nyuuzyou/wikis huggingface dataset:
Bounded Real Function may not be of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a bounded function.
Then $f$ is not necessarily of bounded variation.
== Proof ==
Let $a = 0$, $b = 1$.
Define $f : \closedint 0 1 \to \R$ by: 
:$\map f x = \begin{cases} 1 & x \in \Q \\ 0 & x \not \in \Q \end{cases}$
For each finite subdivision $P$ of $\closedint 0 1$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$0 = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = 1$
For each such subdivision also write:
:$\ds \map {V_f} {P ; \closedint 0 1} = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$
It suffices to find a sequence of finite subdivisions $\sequence {P_n}$ such that: 
:$\ds \lim_{n \mathop \to \infty} \map {V_f} {P_n ; \closedint 0 1} = \infty$
by the definition of bounded variation.
For each $n \in \N$, let $P_n$ be a finite subdivision of size $n + 2$.
Let $\sequence {x_n}_{0 \mathop \le i \mathop \le n + 2}$ be the sequence forming this subdivision.
Note that from Between two Rational Numbers exists Irrational Number:
:it is possible to select an irrational number strictly between two rational numbers.
Similarly from Between two Real Numbers exists Rational Number:
:it is possible to select a rational number strictly between two irrational numbers.
We can therefore define $x_i$ for $1 \le i \le n + 1$ as follows:
:if $i$ is odd, let $x_i$ be an irrational number between $x_{i - 1}$ and $b$
:if $i$ is even, let $x_i$ be a rational number between $x_{i - 1}$ and $b$.
Let $x_0 = 0$ and $x_{n + 2} = 1$ to complete the definition.
Then:
We omit the term $i = n + 2$ for convenience.
Note that:
:if $x_i$ is rational then $x_{i + 1}$ is irrational.
That is:
:if $\map f {x_i} = 1$ then $\map f {x_{i + 1} } = 0$.
Similarly:
:if $x_i$ is irrational then $x_{i + 1}$ is rational.
That is:
:if $\map f {x_i} = 0$ then $\map f {x_{i + 1} } = 1$.
So for all $1 \le i \le n + 1$, we have:
:$\size {\map f {x_i} - \map f {x_{i - 1} } } = 1$
Hence:
:$\ds \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } } = n + 1$
So:
:$\ds \map {V_f} {P_n ; \closedint 0 1} \ge n + 1$
giving:
:$\ds \lim_{n \mathop \to \infty} \map {V_f} {P_n ; \closedint 0 1} = \infty$
as required.
== Also see ==
* Function of Bounded Variation is Bounded: this result shows that the converse of that result is false.
Category:Bounded Variation
Bounded Real Sequence/Examples/1 minus Reciprocals

== Example of Bounded Real Sequence ==
Let $\sequence {s_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$s_n = 1 - \dfrac 1 n$
That is:
:$\sequence {s_n}$ is the sequence of $1$ minus the reciprocals of the strictly positive integers.
Then $\sequence {s_n}$ is bounded.
== Proof ==
We have that $n \ge 1$.
Hence:
:$0 < \dfrac 1 n \le 1$
Hence:
:$0 \le 1 - \dfrac 1 n < 1$
Hence:
:$0$ is a lower bound of $\sequence {s_n}$
:$1$ is an upper bound of $\sequence {s_n}$.
Hence the result by definition of bounded real sequence.
== Sources ==
* 
* 
Category:Examples of Bounded Real Sequences
Bounded Right-Continuous Increasing Function Vanishing at Infimum is Distribution Function of Finite Borel Measure/Corollary

