Mathematical connection in the cartesian product

[V0ODO0CH1LD]

mathematical "connection" in the cartesian product

What is the mathematical connection between elements of a cartesian product $A\times{}B$ and the elements of the sets $A$ and $B$?

In other words, what is the difference between the set $A\times{}B$ and just any set $Z$ with $|A|.|B|$ elements that creates no contradictions if I choose to make a connection in my head between each element of it and one element of $A$ and one element of $B$ the same way the cartesian product of $A$ and $B$ does?

Because if you forget the usual notation for elements of a cartesian product (e.g. $(a,b)$, $a\times{}b$, $ab$) all you have is a set with cardinality $|A|.|B|$ that you as a human connect to the elements of the sets involved in the product in a particular way, usually through notation.

But if I have $|Z|=|A|.|B|$ how can I prove or disprove that the set $Z$ is the cartesian product of $A$ and $B$? Are the elements of $A$ and $B$ set theoretically contained in the elements of $A\times{}B$?

If I choose to say the set $\{1,2,3,4\}$ is the cartesian product of the sets $\{a,b\}$ and $\{c,d\}$ is that incorrect just because of the way that I chose to write the sets down?

 

[jgens]

The Cartesian product of sets X₀ and X₁ actually gives you more information than a set with cardinality |X₀||X₁|. It also comes equipped with canonical projections p_i:X₀×X₁→X_i that satisfy nice mapping properties. Any set Z equipped with projections q_i:Z→X_i satisfying these same properties could equally well be called the product of X₀ and X₁ and in some cases this identification is made. On the other hand, if someone asks whether {1,2,3,4} is the product of {a,b} and {c,d}, then you should probably answer "no" since point-wise these sets differ.

There are other reasons to pay attention to the usual description of Cartesian products by the way. When you have sets with structure (like groups, rings, topological spaces, etc.) often times a product of these objects is constructed by looking at the usual Cartesian product of the underlying sets and then defining the extra structure on that. So it genuinely makes life easier to keep the ordered pair definition in mind.

 

[V0ODO0CH1LD]

I don't know why it occurred to me to google "cartesian product" only after I posted my question, but anyway..

In ZFC everything is a set, which means every element of every set is a set. So even though the ordered pair (a,b) has been defined as (a,b) = {{a},{a,b}} and (a,b) = {{a},ø},{{b}}} for different reasons, the only thing that matters is that your definition makes sure that (a,b) = (c,d) <=> a = b and c = d. In a sense it is okay to think that the element (a,b) in AxB set theoretically contains the elements a in A and b in B.

 

[jgens]

So even though the ordered pair (a,b) has been defined as (a,b) = {{a},{a,b}} and (a,b) = {{a},ø},{{b}}} for different reasons, the only thing that matters is that your definition makes sure that (a,b) = (c,d) <=> a = b and c = d.

This is true for defining ordered pairs. For Cartesian products the important thing is the canonical projections.

In a sense it is okay to think that the element (a,b) in AxB set theoretically contains the elements a in A and b in B.

You hypothetically can but there really is no reason to.