Bounded Sets and Bounded Functions in a Metric Space

Bounded Sets in a Metric Space

Definition: Let $(M,d)$ be a metric space. A subset $S\subseteq M$ is said to be Bounded if there exists a positive real number $r>0$ such that $S\subseteq B(x,r)$ for some $x\in M$. The set $S$ is said to be Unbounded if it is not bounded.

By the definition above, we see that $S$ is bounded if there exists some open ball with a finite radius that contains $S$.

For example, consider the metric space $(M,d)$ where $d$ is the discrete metric defined for all $x,y\in M$ by:

${\begin{aligned}\quad d(x,y)=\left\{{\begin{matrix}0&\mathrm {if} \;x=y\\1&\mathrm {if} \;x\neq y\end{matrix}}\right.\end{aligned}}$

Let $S\subseteq M$. Then by the definition of the discrete metric, for all $x,y\in S$ we have that $d(x,y)\leq 1$. Therefore, if we consider any point $x\in S$ and take $r=2$ then:

${\begin{aligned}\quad S\subseteq B(x,2)\end{aligned}}$

Therefore, $S$ is bounded. This shows that every subset of $M$ is bounded with respect to the discrete metric. In fact, the wholeset $M$ is also bounded and $M\subseteq B(x,2)$ for any $x\in M$.

For another example, consider the metric space $(\mathbb {R} ^{3},d)$ where $d$ is the Euclidean metric. Consider the following set:

${\begin{aligned}\quad S=\{(x,y,z)\in \mathbb {R} ^{3}:x,y,z\geq 0\}\subseteq \mathbb {R} ^{3}\end{aligned}}$

The set $S$ above is the first octant of $\mathbb {R} ^{3}$, and is actually unbounded. To prove this, suppose that instead $S$ is bounded. Then there exists a maximal distance between some pair of points $\mathbf {x} ,\mathbf {y} \in S$, say:

${\begin{aligned}\quad d_{\mathrm {max} }=d(\mathbf {x} ,\mathbf {y} )\end{aligned}}$

Then $S\subseteq B(x,d_{\mathrm {max} })$. For $k>1$ consider the point $k\mathbf {y} =(kx_{1},kx_{2},kx_{3})\in \mathbb {R} ^{3}$. Then:

${\begin{aligned}\quad d(\mathbf {x} ,k\mathbf {y} )={\sqrt {(x_{1}-ky_{1})^{2}+(x_{2}-ky_{2})^{2}+(x_{3}-ky_{3})^{2}}}\end{aligned}}$

Since $(x_{i}-ky_{1})^{2}\geq (x_{i}-y_{i})^{2}$ for each $i\in \{1,2,3\}$ we see that:

${\begin{aligned}\quad d(\mathbf {x} ,k\mathbf {y} )\geq d(\mathbf {x} ,\mathbf {y} )=d_{\mathrm {max} }\end{aligned}}$

But then $\mathbf {y} \not \in B(\mathbf {x} ,d_{\mathrm {max} })$ implies $\mathbf {y} \not \in S$ which is a contradiction. Therefore our assumption that $S$ was bounded is false.

Bounded Functions

A schematic illustration of a bounded function (red) and an unbounded one (blue). Intuitively, the graph of a bounded function stays within a horizontal band, while the graph of an unbounded function does not.

In mathematics, a function f defined on some set X with real or complex values is called bounded if the set of its values is bounded. In other words, there exists a real number M such that

$|f(x)|\leq M$

for all x in X. A function that is not bounded is said to be unbounded.

If f is real-valued and f(x) ≤ A for all x in X, then the function is said to be bounded (from) above by A. If f(x) ≥ B for all x in X, then the function is said to be bounded (from) below by B. A real-valued function is bounded if and only if it is bounded from above and below.

An important special case is a bounded sequence, where X is taken to be the set N of natural numbers. Thus a sequence f = (a₀, a₁, a₂, ...) is bounded if there exists a real number M such that

$|a_{n}|\leq M$

for every natural number n. The set of all bounded sequences forms the sequence space $l^{\infty }$.

The definition of boundedness can be generalized to functions f : X → Y taking values in a more general space Y by requiring that the image f(X) is a bounded set in Y.

Related notions

Weaker than boundedness is local boundedness. A family of bounded functions may be uniformly bounded.

A bounded operator T : X → Y is not a bounded function in the sense of this page's definition (unless T = 0), but has the weaker property of preserving boundedness: Bounded sets M ⊆ X are mapped to bounded sets T(M) ⊆ Y. This definition can be extended to any function f : X → Y if X and Y allow for the concept of a bounded set. Boundedness can also be determined by looking at a graph.

Examples

• The sine function sin : R → R is bounded since $|\sin(x)|\leq 1$ for all $x\in \mathbf {R}$.
• The function $f(x)=(x^{2}-1)^{-1}$, defined for all real x except for −1 and 1, is unbounded. As x approaches −1 or 1, the values of this function get larger and larger in magnitude. This function can be made bounded if one considers its domain to be, for example, [2, ∞) or (−∞, −2].

• The function ${\textstyle f(x)=(x^{2}+1)^{-1}}$, defined for all real x, is bounded.
• The inverse trigonometric function arctangent defined as: y = arctan(x) or x = tan(y)}} is increasing for all real numbers x and bounded with −${\tfrac {\pi }{2}}$ < y < ${\tfrac {\pi }{2}}$ radians
• By the boundedness theorem, every continuous function on a closed interval, such as f : [0, 1] → R, is bounded. More generally, any continuous function from a compact space into a metric space is bounded.
• All complex-valued functions f : C → C which are entire are either unbounded or constant as a consequence of Liouville's theorem. In particular, the complex sin : C → C must be unbounded since it is entire.
• The function f which takes the value 0 for x rational number and 1 for x irrational number (cf.Dirichlet function) is bounded. Thus, a function does not need to be "nice" in order to be bounded. The set of all bounded functions defined on [0, 1] is much larger than the set of continuous functions on that interval. Moreover, continuous functions need not be bounded; for example, the functions $g:\mathbb {R} ^{2}\to \mathbb {R}$ and $h:(0,1)^{2}\to \mathbb {R}$ defined by $g(x,y):=x+y$ and $h(x,y):={\frac {1}{x+y}}$ are both continuous, but neither is bounded. (However, a continuous function must be bounded if its domain is both closed and bounded.)

Licensing

Content obtained and/or adapted from:

• Bounded Sets in a Metric Space, mathonline.wikidot.com under a CC BY-SA license
• Bounded function, Wikipedia under a CC BY-SA license