Mapping Images are Disjoint only if Domains are Disjoint
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Theorem
Let $S$ and $T$ be sets.
Let:
- $f \sqbrk S \cap f \sqbrk T = \O$
where $f \sqbrk S$ denotes the image set of $S$.
Then:
- $S \cap T = \O$
Proof
From Image of Intersection under Mapping:
- $f \sqbrk {S \cap T} \subseteq f \sqbrk S \cap f \sqbrk T$
From Empty Set is Subset of All Sets:
- $f \sqbrk {S \cap T} = \O$
From Image of Subset under Mapping is Subset of Image:
- $S \cap T = \O$
$\blacksquare$
Sources
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.10$: Functions: Exercise $5 \ \text{(c)}$