nyuuzyou/wikis huggingface dataset:
Continuum has Cardinality of Continuum

== Theorem ==
Let $\CC$ be a continuum, including continuum in the topological sense.
Then the cardinality of $\CC$ is equal to the cardinality of the continuum.
That is: all continua have the same cardinality.
== Proof ==
== Sources ==
* 
* 
Category:Continua
Category:Cardinality of Continuum
Contour Integral/Examples/Circulation of Fluid

== Example of Contour Integral ==
Let $\mathbf v$ be the velocity within a body $B$ of fluid as a point-function.
Let $\Gamma$ be a closed contour in $B$.
Let $\d \mathbf l$ be a small element of length of $\Gamma$ at a point $P$.
Then the circulation of $B$ over $\Gamma$ is given by the contour integral:
:$\ds \int_\Gamma \mathbf v \cdot \d \mathbf l$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Contour Integrals
Contour Integral/Examples/Electromotive Force

== Example of Contour Integral ==
Let $\mathbf E$ be an electromagnetic field acting over a region of space $R$.
Let $\Gamma$ be a closed contour in $R$.
Let $\d \mathbf l$ be a small element of length of $\Gamma$ at a point $P$.
Then the electromotive force in $\Gamma$ is given by the contour integral:
:$\ds \int_\Gamma \mathbf E \cdot \d \mathbf l$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Contour Integrals
Contour Integral/Examples/Potential Difference

== Example of Contour Integral ==
Let $\mathbf E$ be an electric field acting over a region of space $R$.
Let $OA$ be a contour in $R$.
Let $\d \mathbf l$ be a small element of length of $OA$ at a point $P$.
Then the potential difference between $O$ to $A$ is given by the contour integral:
:$\ds \int_O^A \mathbf E \cdot \d \mathbf l$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Contour Integrals
Contour Integral/Examples/Work Done

== Example of Contour Integral ==
Let $\mathbf F$ be a force acting as a point-function giving rise to a vector field $\mathbf V$.
Let $OA$ be a contour in $\mathbf V$ along which a particle $P$ is moved by $\mathbf F$.
Let $\d \mathbf l$ be a small element of length of $OA$ at $P$.
Then the work done by $\mathbf F$ moving $P$ from $O$ to $A$ is given by the contour integral:
:$\ds \int_O^A \mathbf F \cdot \d \mathbf l$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Contour Integrals
Contour Integral along Reversed Contour

== Theorem ==
Let $C$ be a contour in the complex plane $\C$.
Let $f: \Img C \to \C$ be a continuous complex functions, where $\Img C$ denotes the image of $C$.
Then the contour integral of $f$ along the reversed contour $-C$ is:
:$\ds \int_{-C} \map f z \rd z = -\int_C \map f z \rd z$
== Proof ==
First, suppose that $C$ is a directed smooth curve in $\C$.
Let $C$ be parameterized by the smooth path $\gamma: \closedint a b \to \C$.
By definition of reversed directed smooth curve, $-C$ is parameterized by a smooth path $\rho: \closedint a b \to \C$ with $\rho = \gamma \circ \psi$.
Here, $\psi: \closedint a b \to \closedint a b$ is defined by $\map \psi t = a + b - t$.
From Derivatives of Function of $a x + b$:
:$\map {\psi'} t = -1$ for all $t \in \closedint a b$
Then:
Next, suppose that $C$ is a contour.
Then $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.
By definition of reversed contour, $-C$ is a concatenation of the finite sequence $-C_n, \ldots, -C_1$ of directed smooth curves.
Then:
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Contour Integral is Independent of Parameterization

== Theorem ==
Let $C$ be a contour defined by a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $f: \Img C \to \C$ be a continuous complex function, where $\Img C$ denotes the image of $C$.
Suppose that $\sigma_k: \closedint {c_k} {d_k} \to \C$ is a reparameterization of $C_k$ for all $k \in \set {1, \ldots, n}$.
Then:
:$\ds \int_C \map f z \rd z = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t = \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \map f {\map {\sigma_k} t} \map {\sigma_k'} t \rd t$
== Proof ==
By definition of parameterization:
:$\gamma_k = \sigma_k \circ \phi_i$
for all $k \in \set {1, \ldots, n}$.
Here, $\phi_k: \closedint {c_k} {d_k} \to \closedint {a_k} {b_k}$ is a bijective differentiable strictly increasing real function.
Then:
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Contour Integral is Well-Defined

