avewright/proofwiki-math huggingface dataset:
Theorem: Points in Product Spaces are Near Open Sets

Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of topological spaces, where $I$ is an arbitrary index set.
Let $X = \ds \prod_{i \mathop \in I} X_i$ be the product space of $\family {X_i}_{i \mathop \in I}$.
Let $U$ be nonempty open subset of $X$.
Let $x$ be a point in $X$.
For each point $y$ in $X$, let $\map K y = \set {i \in I : y_i \ne x_i}$.
Then there exists a point $u$ in $U$ such that $\map K u$ is finite.

Proof:
Let $q$ be any point in $U$.
The topology on the product space $X$ is the product topology which has the natural basis as a (synthetic) basis.
Then:
:$\exists Q \in \BB : q \in Q \subseteq U$
where $\BB$ is the natural basis for product topology on $X$.
From Natural Basis of Product Topology:
:$\ds Q = \prod_{i \mathop \in I} Q_i$ where:
::for all $i \in I : Q_i$ is open in $X_i$
::$J = \set {i \in I : Q_i \ne X_i}$ is finite
Let $u$ be the point defined by:
:$u_i = \begin {cases} q_i & i \in J \\ x_i & i \notin J \end {cases}$
By definition of $u$:
:$\forall i \in I : u_i \in Q_i$
By definition of the Cartesian product:
:$u \in Q \subseteq U$
By definition of $u$:
:$\forall i \in I: i \notin J \implies u_i = x_i$
So by Rule of Transposition:
:$\forall i \in I: u_i \ne x_i \implies i \in J$
Thus:
:$\map K u \subseteq J$ 
From Subset of Finite Set is Finite, and because $J$ is finite, $\map K u$ is also finite.
{{qed}}
Category:Product Topology
870vyj10kjk0mjvzemwtaaj3avev15a
Theorem: Real and Imaginary Part Projections are Continuous

Define the real-valued functions $x, y: \C \to \R$ by:
:$\forall z \in \C: \map x z = \map \Re z$
:$\forall z \in \C: \map y z = \map \Im z$
Equip $\R$ with the usual Euclidean metric.
Equip $\C$ with the usual Euclidean metric.
Then both $x$ and $y$ are continuous functions.

Proof:
Let $z \in \C$, and let $\epsilon \in \R_{>0}$.
Put $\delta = \epsilon$.
For all $w \in \C$ with $\cmod {w - z} < \delta$:
{{begin-eqn}}
{{eqn | l = \cmod {\map \Re w - \map \Re z}
      | r = \cmod {\map \Re w + i \map \Im w - \map \Re z - i \map \Im z + i \map \Im z - i \map \Im w}
}}
{{eqn | o = \le
      | r = \cmod {\map \Re w + i \map \Im w - \map \Re z - i \map \Im z} + \cmod {i \map \Im z - i \map \Im w}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = \le 
      | r = \cmod {\map \Re w + i \map \Im w - \map \Re z - i \map \Im z}
      | c = Complex Modulus is Non-Negative
}}
{{eqn | r = \cmod {w - z}
}}
{{eqn | o = <
      | r = \delta
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \cmod {\map \Im w - \map \Im z}
      | r = \cmod i \cmod {\map \Im w - \map \Im z}
      | c = as $\cmod i = 1$
}}
{{eqn | r = \cmod {i \map \Im w - i \map \Im z}
      | c = Complex Modulus of Product of Complex Numbers
}}
{{eqn | r = \cmod {\map \Re w + i \map \Im w - \map \Re z - i \map \Im z + \map \Re z - \map \Re w}
}}
{{eqn | o = \le
      | r = \cmod {\map \Re w + i \map \Im w - \map \Re z - i \map \Im z} + \cmod {\map \Re z - \map \Re w}
      | c = Triangle Inequality for Complex Numbers
}}
{{eqn | o = \le 
      | r = \cmod {\map \Re w + i \map \Im w - \map \Re z - i \map \Im z}
      | c = Complex Modulus is Non-Negative
}}
{{eqn | r = \cmod {w - z}
}}
{{eqn | o = <
      | r = \delta
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
It follows by definition that $x$ and $y$ are both continuous.
{{qed}}
Theorem: Continuous Complex Function is Complex Riemann Integrable

Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Then $f$ is complex Riemann integrable over $\closedint a b$.

