Section 20: Problem 1 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

(a) In $\mathbb{R}^{n}$ , define $d'(\mathbf{x},\mathbf{y})=|x_{1}-y_{1}|+\cdots+|x_{n}-y_{n}|.$ Show that $d'$ is a metric that induces the usual topology of $\mathbb{R}^{n}$ . Sketch the basis elements under $d'$ when $n=2$ .

(b) More generally, given $p\ge1$ , define $d'(\mathbf{x},\mathbf{y})=\left[\sum_{i=1}^{n}|x_{i}-y_{i}|^{p}\right]^{1/p}$ for $x,у\in\mathbb{R}^{n}$ . Assume that $d'$ is a metric. Show that it induces the usual topology on $\mathbb{R}^{n}$ .

(a) The basis elements when $n=2$ are squares turned by 45 degrees (right angle "rhombuses"). The fact that it is a metric follows from the inequality on page 122. Further, based on Theorem 20.3, to show that $d'$ induces the usual topology on $\mathbb{R}^{n}$ we can show that $d'$ induces the same topology as $\rho$ . Indeed, using Lemma 20.2, every such “rhombus” of size $\epsilon$ contains a “square” of size $\epsilon/n$ , because $d'(\mathbf{x},\mathbf{y})\le n\rho(\mathbf{x},\mathbf{y})$ , and every “square” of size $\epsilon$ contains a “rhombus” of size $\epsilon$ , because $\rho(\mathbf{x},\mathbf{y})\le d'(\mathbf{x},\mathbf{y})$ .

(b) Similarly, it follows from $\rho\le d'\le n^{1/p}\rho$ .

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Section 20: Problem 1 Solution

Section 20: Problem 1 Solution