Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
You should probably use a symbol other than $\times$ for your modified Cartesian product to avoid confusion. – Rahul Mar 1 '13 at 2:39
Maybe $\boxtimes$ or $\otimes$ (\boxtimes, \otimes)? – Rahul Mar 1 '13 at 2:46

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.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.