Section 9: Problem 3 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

Suppose that $A$ is a set and $\{f_{n}\}_{n\in\mathbb{Z}_{+}}$ is a given indexed family of injective functions $f_{n}:\{1,\ldots,n\}\rightarrow A.$ Show that $A$ is infinite. Can you define an injective function $f:\mathbb{Z}_{+}\rightarrow A$ without using the choice axiom?

If it were finite, there would be a bijective function $g:A\rightarrow\{1,...,m\}$ for some $m\in\mathbb{Z}_{+}$ , but then $g\circ f_{m+1}:\{1,\ldots,m+1\}\rightarrow\{1,\ldots,m\}$ would be injective.

We can define $f$ recursively. Let $a_{0}=f_{1}(1)$ , and $\begin{eqnarray*} \rho(h:S_{n+1}\rightarrow A) & = & f_{n+1}(\inf\{k\in\mathbb{Z}_{+}|k\le n+1\\ & & \&\mbox{ }f_{n+1}(k)\notin h(\{1,...,n\})\}). \end{eqnarray*}$ $\rho$ is well-defined as $f_{n+1}$ is injective, so for at least one $1\le k\le n+1$ , $f_{n+1}(k)\notin h(\{1,\ldots,n\})$ . Further, the function $f$ defined by this recursive relation is such that $f(1)=a_{0}=f_{1}(1)$ , and $f(n+1)=\rho(f|\{1,\ldots,n\})$ $\notin f(\{1,\ldots,n\})$ .

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Section 9: Problem 3 Solution

Section 9: Problem 3 Solution