== Theorem ==
Let $C$ be a contour defined by a finite sequence $C_1, \ldots, C_n$ of directed smooth curves in the complex plane $\C$.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $f: \Img C \to \C$ be a continuous complex function, where $\Img C$ denotes the image of $C$.
Let $\sigma_k: \closedint {c_k} {d_k} \to \C$ be a reparameterization of $C_k$ for all $k \in \set {1, \ldots, n}$.
Then:
:$\ds \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t = \sum_{k \mathop = 1}^n \int_{c_k}^{d_k} \map f {\map {\sigma_k} t} \map {\sigma_k'} t \rd t$
and all complex Riemann integrals in the equation are defined.
== Proof ==
Define $g_k: \closedint {a_k} {b_k} \to \C$ by $\map {g_k} t = \map f {\map {\gamma_k} t} \map {\gamma_k'} t$ for all $k \in \set {1, \ldots, n}$.
By definition of smooth path, it follows that $\gamma_k$ and $\gamma_k'$ are continuous for all $k \in \set {1, \ldots, n}$.
From Continuity of Composite Mapping/Corollary and Sum Rule for Continuous Complex Functions, it follows that $g_k$ is continuous.
From Continuous Complex Function is Complex Riemann Integrable, we find that $\ds \int_{a_k}^{b_k} \map {g_k} t \rd t$ is defined.
Similarly, it can be shown that $\ds \int_{c_k}^{d_k} \map f {\map {\sigma_k} t} \map {\sigma_k'} t \rd t$ is defined for all $k \in \set {1, \ldots, n}$.
Hence, all complex Riemann integrals in the theorem are defined.
The equality now follows from Contour Integral is Independent of Parameterization.
== Sources ==
* : $\S 2.2$
Category:Complex Contour Integrals
Category:Examples of Well-Defined Mappings‎
Contour Integral of Closed Contour Split into Two Contours

== Theorem ==
Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $C = \sequence {C_1, \ldots, C_n}$ be a closed contour in $D$.
Let $C'$ be a contour in $D$ with start point $z_1$ and end point $z_2$.
Let $-C'$ denote the reversed contour of $C'$.
Suppose:
:$z_1$ is equal to the end point of $C_{k_1}$
and:
:$z_2$ is equal to the end point of $C_{k_2}$ for some $k_1, k_2 \in \set {1, 2, \ldots, n}$ with $k_1 < k_2$.
Define two contours $C^{\paren 1}, C^{\paren 2}$ by concatenation as 
:$C^{\paren 1} = \sequence {C_1, \ldots, C_{k_1}, C', C_{k_2 + 1}, \ldots, C_n}$
:$C^{\paren 2} = \sequence {C_{k_1 + 1}, \ldots, C_{k_2}, -C'}$
Then:
:$\ds \oint_C \map f z \rd z = \oint_{C^{\paren 1} } \map f z \rd z + \oint_{C^{\paren 2} } \map f z \rd z$
== Proof ==
By definition of closed contour, $C^{\paren 1}$ and $C^{\paren 2}$ are closed contours.
Then:
== Also see ==
* Goursat's Integral Lemma, for an illustration that shows how this theorem is used.
* Cauchy-Goursat Theorem, for another illustration.
== Sources ==
* : $\S 2.2$
Category:Complex Contour Integrals
Contour Integral of Concatenation of Contours

