View source for Bounded Sets and Bounded Functions in a Metric Space

=Bounded Sets in a Metric Space=
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<td><strong>Definition:</strong> Let <span class="math-inline"><math>(M, d)</math></span> be a metric space. A subset <span class="math-inline"><math>S \subseteq M</math></span> is said to be <strong>Bounded</strong> if there exists a positive real number <span class="math-inline"><math>r > 0</math></span> such that <span class="math-inline"><math>S \subseteq B(x, r)</math></span> for some <span class="math-inline"><math>x \in M</math></span>. The set <span class="math-inline"><math>S</math></span> is said to be <strong>Unbounded</strong> if it is not bounded.</td>
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<p>By the definition above, we see that <span class="math-inline"><math>S</math></span> is bounded if there exists some open ball with a finite radius that contains <span class="math-inline"><math>S</math></span>.</p>
<p>For example, consider the metric space <span class="math-inline"><math>(M, d)</math></span> where <span class="math-inline"><math>d</math></span> is the discrete metric defined for all <span class="math-inline"><math>x, y \in M</math></span> by:</p>
<div style="text-align: center;"><math>\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} \; x = y\\ 1 & \mathrm{if} \; x \neq y \end{matrix}\right. \end{align}</math></div>
<p>Let <span class="math-inline"><math>S \subseteq M</math></span>. Then by the definition of the discrete metric, for all <span class="math-inline"><math>x, y \in S</math></span> we have that <span class="math-inline"><math>d(x, y) \leq 1</math></span>. Therefore, if we consider any point <span class="math-inline"><math>x \in S</math></span> and take <span class="math-inline"><math>r = 2</math></span> then:</p>
<div style="text-align: center;"><math>\begin{align} \quad S \subseteq B(x, 2) \end{align}</math></div>
<p>Therefore, <span class="math-inline"><math>S</math></span> is bounded. This shows that every subset of <span class="math-inline"><math>M</math></span> is bounded with respect to the discrete metric. In fact, the wholeset <span class="math-inline"><math>M</math></span> is also bounded and <span class="math-inline"><math>M \subseteq B(x, 2)</math></span> for any <span class="math-inline"><math>x \in M</math></span>.</p>
<p>For another example, consider the metric space <span class="math-inline"><math>(\mathbb{R}^3, d)</math></span> where <span class="math-inline"><math>d</math></span> is the Euclidean metric. Consider the following set:</p>
<div style="text-align: center;"><math>\begin{align} \quad S = \{ (x, y, z) \in \mathbb{R}^3 : x, y, z \geq 0 \} \subseteq \mathbb{R}^3 \end{align}</math></div>
<p>The set <span class="math-inline"><math>S</math></span> above is the first octant of <span class="math-inline"><math>\mathbb{R}^3</math></span>, and is actually unbounded. To prove this, suppose that instead <span class="math-inline"><math>S</math></span> is bounded. Then there exists a maximal distance between some pair of points <span class="math-inline"><math>\mathbf{x}, \mathbf{y} \in S</math></span>, say:</p>
<div style="text-align: center;"><math>\begin{align} \quad d_{\mathrm{max}} = d(\mathbf{x}, \mathbf{y}) \end{align}</math></div>
<p>Then <span class="math-inline"><math>S \subseteq B(x, d_{\mathrm{max}})</math></span>. For <span class="math-inline"><math>k > 1</math></span> consider the point <span class="math-inline"><math>k\mathbf{y} = (kx_1, kx_2, kx_3) \in \mathbb{R}^3</math></span>. Then:</p>
<div style="text-align: center;"><math>\begin{align} \quad d(\mathbf{x}, k\mathbf{y}) = \sqrt{(x_1 - ky_1)^2 + (x_2 - ky_2)^2 + (x_3 - ky_3)^2} \end{align}</math></div>
<p>Since <span class="math-inline"><math>(x_i - ky_1)^2 \geq (x_i - y_i)^2</math></span> for each <span class="math-inline"><math>i \in \{1, 2, 3 \}</math></span> we see that:</p>
<div style="text-align: center;"><math>\begin{align} \quad d(\mathbf{x}, k\mathbf{y}) \geq d(\mathbf{x}, \mathbf{y}) = d_{\mathrm{max}} \end{align}</math></div>
<p>But then <span class="math-inline"><math>\mathbf{y} \not \in B(\mathbf{x}, d_{\mathrm{max}})</math></span> implies <span class="math-inline"><math>\mathbf{y} \not \in S</math></span> which is a contradiction. Therefore our assumption that <span class="math-inline"><math>S</math></span> was bounded is false.</p>

= Bounded Functions=
[[Image:Bounded and unbounded functions.svg|right|thumb|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 
:<math>|f(x)|\le M</math>
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'' = (''a''<sub>0</sub>, ''a''<sub>1</sub>,  ''a''<sub>2</sub>, ...) is bounded if there exists a real number ''M'' such that

:<math>|a_n|\le M</math>
for every natural number ''n''. The set of all bounded sequences forms the sequence space <math>l^\infty</math>.

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 <math>|\sin (x)| \le 1</math> for all <math>x \in \mathbf{R}</math>.
* The function <math>f(x)=(x^2-1)^{-1}</math>, 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 <math display="inline">f(x)= (x^2+1)^{-1}</math>, defined for all real ''x'', ''is'' bounded.
* The inverse trigonometric function arctangent defined as: ''y'' = {{math|arctan(''x'')}} or ''x'' = tan(''y'')}} is increasing for all real numbers ''x'' and bounded with −<math>\tfrac{\pi}{2}</math> < ''y'' < <math>\tfrac{\pi}{2}</math> 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 <math>g:\mathbb{R}^2\to\mathbb{R}</math> and <math>h: (0, 1)^2\to\mathbb{R}</math> defined by <math>g(x, y) := x + y</math> and <math>h(x, y) := \frac{1}{x+y}</math> 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:
* [http://mathonline.wikidot.com/bounded-sets-in-a-metric-space Bounded Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license
* [https://en.wikipedia.org/wiki/Bounded_function Bounded function, Wikipedia] under a CC BY-SA license