Image of Intersection under Mapping/Proof 2

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Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping.

Let $S_1$ and $S_2$ be subsets of $S$.


Then:

$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$


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

$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \subseteq \map {f^\to} {S_1} \cap \map {f^\to} {S_2}$


Proof

As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation:

$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

$\blacksquare$


Sources