# Image of Intersection under Mapping

## Contents

## 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:

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

### 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:

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

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

## Proof 1

\(\displaystyle S_1 \cap S_2\) | \(\subseteq\) | \(\displaystyle S_1\) | 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 |

$\blacksquare$

## 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:

\(\displaystyle S_1\) | \(=\) | \(\displaystyle \set {\tuple {m, 1}: m \in \N}\) | |||||||||||

\(\displaystyle S_2\) | \(=\) | \(\displaystyle \set {\tuple {0, 2 n}: n \in \N}\) |

First note that we have:

\(\displaystyle S_1 \cap S_1\) | \(=\) | \(\displaystyle \set {\tuple {m, 1}: m \in \N} \cap \set {\tuple {0, 2 n}: n \in \N}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \O\) | as $1$ is not an integer of the form $2 n$ |

Then:

\(\displaystyle \pr_1 \sqbrk {S_1}\) | \(=\) | \(\displaystyle \set {m: m \in \N}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \N\) | |||||||||||

and: | |||||||||||||

\(\displaystyle \pr_1 \sqbrk {S_2}\) | \(=\) | \(\displaystyle \set 0\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \pr_1 \sqbrk {S_1} \cap \pr_1 \sqbrk {S_2}\) | \(=\) | \(\displaystyle \N \cap 0\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) |

while:

\(\displaystyle \pr_1 \sqbrk {S_1 \cap S_2}\) | \(=\) | \(\displaystyle \pr_1 \sqbrk {\O}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \O\) |

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

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

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $12.13 \ \text{(a)}$ - 1968: A.N. Kolmogorov and S.V. Fomin:
*Introductory Real Analysis*... (previous) ... (next): $\S 1.3$: Functions and mappings. Images and preimages: Remark $1$ - 1971: Robert H. Kasriel:
*Undergraduate Topology*... (previous) ... (next): $\S 1.10$: Functions: Exercise $5 \ \text{(f)}$ - 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(iii)}$ - 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(g)}$

- 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 $4 \ \text{(i)}$