avewright/proofwiki-math huggingface dataset:
Theorem: Linearly Ordered Space is T1

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a $T_1$ space.

Proof:
Let $p \in S$.
By definition of linearly ordered space, the rays:
{{begin-eqn}}
{{eqn | l = R_1
      | o = :=
      | r = \set {x \in S: x \prec p}
}}
{{eqn | l = R_2
      | o = :=
      | r = \set {x \in S: p \prec x}
}}
{{end-eqn}}
are open in $T$.
Thus their union:
:$R_1 \cup R_2 = \set {x \in S: x \prec p \lor p \prec x}$
is also open in $T$.
Consider the complement in $S$ of $R_1 \cup R_2$:
:$\relcomp S {R_1 \cup R_2} = \set {y \in S: y \notin R_1 \cup R_2}$
By definition of closed set, $\relcomp S {R_1 \cup R_2}$ is closed in $T$
But $\relcomp S {R_1 \cup R_2} = \set p$.
Thus $p$ is a closed point of $T$.
Hence the result, by definition of $T_1$ space.
{{qed}}
Theorem: Linearly Ordered Space is Completely Normal

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a completely normal space.

Proof:
By Linearly Ordered Space is $T_1$, $T$ is a $T_1$ space.
By Linearly Ordered Space is $T_5$, $T$ is a $T_5$ space.
Hence the result, by definition of completely normal space.
{{qed}}
Theorem: Linearly Ordered Space is Compact iff Complete

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a compact space {{iff}} it is complete.

Proof:
=== Necessary Condition ===
Let $T$ be a compact space.
Let $A \subseteq S$.
{{AimForCont}} $A$ has no supremum.
Consider the sets:
{{begin-eqn}}
{{eqn | l = P_\alpha
      | r = \set {x \in S: x \prec \alpha}
      | c = for $\alpha \in A$
}}
{{eqn | l = B_\beta
      | r = \set {x \in S: \beta \prec x}
      | c = for an upper bound $\beta$ of $A$.
}}
{{end-eqn}}
We have that $P_\alpha$ and $B_\beta$ cover $S$ but contain no finite subcover.
Thus $T$ is not a compact space.
Similarly if $A$ has no infimum.
It follows that $A$ has both a supremum and infimum.
As $A$ is arbitrary, it follows that $S$ is a complete ordered set.
{{qed}}
Theorem: Every Point except Endpoint in Connected Linearly Ordered Space is Cut Point

Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A \subseteq S$ be a connected space.
Let $p \in A$ be a point of $A$ which is not an endpoint of $A$.
Then $p$ is a cut point of $A$.

Proof:
We have that $A \setminus \set p$ is separated by $\set {x \in A: x \prec p}$ and $\set {x \in A: p \prec x}$.
If $p$ is an endpoint of $A$, then either:
:$\set {x \in A: x \prec p} = \O$
or:
:$\set {x \in A: p \prec x} = \O$
{{qed}}
Theorem: Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\set \Omega$ is a closed set of $\closedint 0 \Omega$ but not a $G_\delta$ set.

Proof:
The complement relative to $\closedint 0 \Omega$ of $\set \Omega$ is $\hointr 0 \Omega$, which is open in $\closedint 0 \Omega$.
Hence, by definition, $\set \Omega$ is a closed set of $\closedint 0 \Omega$.
Let $G_i$ be a countable set of open sets of $\closedint 0 \Omega$ which contain $\Omega$.
Then we can find a set of basis elements of $\closedint 0 \Omega$ of the form $\hointl {\alpha_i} \Omega \subseteq G_i$ for each $i$.
Each $\hointr 0 {\alpha_i}$ is countable.
We also have Countable Union of Countable Sets is Countable.
Thus the infimum of the $\alpha_i$ is an ordinal $\gamma$ which is less than $\Omega$.
Therefore:
:$\ds \bigcap G_i \supseteq \hointl \gamma \Omega \ne \set \Omega$
That is, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$.
{{qed}}
Theorem: Uncountable Closed Ordinal Space is not First-Countable

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a first-countable space.

