Definition of the ordered triple (a, b, c) according to Kuratowski's Set Theory. Can someone give Kuratowski's definition of the ordered triple $(a,b,c)$ assuming $A \times B \times C$ is rewritten as $(A \times B) \times C$, please? I noticed there is already an answered question for the ordered $n$-tuple, but (as I'm very new to Maths) I didn't understand it, and I only need the definition for the ordered triple. 
 A: So we define $(x,y)$ to be $\{\{x\}, \{x,y\}\}$, for any $x,y$.
Then $(a,b,c)$ should be seen as $((a,b),c)$ per the hint.
$((a,b),c) = \{\{(a,b)\}, \{(a,b), c\}\}$.
Now expand $(a,b)$ as well and substitute.
A: I know I come a litle late but wanted to point something out. It has been said that you can define a triple as
$$(a,b,c) = \{\{a\},\{a,b\},\{a,b,c\}\}$$
but this definition has a big problem as you can see that for any a,b,c it holds that $(a,b,a)=(a,b,b)$, as they're both equal to $\{\{a\},\{a,b\}\}$.
This is because you haven't paid attention to one extremely important property that you want any ordered pair (or n-tuple really) to have, which is
$$(x,y)=(x',y') \Leftrightarrow x=x',y=y'$$
I believe that's why we define n-tuples recursively and not the way the former comment suggested.
A: Ordered triples are defined recursivley, so that 
$(x,y)=\{\{x\},\{x,y\}\}$ and $(x,y,z)=((x,y),z)$. Observe that $((x,y),z)$ only has two elements, $(x,y)$  and $z$, so we can just apply the definition. To make our lives easier, let $q=(x,y)=\{\{x\},\{x,y\}\}$. Then the substitution is simple: 
$$\begin{align}\label{eqn:einstein} (x,y,z) &= ((x,y),z) \\
&=(q,z) \\
&= \big\{\{q\},\{q,z\}\big\} \\
&=   \big\{\{\{\{x\},\{x,y\}\}\},\{\{\{x\},\{x,y\}\},z\}\big\} \\
\end{align}$$
Likewise, $(x,y,z,w)=((x,y,z),w)$, so letting $q=(x,y,z)$, we can write $(x,y,z,w)=(q,w)= \big\{\{q\},\{q,w\}\big\}$, which we can expand as shown above. 
Substitutions are your friend.  
A: Just track the definition of the ordered pair:
$$\begin{align*}
\langle a,b,c\rangle&=\big\langle\color{blue}{\langle a,b\rangle},c\big\rangle\\
&=\left\langle\color{blue}{\big\{\{a\},\{a,b\}\big\}},c\right\rangle\\
&=\Big\{\left\{\color{blue}{\big\{\{a\},\{a,b\}\big\}}\right\},\left\{\color{blue}{\big\{\{a\},\{a,b\}\big\}},c\right\}\Big\}
\end{align*}$$
(I’ve evaluated it from the inside out; I see now that Henno Brandsma has done it from the outside in. You can take your pick; both work fine.)