== Corollary ==
Let $K$ be a closed interval with $\inf K > -\infty$. 
Let $m \ge 0$. 
Let $F : K \to \R$ be a bounded right-continuous increasing function such that:
:$\ds \lim_{x \mathop \to \inf K} \map F x = m$
and:
:$\ds \lim_{x \mathop \to \sup K} \map F x = M \in \hointr 0 \infty$ 
Then there exists a finite Borel measure $\mu$ on $K$ such that the distribution function $F_\mu$ of $\mu$ is equal to $F$. 
Further, $\map \mu K = M$ and $\map \mu {\set {\inf K} } = m$. 
== Proof ==
Define $\widetilde F : K \to \R$ by:
:$\map {\widetilde F} x = \map F x - m$ 
for each $x \in K$.
Then $\widetilde F$ is bounded right-continuous increasing function and is such that:
:$\ds \lim_{x \mathop \to \inf K} \map {\widetilde F} x = 0$
and:
:$\ds \lim_{x \mathop \to \sup K} \map {\widetilde F} x = M - m \in \openint 0 \infty$ 
Hence by Bounded Right-Continuous Increasing Function Vanishing at Infimum is Distribution Function of Finite Borel Measure, there exists a finite Borel measure $\widetilde \mu$ on $K$ such that the distribution function $F_{\widetilde \mu}$ of $\widetilde \mu$ is equal to $F$ and $\map {\widetilde \mu} K = M - m$. 
Let:
:$\mu = \widetilde \mu + m \delta_{\set {\inf K} }$
where $\delta_{\set {\inf K} }$ is the Dirac measure concentrated at $\inf K$.
From Dirac Measure is Measure and Linear Combination of Measures, $\mu$ is a Borel measure.
Further, $\map \mu K = \map {\widetilde \mu} K + m \map {\delta_{\set {\inf K} } } K = M - m + m = M$.
Finally for each $x \in K$ we have:
So $\mu$ is our desired Borel measure.
Category:Bounded Right-Continuous Increasing Function Vanishing at Infimum is Distribution Function of Finite Borel Measure
Bounded Sequence in Euclidean Space has Convergent Subsequence

== Theorem ==
Let $\sequence {x_i}_{i \mathop \in \N}$ be a bounded sequence in the Euclidean space $\R^n$.
Then some subsequence of $\sequence {x_i}_{i \mathop \in \N}$ converges to a limit.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Limits of Sequences
Category:Euclidean Space
Category:Bounded Sequence in Euclidean Space has Convergent Subsequence
Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 1

== Theorem ==
== Proof ==
Denote with $d$ the Euclidean metric on $\R^n$.
Because $\sequence {x_i}_{i \mathop \in \N}$ is bounded, we find $v \in \R^n$ and $\epsilon \in \R_{>0}$ such that:
:$\forall i \in \N: \map d {v, x_i} < \epsilon$
Therefore, all $x_i$ are members of the closed $\epsilon$-ball $S = \map {B_\epsilon^-} v$.
By Closed Ball in Euclidean Space is Compact, $S$ is compact.
Thus $\sequence {x_i}_{i \mathop \in \N}$ can be considered as a sequence in the compact metric space $\struct {S, d \restriction_{S \times S} }$.
By Compact Subspace of Metric Space is Sequentially Compact in Itself, $\sequence {x_i}_{i \mathop \in \N}$ has a convergent subsequence in $S$.
In particular, since $S$ is a metric subspace of $\R^n$, it follows that $\sequence {x_i}_{i \mathop \in \N}$ has a convergent subsequence in $\R^n$ as well.
== Sources ==
* : $3.6b$
Category:Bounded Sequence in Euclidean Space has Convergent Subsequence
Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 2

== Theorem ==
== Proof ==
Let the range of $\sequence {x_i}$ be $S$.
By Closure of Bounded Subset of Metric Space is Bounded $\map \cl S$ is bounded in $\R^n$.
By Topological Closure is Closed, $\map \cl S$ is closed in $\R^n$.
By the Heine-Borel Theorem, $S$ is compact.
The result follows from Compact Subspace of Metric Space is Sequentially Compact in Itself.
== Sources ==
* : $7.2$: Sequential compactness: Corollary $7.2.7$
Category:Bounded Sequence in Euclidean Space has Convergent Subsequence
Bounded Sequence in Euclidean Space has Convergent Subsequence/Proof 3

