Math Is Fun Forum

dchilow

Member

Registered: 2007-03-05

Posts: 27

I need help with Bijections!

Prove that if there is a bijection between two sets A and B, then there is a bijection between P(A) and P(B).

I don't think I can use an example because this is a for all statement.
I am pretty stuck here.
Please someone help me!!!!

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: I need help with Bijections!

If A={1,2,3} the bi-function is  2x+1 , then B={3,5,7}

If P(x)=(x-1)(x-2)(x-3)
Then P(A)={0} , P(B)={0,24,120}

I think this statement is a little weird.

Last edited by Stanley_Marsh (2007-04-01 16:55:23)

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Numbers are the essence of the Universe

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: I need help with Bijections!

Once you construct the bijection, this proof is fairly easy.  The trick is finding the bijection.

Well, first lets take a look at what we got.  We have two sets A and B, and we know there is a bijection between the two.  Because this statement is not true if there isn't a bijection between A and B, we should probably use this bijection in our proof...

So how do we use it?  Well, we can send a single element to another element.  But we need elements of P(A) and P(B), not of A and B.  What are the elements of P(A) and P(B)?  They are subsets of P(A) and P(B).  So how is it that we can take a bijection of elements and turn it into a bijection of subsets...

Well, so I don't hold you in suspense, we just take each element of the subset and use the bijection to make our new subset in P(B):

Let s be a subset of A, f be the bijection from A into B.

g(s) = {f(e) : e is in S}

This must be a subset of B.  It also must be 1-1 and onto.  It's now your job to prove that.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: I need help with Bijections!

If A={1,2,3} the bi-function is  2x+1 , then B={3,5,7}

If P(x)=(x-1)(x-2)(x-3)
Then P(A)={0} , P(B)={0,24,120}

I think this statement is a little weird.

P(A) is not a function. It’s the power set of A, i.e. the set of all subsets of A. Thus, if A = {1,2,3}, P(A) = {∅,{1},{2},{3},{1,2},{1,3},{2,3},A}.

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: I need help with Bijections!

Oh , I see .. lol~

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Numbers are the essence of the Universe

dchilow

Member

Registered: 2007-03-05

Posts: 27

Re: I need help with Bijections!

Thank you Ricky, I appreciate it.