Proof:
Define the real function $x: \closedint a b \to \R$ by:
:$\forall t \in \closedint a b : \map x t = \map \Re {\map f t}$
Define the real function $y: \closedint a b \to \R$ by:
:$\forall t \in \closedint a b : \map y t = \map \Im {\map f t}$
where:
:$\map \Re {\map f t}$ denotes the real part of the complex number $\map f t$
:$\map \Im {\map f t}$ denotes the imaginary part of $\map f t$.
From Real and Imaginary Part Projections are Continuous, it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are continuous functions.
From Composite of Continuous Mappings is Continuous, it follows that $x$ and $y$ are continuous.
From Continuous Real Function is Darboux Integrable, it follows that $x$ and $y$ are Darboux integrable over $\closedint a b$.
By definition, it follows that $f$ is complex Riemann integrable.
{{qed}}
Theorem: Compact Subspace of Metric Space is Sequentially Compact in Itself

Let $M = \struct {A, d}$ be a metric space.
Let $C \subseteq M$ be a subspace of $M$ such that $C$ is compact.
Then $C$ is sequentially compact in itself.

Proof:
Let $C \subseteq M$ be compact.
Let $\sequence {x_n}$ be a sequence in $C$.
Let $S$ be the range of $\sequence {x_n}$.
Thus $S \subseteq C$ and $S$ may be either finite or infinite.
Theorem: Products of Products are Homeomorphic to Collapsed Products

Let $I$ be an index set, and for each $i \in I$ let $J_i$ be an index set.
Let the sets $J_i$ be pairwise disjoint.
Let $\ds J = \bigcup_{i \mathop \in I} J_i$
For each $j \in J$, let $X_j$ be a topological space.
Then $\ds \prod_{j \mathop \in J} X_j$ is homeomorphic to $\ds \prod_{i \mathop \in I} \prod_{j \mathop \in J_i} X_j$.

Proof:
{{proof wanted}}
Category:Product Spaces
0s9l5qbbx5pb7ixmlpcjy0sjn45jpqn
Theorem: Sum of Complex Integrals on Adjacent Intervals

Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Let $c \in \closedint a b$.
Then:
:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$

Proof:
From Continuous Complex Function is Complex Riemann Integrable, it follows that all three complex Riemann integrals are well defined.
From Real and Imaginary Part Projections are Continuous, it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are continuous functions.
{{explain|Revisit the above link -- see if there is a more appropriate one to use so as not to invoke the concept of metric spaces}}
From Composite of Continuous Mappings is Continuous, it follows that $\Re \circ f: \R \to \R$ and $\Im \circ f: \R \to \R$ are continuous real functions.
Then:
{{begin-eqn}}
{{eqn | l = \int_a^b \map f t \rd t
      | r = \int_a^b \map \Re {\map f t} \rd t + i \int_a^b \map \Im {\map f t} \rd t
      | c = {{Defof|Complex Riemann Integral}}
}}
{{eqn | r = \int_a^c \map \Re {\map f t} \rd t + \int_c^b \map \Re {\map f t} \rd t + i \paren {\int_a^c \map \Im {\map f t} \rd t + \int_c^b \map \Im {\map f t} \rd t}
      | c = Sum of Integrals on Adjacent Intervals for Continuous Functions
}}
{{eqn | r = \int_a^c \map \Re {\map f t} \rd t + i \int_a^c \map \Im {\map f t} \rd t + \int_c^b \map \Re {\map f t} \rd t + i \int_c^b \map \Im {\map f t} \rd t
}}
{{eqn | r = \int_a^c \map f t \rd t + \int_c^b \map f t \rd t
}}
{{end-eqn}}
{{qed}}
Category:Complex Analysis
a22sv6klgd10vytqvhtm6xsjy53pztz
Theorem: Equivalence of Definitions of Metric Space Continuity at Point

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$.
Let $a \in A_1$ be a point in $A_1$.
{{TFAE|def = Continuous at Point of Metric Space|continuity at a point}}

Proof:
=== $\epsilon$-$\delta$ Definition iff Definition by Limits ===
This is proved in Metric Space Continuity by Epsilon-Delta.
{{qed|lemma}}
Theorem: Equivalence of Definitions of Limit of Mapping between Metric Spaces

{{TFAE|def = Limit of Mapping between Metric Spaces}}
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $c$ be a limit point of $M_1$.
Let $f: A_1 \to A_2$ be a mapping from $A_1$ to $A_2$ defined everywhere on $A_1$ ''except possibly'' at $c$.
Let $L \in M_2$.