== Theorem ==
Let $C$ and $D$ be contours in $\C$.
Suppose that the end point of $C$ is equal to the start point of $D$, so the concatenation $C \cup D$ is defined.
Let $f: \Img {C \cup D} \to \C$ be a continuous complex function, where $\Img {C \cup D}$ denotes the image of $C \cup D$.
Then:
:$\ds \int \limits_{C \mathop \cup D} \map f z \rd z = \int \limits_C \map f z \rd z + \int \limits_D \map f z \rd z$
== Proof ==
By definition of contour, $C$ is a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Similarly, $D$ is a finite sequence $D_1, \ldots, D_m$ of directed smooth curves.
Let $D_j$ be parameterized by the smooth path $\sigma_j: \closedint {c_j} {d_j} \to \C$ for all $j \in \set {1, \ldots, m}$.
Put $\gamma_k = \sigma_{k - n}, a_k = c_{k - n}$ and $b_k = d_{k - n}$ for all $k \in \set {n + 1, \ldots, n + m}$.
Then $C \cup D$ is a sequence of $n + m$ directed smooth curves which are parameterized by $\gamma_1, \ldots, \gamma_{n + m}$.
Then:
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Contour Integral of Gamma Function

== Theorem ==
Let $\Gamma$ denote the gamma function.
Let $y$ be a (strictly) positive real number.
Then for any (strictly) positive real number $c$:
:$\ds \frac 1 {2 \pi i} \int_{c - i \infty}^{c + i \infty} \map \Gamma t y^{-t} \rd t = e^{-y}$
== Proof ==
Let $L$ be the rectangular contour with the vertices $c \pm i R$, $- N - \dfrac 1 2 \pm i R$.
We will take the Contour Integral of $\map \Gamma t y^{-t}$ about the rectangular contour $L$.
Note from Poles of Gamma Function, that the poles of this function are located at the non-positive integers.
Thus, by Cauchy's Residue Theorem:
:$\ds \oint_L \map \Gamma t y^{-t} \rd z = 2 \pi i \sum_{k \mathop = 0}^N \map {\operatorname{Res} } {-k}$
Thus, we obtain:
:$\ds \lim_{N \mathop \to \infty} \lim_{R \mathop \to \infty} \oint_L \map \Gamma t y^{-t} \rd z = 2 \pi i \sum_{k \mathop = 0}^\infty \map {\operatorname{Res} } {-k}$
From Residues of Gamma Function, we see that:
:$\map {\operatorname{Res} } {-k} = \dfrac {\paren {-1}^k y^k} {k!}$ 
Which gives us: 
We aim to show that the all but the of the rectangular contour go to $0$ as we take these limits, as our result follows readily from this.
The top and bottom portions of the contour can be parameterized by:
:$\map \gamma t = c \pm i R - t$
where $0 < t < c + N + \dfrac 1 2$. 
The modulus of the contour integral is therefore given by:
From Bound on Complex Values of Gamma Function, we have that: 
for all $\cmod R \ge 1$. Because $\cmod R \ge 1$, we have that
:Combining the two inequalities we obtain:
We see that:
:$\ds \int_0^{c + N + \frac 1 2} \cmod {\map \Gamma {c- t + i} y^t} \rd t < \infty$
as the poles of Gamma are at the non-positive integers, which means that the integral is a definite integral of a continuous function. 
 
The above is enough to allow for the interchange of limits by the Dominated Convergence Theorem, thus we have:
 
But using Equation $(1)$ from above we see:
Thus by the Squeeze Theorem for Functions we have:
:$\ds \lim_{R \mathop \to \infty} \cmod {\map \Gamma {c \pm i R - t} } = 0$
Which means we have:
 
Thus we have that the top and bottom of the contour go to $0$ in the limit. 
The of the contour may be parameterized by:
:$\map \gamma t = N - \dfrac 1 2 - it$
where $t$ runs from $-R$ to $R$. 
Thus the absolute value of integral of the is given as:
Thus we have:
which gives us:
Thus we have the left, top, and bottom of the rectangular contour go to 0 in the limit, which gives us:
Category:Gamma Function
Category:Complex Analysis
Contour Integration by Substitution

