A nonempty set T1 is finite if and only if there is a bijection from T1->T2?

phillyolly
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Homework Statement



Prove that a nonempty set T1 is finite if and only if there is a bijection from T1 onto a finite set T2.


The Attempt at a Solution



There are at least two different solutions to this problem that I found online:

http://answers.yahoo.com/question/index?qid=20090319202703AACrlT8

http://www.cramster.com//answers-sep-10/advanced-math/bijection-finite-set-prove-nonempty-set-t1-finite_921466.aspx

Is there any other solution? For a dummy like me these solutions are hard enough to understand them completely. If you can help me get the idea, I will really appreciate it.
 
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Hi Phillyolly! :smile:

What is your definition of finite?
 
Hi Micromass! :-)

Definition of finite:

A set S is said to be finite if it has n elements for some n in N or is empty.
 
Well...

<-

Assume there exists a bijection from T1 to T2. Since there exists a bijection from T1 to T2, this means that |T1| = |T2|. Since T2 is finite...

->

T1 is finite. Define T2 to be T1, so...
 
Well, first you'll need to know what it means that "a set has n elements"...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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