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