== Theorem ==
Let $f$ be a holomorphic function on a simply connected domain $V \subseteq \mathbb C$.
Let $\gamma$ be a contour in $V$ starting at $z_1$ and ending at $z_2$.
Let $U$ be a connected domain.
Let $\phi: U \to V$ be a holomorphic function with $\phi^{-1} \sqbrk {\set {z_1, z_2} } \ne \O$.
Let $\omega$ be a contour in $U$ starting at $u_1$ and ending at $u_2$, such that $\map \phi {u_1} = z_1$ and $\map \phi {u_2} = z_2$.
Then the contour integral of $f$ over $\gamma$ satisfies the following substitution:
:$\ds \int_\gamma \map f z \rd z = \int_\omega \map f {\map \phi u} \map {\phi'} u \rd u$
== Proof ==
Because $V$ is simply connected, $f$ has a primitive.
Let one such primitive $F$ be established.
By the Fundamental Theorem of Calculus for Contour Integrals:
Next, observe that $\map f {\map \phi u} \map {\phi'} u$ also has a primitive, that is $\map F {\map \phi u}$.
Again by the Fundamental Theorem of Calculus for Contour Integrals:
Because $\map \phi {u_1} = z_1$, we have that:
:$\map F {\map \phi {u_1} } = \map F {z_1}$
and likewise for $\map \phi {u_2}$. 
Thus $I_1 = I_2$.
Category:Complex Analysis
Contour Plot/Examples/Arbitrary Example 1

== Example of Contour Plot ==
This is a contour plot of the real function $f$ of $2$ variables defined as:
:$\forall x, y \in \R^2: \map f {x, y} = 4 x^2 - \dfrac {21 x^4} {10} + \dfrac {x^6} 3 + x y - 4 y^2 + 4 y^4$
:
for various values of $f$ between $-0 \cdotp 8$ and $10$.
== Sources ==
* 
* 
Category:Examples of Contour Plots
Contraction of Extension of Contraction of Ideal is Contraction

== Theorem ==
Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b$ be an ideal of $B$.
Let ${\mathfrak b}^c$ be the contraction of $\mathfrak b$ by $f$.
Let ${\mathfrak b}^{ce}$ be the extension of ${\mathfrak b}^c$ by $f$.
Let ${\mathfrak b}^{cec}$ be the contraction of ${\mathfrak b}^{ce}$ by $f$.
Then:
:${\mathfrak b}^c = {\mathfrak b}^{cec}$
== Proof ==
Since $\mathfrak b \supseteq {\mathfrak b}^{ce}$ by Ideal Contains Extension of Contraction:
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
Contraction of Primary Ideal is Primary Ideal

== Theorem ==
Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b$ be a primary ideal of $B$.
Let $\mathfrak b^c$ be the contraction of $\mathfrak b$ by $f$.
Then $\mathfrak b^c$ is a primary ideal of $A$.
== Proof ==
Let $x,y \in A$ such that:
:$xy \in \mathfrak b^c$.
That is:
:$\map f {xy} = \map f x \map f y \in \mathfrak b$
Suppose that $x \ne \mathfrak b^c$.
That is:
:$\map f x \not \in \mathfrak b$
Since $\mathfrak b$ is primary:
:$\exists n \in \N_{>0} : \map f {y^n} = \paren {\map f y}^n \in \mathfrak b$
That is:
:$y^n \in \mathfrak b^c$
Category:Primary Ideals
Contradiction is Negation of Tautology

== Theorem ==
A contradiction implies and is implied by the negation of a tautology:
:$\bot \dashv \vdash \neg \top$
That is, a falsehood can not be true, and a non-truth is a falsehood.
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Proof by Boolean Interpretation ==
That is, the proposition:
:If it's not true, it must be false
is valid only in the context where there are only two truth values.
== Also see ==
* Tautology is Negation of Contradiction
== Sources ==
* : $\S 3.5$: The Classification of Propositions
Category:Contradiction
Category:Tautology
Category:Logical Negation
Category:Contradiction is Negation of Tautology
Contradiction is Negation of Tautology/Proof 3

== Theorem ==
== Proof ==
Let $p$ be a propositional formula.
Let $v$ be any arbitrary boolean interpretation of $p$.
Then $\map v p = F \iff \map v {\neg p} = T$ by the definition of the logical not.
Since $v$ is arbitrary, $p$ is false in all interpretations $\neg p$ is true in all interpretations.
Hence:
:$\bot \dashv \vdash \neg \top$
That is, the proposition:
:If it's not true, it must be false
is valid only in the context where there are only two truth values.
Category:Contradiction is Negation of Tautology
Contradiction is Negation of Tautology/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values in the appropriate columns match.
$\begin{array}{|c||cc|} \hline
\top & \neg & \bot \\
\hline
\T & \T & \F \\
\hline
\end{array}$
Category:Contradiction is Negation of Tautology
Contradictory Antecedent

