# Problem on finite sets

I am sorry for my incompete post, I have updated it.

To prove the finitness of a set one has to establish a bijective correspondence with a finite set, say $z_+$. Let $A$ be a set; let $a_0\in A$. Then there exists a bijective correspondence $f$ of set $A$ with the set $\{1,\ldots,n+1\}$ iff there exists a bijective correspondence $g$ of the set $A-a_0$ with the set $\{1,\ldots,n\}$.

To prove the converse in Topology by James Munkres (2nd edition), Lemma 6.1, the argument made is:

Assume there is a bijective correspondence $f:A\to\{1,\ldots,n+1\}$. If $f$ maps $a_0$ to $n+1$, the restriction $f|A-a_0$ is the desired bijective correspondence of $A-a_0$ with $\{1,\ldots,n\}$.

Otherwise, let $f(a_0)=m$, and let $a_1 \in A$ such that $f(a_1)=n+1$. Then $a_1\neq a_0$. Define a new function $h:A\to\{1,\ldots,n+1\}$ by setting

$h(a_0)=n+1$
$h(a_1)=m$
$h(x)=f(x)$ for $x \in A-\{a_0,a_1\}$, and here $h$ is bijective. The restriction $h|A-\{a_0\}$ is the desired bijection of $A-\{a_0\}$ with $\{1,\ldots,n\}$.

My question is the if someone has examples of these hypothetical $f,h$ functions, I tried few but not really getting .

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If you are new, then I just thought that I should tell you that the way in which you put forth your questions are important.Very few people as Asaf said would be interested in answering if they have to refer a said book just to get what the question is.And if you like an answer for your question, accept it to improve your accept rate.I hope you don't mind me telling you this.Cheers :) –  Vishesh Oct 7 '12 at 10:47
I have retagged this question. Although it arises from a chapter of a topology text, it has nothing whatsoever to do with topology. –  dfeuer Jun 7 '13 at 7:30
Also, if I'm reading this right, it looks like part of the proof that any subset of a finite set is finite. @rafiki: if you're having trouble making head or tail of what this whole thing is about, you may want to step back and ask a different question. –  dfeuer Jun 7 '13 at 7:32
@dfeuer: this is complete in itself. –  Rorschach Jun 7 '13 at 18:12
@rafiki, you haven't actually stated any theorem. –  dfeuer Jun 7 '13 at 18:48

This question probably doesnt warrant a proper answer. But I thought the OP would be looking for one to be satisfied. The result proved in the book is just a restatement of the fact that a subset of a finite set is necessarily finite. Or even better, it says that removal of a finite number of elements from a set A will give you a finite set iff the set A was finite to begin with.The functions $g$ and $h$ are merely constructed in the context of the proof. To give an explicit example, just take a specific finite set A and redo the proof. You will realise what $h$ and $g$ look like.