== Theorem ==
== Proof ==
We have that all norms on $\R^n$ are equivalent.
Choose the Euclidean norm $\norm {\, \cdot \,}_2$.
Proof by induction will be used.
=== Basis for the induction ===
Let $n = 1$.
Then the proof is given by Bolzano-Weierstrass theorem.
=== Induction hypothesis ===
Suppose, a bounded sequence $\sequence {\boldsymbol x_i}_{i \mathop \in \N}$ in $\R^n$ has a convergent subsequence $\sequence {\boldsymbol x_{i_k}}_{k \mathop \in \N}$.
Then we have to show, that the same property holds for $\R^{n + 1}$.
=== Induction step ===
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a bounded sequence in $\R^{n + 1}$.
Split $\mathbf x_n$ into its first $n$ components and the last one:
:$\mathbf x_n := \tuple {\boldsymbol \alpha_n, \beta_n}$
where $\boldsymbol \alpha_n \in \R^n$ and $\beta_n \in \R$
Then:
Hence, $\sequence {\boldsymbol \alpha_n}_{n \mathop \in \N}$ is a bounded sequence in $\R^n$.
By induction hypothesis, it has a convergent subsequence $\sequence {\mathbf \alpha_{n_k} }_{k \mathop \in \N}$.
Denote its limit by $\alpha$.
Furthermore:
Therefore, $\sequence {\beta_n}_{n \mathop \in \N}$ is bounded.
By Bolzano-Weierstrass theorem, $\sequence {\beta_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {\beta_{n_k}}_{k \mathop \in \N}$.
Denote its limit by $\beta$.
Then we have that:
Hence, the bounded sequence $\sequence {\mathbf x_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {\mathbf x_{n_k}}_{k \mathop \in \N}$.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Bounded Sequence in Euclidean Space has Convergent Subsequence
Bounded Subset of Minimally Closed Class under Progressing Mapping has Greatest Element

== Theorem ==
Every bounded subset of $N$ has a greatest element.
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.17 \ (2)$
Category:Minimally Closed Classes under Progressing Mapping
Bounded Subset of Minimally Closed Class under Progressing Mapping has Greatest Element/Proof

== Theorem ==
== Proof ==
Let the hypothesis be assumed.
The proof proceeds by general induction.
For all $x \in N$, let $\map P x$ be the proposition:
:Every subset of $N$ which is bounded by $x$ has a greatest element.
=== Basis for the Induction ===
Let $x = b$.
From Smallest Element of Minimally Closed Class under Progressing Mapping, $b$ is the smallest element of $N$.
That is:
:$\forall x \in N: b \subseteq x$
So the only subset $B$ of $N$ such that $B$ is bounded by $b$ is $B = \set b$.
By definition, $\set b$ trivially has a greatest element, which is $b$.
Thus $\map P b$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true.
So this is the induction hypothesis:
:Every subset of $N$ which is bounded by $x$ has a greatest element.
from which it is to be shown that:
:Every subset of $N$ which is bounded by $\map g x$ has a greatest element.
=== Induction Step ===
This is the induction step:
Let the induction hypothesis be assumed.
Let $B$ be a subset of $N$ which is bounded by $\map g x$.
Thus, by definition, every element of $B$ is a subset of $\map g x$.
It is to be shown that $B$ has a greatest element $m$.
That is, that $m \in B$ and every element of $B$ is a subset of $m$
Let $\map g x \in B$.
By definition of subset bounded by $\map g x$, every element of $B$ is a subset of $\map g x$.
That is, $\map g x$ is itself the greatest element of $B$.
Suppose that $\map g x \notin B$.
Then every element of $B$ is a proper subset of $\map g x$.
Let $y \in B$ be arbitrary.
Because $y$ is a proper subset of $\map g x$, there exists $z \in \map g x$ such that $z \notin y$.
Thus $\map g x$ is not a subset of $y$.
From Minimally Closed Class under Progressing Mapping induces Nest:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
Because it is not the case that $\map g x \subseteq y$, it must be that:
:$y \subseteq x$
Thus, as $y$ is arbitrary, it follows that every element of $B$ is a proper subset of $x$.
Thus $B$ is bounded by $x$.
It follows by the induction hypothesis that $B$ has a greatest element.
So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction.
Therefore:
:For all $x \in N$, every subset of $N$ which is bounded by $x$ has a greatest element.
That is:
:Every bounded subset of $N$ has a greatest element.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Exercise $4.1$
Category:Minimally Closed Classes under Progressing Mapping
Bounded Subset of Real Numbers/Examples/Example 1

== Example of Bounded Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \set {1, 2, 3}$
is bounded both above and below.
Some upper bounds of $S$ are:
:$100, 10, 4, 3$
Some lower bounds of $S$ are:
:$-27, 0, 1$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.4$ Examples: $\text {(i)}$
Category:Examples of Bounded Sets of Real Numbers
Bounded Subset of Real Numbers/Examples/Example 2

