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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

$h(x)=f(x)$ 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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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
@dfeuer:I checked in book but they have not specified any theorem prior to this and I think lemma and theorem and some what same. –  Rorschach Jun 8 '13 at 6:30

1 Answer 1

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.

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