Question about ordered pairs While reading through a book on discrete mathematics, it never occured to me to think of a 1-tuple. That is, an ordered singleton (or ordered unit set). Now, do such sets exist? Using the Kuratowski Definition, the ordered singleton (a) is equal to {{a}}. But as so is the ordered pair (a,a) equal to {{a}}. So is (a,a) = (a)? Could someone point out where my logic is flawed, please. Thank you in advance.
 A: The Kuratowski definition is used for ordered pairs, so $(a)$ is not defined as $\{\{a\}\}$.
$1$-tuples are not defined, since they carry the same amount of information as a single element. The point of defining tuples was to introduce an order into the set. When the set consists of a single element, such orders are useless.
Finally, it is true that $(a,a)=\{\{a\}\}$, which is yet another reason why one wouldn't want to define $1$-tuple the way you mentioned.
A: Having defined ordered pairs (e.g., using the Kuratowski definition $(x, y) = \{\{x\}, \{x, y\}\}$), we define functional relations (as certain sets of ordered pairs) and then define $n$-tuples as functions whose domain is $\{1, \ldots, n\}$. $1$-tuples then show up as functions with domain $\{1\}$, but the notation $(a)$ for the $1$-tuple $\{(1, a)\}$ is never used in practice in my experience.
As suggested in my comment below, you can use angle brackets as a (reasonably standard) substitute for round brackets so that you can write $\langle a \rangle$ to make it clear that that your mean the $1$-tuple with one element $a$.