== Example of Bounded Subset of Real Numbers ==
The subset $T$ of the real numbers $\R$ defined as:
:$T = \set {x \in \R: 1 \le x \le 2}$
is bounded both above and below.
Some upper bounds of $T$ are:
:$100, 10, 4, 2$
Some lower bounds of $T$ are:
:$-27, 0, 1$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.4$ Examples: $\text {(ii)}$
Category:Examples of Bounded Sets of Real Numbers
Bounded Subset of Real Numbers/Examples/Reciprocals of Positive Integers

== Example of Bounded Subset of Real Numbers ==
The subset $T$ of the real numbers $\R$ defined as:
:$T = \set {\dfrac 1 n: n \in \Z_{>0} }$
is bounded both above and below.
We have that:
where $\sup T$ and $\inf T$ denote the supremum and infimum of $T$ respectively.
We also have:
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.35$. Example $\text{(a)}$
Category:Examples of Bounded Sets of Real Numbers
Bounded Subspace of Euclidean Space is Totally Bounded

== Theorem ==
Let $\struct {\R^n, \norm {\, \cdot \,} }$ be a Euclidean space, where $\norm {\, \cdot \,}$ denotes the usual metric.
Let $M$ be a metric subspace of $\struct {\R^n, \norm {\, \cdot \,} }$.
Let $M$ be bounded.
Then $M$ is totally bounded.
== Proof ==
As $M$ is bounded, it has a finite diameter $R$.
Consider arbitary $x, y \in \R^n$, expressed in their usual coordinates.
Thus, up to a positive constant multiple, the inequality:
:$\ds \sup_{x, y \mathop \in M} \max_i \size {y_i - x_i} \le R$
implies that $M$ is bounded with radius $R$.
But this inequality is a characterization of an $n$-cube.
Because $R \le 2 R \sqrt n$ for all $R > 0$, $M$ can be placed inside an $n$-cube of the form $\closedint {c - R} {c + R}^n$, by diameter of $n$-cube.
For any $\epsilon > 0$, pick an integer $k > \sqrt n R \epsilon^{-1}$.
Consider a normal subdivision of $\closedint {c - R} {c + R}$ into $k$ pieces.
The length of each subinterval is then:
:$\dfrac {\paren {c + R - \paren {c - R} } } k = 2 R k^{-1}$
For any subinterval chosen this way, the product of $n$ copies of such an interval will itself be a cube, because the subintervals were chosen as to all be the same length.
By diameter of $n$-cube, each sub-cube has diameter $2 R k^{-1} \sqrt n$
By the choice of $k$, we have $2 R k^{-1} \sqrt n < 2 \epsilon$
That means that each sub-cube can fit inside an open ball $n$-ball of radius $\epsilon$.
This in turn implies that $M$ can be covered by a finite number of balls with radius $\epsilon$.
Thus $M$ is totally bounded by definition, as $\epsilon$ was arbitrary.
== Sources ==
* : $\S \text {P}.6$
Category:Bounded Metric Spaces
Bounded Summation is Primitive Recursive

== Theorem ==
Let the function $f: \N^{k + 1} \to \N$ be primitive recursive.
Then so is the function $g: \N^{k + 1} \to \N$ defined as:
:$\ds \map g {n_1, n_2, \ldots, n_k, z} = \begin{cases}
0 & : z = 0 \\
\sum_{y \mathop = 1}^z \map f {n_1, n_2, \ldots, n_k, y} & : z > 0
\end{cases}$
== Proof ==
The function $g$ satisfies:
:$\map g {n_1, n_2, \ldots, n_k, 0} = 0$
:$\map g {n_1, n_2, \ldots, n_k, z + 1} = \map g {n_1, n_2, \ldots, n_k, z} + \map f {n_1, n_2, \ldots, n_k, z + 1}$.
Hence $g$ is defined by primitive recursion from:
:the primitive recursive function $\operatorname{add}$
:$f$, which is primitive recursive
:constants, which are primitive recursive.
Hence the result.
== Note ==
The sum $\ds \sum_{y \mathop = 1}^z$ is referred to as a bounded sum to distinguish it from $\ds \sum_{y \mathop = 1}^\infty$ which is not. 
Category:Primitive Recursive Functions
Bounded Summation of Integers is Primitive Recursive