Proof:
From Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$.
Thus, by definition, $\Omega$ does not have a countable local basis.
Hence the result by definition of first-countable space.
{{qed}}
Theorem: Uncountable Open Ordinal Space is not Separable

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a separable space.

Proof:
Because $\Omega$ is the first uncountable ordinal, any ordinal which strictly precedes $\Omega$ is countable.
Let $H \subseteq \hointr 0 \Omega$ be a countable subset of $\hointr 0 \Omega$.
Let $\sigma$ be the supremum of $H$.
As $H$ by definition strictly precedes $\Omega$, $H$ itself is countable.
Thus $\sigma$ strictly precedes $\Omega$.
The closed interval $\closedint 0 \sigma$ is such that $H \subseteq \closedint 0 \sigma$.
By definition of the closure $H^-$ of $H$ as the smallest closed set of $\hointr 0 \Omega$ containing $H$, it follows that $H^- \subseteq \closedint 0 \sigma$.
Therefore, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$.
Thus the closure of $H$ does not equal $\hointr 0 \Omega$.
Thus $H$ is not everywhere dense in $\hointr 0 \Omega$.
Hence, by definition, $\hointr 0 \Omega$ is not separable.
{{qed}}
Theorem: Uncountable Closed Ordinal Space is not Separable

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a separable space.

Proof:
Let $H \subseteq \closedint 0 \Omega$ be a countable subset of $\closedint 0 \Omega$.
From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$, and so also in $\closedint 0 \Omega$.
Thus the closure of $H$ in $\closedint 0 \Omega$ does not equal $\closedint 0 \Omega$.
Thus $H$ is not everywhere dense in $\closedint 0 \Omega$.
Hence, by definition, $\closedint 0 \Omega$ is not separable.
{{qed}}
Theorem: Uncountable Open Ordinal Space is First-Countable

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a first-countable space.

Proof:
{{ProofWanted|It is to be shown that the only point of $\closedint 0 \Omega$ that does not have a countable local basis is $\Omega$. So all the others have, and that means all the points in $\hointr 0 \Omega$ have a countable local basis.}}
Theorem: Ordinal Space is Completely Normal

Let $\Gamma$ denote a limit ordinal.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both completely normal.

Proof:
By definition, $\hointr 0 \Gamma$ and $\closedint 0 \Gamma$ are both linearly ordered spaces.
The result follows from Linearly Ordered Space is Completely Normal.
{{qed}}
Theorem: Uncountable Closed Ordinal Space is not Perfectly Normal

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a perfectly normal space.

Proof:
From Omega is Closed in Uncountable Closed Ordinal Space but not $G_\delta$ Set, $\set \Omega$ is not a $G_\delta$ set.
From Ordinal Space is Completely Normal, $\closedint 0 \Omega$ is a $T_1$ space.
Thus by definition $\set \Omega$ is closed in $\closedint 0 \Omega$.
Thus we have that $\set \Omega$ is a closed set of $\closedint 0 \Omega$ which is not a $G_\delta$ set.
The result follows by definition of perfectly normal space.
{{qed}}
Theorem: Uncountable Closed Ordinal Space is not Second-Countable

Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a second-countable space.

Proof:
{{ProofWanted|Show it does not have a countable basis}}
Theorem: Uncountable Open Ordinal Space is not Second-Countable

Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a second-countable space.

Proof:
{{ProofWanted|Show it does not have a countable basis}}
Theorem: Ring Element is Unit iff Unit in Integral Extension

let $A$ be a commutative ring with unity.
Let $a \in A$.
Let $B$ be an integral ring extension of $A$.
{{TFAE}}
:$(1): \quad a$ is a unit of $A$
:$(2): \quad a$ is a unit of $B$

Proof:
=== 1 implies 2 ===
Follows from Ring Homomorphism Preserves Units.
{{qed|lemma}}
Theorem: Separable Elements Form Field

Let $E / F$ be an algebraic field extension.
Let $K$ be the relative separable closure of $F$ in $E$.
Then $K$ is an intermediate field of $E / F$.