== Theorem ==
A conditional with a contradiction as antecedent:
:$\bot \implies p \dashv \vdash \top$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Also see ==
* Rule of Explosion
== Sources ==
* : $\S 2.3.3$
Category:Contradictory Antecedent
Category:Contradiction
Category:Conditional
Contradictory Antecedent/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|} \hline
p & \bot & \implies & p & \top \\
\hline
\F & \F & \T & \F & \T \\
\T & \F & \T & \T & \T \\
\hline
\end{array}$
Category:Contradictory Antecedent
Contradictory Consequent

== Theorem ==
A conditional with a contradiction as consequent:
:$p \implies \bot \dashv \vdash \neg p$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Also see ==
* Proof by Contradiction
== Sources ==
* : $\S 2.3.3$
Category:Contradictory Consequent
Category:Contradiction
Category:Conditional
Contradictory Consequent/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||cc|} \hline
p & p & \implies & \bot & \neg & p \\
\hline
\F & \F & \T & \F & \T & \F \\
\T & \T & \F & \F & \F & \T \\
\hline
\end{array}$
Category:Contradictory Consequent
Contravariant Hom Functor maps Colimits to Limits

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Let $\mathbf C$ be a locally small category.
Let $C$ be an object of $\mathbf C$.
Let $\map \hom {\cdot, C}: \mathbf C \to \mathbf{Set}$ be the contravariant hom functor based at $C$.
Then $\map \hom {\cdot, C}$ maps every colimit to a limit, in that:
:$\map \hom { {\varinjlim \,}_j \,D_j, C} \cong {\varprojlim \,}_j \, \map \hom {D_j, C}$
for every diagram $D: \mathbf J \to \mathbf C$.
== Proof ==
== Sources ==
* : $\S 5.5$: Corollary $5.27$
Category:Category Theory
Contravariant Tensor/Examples/Rank 1

== Example of Contravariant Tensor ==
A contravariant tensor of rank $1$ is a set satisfying, for all $i$:
:$\overline F^i = \ds \sum_{r \mathop = 1}^l \dfrac {\partial \overline x^i} {\partial x^r} F^r$
== Sources ==
* 
* 
Category:Examples of Contravariant Tensors
Contravariant Tensor/Examples/Rank 2

== Example of Contravariant Tensor ==
A contravariant tensor of rank $2$ is a set satisfying, for all $i$ and $j$:
:$\overline F^{ij} = \ds \sum_r \sum s \dfrac {\partial \overline x^i} {\partial x^r} \dfrac {\partial \overline x^j} {\partial x^s} F^{rs}$
== Sources ==
* 
* 
Category:Examples of Contravariant Tensors
Control Path in Flow Chart is Unique

