Set Theory Q: Is |P(A)| = |P(B)| iff A=B? Hints Needed

In summary, the statement "If |P(A)| = |P(B)|, then |A| = |B|" is not true for infinite sets without the Continuum Hypothesis. A counterexample can be found by considering the sets A = {0,1} and B = {0,1,2}, which have the same power set cardinality but different cardinalities.
  • #1
JonF
621
1
If P(X) denotes the power set of X. Is |P(A)| = |P(B)| iff A=B true? If so, I have no idea how to prove the |P(A)| => |P(B)| iff A=B direction, so any hints would be great.
 
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  • #2
does |A| mean the cardinality of A? If so, it is clearly false. For example: the sets {0,1,2} and {3,4,5} have power sets with the same cardinality but they are not equal.
 
  • #3
No.

If A = B, then P(A) = P(B) so of course we have |P(A)| = |P(B)|.

But if |P(A)| = |P(B)|, A need not be equal to B.
Just take say A = {1}, B = {2}, then P(A) = {{},{1}} and P(B) = {{}, {2}}, so |P(A)| = |P(B)| but A != B.
 
  • #4
Does |P(A)| = |P(B)| imply |A| = |B| for infinite sets without the CH?
 
  • #5
Ok never mind on help trying to prove this, is it even true? I suspect it might be equivalent to the SCH.
 
  • #6
CRGreathouse said:
Does |P(A)| = |P(B)| imply |A| = |B| for infinite sets without the CH?

Yes.

Any set [tex]A[/tex] is equinumerous to the set [tex]A_s = \{ \{x\} \in P(A) | x \in A \}[/tex]

For any set [tex]A[/tex] Let [tex]f_A[/tex] denote the bijective map from [tex]A[/tex] to [tex]A_s[/tex].

If [tex]|P(A)| = |P(B)|[/tex] then there is a bijection [tex] g [/tex] from [tex]P(A)[/tex] to [tex]P(B)[/tex] and the restriction of [tex] g [/tex] to the set [tex]A_s[/tex], is a bijection from [tex]A_s[/tex] to [tex]B_s[/tex]. Thus a bijection from [tex]A[/tex] to [tex]B[/tex] is given by the composition of functions

[tex]f_B^{-1} \circ g|A_s \circ f_A[/tex]

Where [tex] g|A_s [/tex] is the restriction of g to the set [tex]A_s[/tex] and [tex]f_B^{-1}[/tex] is the inverse of the function mapping [tex]B[/tex] onto [tex]B_s[/tex].
 
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  • #7
CrankFan said:
the restriction of [tex] g [/tex] to the set [tex]A_s[/tex], is a bijection from [tex]A_s[/tex] to [tex]B_s[/tex]
Why do you know that? That's false for most bijections [itex]\mathcal{P}(A) \rightarrow \mathcal{P}(B)[/itex].

For example, suppose that A and B actually are both equal to the set {0, 1}. Then, one example of a bijection [itex]\mathcal{P}(A) \rightarrow \mathcal{P}(B)[/itex] is:

g({}) = {0}
g({0}) = {}
g({1}) = {0, 1}
g({0, 1}) = {1}

Here, the image of [itex]g \circ f_A[/itex] never takes values in [itex]B_s[/itex].
 
  • #8
I guess I said that because I wasn't thinking clearly. Thanks for the correction.
 

Related to Set Theory Q: Is |P(A)| = |P(B)| iff A=B? Hints Needed

1. Is set A equal to set B if and only if the power set of A is equal to the power set of B?

Yes, if the power set (the set of all subsets) of set A is equal to the power set of set B, then set A and set B must be equal.

2. Can two sets have the same power set without being equal?

Yes, it is possible for two different sets to have the same power set. For example, the sets {1,2} and {3,4} have the same power set of {{}, {1}, {2}, {1,2}} even though they are not equal.

3. Is the power set of an empty set equal to the empty set?

Yes, the power set of the empty set is equal to the set containing only the empty set. In other words, the power set of an empty set is {{}}.

4. Can the power set of a set be smaller than the set itself?

No, the power set of a set will always have more elements than the set itself. This is because the power set contains all possible subsets of the original set, including the empty set and the set itself.

5. What is the relationship between the cardinality of a set and its power set?

The cardinality (number of elements) of a set is always smaller than the cardinality of its power set. This is because the power set contains all possible subsets, including the empty set and the set itself, which adds at least two elements to the power set.

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