, d _ Y)$, then the product space $(X \times Y, d)$ given by the metric
\[d((x_1, y_1), (x_2, y_2)) = \sqrt{d_X(x_1, x_2)^2 + d_Y(y_1, y_2)^2}\]
defines a metric space.
Symmetry and positive definiteness is easy. We want to show
\[d_{X \times Y} ((x_1, y_1), (x_2, y_2)) \le d_{X \times Y} ((x_1, y_1), (x_3, y_3)) + d_{X \times Y} ((x_3, y_3), (x_2, y_2))\]Expanded out:
\[\sqrt{d_X(x_1, x_3)^2 + d_Y(y_1, y_3)^2} + \sqrt{d_X(x_3, x_2)^2 + d_Y(y_3, y_2)^2} \ge \sqrt{d_X(x_1, x_2)^2 + d_Y(y_1, y_2)^2}\]Then write:
\[\begin{aligned} a_1 &= d(x_2, x_3) \\\\ a_2 &= d(x_1, x_3) \\\\ a_3 &= d(x_1, x_2) \\\\ b_1 &= d(y_2, y_3) \\\\ b_2 &= d(y_1, y_3) \\\\ b_3 &= d(y_1, y_2) \end{aligned}\]So it becomes:
\[\sqrt{a_2^2 + b_2^2} + \sqrt{a_1^2 + b_1^2} \ge \sqrt{a_3^2 + a_3^2}\]By the ordinary triangle inequality, we have
\[\begin{aligned} &a_3 \le a_1 + a_2 \\\\ &b_3 \le b_1 + b_2 \end{aligned}\]Squaring these gives
\[\begin{aligned} &a_3^2 \le a_1^2 + a_2^2 + 2a_1a_2 \\\\ &b_3^2 \le b_1^2 + b_2^2 + 2b_1 b_2 \end{aligned}\]Adding these
\[a_3^2 + b_3^2 + 2(a_1 a_2 + b_1 b_2) + a_1^2 + b_1^2 \le a_2^2 + b_2^2\]Then by Cauchy-Schwarz,
\[a_1 a_2 + b_1 b_2 \le \sqrt{a_1^2 + b_1^2} \sqrt{a_2^2 + b_2^2}\]Subsituting this into
\[a_3^2 + b_3^2 + 2(a_1 a_2 + b_1 b_2) + a_1^2 + b_1^2 \le a_2^2 + b_2^2\]gives the square of
\[\sqrt{a_2^2 + b_2^2} + \sqrt{a_1^2 + b_1^2} \ge \sqrt{a_3^2 + a_3^2}\]and hence we are done.