Preimage of Image under Left-Total Relation is Superset

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Let $\RR \subseteq S \times T$ be a left-total relation.


$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$


$\RR \sqbrk A$ denotes the image of $A$ under $\RR$
$\RR^{-1} \sqbrk A$ denotes the preimage of $A$ under $\RR$
$\RR^{-1} \circ \RR$ denotes composition of $\RR^{-1}$ and $\RR$.

This can be expressed in the language and notation of direct image mappings and inverse image mappings as:

$\forall A \in \powerset S: A \subseteq \map {\paren {\RR^\gets \circ \RR^\to} } A$


Suppose $A \subseteq S$.

We have:

\(\ds x\) \(\in\) \(\ds A\)
\(\ds \leadsto \ \ \) \(\ds \set x\) \(\subseteq\) \(\ds A\) Singleton of Element is Subset
\(\ds \leadsto \ \ \) \(\ds \RR \sqbrk {\set x}\) \(\subseteq\) \(\ds \RR \sqbrk A\) Image of Subset under Relation is Subset of Image
\(\ds \leadsto \ \ \) \(\ds \RR^{-1} \sqbrk {\RR \sqbrk {\set x} }\) \(\subseteq\) \(\ds \RR^{-1} \sqbrk {\RR \sqbrk A}\) Image of Subset under Relation is Subset of Image: Corollary 1

Then we have that $\RR$ is a left-total relation.


$\exists t \in T: \tuple {x, t} \in \RR$

Hence by definition of inverse relation:

$\exists t \in T: \tuple {t, x} \in \RR^{-1}$


\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \RR^{-1} \sqbrk {\RR \sqbrk A}\) Definition of Relation
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \paren {\RR^{-1} \circ \RR} \sqbrk A\) Definition of Composition of Relations

So by definition of subset:

$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$


Also see