Proof:
We need to show that $K$ is a field.
By Transitivity of Separable Field Extensions, an algebraic extension generated by a family of separable elements is separable.
{{proof wanted|proof of Theorem 4.5 in Lang's Algebra}}
Theorem: Decomposition of Field Extension as Separable Extension followed by Purely Inseparable

Let $E / F$ be an algebraic field extension.
Then the relative separable closure $K = F^{sep}$ in $E$ is the unique intermediate field with the following properties:
:$K / F$ is separable
:$E / K$ is purely inseparable.

Proof:
Let $K = F^{sep}$ be the set of elements of $E$ which are separable over $F$.
This is a subextension by Separable Elements Form Field.
Then elements of $E \setminus K$ are not separable over $F$, since all elements that are separable over $F$ are in $K$.
Then elements of $E \setminus K$ are not separable over $K$, since if an element of $E$ is separable over $K$, then it is separable over $F$, by Transitivity of Separable Field Extensions.
{{explain|$E \setminus K$ in this context}}
By definition of purely inseparable field extension, $E/K$ is purely inseparable.
{{qed}}
Theorem: Subextensions of Separable Field Extension are Separable

Let $E / K / F$ be a tower of fields.
Let $E / F$ be separable.
Then $E / K$ and $K / F$ are separable.

Proof:
=== Upper extension ===
We prove that $E / K$ is separable.
Let $\alpha \in E$.
Let $f$ be its minimal polynomial over $F$.
Let $g$ be its minimal polynomial over $K$.
Then {{hypothesis}}, $f$ is separable.
On the other hand:
:$f \in K \sqbrk x$
and:
:$\map f \alpha = 0$
Hence by definition $g$ divides $f$.
{{ExtractTheorem|The following proves Divisor of Separable Polynomial is Separable}}
Consider the roots of $f$ and $g$ in the algebraic closure of $K$.
Since $f$ is separable, it has no repeated roots.
Since $g$ divides $f$, the roots of $g$ are included in the roots of $f$,
we see that $g$ has no repeated roots in the algebraic closure of $K$,
so $g$ is separable.
{{qed|lemma}}
Theorem: Finite Orbit under Group of Automorphisms of Field implies Separable over Fixed Field

Let $E$ be a field.
Let $G \le \Aut E$ be a subgroup of its automorphism group.
Let $F = \map {\operatorname {Fix}_E} G$ be its fixed field.
Let $\alpha \in E$ have a finite orbit under $G$.
Then $\alpha$ is separable over $F$.

Proof:
Let $\Lambda$ be the orbit of $\alpha$ under $G$.
By:
:Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit
the product:
:$\map p x = \ds \prod_{\beta \in \Lambda} \paren {x - \beta}$
is the minimal polynomial of $\alpha$ over $F$.
By Product of Distinct Monic Linear Polynomials is Separable, $p$ is separable.
Thus $\alpha$ is separable over $F$.
{{qed}}
Category:Field Extensions
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Theorem: Countable Closed Ordinal Space is Second-Countable

Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a second-countable space.

Proof:
From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\closedint 0 \Gamma$ of the form:
:$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$
for $\alpha, \beta \in \hointr 0 \Gamma$, forms a basis for $\closedint 0 \Gamma$.
As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\closedint 0 \Gamma$.
Each point in $\closedint 0 \Gamma$ has a countable basis.
From Countable Union of Countable Sets is Countable, it follows that $\closedint 0 \Gamma$ has a countable basis.
{{qed}}
Theorem: Countable Open Ordinal Space is Second-Countable

Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Then $\hointr 0 \Gamma$ is a second-countable space.

