Consider the following subsets of the plane: $$C_{1} = \left\{(x, y) : x > 0,\ y = \frac{1}{x}\right\}$$ and $$C_{2} = \left\{(x, y) : x < 0,\ y = -1 + \frac{1}{x}\right\}$$ Given any two points $P = (x, y)$ and $Q = (u, v)$ of the plane, their distance $d(P, Q)$ is defined by $$d(P, Q) = \sqrt{(x - u)^{2} + (y - v)^{2}}$$ Show that there exists a unique choice of points $P_{0} \in C_{1}$ and $Q_{0} \in C_{2}$ such that $$d(P_{0}, Q_{0}) \leq d(P, Q) \quad \text{for all } P \in C_{1} \text{ and } Q \in C_{2}.$$
Consider the following subsets of the plane:
$$C_{1} = \left\{(x, y) : x > 0,\ y = \frac{1}{x}\right\}$$
and
$$C_{2} = \left\{(x, y) : x < 0,\ y = -1 + \frac{1}{x}\right\}$$
Given any two points $P = (x, y)$ and $Q = (u, v)$ of the plane, their distance $d(P, Q)$ is defined by
$$d(P, Q) = \sqrt{(x - u)^{2} + (y - v)^{2}}$$
Show that there exists a unique choice of points $P_{0} \in C_{1}$ and $Q_{0} \in C_{2}$ such that
$$d(P_{0}, Q_{0}) \leq d(P, Q) \quad \text{for all } P \in C_{1} \text{ and } Q \in C_{2}.$$