# Image of Intersection under Mapping

## 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:

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

That is, the image of the intersection of subsets of a mapping is a subset of the intersection of their images.

### General Result

Let $S$ and $T$ be sets.

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

Let $\powerset S$ be the power set of $S$.

Let $\mathbb S \subseteq \powerset S$.

Then:

$\ds f \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$

### Family of Sets

Let $S$ and $T$ be sets.

Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.

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

Then:

$\ds f \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} f \sqbrk {S_i}$

where $\ds \bigcap_{i \mathop \in I} S_i$ denotes the intersection of $\family {S_i}_{i \mathop \in I}$.

## Proof 1

 $\ds S_1 \cap S_2$ $\subseteq$ $\ds S_1$ Intersection is Subset $\ds \leadsto \ \$ $\ds f \sqbrk {S_1 \cap S_2}$ $\subseteq$ $\ds f \sqbrk {S_1}$ Image of Subset under Mapping is Subset of Image

 $\ds S_1 \cap S_2$ $\subseteq$ $\ds S_2$ Intersection is Subset $\ds \leadsto \ \$ $\ds f \sqbrk {S_1 \cap S_2}$ $\subseteq$ $\ds f \sqbrk {S_2}$ Image of Subset under Mapping is Subset of Image

 $\ds \leadsto \ \$ $\ds f \sqbrk {S_1 \cap S_2}$ $\subseteq$ $\ds f \sqbrk {S_1} \cap f \sqbrk {S_2}$ Intersection is Largest Subset

$\blacksquare$

## Proof 2

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

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

$\blacksquare$

## Examples

### Square Function

Let:

$S_1 = \set {x \in \Z: x \le 0}$
$S_2 = \set {x \in \Z: x \ge 0}$
$f: \Z \to \Z: \forall x \in \Z: \map f x = x^2$

We have:

$f \sqbrk {S_1} = \set {0, 1, 4, 9, 16, \ldots} = f \sqbrk {S_2}$

Then:

$f \sqbrk {S_1} \cap f \sqbrk {S_2} = \set {0, 1, 4, 9, 16, \ldots}$

but:

$f \sqbrk {S_1 \cap S_2} = f \sqbrk {\set 0} = \set 0$

As can be seen, the inclusion is proper, that is:

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

### First Projection on Subsets of $\N \times \N$

Let $\pr_1: \N \times \N \to \N$ denote the first projection from the cartesian space $\N \times \N$ of the natural numbers.

Let:

 $\ds S_1$ $=$ $\ds \set {\tuple {m, 1}: m \in \N}$ $\ds S_2$ $=$ $\ds \set {\tuple {0, 2 n}: n \in \N}$

First note that we have:

 $\ds S_1 \cap S_1$ $=$ $\ds \set {\tuple {m, 1}: m \in \N} \cap \set {\tuple {0, 2 n}: n \in \N}$ $\ds$ $=$ $\ds \O$ as $1$ is not an integer of the form $2 n$

Then:

 $\ds \pr_1 \sqbrk {S_1}$ $=$ $\ds \set {m: m \in \N}$ $\ds$ $=$ $\ds \N$ and: $\ds \pr_1 \sqbrk {S_2}$ $=$ $\ds \set 0$ $\ds$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \pr_1 \sqbrk {S_1} \cap \pr_1 \sqbrk {S_2}$ $=$ $\ds \N \cap 0$ $\ds$ $=$ $\ds 0$

while:

 $\ds \pr_1 \sqbrk {S_1 \cap S_2}$ $=$ $\ds \pr_1 \sqbrk {\O}$ $\ds$ $=$ $\ds \O$

As can be seen, the inclusion is proper, that is:

$\pr_1 \sqbrk {S_1 \cap S_2} \ne \pr_1 \sqbrk {S_1} \cap \pr_1 \sqbrk {S_2}$