# Cartesian products and power sets

another discrete math question. I was watching this video about Cartesian products, power sets and cardinality when a thought occurred to me. If the set S were arbitrary, would its Cartesian product be a subset of the power set? In other words would this statement be true:

$S \times S \subseteq \mathcal{P}(S)$

I've looked around on the site to find a similar question, but any questions I found involved the Cartesian product of two or more sets, not one. I think the statement is false because with an arbitrary set, we don't know what the elements are, so there's a chance the statement is false by default. Any thoughts?

It is not true for most sets $A$, because the elements of $A\times A$ are ordered pairs, and an ordered pair of elements from $A$ is not a usually set of elements of $A$ (which it would have to be in order to be a member of $\mathcal P(A)$.

In axiomatic set theory, where sets are the only things that exist, an ordered pair must be represented by a particular set -- the most common convention is to consider $(a,b)$ to "really" be an abbreviation for $\{\{a\},\{a,b\}\}$.

When this convention is used, there are a few particular cases where $A\times A$ is indeed a subset of $\mathcal P(A)$.

One example of this is $A=\varnothing$, in which case $A\times A=\varnothing$ which is a subset of anything. A different example is $A=H_\omega$, the set of hereditarily finite sets (which means basically the sets that can be written in finite space with only the symbols {, }, and ,, such as "$\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}$").

But in both of those cases $A\times A\subseteq \mathcal P(A)$ is more of an accident than something that tells us something interesting and useful about the sets in question.

• So I was right that the statement was false because the set was arbitrary?
– 009
Jan 23 '17 at 4:34
• @009: You can't be sure that the statement is false -- its truth value depends on which set it is about. After all an "arbitrary" set may turn out to be the empty set. Jan 23 '17 at 12:14

That is not true. To understand why you have to remember what a Cartesian product spits out. It spits out an ordered Tuple $(a,b)$ so $(a,b)$ and $(b,a)$ are not the same thing. But as subsets of $S$ they are the same.

Try thinking of it this way: What does it mean for two sets to be the same? They must have all of the same elements. But what does it mean for a set to be ordered? The have to have the same elements and in the same order. Two sets being the same as subsets of $S$ doesn't mean that they have the same ordering so $S members of the cartesian product cannot be members of the power set. Basically you're trying to compare things that aren't the same type. One way to look at it is as asking is carrot a subset of 12? The question just doesn't make any sense. • To be fair, in set theory$5 \subset 12$and that makes about as much sense as carrot$\subset 12$. Jan 22 '17 at 23:57 • I wanted to avoid that statement because that might be true depending on the context. Eg. that's true for dedekind cuts. It's better to be completly unambiguous. Jan 22 '17 at 23:59 • I wasn't being overly serious. But I did want to make the point that "making sense" is not necessarily a guide to mathematical truth. Jan 23 '17 at 0:06 Take a specific example. Say$S = \{1, 2, 3\}$. Now$S \times S = \{(1,1), (1, 2), (1,3), (2, 1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$. And$\mathscr P(S) = \{ \{\emptyset\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\},\{2,3\}, \{1,2,3\}\}$. The Cartesian product consists of ordered pairs, while the power set consists of subsets of$S\$. Clearly the Cartesian product contains elements that are not in the power set, so the subset relationship does not hold.

(To be technically accurate, ordered pairs are themselves sets. But they're not any of the sets in the power set, so the conclusion still holds).

Frederic Schuller in his notes on geometric anatomy of theoretical physics describes axiomatic set theory (ZFC) and comes up with that definition.

If you apply it to the example shown in the last answer, you will come up in that it is correct.