== Theorem ==
Let $C = \struct {F, P, V, E}$ be a flow chart.
Let $\struct {X, \set {f_g}, \set {p_q}}$ be an interpretation for $C$.
Let $b \in V$ and $x \in X$, and $N \in \N_{> 0}$.
Then, there is at most one length $N$ control path in $\struct {C, X}$, starting from $\tuple {b, x}$.
== Proof ==
We will proceed by induction on $N$.
=== Basis for the Induction ===
Suppose $N = 1$.
Then, the control path consists of:
:$\sequence {\tuple {b_1, x_1}}$
But the control path starts from $\tuple {b, x}$ by assumption.
Therefore:
:$b_1 = b$
:$x_1 = x$
and the control path is completely determined.
=== Induction Hypothesis ===
Now, suppose that there is at most one control path of length $N$, starting from $\tuple {b, x}$.
We need to show that there is at most one control path of length $N + 1$, starting from $\tuple {b, x}$.
=== Induction Step ===
If there is no length $N$ control path starting from $\tuple {b, x}$, then clearly there is no length $N + 1$ control path.
Otherwise, let $\sequence {\tuple {b_j, x_j}}_{1 \mathop \le j \mathop \le N}$ be the unique control path.
If $b_N \in V_F \cup V_J \cup V_I$, then it has out-degree $1$.
Thus:
:$b_{N + 1}$ is uniquely incident from the only $e \in E$ which is itself incident from $b_N$.
If $b_N \in V_O$, then it has out-degree $0$.
Then, there is no $b_N b_{N + 1} \in E$, and thus there is no length $N + 1$ control path.
If $b_N \in V_P$, but $x_N \notin X_{b_N}$, then any extension to $\sequence {\tuple {b_j, x_j}}_j$ will violate the property:
:$j < N \implies x_j \in X_{b_j}$
so there is no length $N + 1$ control path.
If $b_N \in V_P$ and $x_N \in X_{b_N}$, then $b_{N + 1}$ is completely determined by whether $x_N \in p_{b_N}$.
:If $x_N \in p_{b_N}$, then $b_{N + 1}$ is unique such that $b_N b_{N + 1} \in E$ is labeled $\top$.
:If $x_N \notin p_{b_N}$, then $b_{N + 1}$ is unique such that $b_N b_{N + 1} \in E$ is labeled $\bot$.
The above cover all cases for $b_{N + 1}$, so it is unique if it exists.
If $b_N \notin V_F$, then $x_{N + 1} = x_N$, completely determining it.
If $b_N \in V_F$, then $x_{N + 1} = \map {f_{b_N}} {x_N}$, if it exists.
:Uniqueness follows from the definition of partial mapping.
Since both $b_{N + 1}$ and $x_{N + 1}$ are uniquely determined or do not exist, then $\tuple {b_{N + 1}, x_{N + 1}}$ is either unique, or fails to exist.
By Principle of Mathematical Induction, the result follows.
Category:Algorithms
Convergence a.u. Implies Convergence a.e.

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f_n: D \to \R$ be a sequence of $\Sigma$-measurable functions for $D \in \Sigma$.
Let $f_n$ converge a.u. to a function $f$ on $D$.
Then $f_n$ converges a.e. to $f$.
== Proof ==
Assume $f_n$ converge a.u. to $f$ on $D$.
Then for each $\epsilon > 0$ there is a $B_\epsilon \subseteq D$ with $\map \mu {B_\epsilon} < \epsilon$ outside of which $f_n$ converges uniformly to $f$.
Thus, $f_n$ converges pointwise to $f$ outside of each $B_\epsilon$.
Next, define $\ds B \equiv \bigcap_{n \mathop \in \N} B_{\frac 1 n}$.
First, $B \in \Sigma$ since $\Sigma$ is a sigma-algebra, so that $\map \mu B$ is defined.
Second, note that since the measure is monotonic:
:$\forall n \in \N: \map \mu B \le \map \mu {B_{\frac 1 n} } < \frac 1 n$
Hence:
:$\map \mu B = 0$
But $x \in D - B$ only if $x \notin B_{\frac 1 k}$ for some $k \in \N$.
So by the above:
:$\map {f_n} x \to \map f x$
It follows by definition that $f_n$ converges a.e. to $f$.
== Sources ==
* : $\S 7.2$: Almost Uniform Convergence: Theorem $8$
Category:Measure Theory
Convergence by Multiple of Error Term

== Theorem ==
Let $\sequence {s_n}$ be a real sequence.
Suppose that $\exists \epsilon \in \R, \epsilon > 0$ such that:
:$\exists N \in \N: \forall n \ge N: \size {s_n - l} < K \epsilon$
for any $K \in \R, K > 0$, independent of both $\epsilon$ and $N$.
Then $\sequence {s_n}$ converges to $l$.
== Proof ==
Let $\epsilon > 0$.
Then $\dfrac \epsilon K > 0$.
If the condition holds as stated, then:
:$\exists N \in \N: \forall n \ge N: \size {s_n - l} < K \paren {\dfrac \epsilon K}$
Hence the result by definition of a convergent sequence.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Lemma $1.2.4$
Category:Limits of Sequences