Equality of Relations

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Theorem

Let $\mathcal R_1$ and $\mathcal R_2$ be relations on $S_1 \times T_1$ and $S_2 \times T_2$ respectively.

Then $\mathcal R_1$ and $\mathcal R_2$ are equal if and only if:

$S_1 = S_2$
$T_1 = T_2$
$\tuple {s, t} \in \mathcal R_1 \iff \tuple {s, t} \in \mathcal R_2$


It is worth labouring the point that for two relations to be equal, not only must their domains be equal, but so must their codomains.


Proof

This follows from set equality and Equality of Ordered Pairs.

A relation on $S \times T$ is an ordered triple, thus:

\(\displaystyle \mathcal R_1\) \(=\) \(\displaystyle \tuple {S_1, T_1, R_1}\) where $R_1 \subseteq S_1 \times T_1$
\(\displaystyle \mathcal R_2\) \(=\) \(\displaystyle \tuple {S_2, T_2, R_2}\) where $R_2 \subseteq S_2 \times T_2$


From Equality of Ordered Tuples, $\mathcal R_1 = \mathcal R_1$ if and only if:

\(\displaystyle S_1\) \(=\) \(\displaystyle S_2\)
\(\displaystyle T_1\) \(=\) \(\displaystyle T_2\)
\(\displaystyle R_1\) \(=\) \(\displaystyle R_2\)


Then we have that $R_1$ and $R_2$ are sets of ordered pairs.

Thus by definition of set equality:

$R_1 = R_2 \iff \paren {\forall \tuple {x, y} \in \S_1 \times T_2: \tuple {x, y} \in R_1 \iff \tuple {x, y} \in R_2}$

Hence the result.

$\blacksquare$


Also known as

Some sources refer to this concept between two relations as being equivalence, rather than equality.


Sources