Image of Intersection under Mapping/Examples/Square Function

Example of Image of Intersection under Mapping

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:

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

Also see

• Image of Intersection under Injection: equality holds if and only if $f$ is an injection.

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 21$: The image of a subset of the domain; surjections: Postscript to $\S 21.4 \ \text{(i)}$