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 codomain 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.

Sequence
A special case of a bounded mapping is a bounded sequence, where the domain of the mapping is $$\N$$.

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

Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $T$.

Then $$\left \langle {x_n} \right \rangle$$ is bounded iff $$\forall i \in \N$$:
 * $$\exists m \in T: m \preceq x_i$$;
 * $$\exists M \in T: x_i \preceq M$$.

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 $$x \in S$$.

See Bounded Set of Real Numbers‎ for a demonstration that this definition is compatible with boundedness on an ordered set.

Function Attaining its Bounds
If a 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 $$S$$.

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

From the definition of a metric, $$d: X \times X \to \R$$ is a real-valued function.

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

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

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

Metric Subspace
Let $$M = \left({X, d}\right)$$ be a metric space.

Let $$M' = \left({Y, d_Y}\right)$$ be a subspace of $$M$$.

Then $$M'$$ is bounded (in $$M$$) if $$M'$$ is bounded with respect to the subspace metric $$d_Y$$.

Mapping into Metric Space
Let $$M$$ be a metric space.

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

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

Mapping into Real Number Line
Note that as the real numbers form a metric space, we can in theory consider defining boundedness on a real-valued function in terms of boundedness of a mapping into a metric space.

However, as a metric space is itself defined in terms of a real-valued function in the first place, this concept can be criticised as being a circular definition.

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 $$\left|{f \left({z}\right)}\right| \le K$$ 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.

Unbounded
Any space which is not bounded is described as unbounded.

Also see

 * Totally Bounded
 * Uniformly Bounded