Section 7: Problem 2 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 maps $f$ and $g$ of Examples 1 and 2 are bijections.

Example 1. $f$ is surjective as for every $m\in\mathbb{Z}_{+}$ , either $m=2k$ , $k\ge1$ , and $f(k)=2k$ , or $m=2k-1$ , $k\ge1$ , and $f(-k+1)=2k-1$ , and $f$ is injective, as for every $m,n\in\mathbb{Z}$ , $2m\neq-2n+1$ , $2m=2n\Rightarrow m=n$ , and $-2m+1=-2n+1\Rightarrow m=n$ .

Example 2. $f$ is surjective as if $a,b\in\mathbb{Z}_{+}$ and $a\ge b$ then $f(a-b+1,b)=(a,b)$ where $a-b+1\in\mathbb{Z}_{+}$ , and $f$ is injective as for every $x,y\in\mathbb{Z}_{+}$ , if $x+y-1=x'+y'-1$ and $y=y'$ then $x=x'$ .

$g$ is injective, as for every $x,y\in\mathbb{Z}_{+}$ such that $1\le y\le x$ , $g(x,1)\le g(x,y)$ $<g(x,x)+1=g(x+1,1)$ , so that for $x,y,x',y'\in\mathbb{Z}_{+}$ such that $1\le y\le x$ and $1\le y'\le x'$ , $g(x,y)=g(x',y')$ $\Rightarrow(x,y)=(x',y')$ , and $g$ is surjective as for every $n\in\mathbb{Z}_{+}$ , we can let $x=\sup\{k\in\mathbb{Z}_{+}|g(k,1)\le n\}$ (note that $g(1,1)=1\le n$ $<g(n+1,1)$ , hence, the supremum exists and $x\ge1$ ) and $y=n-\frac{(x-1)x}{2}$ (note that $y\in\mathbb{Z}$ , and, by definition, $y=n-g(x,1)+1\ge1$ and $y=n-g(x+1,1)+x+1<x+1$ ), so that $g(x,y)=n$ .

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Section 7: Problem 2 Solution

Section 7: Problem 2 Solution