Proof:
From Basis for Open Ordinal Topology, the set $\BB$ of subsets of $\hointr 0 \Gamma$ of the form:
:$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$
for $\alpha, \beta \in \hointr 0 \Gamma$, forms a basis for $\hointr 0 \Gamma$.
As $\Gamma$ strictly precedes $\Omega$, there are a countable number of points of $\hointr 0 \Gamma$.
Each point in $\hointr 0 \Gamma$ has a countable basis.
From Countable Union of Countable Sets is Countable, it follows that $\hointr 0 \Gamma$ has a countable basis.
{{qed}}
Theorem: Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit

Let $E$ be a field.
Let $G \le \Aut E$ be a subgroup of its automorphism group.
Let $F = \map {\operatorname {Fix_E} } G$ be its fixed field.
Let $\alpha \in E$ have a finite orbit under $G$.
Then $\alpha$ is algebraic over $F$ and the product of polynomials
:$\ds \map p x = \prod_{\beta \mathop \in \Lambda} \paren {x - \beta}$
is the minimal polynomial of $\alpha$ over $F$.

Proof:
By Product over Finite Set with Zero Factor, we have $\map p \alpha = 0$.
By definition, $p \in E \sqbrk x$.
Theorem: Countable Closed Ordinal Space is Metrizable

Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a metrizable space.

Proof:
From Countable Closed Ordinal Space is Second-Countable, $\closedint 0 \Gamma$ has a basis which is $\sigma$-locally finite.
From Ordinal Space is Regular, $\closedint 0 \Gamma$ is a regular space.
The result follows from Metrizable iff Regular and has Sigma-Locally Finite Basis.
{{qed}}
Theorem: Countable Open Ordinal Space is Metrizable

Let $\Omega$ denote the first uncountable ordinal.
Let $\Gamma$ be a limit ordinal which strictly precedes $\Omega$.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Then $\hointr 0 \Gamma$ is a metrizable space.

Proof:
From Countable Open Ordinal Space is Second-Countable, $\hointr 0 \Gamma$ has a basis which is $\sigma$-locally finite.
From Ordinal Space is Regular, $\hointr 0 \Gamma$ is a regular space.
The result follows from Metrizable iff Regular and has Sigma-Locally Finite Basis.
{{qed}}
Theorem: Closed Ordinal Space is Complete Order Space

Let $\Gamma$ be a limit ordinal.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a complete order space.

Proof:
Let $H$ be a subset of an ordinal space.
Then $H$ has an infimum: its first element.
Let $H$ be a subset of $\closedint 0 \Gamma$.
Then $H$ has a supremum.
Therefore $\closedint 0 \Gamma$ is a complete order space.
{{qed}}
Theorem: Closed Ordinal Space is Compact

Let $\Gamma$ be a limit ordinal.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a compact space.

Proof:
By definition, $\closedint 0 \Gamma$ is a linearly ordered space.
The result follows from Linearly Ordered Space is Compact iff Complete.
{{qed}}
Theorem: Ordinal Space is Strongly Locally Compact

Let $T$ denote an ordinal space on a limit ordinal $\Gamma$.
Then $T$ is a strongly locally compact space.

Proof:
{{ProofWanted|Demonstrated by showing that the closure of each basis neighborhood is compact.}}
Theorem: Open Ordinal Space is not Compact Space

Let $\Gamma$ be a limit ordinal.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Consider the compact subspace $\hointr 0 \Gamma$.
Then $\hointr 0 \Gamma$ is not compact in $\closedint 0 \Gamma$.

Proof:
Consider the set:
:$\set {\hointr 0 \Gamma: \alpha < \Gamma}$
This is an open cover of $\hointr 0 \Gamma$.
But because $\Gamma$ is a limit ordinal, it has no finite subcover.
Hence the result by definition of compact.
{{qed}}
Theorem: Integral Transform is Mapping

Let $\map F p$ be an integral transform:
:$\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$
Let $T$ be the integral operator associated with $\map F p$.
Then $T$ is a mapping from the domain of $T$ to its image.
That is, for every $\map f x$ there exists a unique $\map F p$.

