Why is the Cartesian Product not {《x,y 》,z 》}? 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 $c\in B$.
 A: 
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.