== Theorem ==
Let the function $f : \N^{k + 1} \to \N$ be primitive recursive.
Let the function $g : \N^{k + 1} \to \N$ be defined as:
:$\map g {n_1, \dotsc, n_k, z} = \paren {\sum_{y = 0}^{z - 1} \ell_y}_\Z$
where $\map f {n_1, \dotsc, n_k, y}$ codes the integer $\ell_y$.
Then, $g$ is primitive recursive.
== Proof ==
We can equivalently write $g$ as:
:$\map g {n_1, \dotsc, n_k, z} = \begin{cases} 0 & : z = 0 \\ \map g {n_1, \dotsc, n_k, z - 1} +_\Z \map f {n_1, \dotsc, n_k, z - 1} & : z > 0 \end{cases}$
which is clearly obtained by primitive recursion from:
* Addition of Integers is Primitive Recursive
primitive recursive functions.
Bounded Universal Quantification of Provable Arithmetic Formula is Provable

== Theorem ==
Let $\map \phi x$ be a WFF in the language of arithmetic with $1$ free variable.
Let $\tau$ be a term in the language of arithmetic with no variables.
Let $\sqbrk a$ denote the unary representation of $a$.
Suppose that, for every $n \in \N$ such that $\N \models \map \phi {x \gets \sqbrk n}$:
:$\map \phi {x \gets \sqbrk n}$
is a theorem of minimal arithmetic.
Suppose also that:
:$\N \models \forall x < \tau: \map \phi x$
Then:
:$\forall x < \tau: \map \phi x$
is a theorem of minimal arithmetic.
== Proof ==
Let $t = \map {\operatorname{val}_\N} \tau$ be the value of $\tau$ under the natural numbers.
By definition of value of formula:
:$\N \models \map \phi {x \gets x_0}$
for every $x_0 < t$.
Therefore, :
:$\map \phi {x \gets \sqbrk {x_0} }$
is a theorem, for every $x_0 < t$.
Additionally, by Lower Section of Natural Number is Provable:
:$\forall x: x = 0 \lor x = \map s 0 \lor \dotso \lor x = \sqbrk {t - 1} \lor \neg {x < \sqbrk t}$
is a theorem.
The result follows from False Statement implies Every Statement and Proof by Cases.
Category:Natural Numbers
Category:Proof Theory
Boundedness is not Topological Property

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be homeomorphic.
Then it is not necessarily the case that:
:$M_1$ is bounded $M_2$ is bounded.
That is, boundedness is not a topological property.
== Proof ==
Proof by Counterexample:
Let the metric space $M_1 = \struct {S_1, d}$ such that:
:$S_1 = \openint 0 1$ is the open unit interval
:$d$ is the usual (Euclidean) metric on $S_1$.
Let the metric space $M_2 = \struct {\R, d}$ such that:
:$\R$ is the set of real numbers
:$d$ is again the usual (Euclidean) metric on $\R$.
Then $M_1$ is bounded by, for example, $1$.
However, $M_2$ is not bounded.
But from Open Real Interval is Homeomorphic to Real Number Line, we have that $M_1$ and $M_2$ are homeomorphic.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.4$
Category:Examples of Topological Properties
Boundedness of Sine X over X

== Theorem ==
Let $x \in \R$.
Then:
:$\size {\dfrac {\sin x} x} \le 1$
== Proof ==
From Derivative of Sine Function, we have:
:$D_x \paren {\sin x} = \cos x$
So by the Mean Value Theorem, there exists $\xi \in \R$ between $0$ and $x$ such that:
:$\dfrac {\sin x - \sin 0} {x - 0} = \cos \xi$
From Real Cosine Function is Bounded we have that:
:$\size {\cos \xi} \le 1$
== Sources ==
* : $\S 16.3 \ (5)$
Category:Sine Function
Bounds for Complex Exponential

== Theorem ==
Let $\exp$ denote the complex exponential.
Let $z \in \C$ with $\cmod z \le \dfrac 1 2$.
Then
:$\dfrac 1 2 \cmod z \le \cmod {\exp z - 1} \le \dfrac 3 2 \cmod z$
== Proof ==
By definition of complex exponential:
:$\exp z = \ds \sum_{n \mathop = 1}^\infty \frac {z^n} {n!}$
Thus
By the Triangle Inequality:
:$\dfrac 1 2 \cmod z \le \cmod {\exp z - 1} \le \dfrac 3 2 \cmod z$
== Also see ==
* Bounds for Complex Logarithm
Category:Exponential Function
Bounds for Complex Logarithm