Proof:
Let $p$ be fixed.
In this context, $\map f x \map K {p, x}$ is the pointwise product of the functions $\map f x$ and $\map K {p, x}$.
From Pointwise Operation is Well-Defined, it follows that $\map f x \map K {p, x}$ is a real function on $x$.
We have that both $\map f x$ and $\map K {p, x}$ are integrable.
It follows from Pointwise Product of Integrable Real Functions is Integrable that $\map f x \map K {p, x}$ is an integrable function.
From Definite Integral is Unique, for a given $p$ there is exactly one real number $\map F p$ for which:
:$\map F p = \ds \int_a^b \map f x \map K {p, x} \rd x$
{{qed}}
{{explain|A lot of stuff in analysis and calculus has been glossed over, as can be seen from the redlinks above. This will need to be tightened up.}}
Theorem: Integral Operator is Linear

Let $T$ be an integral operator.
Let $f$ and $g$ be integrable real functions on a domain appropriate to $T$.
Then $T$ is a linear operator:
:$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$

Proof:
Let $T$ be expressed in its full form as an integral fransform:
:$\map T f := \ds \int_a^b \map f x \map K {p, x} \rd x$
for some integrable function $\map K {p, x}$.
Then:
{{begin-eqn}}
{{eqn | l = \map T {\alpha f + \beta g}
      | r = \int_a^b \paren {\alpha \map f x + \beta \map g x} \map K {p, x} \rd x
      | c = 
}}
{{eqn | r = \int_a^b \paren {\alpha \map f x \map K {p, x} + \beta \map g x \map K {p, x} } \rd x
      | c = Real Multiplication Distributes over Addition
}}
{{eqn | r = \alpha \int_a^b \map f x \map K {p, x} \rd x + \beta \int_a^b \map g x \map K {p, x} \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \alpha \map T f + \beta \, \map T g
      | c = {{Defof|Integral Operator}}
}}
{{end-eqn}}
{{Qed}}
Category:Integral Transforms
Category:Integral Operator is Linear
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Theorem: Integral Operator is Linear/Corollary 1

:$\forall \alpha, \beta \in \R: \map T {f + g} = \map T f + \map T g$

Proof:
From Integral Operator is Linear:
:$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$
The result follows by setting $\alpha = \beta = 1$.
{{Qed}}
Theorem: Integral Operator is Linear/Corollary 2

:$\forall \alpha \in \R: \map T {\alpha f} = \alpha \map T f$

Proof:
From Integral Operator is Linear:
:$\forall \alpha, \beta \in \R: \map T {\alpha f + \beta g} = \alpha \map T f + \beta \map T g$
The result follows by setting $\beta = 0$.
{{Qed}}
Theorem: Inverse Integral Operator is Linear if Unique

Let $T$ be an integral operator.
Let $f$ be an integrable real function on a domain appropriate to $T$.
Let $F = \map T f$ and $G = \map T g$.
Let $T$ have a unique inverse $T^{-1}$.
Then $T^{-1}$ is a linear operator:
:$\forall p, q \in \R: \map {T^{-1} } {p F + q G} = p \map {T^{-1} } F + q \map {T^{-1} } G$

Proof:
Let:
:$x_1 = \map {T^{-1} } F$
:$x_2 = \map {T^{-1} } G$
Thus:
:$F = \map T {x_1}$
:$G = \map T {x_2}$
Then for all $p, q \in \R$:
{{begin-eqn}}
{{eqn | l = \map T {p x_1 + q x_2}
      | r = p \map T {x_1} + q \map T {x_2}
      | c = Integral Operator is Linear
}}
{{eqn | r = p F + q G
      | c = 
}}
{{end-eqn}}
and so $x = p F + q G$ is a solution to the equation:
:$\map T x = p F + q G$
But this equation has only one solution:
:$x = \map {T^{-1} } {p F + q G}$
Thus $p F + q G$ must coincide with the above:
:$p \map {T^{-1} } F + q \map {T^{-1} } G = \map {T^{-1} } {p F + q G}$
which proves that $T^{-1}$ is a linear operator.
{{qed}}
Theorem: Trigonometric Series is Convergent if Sum of Absolute Values of Coefficients is Convergent

