Definition of cartesian product

Question

I have some problem understanding some aspects of the definition of an arbitrary cartesian product. What I have so far:

Let $\mathscr{A} = (A_s)_{s \in S}$ be a collection of sets. The cartesian product of the collection $\mathscr{A}$ is: $$\begin{align} \prod\mathscr{A} = \prod_{s \in S}A_s &:= \left\{ f: \mathscr{A} \to \bigcup_{s \in S}A_s \mid f(A) \in A, \, \forall\, A \in \mathscr{A} \right\} \\ &= \left\{ f: S \to \bigcup_{s \in S}A_s \mid f(s) \in A_s, \forall\, s \in S \right\}\end{align}. $$If $\mathscr{A} =\{A_1, \ldots A_n\}$, we will write $\prod\mathscr{A} = A_1 \times \cdots \times A_n$.

By my understanding, I get that if $A$ and $B$ are sets, we have: $$A \times B = \{f: \{A,B\} \to A \cup B\mid f(A) \in A \text{ and }f(B) \in B\}$$

I am asked, then, to prove that given sets $A,B$ and $C$, we have $A \times(B \cup C) = (A \times B)\cup(A \times C)$.

Doubt 1: How can I make the comparison if the domains of the elements on both sides are different? It does not make any sense at all.

Suppose I don't care and go on: If $f \in A \times(B \cup C)$, then $f(A) \in A$ and $f(B \cup C) \in B \cup C$. I would think then, that if $f(B \cup C) \in B$, then $f \in A \times B$, and if $f(B \cup C)\in C$, we have $f \in A \times C$, so anyway we have the inclusion $\subseteq$. On the other hand, it suffices to prove that $A \times B \subseteq A \times(B \cup C)$ (WLOG). If $f \in A \times B$, then $f(A) \in A$ and $f(B) \in B \subset B \cup C$ and so $f \in A \times(B \cup C)$. I am not comfortable at all with this, since I do not have a rigorous definition of a functions, and the most formal one I know is that $f : A \to B$ is a function if $f \in A \times B$ satisfies blah blah blah. It is circular.

Up next I am asked to prove that $A \times B = A \times C$ and $A \neq \varnothing$, then $B = C$. My idea consists mainly of the following: since $A \neq \varnothing$, then exists $a \in A$. Take $x \in B$, if I prove that $x \in C$ then I am done. Define $f:\{A, B\}\to A \cup B$ by setting $f(A) = a \in A$ and $f(B) = x \in B$. This way $f \in A \times B$, and by hypothesis $f \in A \times C$. I don't know how to give the next step in this argument.

Doubt 2: Can't I just say that if $f \in A \times B = A \times B$ implies $\{A,B\}=\{A,C\}$ and $A \cup B = A \cup C$, based on domains and codomains, and from there try to get $B = C$?

I probably can tackle these problems on my on, if I can understand the definitions correctly, but all of this is very cloudy and I need some help understanding. Thanks.

Answer 1

The first problem that we encounter is in the first definition. Namely, it does not work if $(A_s)_{s\in S}$ contains duplicates. It therefore cannot be used to define one of the most common examples of the Cartesian product, namely $A^n$.

As to your problem with the circularity in the definition of a function, I think it's mostly a matter of recognising that the definition of the binary Cartesian product is a case treated separately—defined in the usual way: ordered pairs via Kuratowski's definition, then setting:

$$A \times B := \{(a,b): a \in A, b \in B\}$$

which is proved to exist in $\sf ZF$ in many, many places. It is correct that this is not equal to $\prod \{A, B\}$, and this obfuscation of the distinction between binary and arbitrary Cartesian product is a weak point in the exposition.

Next, we also have the definition of a function, $f$ is a function iff:

• $f \subseteq A \times B$;
• $(a,b) \in f$ and $(a,c) \in f$ imply $b= c$.

Then, the notation "$f: A \to B$" means:

• $\forall a: (a \in A \iff \exists b \in B: (a,b) \in f)$;
• $\forall b: \exists a \in A: (a,b) \in f \implies b \in B$.

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With that all out of the way, we can get to the definition of the Cartesian product proper. To avoid confusion, let us write $\prod \{A, B\}$ instead of $A \times B$ in addressing "binary arbitrary Cartesian products". Let us also agree that for such finite products, the indexing set is $\{1\ldots n\}$.

These presuppositions having been specified, we can see that:

\begin{align*} \prod \{A,B \cup C\} &= \left\{f: \{1,2\} \to A \cup B \cup C \middle\vert f(1) \in A, f(2) \in B \cup C \right\}\\ \prod \{A,B\} &= \left\{f: \{1,2\} \to A \cup B \middle\vert f(1) \in A, f(2) \in B \right\}\\ \prod \{A,C\} &= \left\{f: \{1,2\} \to A \cup C \middle\vert f(1) \in A, f(2) \in C \right\}\\ \end{align*}

Now that we have cut through the complicated language, I trust that you can see that the asserted equation does make sense, and indeed can prove it yourself.

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Lastly, your second doubt concerns the first definition, which as said is problematic because it is not general enough.

Your proof is good, and it can be concluded from $f \in A \times C$ that $f(2) \in C$, and by construction $f(2) = x$; hence $x \in C$. Swapping the roles of $B$ and $C$ yields the reverse inclusion, proving equality.