Real-Valued Function on Finite Set is Bounded
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Theorem
Let $S$ be a finite set.
Let $f: S \to \R$ be a real-valued function on $S$.
Then $f$ is bounded.
Proof
Let $K$ be defined as:
- $K = \ds \max_{x \mathop \in S} \size {\map f x}$
where $\size {\map f x}$ denotes the absolute value of $\map f x$.
Then trivially:
- $\exists K \in \R_{\ge 0}: \forall x \in S: \size {\map f x} \le K$
This is the definition of a bounded real-valued function.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.1$: Motivation: Step $1$