A f:K→Rf:K\to\mathbb R on a KK is bounded: there exists M≥0M\geq0 such that

∣f(x)∣≤M(x∈K).|f(x)|\le M\qquad(x\in K).
Proof

If KK is empty the assertion is vacuous. Otherwise the gives a minimum mm and maximum M+M_+; take M=max⁡{∣m∣,∣M+∣}M=\max\{|m|,|M_+|\}.

Scope

No metric on the domain is required. Applied to a compact subset of another space, continuity means continuity for the subspace topology.