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