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The power set can be constructed as a "Cartesian product" as in the following example. Suppose $S=\{a, b, c\}$ and that $\times$ is a modified Cartesian product operator (see following paragraph), then:

$$Pset(S)=\{\emptyset,\{a\}\}\boxtimes \{\emptyset,\{b\}\} \boxtimes \{\emptyset,\{c\}\}$$

The difference being that the pairs of the product are actually unified together to make one set. For example, the pair $(\emptyset, \emptyset)$ yields $\emptyset$, the pair $(\emptyset, \{a\})$ yields $\{a\}$.

Does this operation have a standard name/notation? I want to prove that the resulting set is actually the power set but I need notation or try to introduce appropriate notation.

Note: This construction tells us that the cardinality is $2^{|S|}$ right away which is true.

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How is what you do any different from $2^S = \{ S \to \{\mathbf{0},\mathbf{1}\}\}$? –  Willie Wong Mar 1 '13 at 2:08
    
@WillieWong That's what I'm trying to prove! –  saadtaame Mar 1 '13 at 2:12
    
What do you mean that's what you are trying to prove? What I wrote is a definition. –  Willie Wong Mar 1 '13 at 2:14
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You should probably use a symbol other than $\times$ for your modified Cartesian product to avoid confusion. –  Rahul Mar 1 '13 at 2:39
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Maybe $\boxtimes$ or $\otimes$ (\boxtimes, \otimes)? –  Rahul Mar 1 '13 at 2:46

1 Answer 1

The set you describe is not the power set of $\{a,b,c\}$. It is a set that has the same cardinality as the power set, but its elements $x$ do not satisfy $x\subseteq S$.

EDIT: after the clarifications and modifications it all makes sense now. I never saw the definition $A\boxtimes B=\{a\cup b\mid a\in A,b\in B\}$, which in the context of axiomatic set theory makes sense. It does not make sense universally in naive set theory though. The equality $\mathcal P(S)=\{\emptyset ,\{s_1\}\}\boxtimes \cdots \{ \emptyset, \{s_n\}\}$ makes sense and is true when $S=\{s_1,\cdots , s_n\}$. With a bit of care it can be extended to infinite sets and infinite box products.

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It is the power set! It's not the Cartesian product that's way I put it in quotes. –  saadtaame Mar 1 '13 at 2:04
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what makes it the power set? –  Ittay Weiss Mar 1 '13 at 2:04
    
Try and list all the components of the set. You can try with a set that has 1 or two elements. –  saadtaame Mar 1 '13 at 2:05
    
if the context of the question is axiomatic set theory (say ZF) then the existence of power sets is an axiom, not a definition. If the context is naive set theory, then the power set of $S$ is the set of all subsets of $S$. What you describe is not the power set, but rather a set of the same cardinality. –  Ittay Weiss Mar 1 '13 at 2:16
    
The OP is using $\times$ to denote something different from the usual Cartesian product. In their notation, $A\times B=\{a\cup b:a\in A,b\in B\}$. –  Rahul Mar 1 '13 at 2:41

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