TITLE: How to write 1 tuple as sets?
QUESTION [5 upvotes]: A 2-tuple is $(a, b) = \{\{a\}, \{a, b\}\}$
What about a 1 tuple? Is it $(a) = \{\{a\}\}$?
Then if $a = b$: $(a, b) = (a, a) = \{\{a\}, \{a, a\}\} = \{\{a\}, \{a\}\} = \{\{a\}\}$
So $(a, a) = (a)$?

REPLY [5 votes]: In set theory, we usually (see page 18 in Hrbacek, K. and Jech, T., 1999 and page 7 in Jech, T., 2000) define $n$ tuples as follows.
Definition 1. Suppose $n$ is a natural number and $x_0,\cdots,x_{n-1}$ are sets. Set
\begin{align*}
(\,)&:=\varnothing,\\
(x_0)&:=x_0,\\
(x_0,x_1)&:=\{\{x_0\},\{x_0,x_1\}\},\\
\vdots\quad\qquad&\qquad\qquad~\vdots\\
(x_0,\cdots,x_{n-2},x_{n-1})&:=((x_0,\cdots,x_{n-2}),x_{n-1}).
\end{align*}
Remark 2. As you see, $0$ tuples and $1$ tuples has nothing to do with Kuratowski's definition for ordered pairs$^*$, while $n$ tuples with $n\geq 2$ are based on Kuratowski's definition for ordered pairs. The reason that we set $0$ tuples to be the empty set is clear, and the reason that we set $1$ tuples to be the element laying in itself is to coincide with definitions for unary relations (any subset of $X$ is a unary relation on $X$) and with definitions for unary Cartesian products ($X^1=X$ which selects elements of itself as its elements).
Furthermore, keeping the definitions for $0$ tuples, $1$ tuples and $2$ tuples unchanged, we can also define $n$ tuples with $n\geq 3$ to be functions with domain $n$.
Definition 3. Suppose $n$ is a natural number no less than $3$ and $x_0,\cdots,x_{n-1}$ are sets. Set
$$(x_0,\cdots,x_{n-1}):=\{(0,x_0),\cdots,(n-1,x_{n-1})\}.$$
Remark 4. Note that $2$ tuples (ordered pairs$^\dagger$) can't be defined as functions with domain $2$, since the definitions for functions depends on relations while the definitions for relations depends on $2$ tuples, and so the definition for $2$ tuples should only be Kuratowski's or similar (for example, Hausdorff's or Wiener's) way.
Remark 5. Both of the two definitions for $n$ tuples with $n\geq 3$ have their own advantages: (1) Definition 1 has real order while Definition 3 has imaginary order since the elements of functions have no order and the imaginary order follows from the order of ordinals (natural numbers); (2) Definition 3 is clearly simpler than the first one. In fact, since we also define sequences as functions with ordinals as domains, if we adopt Definition 1, then tuples can only be regarded as finite sequences while in fact not, and if we adopt Definition 3 then $n$ tuples are indeed $n$ sequences (finite sequences).
Anyway, $(a,a)=\{\{a\}\}\neq a=(a)$. Note that by Axiom of Foundation/Regularity, we can by induction show that: if $n$ is a non-zero natural number, then $a\neq \underbrace{\{\cdots\{}_{n\text{ many}}a\underbrace{\}\cdots\}}_{n\text{ many}}$.

$*$ By the way, Kuratowski's definition for ordered pairs is great although it seems to be very simple at first glance since it transforms the order of elements laying in the ordered pairs into the difference of sets (not difference operation of sets) of elements laying in the ordered pairs,  and based on it we can define other basic mathematical concepts such as relations, functions and so on.
$\dagger$ Considering of no mathematical definition, there are many equivalent names to $2$ tuples: couple, double, ordered pair, two-ple, twin, dual, duad, dyad, twosome. So it's better to define $2$ tuples and ordered pairs to be the same object, while @Gregory Nisbet 's new definition for $(a,b)$ (i.e., 2 tuples) and definition for $(a,b)_K$ (i.e., Kuratowski's definition for ordered pairs) are different objects. Without this, it's also ok to define $2$ tuples in @Gregory Nisbet 's new way. By the way, in fact @Gregory Nisbet 's new way is the same as Definition 3 for $n$ tuples with $n\geq 3$ as above.