Let $\map S x$ be a trigonometric series:
:$\map S x = \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
Let the series:
:$\ds \sum_{n \mathop = 1}^\infty \paren {\size {a_n} + \size {b_n} }$
be convergent.
Then $\map S x$ is a convergent series for each $x \in \R$.

Proof:
For all $n \in \N_{\ge 1}$ and $x \in \R$, we have:
{{begin-eqn}}
{{eqn | o = \le
      | l = \size {a_n \cos n x + b_n \sin n x} 
      | r = \size {a_n \cos n x} + \size {b_n \sin n x}
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \size {a_n} \size {\cos n x} + \size {b_n} \size {\sin n x}
      | c = Absolute Value Function is Completely Multiplicative
}}
{{eqn | o = \le
      | r = \size {a_n} + \size {b_n}
      | c = Real Cosine Function is Bounded and Real Sine Function is Bounded
}}
{{end-eqn}}
We have {{hypothesis}}, the series $\ds \sum_{n \mathop = 1}^\infty \paren {\size {a_n} + \size {b_n} }$ is convergent.
By the Comparison Test, it follows that:
:$\ds \sum_{n \mathop = 1}^\infty \paren {\size {a_n} + \size {b_n} }$
is absolutely convergent for all $x \in \R$.
From Absolutely Convergent Real Series is Convergent, it follows that $\map S x$ is convergent for all $x \in \R$.
{{qed}}
Theorem: Convergent Trigonometric Series is Periodic

Let $\map S x$ be a trigonometric series:
:$\map S x = \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
Let $S$ be convergent.
Then $S$ is periodic:
:$\forall r \in \Z: \map S {x + 2 r \pi} = \map S x$

Proof:
Let $\map S x$ converge to some $L \in \R$.
Let $r \in \Z$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map S {x + 2 r \pi}
      | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n \paren {x + 2 r \pi} + b_n \sin n \paren {x + 2 r \pi} }
      | c = Definition of $\map S {x + 2 r \pi}$
}}
{{eqn | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \map \cos {n x + 2 r n \pi} + b_n \map \sin {n x + 2 r n \pi} }
      | c = 
}}
{{eqn | r = \dfrac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}
      | c = Sine and Cosine are Periodic on Reals
}}
{{eqn | r = \map S x
      | c = Definition of $\map S x$
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Infinite Series of Product of Power and Sine

Let $r \in \R$ such that $\size r < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^\infty r^k \sin k x
      | r = r \sin x + r^2 \sin 2 x + r^3 \sin 3 x + \cdots
      | c = 
}}
{{eqn | r = \dfrac {r \sin x} {1 - 2 r \cos x + r^2}
      | c = 
}}
{{end-eqn}}

Proof:
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^\infty r^k \sin k x
      | r = \map \Im {\sum_{k \mathop = 1}^\infty r^k e^{i k x} }
      | c = 
}}
{{eqn | r = \map \Im {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}
      | c = as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$
}}
{{eqn | r = \map \Im {\frac 1 {1 - r e^{i x} } }
      | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$
}}
{{eqn | r = \map \Im {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }
      | c = 
}}
{{eqn | r = \map \Im {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }
      | c = 
}}
{{eqn | r = \map \Im {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + r^2} }
      | c = {{Corollary|Euler's Formula}}
}}
{{eqn | r = \dfrac {r \sin x} {1 - 2 r \cos x + r^2}
      | c = after simplification
}}
{{end-eqn}}
{{qed}}