Prove that $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$, where $D_a = \{ (x,y) \in \mathbb{R}^2\ | \ |x-y| < a \}$

Question

Suppose $r$ and $s$ are two positive real numbers. Let $D_r = \{ (x,y) \in \mathbb{R}^2\ | \ |x-y| < r \}$ and $D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y| < s \}$. Prove that $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$. Well, I only proved that $D_r \circ D_s \subseteq \{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \}$ by letting $(x,y)$ be an arbitrary element of $D_r \circ D_s$ and using the triangle inequality. How can I prove that $\{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \} \subseteq D_r \circ D_s$?

Answer 1

Let $(x,y) \in D_{r+s}$. Let

$$z = \frac{r}{r+s}x + \frac{s}{r+s}y.$$

Then

$$\begin{align} \lvert z - x\rvert &= \left\lvert\frac{s}{r+s}y + \left(\frac{r}{r+s}-1\right)x \right\rvert = \frac{s}{r+s}\lvert y-x\rvert < s\\ \lvert z - y\rvert &=\left\lvert \frac{r}{r+s}x + \left( \frac{s}{r+s}-1\right)y\right\rvert = \frac{r}{r+s}\lvert x-y\rvert < r \end{align}$$

So $(x,z) \in D_s$ and $(z,y) \in D_r$, hence $(x,y) \in D_r \circ D_s$.