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How many relations in ${\mathscr P}(A \times A)$ defined on $A$ are symmetric?

$A$ = $\{1,2,...,n\}$ and ${\mathscr P}(A \times A)$ is the power set of the cartesian product of $A$ with itself.

A symmetric relation in this case would be $\forall x \forall y, \{x,y\} \in X \implies \{y,x\} \in X$ for $X$ in ${\mathscr P}(A \times A)$.

EDIT: I think the answer is $2^{\frac{n(n+1)}{2}}$, because you have $2^{\frac{n(n-1)}{2}}$ number of sets containing only $\{x, y\}$ where $x < y$ and $2^n$ number of sets containing only $\{x, y\}$ where $x = y$

Jack Black
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1 Answers1

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Let $\mathscr{S}$ be the set of all symmetric relations on $A=\{1,2,\dots ,n\},$ and define $V=\{\langle x, y\rangle \in A\times A \mid x\le y\}.$

Define $f\colon \mathscr{S} \to\mathscr{P}(V)$ by setting $f(R)=R\cap V$ for any symmetric relation $R$ on $A.$ Then $f$ is a bijection between $\mathscr{S}$ and $\mathscr{P}(V).$

Since $\lvert V \rvert=\dfrac{n(n+1)}{2},$ it follows that \begin{align}\lvert \mathscr{S} \rvert &= \lvert \mathscr{P}(V) \rvert\\&=2^{\lvert V \rvert} \\&=2^{n(n+1)/2}. \end{align}

Mitchell Spector
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