# Can only one ordered pair be a relation?

I'm sorry, but I really can't find an answer to this no matter how deep I dig.

A relation is defined as any set of ordered pairs.

But what about a set of only one ordered pair? Is it still a relation? Is it a special kind of relation? I get Binary Relation in some pages but is a Single,Singular,Just-one-ordered-pair relation an actual thing?

I know it's quite elementary but thanks!

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It is. Just consider the case of partial order relation. Not all elements are required to be comparable. –  Troy Woo Aug 25 '14 at 10:23
A single ordered pair is not a relation. But a set that contains a single ordered pair is a relation. –  MJD Aug 25 '14 at 13:32
As a trivial example (but not maximally trivial), consider $A=B=\{0,1\}$. Then the $<$ relation is the set containing the single ordered pair $(0,1)$, and the $>$ relation is $\{(1,0)\}$. –  Scott Aug 25 '14 at 16:52

A binary relation (or relation, means the same) from a set $A$ to a set $B$ is any subset $R\subseteq A\times B$. We take any here seriously so in particular, if $A$ contains some element $a$, and $B$ contains some element $b$, then $R=\{(a,b)\}$, being a subset of $A\times B$ is a relation from $A$ to $B$. For that matter, given any two sets $A$ and $B$, the empty relation $\emptyset \subseteq A\times B$ is always a relation from $A$ to $B$. So not only can relations consists of just one single pair, they can also consists of no pairs at all.

If this seems useless to you, and in a sense we hardly ever really care about such simple relations, then you are, in a sense, correct. However, just because something is not particularly complicated or interesting does not mean we should discard it. For instance, you may argue that the empty set is kinda useless. True, we will never study it since there is not much we can say. but it is useful, for instance to express that two sets are disjoins by saying $A\cap B=\emptyset$. Categorically and notationally, it would be a disaster to discard of it since it will force upon you very cumbersome formulations of results. For roughly the same reasons we like to have all relations, even trivial-looking ones.

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just trust the definition! a relation on a set $A$ is any element of $\mathfrak{P}(A \times A)$. a singleton is such an element. don't confuse with a function $A \to A$, which is a relation which must satisfy two further conditions. what about the empty relation?

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An ordered pair is defined as the following by Kuratowski $(x,y):=\{ \{x\},\{x,y\} \}.$ If $X,Y$ are sets, we define their Cartesian product as $X\times Y = \{\,(x,y)\mid x\in X, \ y\in Y\,\}.$

A set called binary relation if all of its elements are ordered pair. If $R$ is a binary relation we say $(x,y) \in R$ or $xRy$. Sometimes we also speak about the graph of the relation.

There are two extremal situation of a binary relation: the empty set, and Cartesian product of two sets. We can define equity on a set $X$ as $\mathbb{I}_{X} = \{ (x,x) \in X \times X \mid x \in X)$. We also call it the digaonal of $X \times X$.

The answer for your question is yes. Let $X=\{1\}$. Then the equity on $X$ is a binary relation on $X$ which contains only one element. But moreover the equity on empty set is also a binary relation which is empty, since a Cartesian product is empty if and only if one of its factor is empty. Its called empty product on sets.

And last but not least: the name of a set with exactly one element is singleton.

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