# Why is the inverse of a relation defined that way?

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?

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.

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 $$R^{-1}\subseteq B\times A$$.

• OK, I will put it differently: are these two definitions the same? $R^{-1} = \{(b,a) \in B \times A \mid (a,b) \in R\}$ and $R^{-1} = \{(b,a) \mid (a,b) \in R\}$ – Nelver Oct 8 '19 at 8:02
• Indeed, they are the same. But the first one is preferable. – Wuestenfux Oct 8 '19 at 8:06
• Why is the first one preferable? – Nelver Oct 8 '19 at 8:07