== Theorem ==
Let $\ln$ denote the complex logarithm.
Let $z \in \C$ with $\cmod z \le \dfrac 1 2$.
Then:
:$\dfrac 1 2 \cmod z \le \cmod {\map \ln {1 + z} } \le \dfrac 3 2 \cmod z$
== Proof ==
By definition of complex logarithm:
:$-\map \ln {1 + z} = \ds \sum_{n \mathop = 1}^\infty \frac {\paren {-z}^n} n$
Thus
By the Triangle Inequality:
:$\dfrac 1 2 \le \cmod {\dfrac {\map \ln {1 + z} } z} \le \dfrac 3 2$
== Also see ==
* Bounds of Natural Logarithm
* Bounds for Complex Exponential
== Sources ==
* $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: $5.3$
Category:Logarithms
Bounds for Cumulative Distribution Function

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $F_X$ be the cumulative distribution function.
Then:
:$0 \le \map {F_X} x \le 1$ for each $x \in \R$
== Proof ==
From the definition of the cumulative distribution function, we have: 
:$\map {F_X} x = \map \Pr {X \le x}$
for each $x \in \R$.
We have: 
:$\O \subseteq \set {\omega \in \Omega : \map X \omega \le x} \subseteq \Omega$
So, from Measure is Monotone, we have: 
:$\map \Pr \O \le \map \Pr {X \le x} \le \map \Pr \Omega$
From the definition of a probability measure, we have: 
:$\map \Pr \O = 0$
and:
:$\map \Pr \Omega = 1$
so:
:$0 \le \map \Pr {X \le x} \le 1$
for each $x \in \R$.
Category:Cumulative Distribution Functions
Bounds for Finite Product of Real Numbers

== Theorem ==
Let $a_1, a_2, \ldots, a_n$ be positive real numbers.
Then:
:$\ds \sum_{k \mathop = 1}^n a_k \le \prod_{k \mathop = 1}^n \paren {1 + a_k} \le \map \exp {\sum_{k \mathop = 1}^n a_k}$
== Proof ==
=== Lower bound ===
Follows by expanding.
=== Upper Bound ===
==== Proof 1 ====
By Exponential of x not less than 1+x:
:$\ds \prod_{k \mathop = 1}^n \paren {1 + a_k} \le \prod_{k \mathop = 1}^n \exp a_k = \map \exp {\sum_{k \mathop = 1}^n a_k}$
==== Proof 2 ====
By the AM-GM Inequality:
:$\ds \prod_{k \mathop = 1}^n \paren {1 + a_k} \le \paren {\frac {n + \sum_{k \mathop = 1}^n a_k} n}^n$
== Also see ==
* Weierstrass Product Inequality
Category:Inequalities
Bounds for Integer Expressed in Base k

== Theorem ==
Let $n \in \Z$ be an integer.
Let $k \in \Z$ such that $k \ge 2$.
Let $n$ be expressed in base $k$ notation:
:$n = \ds \sum_{j \mathop = 1}^s a_j k^j$
where each of the $a_j$ are such that $a_j \in \set {0, 1, \ldots, k - 1}$.
Then:
:$0 \le n < k^{s + 1}$
== Proof ==
As none of the coefficients $a_j$ in $\ds \sum_{j \mathop = 1}^s a_j k^j$ is (strictly) negative, the summation itself likewise cannot be negative
Thus:
:$0 \le n$
The equality is satisfied when $a_j = 0$ for all $j$.
We then have:
== Sources ==
* : $\text {1-2}$ The Basis Representation Theorem: Exercise $6$
Category:Number Bases
Bounds for Modulus of e^z on Circle x^2 + y^2 - 2x - 2y - 2 = 0

== Theorem ==
Consider the circle $C$ embedded in the complex plane defined by the equation:
:$x^2 + y^2 - 2 x - 2 y - 2 = 0$
Let $z = x + i y \in \C$ be a point lying on $C$.
Then:
:$e^{-1} \le \cmod {e^z} \le e^3$
== Proof ==
This defines a circle whose center is at $1 + i$ and whose radius is $2$.
From Modulus of Exponential is Exponential of Real Part:
:$\cmod {e^z} = e^x$
If $z \in C$ then from the geometry of the circle $C$:
:$-1 \le x \le 3$
Then from Exponential is Strictly Increasing:
:$e^{-1} \le e^x \le e^3$
Hence the result.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $5$
Category:Circles
Category:Geometry of Complex Plane
Category:Equation of Circle