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Finding the Power Set of a Power Set

Date: 02/17/2005 at 04:45:48
From: Richard
Subject: "The second power set", or the power set of a power set 

What is "the second power set", or the power set of the power set of 
a set, say set <1,2,3>?
 
As we know, the power set of set <1,2,3> is

  << >,<1>,<2>,<3>,<1,2>,<1,3>,<2,3>,<1,2,3>>. 

But I do not know the power set of that set.



Date: 02/22/2005 at 09:47:35
From: Doctor Ian
Subject: Re: 

Hi Richard, 

Suppose our original set is 

  A = {a,b}

The power set of this is the set of all subsets:

  P(A) = { {},                No elements
           {a}, {b},          One element
           {a,b}              Two elements
         }

To make subsequent expansion less of a notational nightmare, I'm going
to use variables to represent these subsets, i.e., 

  P(A) = { W, X, Y, Z } 

Now we can take the power set of the power set, which is the set of
all subsets of P(A):

  P(P(A)) = { {},          No elements
              {W},         One element
              {X},             
              {Y}, 
              {Z}, 
              {W,X},       Two elements
              {W,Y},
              {W,Z},
              {X,Y},
              {X,Z},
              {Y,Z},
              {W,X,Y},     Three elements
              {W,X,Z},
              {W,Y,Z},
              {X,Y,Z},
              {W,X,Y,Z}    Four elements
            }

This is 2^4 = 16 elements, as we would expect.  To express them in
terms of the original elements, we can substitute as needed:

  W -> {}
  X -> {a} 
  Y -> {b}
  Z -> {a,b}

  P(P(A)) = { {},          
              {{}},         
              {{a}},             
              {{b}}, 
              {{a,b}}, 
              {{},{a}},       
              {{},{b}},
              {{},{a,b}},
              {{a},{b}},
              {{a},{a,b}},
              {{b},{a,b}},
              {{},{a},{b}},    
              {{},{a},{a,b}},
              {{},{b},{a,b}},
              {{a},{b},{a,b}},
              {{},{a},{b},{a,b}}   
            }

If I wanted to find P(P(P(A))), I would use the same trick again,
i.e., assign a variable to each element of P(P(A)), expand using those
variables, and substitute.  Otherwise, it's too easy to get lost in a
blizzard of brackets.  

Does this help?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Logic
High School Logic

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