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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 $\{ (x,y) \in \mathbb{R}^2 \ | \ |x-y|<r+s \} \subseteq D_r \circ D_s$?

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What is the definition of $A\circ B$? – Pocho la pantera Sep 1 '13 at 12:50
Suppose $D_r$ and $D_s$ are both relations on $\mathbb{R}$. Then $D_r \circ D_s = \{ (x,y) \in \mathbb{R}^2 \ | \ \exists z \in \mathbb{R}((x,z) \in D_s \text{ and } (z,y) \in D_r) \}$ – Stavros Sep 1 '13 at 12:56

1 Answer 1

up vote 4 down vote accepted

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

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


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

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

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Nice answer! Any tip on how you came up with this $z$? – Stavros Sep 1 '13 at 13:13
You want a point between $x$ and $y$, so that's a convex combination of $x$ and $y$, $z_t = x + t\cdot(y-x) = y - (1-t)(x-y),\: t \in [0,1]$. You want the point close enough to $x$, so $t\lvert y-x\rvert < s$ and close enough to $y$, so $(1-t)\lvert x-y\rvert < r$. You know $\lvert x-y\rvert < r+s$, so trying $t(r+s) = s$ is a natural choice. – Daniel Fischer Sep 1 '13 at 13:28
Really elegant solution! This part of the problem got me stumped way more than any other exercise preceding this one (in Velleman's book)... Nice to find help here. – Amjad Jan 23 at 17:20

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