Suppose S is finite of cardinality n. If f: S ---> T is a bijection, then T is finite and has cardinality n.

Any advice on how to give a solid proof of this? THanks.

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- February 16th 2009, 11:53 PMjzelltSet Theory
Suppose S is finite of cardinality n. If f: S ---> T is a bijection, then T is finite and has cardinality n.

Any advice on how to give a solid proof of this? THanks. - February 17th 2009, 02:59 AMPlato
- February 17th 2009, 06:18 AMInvisibleMan
There's nothing to prove here, two sets have the same cardinality iff there's a bijection between the two sets.

This is the definition. - February 17th 2009, 02:08 PMarchidi

From the book: General Topology,from Schaum's Outline Series i quote from page 32 the part that solves your problem:

" A set A is called equivalent to a set B,written A~B,if there exists a function f: A --->B which is one to one and onto.The function f is then said to define a one to one correspondence between the set A and B.

A set is finite iff it is empty or equivalent to {1,2,.........,n} for some nεN,otherwise it is said to be infinite.

Clearly two finite sets are equivalent iff they contain the same number of elements.

According to the above quotation ,since S is equivalent to T ,THEY have the same number of elements,and they have the same cardinality n