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February 13th, 2008, 07:55 PM   #1
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An error on functions and powersets, in a Dover book?

Hey all,

Taking discrete mathematics this semester and I'm really enjoying the course material. I'm digging a bit farther into it because I have an interest in these topics, and, of course, I want to get an A in the class.

So I'm working on a proof provided in the Dover book "Elements of Abstract Algebra" by Allan Clark. In it, there is a theorem which states:

There is no one-to-one-correspondence f: X -> 2^X for any set X.

(Note that 2^X in this context refers to the powerset of X)

For the vast majority of sets, this is true.. But what about the empty set ({}) ? If I'm not mistaken

2^{} = {{}}

and a mapping:

{} -> {{}}

Is both one-to-one and onto. So, is this an error in the book? Or am I doing something wrong?

Although the book provided a proof by contradiction, I gave it a try without. I'm still learning proofs, so I'd like advice if it's wrong, or could be improved:

Theorem: There is no one-to-one correspondence f: X -> 2^X for any set X

Let n(X) := the number of elements in a set X

In order for f to have one-to-one correspondence, the Domain(f) and Codomain(f) must have equal number of elements.

However, for any set X, the 2^X has 2^n(X) elements, therefore f: X-> 2^X cannot be a one-to-one correspondence.

That was my first try. If the theorem was changed to not include the empty set, it would be correct.. if I'm not already mistaken.

- Rob

Update: Correction, thanks to CRGreathouse
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February 13th, 2008, 08:00 PM   #2
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Ah, but you did make a mistake early on. You wrote that 2^{} = {}, but in fact the power set of {} is not {} but {{}}, a set which has one element instead of the zero elements in {}.

Your proof works, though, and does tell you that the power set of a 0-element set will have 1 element. You almost figured it out... just like Saccheri nearly discovered non-Euclidean geometry. :P
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February 15th, 2008, 06:22 PM   #3
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Greetings:

In order that a 1-1 correspondence may exist between two sets, they must both share the same cardinality. But for any set, S, n(P(S)) = 2^n(S), where P(S)==pwr set of S, and n(S
i) refers to the cardinality of set Si. That is, the cardinality of P(S) is equal to 2 raised to the cardinality of S. Because 2^x > x for all non-negative integers, x, it is not possible for S and P(S) to have like cardinality. Hence no such 1-1 correspondence exists and the proof is therefore complete.

Regards,

Rich B.
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