Proof:
=== $\epsilon$-$\delta$ Condition implies $\epsilon$-Ball Condition ===
Suppose that $f$ satisfies the $\epsilon$-$\delta$ condition:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \map {d_1} {x, c} < \delta \implies \map {d_2} {\map f x, L} < \epsilon$
Let $y \in f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c}$.
By definition of open $\epsilon$-ball, this means:
:$\exists x \in A_1: 0 < \map {d_1} {x, c} < \delta: y = \map f x$
By hypothesis, it follows that $\map {d_2} {y, L} < \epsilon$
That is:
:$y \in \map {B_\epsilon} {L; d_2}$
By definition of subset:
:$f \sqbrk {\map {B_\delta} {c; d_1} \setminus \set c} \subseteq \map{B_\epsilon} {L; d_2}$
Thus it follows that $f$ satisfies the $\epsilon$-ball condition.
{{qed|lemma}}
Theorem: Ordering Principle

Let $S$ be a set.
Then there exists a total ordering on $S$.

Proof:
From Zermelo's Well-Ordering Theorem, $S$ has a well-ordering.
The result follows from Well-Ordering is Total Ordering.
{{qed}}
Theorem: Ordering Principle

Let $S$ be a set.
Then there exists a total ordering on $S$.

Proof:
This theorem follows trivially from the Order-Extension Principle.
{{qed}}
Theorem: Ordering Principle

Let $S$ be a set.
Then there exists a total ordering on $S$.

Proof:
Let $S$ be a non-empty subset of the set of natural numbers $\N$.
We take as axiomatic that $\N$ is itself a subset of the set of real numbers $\R$.
Thus $S \subseteq \R$.
By definition:
:$\forall n \in \N: n \ge 0$
and so:
:$\forall n \in S: n \ge 0$
Hence $0$ is a lower bound of $S$.
This establishes the fact that $S$ is bounded below.
By the Continuum Property, we have that $S$ admits an infimum.
Hence let $b = \inf S$.
Because $b$ is the infimum of $S$, it follows that $b + 1$ is not a lower bound of $S$.
So, for some $n \in S$:
:$n < b + 1$
{{AimForCont}} $n$ is not the smallest element of $S$.
Then there exists $m \in S$ such that:
:$b \le m < n < b + 1$
from which it follows that:
:$0 < n - m < 1$
But there exist no natural numbers $k$ such that $0 < k < 1$.
Hence $n = b$ is the smallest element of $S$.
{{qed}}
Theorem: Ordering Principle

Let $S$ be a set.
Then there exists a total ordering on $S$.

Proof:
Consider the natural numbers $\N$ defined as the naturally ordered semigroup $\struct {S, \circ, \preceq}$.
From its definition, $\struct {S, \circ, \preceq}$ is well-ordered by $\preceq$.
The result follows.
As $\N_{\ne 0} = \N \setminus \set 0$, by Set Difference is Subset $\N_{\ne 0} \subseteq \N$.
As $\N$ is well-ordered, by definition, every subset of $\N$ has a smallest element.
{{qed}}
Theorem: Ordering Principle

Let $S$ be a set.
Then there exists a total ordering on $S$.

Proof:
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is a minimally inductive class under the successor mapping.
From Successor Mapping on Natural Numbers is Progressing, this successor mapping is a progressing mapping.
The result is a direct application of Minimally Inductive Class under Progressing Mapping is Well-Ordered under Subset Relation.
{{qed}}
Theorem: Number Smaller than Lebesgue Number is also Lebesgue Number

Let $M = \struct {A, d}$ be a metric space.
Let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$.
Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.
Then $\epsilon'$ is also a Lebesgue number for $M$.

Proof:
By hypothesis, let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$.
Then by definition:
:$\forall x \in A: \exists \map U x \in \UU: \map {B_\epsilon} x \subseteq \map U x$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$.
Let $y \in \map {B_{\epsilon'} } x$.
{{begin-eqn}}
{{eqn | l = \map d {y, x}
      | o = <
      | r = \epsilon'
      | c = {{Defof|Open Ball of Metric Space}}
}}
{{eqn | ll= \leadsto
      | l = \map d {y, x}
      | o = <
      | r = \epsilon
      | c = as $\epsilon' < \epsilon$
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \map {B_\epsilon} x
      | c = {{Defof|Open Ball of Metric Space}}
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \map U x
      | c = {{Defof|Subset}}: $\map {B_\epsilon} x \subseteq \map U x$
}}
{{end-eqn}}
That is:
:$\map {B_{\epsilon'} } x \subseteq \map U x$
The result follows by definition of Lebesgue number.
{{qed}}
Theorem: Open Cover may not have Lebesgue Number

Let $M = \struct {A, d}$ be a metric space.
Let $\CC$ be an open cover of $M$.
Then it may not necessarily be the case that $\CC$ has a Lebesgue number.

