# Set Theory

• Feb 16th 2009, 11:53 PM
jzellt
Set 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.
• Feb 17th 2009, 02:59 AM
Plato
Quote:

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

HINT: The composition of two bijections is a bijection.
• Feb 17th 2009, 06:18 AM
InvisibleMan
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.
• Feb 17th 2009, 02:08 PM
archidi
Quote:

Originally Posted by jzellt
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.

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