Difference of Images under Mapping not necessarily equal to Image of Difference

Theorem

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

The image of the set difference of two subsets of $S$ is not necessarily equal to the set difference of the images.

That is:

Let $S_1$ and $S_2$ be subsets of $S$.

Then it is not always the case that:

$f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$

where $\setminus$ denotes set difference.

Proof

Note that from Image of Set Difference under Mapping:

$f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$

By Proof by Counterexample it is demonstrated that the inclusion does not necessarily apply in the other direction.

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:

$S_1 \setminus S_2 = \set {-1, -2, -3, \ldots}$

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

Then from Set Difference with Self is Empty Set:

$f \sqbrk {S_1} \setminus f \sqbrk {S_2} = \O$

but:

$f \sqbrk {S_1 \setminus S_2} = f \sqbrk {\set {x \in \Z: x > 0} } = \set {1, 4, 9, 16, \ldots}$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $1$
• 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$
• 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 $4 \ \text{(i)}$

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• 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(h)}$