Proof:
Let $M := \openint 0 1 \subseteq \R$ under the Euclidean metric.
Let $\CC := \set {\openint {\dfrac 1 n} 1: n \ge 2}$.
For any $\epsilon \in \R_{>0}$, take $n > \dfrac 1 \epsilon$ and $x = \dfrac 1 n$.
There is no $\openint {\dfrac 1 m} 1$ such that $\map {B_\epsilon} x \subseteq \openint {\dfrac 1 m} 1$, since $\map {B_\epsilon} x = \map {B_\epsilon} {\dfrac 1 n} = \openint 0 {\dfrac 1 n + \epsilon}$.
{{qed}}
Theorem: Burnside's Theorem

Let $G$ be a finite group.
Let the order of $G$ be $p^m q^n$ where:
:$p, q$ are prime
:$m, n \in \N$
Then $G$ is solvable.

Proof:
{{ProofWanted}}
{{Namedfor|William Burnside|cat = Burnside}}
Theorem: Constant Function is Uniformly Continuous/Metric Space

Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.
Let $f_c: A_1 \to A_2$ be the constant mapping from $A_1$ to $A_2$:
:$\exists c \in A_2: \forall a \in A_1: f_c \left({a}\right) = c$
That is, every point in $A_1$ maps to the same point $c$ in $A_2$.
Then $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.

Proof:
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$.
Let $\epsilon \in \R: \epsilon > 0$.
Pick any $\delta \in \R$ such that $\delta > 0$.
Let $x, y \in A_1$ such that $d_1 \left({x, y}\right) < \delta$.
Now we have:
:$f_c \left({x}\right) = c = f_c \left({y}\right)$
Hence:
:$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$
By definition of a metric:
:$d_2 \left({c, c}\right) = 0 < \epsilon$.
Thus we have that:
:$\forall \epsilon > 0: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$
Hence by definition of uniform continuity on a metric space, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$.
{{qed}}
Category:Constant Mappings
Category:Metric Spaces
Category:Uniformly Continuous Mappings
dwll4poxn4ev0ljeemh7zgf578t7aky
Theorem: Constant Function is Uniformly Continuous/Real Function

Let $f_c: \R \to \R$ be the constant mapping:
:$\exists c \in \R: \forall a \in \R: \map {f_c} a = c$
Then $f_c$ is uniformly continuous on $\R$.

Proof:
Follows directly from:
:Constant Function is Uniformly Continuous: Metric Space
:Real Number Line is Metric Space.
{{qed}}
Theorem: Pointwise Difference of Simple Functions is Simple Function

Let $\struct {X, \Sigma}$ be a measurable space.
Let $f,g : X \to \R$ be simple functions.
Let $f - g: X \to \R$ be the pointwise difference of $f$ and $g$:
:$\forall x \in X: \map {\paren {f - g} } x := \map f x - \map g x$
Then $f - g$ is also a simple function.

Proof:
By Scalar Multiple of Simple Function is Simple Function, $-g = -1 \cdot g$ is a simple function.
By Pointwise Sum of Simple Functions is Simple Function, so is $f - g$.
{{qed}}
Category:Simple Functions
7evs65fh65472uutz4nfnefrd9atbrp
Theorem: Modulus and Argument of Complex Exponential

Let $z \in \C$ be a complex number.
Let $\hointr a {a + 2 \pi}$ be a half open interval of length $2 \pi$.
Let $r \in \hointr 0 {+\infty}$ and $\theta \in \hointr a {a + 2 \pi}$.
Then:
:$r = \cmod z$ and $\theta = \map \arg z$
{{iff}}:
:$z = r e^{i \theta}$
where:
:$\cmod z$ denotes the modulus of $z$
:$\map \arg z$ denotes the argument of $z$
:$x \mapsto e^x$ is the complex exponential function.
If $z = 0$ or $r = 0$, then $\theta$ may be any number in $\hointr a {a + 2 \pi}$.

