Non associativity of Cartesian product

Question

I am trying to prove that if

$$(A \times B) \times C = A \times (B \times C)$$

Then either $A = \emptyset, B = \emptyset$ or $C = \emptyset$.

So far I start of by assuming the equivalence and that $A \neq \emptyset, B \neq \emptyset$ and $C \neq \emptyset$. This gives $a \in A, b \in B, c \in C$ such that $((a,b), c) \in (A \times B) \times C $ and $((a,b), c)\in A \times (B \times C)$, whence $(a, b) \in A \times B$ and $(a, b) \in A$ and $c \in C$ and $c \in B \times C$, so that $A \times B = A$ and $B \times C = C$.

I understand on an intuitive level that $A \times B \neq A$, unless they are both empty (and likewise for $B \times C = C$, but what rules it out formally (in particular, what axiom or chain of reasoning in ZFC would rule it out)?

Observe that a general element of $;(A\times B)\times C;$ is of the form $;((a,b)\times c);$, whereas a general element in $;A\times(B\times C);$ is of the form $;(a,(b,c));$ ... — DonAntonio, Sep 22 '22 at 12:09
Yes, thanks, that's what I observed. But I want to know what axiom of ZFC (or reasoning from those axioms) rules it out. — Edward.Lin, Sep 22 '22 at 12:10
If there's a least one element in either cartesian product, say $;(a,(b,c))\in A\times(B,C);$ , then it can NOT be in $;(A\times B)\times C;$ ...it is basically a different ordered couple — DonAntonio, Sep 22 '22 at 12:12
Does this answer your question? Associativity of Cartesian Product — Anne Bauval, Sep 22 '22 at 12:16
It didn't. It took the answer for granted. But then I realised they were distinct by the Kuratowski definition and the axiom of extensionality. — Edward.Lin, Sep 22 '22 at 12:25
@AnneBauval I'm not sure that is a duplicate of this question. — Suzu Hirose, Sep 26 '22 at 01:02
While they aren't equal, there's such a trivial bijection between the two that leads us to work with "ordered triples" of $A\times B \times C$ knowing that underneath it's one of those two — Alan, Sep 27 '22 at 05:24

David Perrella · Answer 1 · 2022-09-27T05:12:40.030

You're right that it boils down to showing that if $A, B$ are sets, then $A = A\times B$ only if $A$ is empty. This holds without choice. Suppose $A = A\times B$ where $A$ is nonemoty.

We will construct an infinite sequence which contradicts a consequence of the axiom of regularity and pairing mentioned on Wikipedia.

For each $a$ in $A$, there exists a unique $a' \in A, b \in B$ such that $a = (a', b)$. Moreover $(a', b) = \{\{a'\}, \{a', b\} \}$. So we have that $\{a'\} \in a$.

Then, fixing $a\in A$ consider the infinite sequence $a(1),...$ given by \begin{align} a(1) &= a\\ a(2) &= \{a'\}\\ a(3) &= a'\\ a(4) &= \{(a')'\} \\ a(5) & = (a') '\\ &... \end{align} This sequence satisfies $a(n+1) \in a(n)$ for each $n \in \mathbb{N} $. This is a contradiction.

Eric Wofsey · Accepted Answer · 2022-09-22T14:23:31.320

1

To complement David Perrella's answer, let me point out that the axiom of regularity is crucial for this statement. Without regularity, it is possible to have a set $A$ such that $A=\{A\}$. This set then satisfies $(A,A)=\{\{A\},\{A,A\}\}=\{\{A\}\}=\{A\}=A$ and so $A\times A=\{A\}=A$ and thus $(A\times A)\times A=A=A\times (A\times A)$. So, the axiom of regularity is the only axiom of ZFC that really rules out some sort of coincidence in the structures of the sets of $A, B,$ and $C$ that would result in $A\times (B\times C)=(A\times B)\times C$.

edited Sep 22 '22 at 14:23

answered Sep 22 '22 at 13:35

Eric Wofsey

330,363

Indeed, and one must assume the axiom of pairing here – David Perrella Sep 22 '22 at 13:49

Non associativity of Cartesian product

2 Answers2