Section 8*: Problem 7 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

Prove Theorem 8.4.

First, similar to Lemma 8.1, by induction, we prove that for $n\in\mathbb{Z}_{+}$ there exists a function $h_{n}:\{1,\ldots,n\}\rightarrow A$ satisfying the equations $(\ast)$ of Theorem 8.4 for all $i$ in its domain. Namely, this is true for $n=1$ , $h_{1}(1)=a_{0}$ , and, given this is true for $n$ , i.e. there is $h_{n}$ , we define $h_{n+1}(i)=h_{n}(i)$ for $i\le n$ , and $h_{n+1}(n+1)=\rho(h_{n})$ . Note, that $h_{n+1}(1)=h_{n}(1)=\cdots$ $=h_{1}(1)=a_{0}$ , and $h_{n}=h_{n+1}|\{1,\ldots,n\}$ , hence, the equations $(\ast)$ hold for $h_{n+1}$ .

Then, similar to Lemma 8.2, we prove that for every $m\le n$ , and $g_{m}$ and $h_{n}$ satisfying $(\ast)$ on their domains, respectively, $h_{n}|\{1,\ldots,m\}=g_{m}$ . Again, we argue that $g_{m}(1)=h_{n}(1)$ , and for every $1\le k<m$ , $h_{n}|\{1,\ldots,k\}=g_{m}|\{1,\ldots,k\}$ implies $h_{n}|\{1,\ldots,k+1\}=g_{m}|\{1,\ldots,k+1\}$ as both satisfy $(\ast)$ .

Finally, as in Theorem 8.3, we construct $h$ for all positive integers by taking the union of the rules of all $h_{n}$ ’s. The argument is, in fact, almost identical to the proof of Theorem 8.3, as the latter mostly operates in general terms without even using the particular definition of $\rho$ it is proved for.

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Section 8*: Problem 7 Solution

Section 8*: Problem 7 Solution