Image of Element under Composite Relation with Common Codomain and Domain

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Theorem

Let $\RR_1 \subseteq S \times T$ and $\RR_2 \subseteq T \times U$ be relations.

Let $\RR_2 \circ \RR_1 \subseteq S \times U$ be the composition of $\RR_1$ and $\RR_2$.

Let $x \in S$.

Then:

$\RR_2 \sqbrk {\map {\RR_1} x} = \map {\paren{\RR_2 \circ \RR_1}} x$

Proof

We have:

\(\ds \RR_2 \sqbrk {\map {\RR_1} x}\)

\(=\)

\(\ds \RR_2 \sqbrk {\RR_1 \sqbrk {\set x} }\)

Image of Singleton under Relation

\(\ds \)

\(=\)

\(\ds \paren{\RR_2 \circ \RR_1} \sqbrk {\set x}\)

Image of Subset under Composite Relation with Common Codomain and Domain

\(\ds \)

\(=\)

\(\ds \map {\paren{\RR_2 \circ \RR_1} } x\)

Image of Singleton under Relation

$\blacksquare$

Also see

• Image of Subset under Composite Relation with Common Codomain and Domain

• Image of Element under Composite Relation

• Image of Subset under Composite Relation