0
$\begingroup$

I am only starting to study Set Theory and I was confused as to why the following is not true.

I was wondering why the Cartesian product of sets whose elements are ordered pairs is not a set of ordered pairs of ordered pairs.

Cartesian Product of sets $A$ and $B$ is defined to be a set $S$ such that $\{(x,y):x \in A\ \text{and}\ y \in B\}$. For set $A := \{\langle a,b\rangle\}$ and set $B := \{c\}$, why is the following not true:

$$A \times B = \{\langle \langle a,b \rangle, c \rangle\}?$$

Since, $\langle a,b \rangle \in A$ and $c\in B$.

$\endgroup$

closed as unclear what you're asking by Lord Shark the Unknown, Xander Henderson, Siong Thye Goh, Shailesh, user91500 Sep 3 '17 at 8:44

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ What does $\langle a,b\rangle$ mean? $\endgroup$ – Ennar Sep 2 '17 at 12:53
  • $\begingroup$ it can only take one element per set ... $\endgroup$ – user451844 Sep 2 '17 at 12:58
  • $\begingroup$ @Ennar sorry about that i meant an ordered pair (a,b) $\endgroup$ – chittychitty Sep 2 '17 at 13:04
  • $\begingroup$ Then it is true, who told you it isn't? $\endgroup$ – Ennar Sep 2 '17 at 13:07
  • 1
    $\begingroup$ What you have written seems to be true. $\endgroup$ – астон вілла олоф мэллбэрг Sep 2 '17 at 13:10
1
$\begingroup$

I was wondering why the Cartesian product of sets whose elements are ordered pairs is not a set of ordered pairs of ordered pairs.

Let $X\subseteq A\times B$ and $Y\subseteq C\times D$. Then, \begin{align}X\times Y\subseteq (A\times B)\times (C\times D) &= \{(x,y)\mid x\in A\times B,\ y\in C\times D\}\\ &= \{((a,b),(c,d))\mid a\in A,\ b\in B,\ c\in C,\ d\in D\}.\end{align}

What you actually seem to be confused about is why $A\times B\times C \neq (A\times B)\times C$ as sets.

Now, this really depends on how you define Cartesian product of multiple sets. One way actually is just to define $A\times B\times C = (A\times B)\times C$ and then the question is moved to why $$(A\times B)\times C\neq A\times (B\times C).$$

Another way can be $$A_1\times A_2\times A_3 := \{f\colon \{1,2,3\}\to A_1\cup A_2\cup A_3\mid f(i)\in A_i,\ i = 1,2,3\}$$ and then indeed it is not true that $A\times B\times C = (A\times B)\times C.$

You should clarify the definitions you are using.

However, there are natural bijections between sets $$\{(a,b,c)\mid a\in A,\ b\in B,\ c\in C\},\ \{((a,b),c)\mid a\in A,\ b\in B,\ c\in C\}\ \text{and}\ \{(a,(b,c))\mid a\in A,\ b\in B,\ c\in C\}.$$ Can you find them?

Because of those natural bijections, we tend to just write $$A\times B\times C = (A\times B)\times C = A\times (B\times C)$$ even though it is not technically correct.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.