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
A metric space $$\left({S, d}\right)$$ is called bounded if there exists $$a \in S$$ and $$K \in \R$$ such that $$d \left({x, a}\right) \le K$$ for all $$x \in S$$.

It follows immediately that, if $$S$$ satisfies this condition for one $$a \in S$$, then it does so for all $$a'$$, with $$K$$ replaced by $$K^{\prime} = K + d \left({a, a^{\prime}}\right)$$.

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

If $$M$$ is a metric space and $$S$$ is a subset of $$M$$, then we say that $$S$$ is bounded (in $$M$$) if $$S$$ is bounded with respect to the subspace metric.

Mapping into Metric Space
Let $$M = \left({A, d}\right)$$ be a metric space, and suppose that $$f: X \to M$$ is a mapping from any set $$X$$ into $$M$$.

Then $$f$$ is called bounded if $$f \left({X}\right)$$ is bounded in $M$.

Totally Bounded Metric Space
A metric space $$\left({S, d}\right)$$ is called totally bounded if, for every $$\varepsilon > 0$$, there exist finitely many points $$x_0, \dots, x_n \in S$$ such that
 * $$\inf_{0 \leq i \leq n} d \left({x_i, x}\right) \leq \varepsilon$$

for all $$x\in S$$.

Note:
 * Any totally bounded metric space is also bounded, but the converse is not true. (The simplest example is a countable set with the discrete metric.)


 * A metric space is compact if and only if it is complete and totally bounded.

Real-valued Function
A real-valued function $$f: S \to \R$$ is bounded if there is a number $$K \ge 0$$ such that $$\left|{f \left({x}\right)}\right| \le K$$ for all $$x \in S$$.

Note that this coincides with the above definitions of bounded functions (using the standard ordering / metric on $$\R$$).

See Bounded Set of Real Numbers‎ and Real Number Line is Metric Space.

Function Attaining its Bounds
If a map real-valued function $$f: S \to \R$$ is bounded, then $$f \left({S}\right)$$ is by definition a bounded subset of $$\R$$, and hence has a supremum and infimum.

These are the bounds on $$f \left({S}\right)$$, which may or may not be in $$f \left({S}\right)$$.

If $$\inf \left({f \left({S}\right)}\right) \in f \left({S}\right)$$ and $$\sup \left({f \left({S}\right)}\right) \in f \left({S}\right)$$, then $$f$$ attains its bounds on $$C$$.

Complex-Valued Function
A complex-valued function $$f: S \to \C$$ is called bounded if the real-valued function $$\left|{f}\right|: S \to \R$$ is bounded, where $$\left|{f}\right|$$ is the modulus of $$f$$.

That is, $$f$$ is bounded if there is a constant $$K \ge 0$$ such that $$\left|{f \left({z}\right)}\right| \leq K$$ for all $$z \in S$$.

This coincides with the definition of a bounded mapping into a metric space, using the standard metric on $$\C$$.

See Complex Plane is Metric Space.