Definition:Bounded

Ordered Set
Let $$\left({S; \preceq}\right)$$ be a poset.

Let $$T \subseteq S$$ be both bounded below and bounded above in $$S$$.

Then $$T$$ is bounded in $$S$$.

Mapping
Let $$\left({T; \preceq}\right)$$ be a poset.

Let $$f: S \to T$$ be a mapping.

Let the range of $$f$$ be bounded.

Then $$f$$ is defined as being bounded.

That is, $$f$$ is bounded if it is both bounded above and bounded below.

Metric Space
Let $$M = \left\{{A, d}\right\}$$ be a metric space.

Let $$S \subseteq M$$.

Then $$S$$ is bounded in $$M$$ iff $$\exists a \in A, K \in \reals: \forall x \in S: d \left({x, a}\right) \le K$$.

It follows immediately that if $$S$$ satisfies the definition for $$a \in \reals$$, it also satisfies it for $$a^{\prime} \in \reals$$, with $$K$$ replaced by $$K^{\prime} = K + d \left({a, a^{\prime}}\right)$$.

This is because $$d \left({x, a}\right) \le K \Longrightarrow d \left({x, a^{\prime}}\right) \le d \left({x, a}\right) + d \left({a, a^{\prime}}\right) \le K + d \left({a, a^{\prime}}\right)$$.

Mapping into Metric Space
Let $$M = \left\{{A, d}\right\}$$ be a metric space.

Let $$f: X \to M$$ be a mapping from any set $$X$$ into $$M$$

Let $$f \left({X}\right)$$ be bounded in $M$.

Then $$f$$ is bounded.

Real Function
The definition for bounded as applied to a real function is the same as for a mapping.

Note that it follows from Bounded Set of Real Numbers‎ that $$f$$ is bounded on a set $$S \subseteq \reals$$ iff $$\exists K \in \reals: \forall x \in S: \left|{f \left({x}\right)}\right| \le K$$.

This also follows directly from Real Number Line is Metric Space.

Complex Function
The definition for bounded as applied to a complex function follows from the definition for a bounded mapping into a metric space.

Let $$f: \mathbb{C} \to \mathbb{C}$$ be a complex function.

From Complex Plane is Metric Space, it follows that $$f$$ is bounded on a set $$S \subseteq \mathbb{C}$$ iff $$\exists K \in \reals: \forall x \in S: \left|{f \left({x}\right)}\right| \le K$$.