Image of Intersection under Mapping/Examples

Examples of Image of Intersection under Mapping

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:

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