Null Relation is Mapping iff Domain is Empty Set
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Theorem
Let $S$ and $T$ be sets.
The null relation $\RR = \O \subseteq S \times T$ is a mapping if and only if $S = \O$.
Proof
Sufficient Condition
Let $S = \O$.
Then the null relation $\RR = \O \subseteq S \times T$ is a mapping from Empty Mapping is Mapping.
$\Box$
Necessary Condition
Suppose $S \ne \O$.
From the definition of an empty set, $S \ne \O \implies \exists x \in S$.
Thus:
\(\ds \RR\) | \(=\) | \(\ds \O\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in S: \neg \exists y \in T: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds \RR\) | Definition of Empty Set | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \RR\) | \(\) | \(\ds \text{ is not a mapping}\) | Definition of Mapping |
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Functions