# 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

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 21.4 \ \text{(i)}$: The image of a subset of the domain; surjections

- 1955: John L. Kelley:
*General Topology*... (previous) ... (next): Chapter $0$: Functions - 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $\S 2$ - 2005: René L. Schilling:
*Measures, Integrals and Martingales*... (previous) ... (next): $\S 2$: Problem $4 \ \text{(i)}$