Section 20: Problem 9 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Show that the euclidean metric $d$ on $\mathbb{R}^{n}$ is a metric, as follows: If $\mathbf{x},\mathbf{y}\in\mathbb{R}^{n}$ and $c\in\mathbb{R}$ , define $\begin{eqnarray*} \mathbf{x}+\mathbf{y} & = & (x_{1}+y_{1},\ldots,x_{n}+y_{n}),\\ c\mathbf{x} & = & (cx_{1},\ldots,cx_{n}),\\ \mathbf{x}\cdot\mathbf{y} & = & x_{1}y_{1}+\cdots+x_{n}y_{n}. \end{eqnarray*}$

(a) Show that $\mathbf{x}\cdot(\mathbf{y}+\mathbf{z})=(\mathbf{x}\cdot\mathbf{y})+(\mathbf{x}\cdot\mathbf{z})$ .

(b) Show that $|\mathbf{x}\cdot\mathbf{y}|\le\Vert\mathbf{x}\Vert\Vert\mathbf{y}\Vert$ . [Hint: If $\mathbf{x},\mathbf{y}\neq0$ , let $a=1/\Vert\mathbf{x}\Vert$ and $b=1/\Vert\mathbf{y}\Vert$ , and use the fact that $\Vert a\mathbf{x}\pm b\mathbf{y}\Vert\ge0$ .]

(c) Show that $\Vert\mathbf{x}+\mathbf{y}\Vert\le\Vert\mathbf{x}\Vert+\Vert\mathbf{y}\Vert$ . [Hint: Compute $(\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}+\mathbf{y})$ and apply (b).]

(d) Verify that $d$ is a metric.

(a) Just use the distributive law (and, of course, commutative and associative laws) for the expression on the left hand side.

(b) If $\mathbf{x}$ or $\mathbf{y}$ is zero, then the equality holds, otherwise, using the hint, as well as $\Vert\mathbf{x}\Vert^{2}=\mathbf{x}\cdot\mathbf{x}$ , $\mathbf{x}\cdot\mathbf{y}=\mathbf{y}\cdot\mathbf{x}$ , $(a\mathbf{x})\cdot(b\mathbf{y})=ab(\mathbf{x}\cdot\mathbf{y})$ and (a),

$a^{2}\|\mathbf{x}\|^{2}+b^{2}\|\mathbf{y}\|^{2}\pm2ab(\mathbf{x}\cdot\mathbf{y})=$ $2\pm2ab(\mathbf{x}\cdot\mathbf{y})$ . Therefore, $|\mathbf{x}\cdot\mathbf{y}|\le\tfrac{1}{ab}=\|\mathbf{x}\|\|\mathbf{y}\|$ .

(c) Using the hint and (b), $0\le\|\mathbf{x}+\mathbf{y}\|^{2}=$ $\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}+2(\mathbf{x}\cdot\mathbf{y})\le$ $\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}+2|\mathbf{x}\cdot\mathbf{y}|\le$ $(\|\mathbf{x}\|+\|\mathbf{y}\|)^{2}$ and both are positive.

(d) $d(\mathbf{x},\mathbf{y})\ge0$ and is $0$ iff $\mathbf{x}=\mathbf{y}$ . The triangle inequality holds due to (c) because $d(\mathbf{x},\mathbf{y})=\Vert\mathbf{x}-\mathbf{y}\Vert$ .

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Section 20: Problem 9 Solution

Section 20: Problem 9 Solution