Is this a rigorous mathematical proof that $P(A\times B)\neq P(A)\times P(B)$?

Question

I have to prove that $P(A\times B)\neq P(A)\times P(B)$

My approach:

I am showing that the cardinality of $P(A\times B)$ and $P(A)\times P(B)$ is not the same. I think this method works because powerset of a set is a set and for equality of two sets, it is mandatory for the sets to have the same elements.

Let, $A, B$ be two finite sets with $x,y$ elements respectively.

Thus no of elements in $P(A\times B)$ is $2^{xy}$ and no of elements in $P(A)\times P(B)=2^{x+y}$.

Let,$x=1,y=2$. Clearly, $2^{x+y}\neq 2^{xy}$.

Thus the two sets don't have the same number of elements. So we have successfully built a counterexample that the two sets contain some dissimilar elements (since their cardinality is different).

Is this a proper rigorous mathematical proof?

Thanks for any help!!

Answer 1

As stated the problem is too obvious. An element of $P(A)\times P(B)$ is an ordered pair $(E,F)$ where $E\subset A$ and $F\subset B$, and that ordered pair is simply not a subset of $A\times B$. If the problem wasn't supposed to be that trivial, surely you're actually supposed to show that $$P(A\times B)\ne\{E\times F: E\in P(A), F\in P(B)\}.$$(It's very easy to simply write down an element of one of those sets that's not an element of the other...)

Answer 2

As said by D.C. Ullrich, the ''data structures'' are different and so equality cannot be proved. But one can show that there is no bijection between these sets (finite case only!) as they have different cardinality which you already indicated.

For instance, if $A=\{a\}$ and $B=\{b,c\}$, then $$P(A\times B) = P(\{(a,b),(a,c)\} = \{\emptyset, \{(a,b)\}, \{(a,c)\}, \{(a,b),(a,c)\}\}$$ while $P(A)=\{\emptyset,\{a\}\}$ and $P(B)=\{\emptyset,\{b\}, \{c\},\{b,c\}\}$ and so $P(A)\times P(B) = \{(x,y)\mid x\in P(A),y\in P(B)\}$ contains for instance the pair $(\{a\},\{b,c\})$.

Answer 3

$\emptyset \in P(A\times B)$ because $\emptyset \subset M$ for any set $M$ and so $\emptyset \in P(M)$ for any set $M$.

And $\emptyset \not \in P(A) \times P(B)$ because $\emptyset \not \in J \times K$ for any sets $J$ and $K$ since $\emptyset$ is not an ordered pair of anything.

Now $\emptyset = \{\}$ and $\emptyset \ne (j,k)$ for any $j,k$s in any set whatsoever. $(j,k)$ is an ordered pair and the empty set is not an ordered pair. It simply isn't. The empty set is not a number; it is not a pair of numbers; it's not a pair of elephants; it's not a pair of an elephant and a marble. The empty set is not a pair of anything.

So $\emptyset \ne (j,k)$ for any $j$ in any set $J$ or any $k$ in any set $K$. Even if $J = K =\emptyset$ we still do not have $\emptyset \in \emptyset \times \emptyset$. $\emptyset \times \emptyset = \{(a,b)|a\in \emptyset; b\in \emptyset\}$ but there are no such $(a,b)$ because there is no $a\in \emptyset$ and there is no $b\in \emptyset$ so $\emptyset \times \emptyset = \{(a,b)|a\in \emptyset; b\in \emptyset\}=\emptyset$. And $\emptyset \not \in \emptyset$.

So $\emptyset \in P(A\times B)$ but $\emptyset \not\in P(A)\times P(B)$. So $P(A\times B)$ can not ever equal $P(A)\times P(B)$.

.....

Also as $P(M)$ is a set of sets. And $J\times K$ is a set of ordered pairs. And sets are not ordered pairs and ordered pairs are not sets so these are two completely different types of things.