# Image of Intersection under Mapping/Proof 1

## 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

$\blacksquare$