Image under Subset of Relation is Subset of Image under Relation
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Theorem
Let $S$ and $T$ be sets.
Let $\RR_1 \subseteq S \times T$ be a relation in $S \times T$.
Let $\RR_2 \subseteq \RR_1$.
Let $A \subseteq S$.
Then:
- $\RR_2 \sqbrk A \subseteq \RR_1 \sqbrk A$
where $\RR_1 \sqbrk A$ denotes the image of $A$ under $\RR_1$.
Corollary
Let $x \in S$.
Then:
- $\map {\RR_2} x \subseteq \map {\RR_1} x$
where $\map {\RR_1} x$ denotes the image of $x$ under $\RR_1$.
Proof
\(\ds y\) | \(\in\) | \(\ds \RR_2 \sqbrk A\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in A: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR_2\) | Definition of Image of Subset under Relation | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in A: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR_1\) | Definition of Subset | |||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds \RR_1 \sqbrk A\) | Definition of Image of Subset under Relation |
$\blacksquare$
Sources
- 1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Relations