Image of Intersection under Mapping/Proof 1

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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

\(\displaystyle S_1 \cap S_2\) \(\subseteq\) \(\displaystyle S_1\) Intersection is Subset
\(\displaystyle \leadsto \ \ \) \(\displaystyle f \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\displaystyle f \sqbrk {S_1}\) Image of Subset under Mapping is Subset of Image


\(\displaystyle S_1 \cap S_2\) \(\subseteq\) \(\displaystyle S_2\) Intersection is Subset
\(\displaystyle \leadsto \ \ \) \(\displaystyle f \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\displaystyle f \sqbrk {S_2}\) Image of Subset under Mapping is Subset of Image


\(\displaystyle \leadsto \ \ \) \(\displaystyle f \sqbrk {S_1 \cap S_2}\) \(\subseteq\) \(\displaystyle f \sqbrk {S_1} \cap f \sqbrk {S_2}\) Intersection is Largest Subset

$\blacksquare$


Sources