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.