Proof:
=== Necessary condition ===
Let $r = \cmod z$.
If $z = 0$, we have:
:$z = 0e^{i \theta} = re^{i \theta}$
Suppose $z \ne 0$ and $\theta = \map \arg z$.
By definition of argument, the following two equations hold:
:$(1): \quad \dfrac {\map \Re z} r = \cos \theta$
:$(2): \quad \dfrac {\map \Im z} r = \sin \theta$
where:
:$\map \Re z$ denotes the real part of $z$
:$\map \Im z$ denotes the imaginary part of $z$.
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = \map \Re z + i \map \Im z
      | c = {{Defof|Complex Number}}
}}
{{eqn | r = r \cos \theta + i r \sin \theta
      | c = from $(1)$ and $(2)$
}}
{{eqn | r = r \paren {\cos \theta + i \sin \theta}
}}
{{eqn | r = re^{i \theta}
      | c = Euler's Formula
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Bases of Vector Space have Equal Cardinality

Let $R$ be a division ring.
Let $V$ be a vector space over $R$.
Let $X$ and $Y$ be bases of $V$.
Then $X$ and $Y$ are equivalent.
That is, $X$ and $Y$ have the same cardinality.

Proof:
We will first prove that there is an injection from $X$ to $Y$.
Let $x \in X$.
By Expression of Vector as Linear Combination from Basis is Unique: General Result, there is a unique finite subset $C_x$ of $R \times Y$ such that:
:$\ds x = \sum_{\tuple {r, v} \mathop \in C_x} r \cdot v$ and
:$\forall \tuple {r, v} \in C_x: r \ne 0_R$
Define $\Phi: X \to \powerset Y$ by:
:$\map \Phi x := \Img {C_x}$
We next show that $\sequence {\map \Phi x}_{x \mathop \in X}$ satisfies the '''marriage condition'''. 
That is, for every finite subset $F$ of $X$:
:$\ds \card F \le \card {\bigcup \map \Phi F}$
Since $X$ is a basis, it is linearly independent.
By Subset of Linearly Independent Set is Linearly Independent, $F$ is also linearly independent.
By the definition of $\Phi$:
:$\ds F \subseteq \span \bigcup \map \Phi F$.
By Finite Union of Finite Sets is Finite, $\ds \bigcup \map \Phi F$ is finite.
Thus by Size of Linearly Independent Subset is at Most Size of Finite Generator:
:$\ds \size F \le \size {\bigcup \map \Phi F}$
By Hall's Marriage Theorem, there is an injection from $X$ into $Y$.
Precisely the same argument with $X$ and $Y$ interchanged shows that there is an injection from $Y$ into $X$ as well.
Thus by the Cantor-Bernstein-Schröder Theorem, $X$ and $Y$ are equivalent.
{{qed}}
{{BPI|Hall's Marriage Theorem/General Set|3}}
Theorem: Equivalence of Definitions of Symmetric Difference/(1) iff (2)

Let $S$ and $T$ be sets.
{{TFAENocat|def = Symmetric Difference|view = symmetric difference $S \symdif T$ between $S$ and $T$}}

Proof:
{{begin-eqn}}
{{eqn | l = S \symdif T
      | r = \paren {S \setminus T} \cup \paren {T \setminus S}
      | c = {{Defof|Symmetric Difference|index = 1}}
}}
{{eqn | r = \paren {\paren {S \cup T} \setminus T} \cup \paren {\paren {S \cup T} \setminus S}
      | c = Set Difference with Union is Set Difference
}}
{{eqn | r = \paren {S \cup T} \setminus \paren {T \cap S}
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {S \cup T} \setminus \paren {S \cap T}
      | c = Intersection is Commutative
}}
{{end-eqn}}
Theorem: Equivalence of Definitions of Symmetric Difference/(1) iff (3)

Let $S$ and $T$ be sets.
{{TFAENocat|def = Symmetric Difference|view = symmetric difference $S \symdif T$ between $S$ and $T$}}

Proof:
{{begin-eqn}}
{{eqn | l = S \symdif T
      | r = \paren {S \setminus T} \cup \paren {T \setminus S}
      | c = {{Defof|Symmetric Difference|index = 1}}
}}
{{eqn | r = \paren {S \cap \overline T} \cup \paren {\overline S \cap T}
      | c = Set Difference as Intersection with Complement
}}
{{end-eqn}}
Theorem: Equivalence of Definitions of Symmetric Difference/(2) iff (4)

