I have come up with the following conjecture which I am trying to prove:
Let $A$ and $B$ be sets such that $|A|=|B|= 3$ if $R$ is a relation from A to B where $|R|=9$ and $R^{-1}=R$ then $A=B$.
The following is my attempt at a proof:
Proof: Since $|R|=9$, $|A \times B|=|A||B| = 3\cdot3 = 9 $ and since $R \subseteq A\times B$, $R = A \times B$.
First we show that $A \subseteq B$. Let $x \in A$. If $(x, y) \in R$ for some $y \in B$, as $R = R^{-1}$, $(x, y) \in R^{-1}$ and so it follows that $(y, x) \in R$. Since $R = A \times B$, we have that $x \in B$, and as x was arbitrary $A \subseteq B$.
Next we show that $B \subseteq A$. Let $y \in B$. If $(x, y) \in R$ for some $x \in A$, as $R = R^{-1}$, $(x, y) \in R^{-1}$ and so it follows that $(y, x) \in R$. Since $R = A \times B$, it follows that $y \in A$, hence as y was arbitrary $B \subseteq A$.
Since $A \subseteq B$ and $B \subseteq A$, $A = B\space\space\square$
Is this argument correct? My main assumption is that if we have two sets of cardinality 3 then the relation must be the entire Cartesian product $A \times B$. By assuming this I can then work with elements being confident that each $(x, y)$ has $x \in A$ and $y \in B$.
This is correct. You also don't need reverse part separately - if $A \subseteq B$ and $|A| = |B|$ is finite, then $A = B$. The same argument works to show $R = A \times B$: $R \subseteq A \times B$, and $|A \times B| = |A| \cdot |B| = 9 = |R|$.