Why is the inverse of a relation defined that way?

Question

Suppose we have three sets $A,B$, $R$, and $R \subseteq A \times B$. $R$ is a relation. According to the book I'm currently studying, inverse of the relation is defined as follows:

$R^{-1} = \{(b,a) \in B \times A \mid (a,b) \in R\}$

Why do you need to specify $B \times A$ part?

If I'm getting this correctly: if condition $(a,b) \in R$ holds, then $(b,a) \in B \times A$ will hold regardless.

In an attempt to be more rigorous, I decided to proof this conjecture.

Suppose $(a,b) \in R$. Then $(a,b) \in A \times B$, which means $a \in A$ and $b \in B$. And since $A,B$ are non-empty, then $B \times A$ is non-empty and $(b,a) \in B \times A$.

Hence $(a,b) \in R \implies (b,a) \in B \times A$. $\Box$

So provided my proof is correct, I think it would be more concise (and equally accurate) if the definition went like this:

$$R^{-1} = \{(b,a) \mid (a,b) \in R\}$$

So repeating my question, is it necessary to mention $(b,a) \in B \times A$ when defining inverse?

Answer 1

Yes, you are right: given $R \subseteq A \times B$, $\{(b,a) \in B \times A \mid (a,b) \in R\}$ and $\{(b,a) \mid (a,b) \in R\}$ denote the same set $R^{-1}$, i.e. $\{(b,a) \in B \times A \mid (a,b) \in R\} = \{(b,a) \mid (a,b) \in R\}$. To prove that you have to show:

• $\{(b,a) \in B \times A \mid (a,b) \in R\} \subseteq \{(b,a) \mid (a,b) \in R\}$, which is trivial.
• $\{(b,a) \mid (a,b) \in R\} \subseteq \{(b,a) \in B \times A \mid (a,b) \in R\}$, which is almost trivial, and your proof is essentially correct. Note that in your argument, the fact that $A$ and $B$ are not empty and hence $B \times A$ is not empty are irrelevant. The argument works even in the case where $A$ and $B$ (and hence $R$ and $R^{-1}$) are empty.

Answer 2

What do you want to prove here?

For any pair of sets $A,B$ one can define the cartesian products $A\times B$ and $B\times A$.

If $R$ is a relation with $R\subseteq A\times B$, the by definition $R^{-1}$ is a relation with $R^{-1}\subseteq B\times A$.