Let $S$ and $T$ be sets.
{{TFAENocat|def = Symmetric Difference|view = symmetric difference $S \symdif T$ between $S$ and $T$}}

Proof:
{{begin-eqn}}
{{eqn | l = S \symdif T
      | r = \paren {S \cup T} \setminus \paren {S \cap T}
      | c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \paren {S \cup T} \cap \paren {\overline {S \cap T} }
      | c = Set Difference as Intersection with Complement
}}
{{eqn | r = \paren {S \cup T} \cap \paren {\overline S \cup \overline T}
      | c = De Morgan's Laws: Complement of Intersection
}}
{{end-eqn}}
Theorem: Subset of Countable Set is Countable

A subset of a countable set is countable.

Proof:
Let $S$ be a countable set.
Let $T \subseteq S$.
By definition, there exists an injection $f: S \to \N$.
Let $i: T \to S$ be the inclusion mapping.
We have that $i$ is an injection.
Because the composite of injections is an injection, it follows that $f \circ i: T \to \N$ is an injection.
Hence, $T$ is countable.
{{qed}}
Theorem: Euler's Sine Identity

:$\sin z = \dfrac {e^{i z} - e^{-i z} } {2 i}$

Proof:
Recall the definition of the sine function:
{{begin-eqn}}
{{eqn | l = \sin z
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}
      | c = 
}}
{{eqn | r = z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \cdots + \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} + \cdots
      | c = 
}}
{{end-eqn}}
Recall the definition of the exponential as a power series:
{{begin-eqn}}
{{eqn | l = e^z
      | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}
      | c = 
}}
{{eqn | r = 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots
      | c = 
}}
{{end-eqn}}
Then, starting from the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \frac {e^{i z} - e^{-i x} } {2 i}
      | r = \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }
      | c = 
}}
{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n - \paren {-i z}^n} {n!} }
      | c = 
}}
{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} - \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }
      | c = split into even and odd $n$
}}
{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!}
      | c = as $\paren {-i z}^{2 n} = \paren {i z}^{2 n}$
}}
{{eqn | r = \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}
      | c = as $\paren {-1}^{2 n + 1} = -1$
}}
{{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}
      | c = cancel $2$
}}
{{eqn | r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n z^{2 n + 1} } {\paren {2 n + 1}!}
      | c = as $i^{2 n + 1} = i \paren {-1}^n $
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1!} }
      | c = cancel $i$
}}
{{eqn | r = \sin z
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Sine Identity

:$\sin z = \dfrac {e^{i z} - e^{-i z} } {2 i}$

Proof:
Recall Euler's Formula:
:$e^{i z} = \cos z + i \sin z$
Then, starting from the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \frac {e^{i z} - e^{-i z} } {2 i}
      | r = \frac {\paren {\cos z + i \sin z} - \paren {\map \cos {-z} + i \map \sin {-z} } } {2 i}
}}
{{eqn | r = \frac {\paren {\cos z + i \sin z - \cos z - i \map \sin {-z} } } {2 i}
      | c = Cosine Function is Even
}}
{{eqn | r = \frac {i \sin z - i \map \sin {-z} } {2 i}
      | c = 
}}
{{eqn | r = \frac {i \sin z - i \paren {-\sin z} } {2 i}
      | c = Sine Function is Odd
}}
{{eqn | r = \frac {2 i \sin z} {2 i}
      | c = 
}}
{{eqn | r = \sin z
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Euler's Sine Identity

:$\sin z = \dfrac {e^{i z} - e^{-i z} } {2 i}$

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | l = e^{i z}
      | r = \cos z + i \sin z
      | c = Euler's Formula
}}
{{eqn | n = 2
      | l = e^{-i z}
      | r = \cos z - i \sin z
      | c = {{Corollary|Euler's Formula}}
}}
{{eqn | ll= \leadsto
      | l = e^{i z} - e^{-i z}
      | r = \paren {\cos z + i \sin z} - \paren {\cos z - i \sin z}
      | c = $(1) - (2)$
}}
{{eqn | r = 2 i \sin z
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \frac {e^{i z} - e^{-i z} } {2 i}
      | r = \sin z
      | c = 
}}
{{end-eqn}}
{{qed}}