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I am having troubles with the following excercise:

$P(A\times B) = Q$ and $Q = \lbrace V\times W \ \vert \ V\in P(A), W\in P(B)\rbrace$

So I have to prove or disprove. I know that $P(A\times B) \neq Q$ and being specific $P(A\times B) \not\subset Q$ and $Q \subset P(A\times B)$. In addition;

$\supseteq \rbrack \ X\in Q \rightarrow X\in V\times W$, but $V\subset A$ and $W\subset B$, $\rightarrow $ $X\subset A\times B \rightarrow X\in P(A\times B)$.

But I am not able to disprove $\subseteq \rbrack$. I know they have diferent sizes but I want to make a formal disprove.

I am sorry for grammar mistakes, but English is not my native language.

Kind regards,

Phi.

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    See e.g. the post [cartesian-products-and-power-sets](http://math.stackexchange.com/questions/2109593/cartesian-products-and-power-sets) – 2017-02-05
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    Set $A=\{(1,a),(2,b)\} \in P(\{1,2\} \times \{a,b\})$ but $A \notin Q$. – 2017-02-05
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    Also $A = \lbrace (1,b),(2,a) \rbrace \in P(\lbrace 1,2 \rbrace \times \lbrace a,b \rbrace )$ but $A \not\in Q$ but I am not able to see why it happens. Some ideas for a formal disprove? – 2017-02-05
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    The way to disprove it is by showing a counter-example; any one would do, however simple. – 2017-02-05
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    If $A$ has $x$ members and $B$ has $y$ members then $P( A\times B)$ has $2^{xy}$ members but $Q$ has $2^{x+y}$ members. – 2017-02-05
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    That should have said AT MOST $2^{x+y}$ members. – 2017-02-05

2 Answers 2

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If one of the sets $A$ or $B$ is a singleton, then $Q=P(A\times B)$.

But, if both $A$ and $B$ contain at least two different elements, say $a_1,a_2 \in A$ and $b_1,b_2 \in B$, where $a_1 \neq a_2$ and $b_1 \neq b_2$, then for $$E=\{(a_1,b_1),(a_2,b_2)\}$$ we have $E \in P(A \times B)$ but $E \notin Q$. A proof by counter-example is as formal as one can get.

P.S. If you ask for a proof why $E \notin Q$, suppose that $E=A_1 \times B_1$ for some $A_1 \in A$ and $B_1 \in B$; then $$(a_1,b_1),(a_2,b_2) \in E \implies a_1,a_2 \in A_1 \wedge b_1,b_2 \in B_1.$$ But, then also $(a_1,b_2) \in E$ which is actually not the case.

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It's usually a good idea to look at small examples to look for a typical pattern for a counter-example. In your case the two sets have different cardinalities, so it's a good approach.

Let's take $A=B=\{0,1\}$. Then $P(A)=P(B)=\{\emptyset, \{0\}, \{1\}, \{0, 1\}\}$ and $A\times B=\{(0,0), (0,1), (1,0), (1,1)\}$.

The following invariant for the elements of $Q$ turns out to be interesting: for all $A \in Q$, if $(x_1, y_1), (x_2,y_2) \in A$, then $(x_1, y_2), (x_2,y_1) \in A$. Thus you won't get subsets of $A \times B$ not closed by such operation. For example $\{(0,1),(1,0)\} \notin Q$.

If $A$ and $B$ have $\ge 2$ elements, this counter-example works.