avewright/proofwiki-math huggingface dataset:
Theorem: Coset Product of Normal Subgroup is Consistent with Subset Product Definition

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Let $a \circ N$ and $b \circ N$ be the left cosets of $a$ and $b$ by $N$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is consistent with the definition of the coset product as the subset product of $a \circ N$ and $b \circ  N$:
:$\paren {a \circ N} \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$

Proof:
Consider the set:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
As $e \in N$, have:
:$\paren {a \circ b} \circ N = \paren {a \circ e} \circ \paren {b \circ N} \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
by {{Group-axiom|1}} of $\circ$.
Hence $\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$.
Now let $x \in a \circ N$ and $y \in b \circ N$.
Then by the definition of subset product:
:$\exists n_1 \in N: x = a \circ n_1$
:$\exists n_2 \in N: y = b \circ n_2$
It follows that:
{{begin-eqn}}
{{eqn | l = x \circ y
      | r = \paren {a \circ n_1} \circ \paren {b \circ n_2}
      | c = 
}}
{{eqn | r = \paren {a \circ n_1} \circ \paren {n_3 \circ b}
      | c = for some $n_3 \in N$: {{Defof|Normal Subgroup}}
}}
{{eqn | r = a \circ \paren {\paren {n_1 \circ n_3} \circ b}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = a \circ \paren {n_4 \circ b}
      | c = for some $n_4 \in N$: {{Defof|Subgroup}}
}}
{{eqn | r = a \circ \paren {b \circ n_5}
      | c = for some $n_5 \in N$: {{Defof|Normal Subgroup}}
}}
{{eqn | r = \paren {a \circ b} \circ n_5
      | c = {{Group-axiom|1}}
}}
{{eqn | o = \in
      | r = \paren {a \circ b} \circ N
      | c = {{Defof|Subset Product}}
}}
{{end-eqn}}
So the definition by subset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \set {x \circ y: x \in a \circ N, y \in b \circ N}$
leads to the definition of coset product as:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
{{qed}}
Category:Coset Product
2hi12gt9tgfp6c8s10quk1u9lj0h5iz
Theorem: Equivalence Relation is Congruence iff Compatible with Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be an equivalence relation on $S$.
Then $\RR$ is a congruence relation for $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = x \mathrel \RR y
      | o = \implies
      | r = \paren {x \circ z} \mathrel \RR \paren {y \circ z}
}}
{{eqn | l = x \mathrel \RR y
      | o = \implies
      | r = \paren {z \circ x} \mathrel \RR \paren {z \circ y}
}}
{{end-eqn}}
That is, {{iff}} $\RR$ is compatible with $\circ$.

Proof:
=== Necessary Condition ===
Let $\RR$ be a congruence relation for $\circ$.
That is:
:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \mathrel \RR x_2 \land y_1 \mathrel \RR y_2 \implies \paren {x_1 \circ y_1} \mathrel \RR \paren {x_2 \circ y_2}$
As $\RR$ is an equivalence relation it is by definition reflexive.
That is:
:$\forall z \in S: z \mathrel \RR z$
Make the substitutions:
{{begin-eqn}}
{{eqn | l = x_1
      | o = \to
      | r = x
}}
{{eqn | l = x_2
      | o = \to
      | r = y
}}
{{eqn | l = y_1
      | o = \to
      | r = z
}}
{{eqn | l = y_2
      | o = \to
      | r = z
}}
{{end-eqn}}
It follows that:
:$\forall x, y, z \in S: x \mathrel \RR y \implies \paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Similarly, make the substitutions:
{{begin-eqn}}
{{eqn | l = x_1
      | o = \to
      | r = z
}}
{{eqn | l = x_2
      | o = \to
      | r = z
}}
{{eqn | l = y_1
      | o = \to
      | r = x
}}
{{eqn | l = y_2
      | o = \to
      | r = y
}}
{{end-eqn}}
It follows that:
:$\forall x, y, z \in S: x \mathrel \RR y \implies \paren {z \circ x} \mathrel \RR \paren {z \circ y}$
{{qed|lemma}}
=== Sufficient Condition ===
Now let $\RR$ have the nature that:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = x \mathrel \RR y
      | o = \implies
      | r = \paren {x \circ z} \mathrel \RR \paren {y \circ z}
}}
{{eqn | l = x \mathrel \RR y
      | o = \implies
      | r = \paren {z \circ x} \mathrel \RR \paren {z \circ y}
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x_1, x_2, y_1, y_2 \in S
      | l = x_1 \mathrel \RR y_1
      | o = \implies
      | r = \paren {x_1 \circ x_2} \mathrel \RR \paren {y_1 \circ x_2}
}}
{{eqn | l = x_2 \mathrel \RR y_2
      | o = \implies
      | r = \paren {y_1 \circ x_2} \mathrel \RR \paren {y_1 \circ y_2}
}}
{{end-eqn}}
As $\RR$ is an equivalence relation it is by definition transitive.
Thus it follows that:
:$\paren {x_1 \circ x_2} \mathrel \RR \paren {y_1 \circ y_2}$
{{qed|lemma}}
The result follows.
{{qed}}
Theorem: Equivalence Relation is Congruence iff Compatible with Operation

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be an equivalence relation on $S$.
Then $\RR$ is a congruence relation for $\circ$ {{iff}}:
{{begin-eqn}}
{{eqn | q = \forall x, y, z \in S
      | l = x \mathrel \RR y
      | o = \implies
      | r = \paren {x \circ z} \mathrel \RR \paren {y \circ z}
}}
{{eqn | l = x \mathrel \RR y
      | o = \implies
      | r = \paren {z \circ x} \mathrel \RR \paren {z \circ y}
}}
{{end-eqn}}
That is, {{iff}} $\RR$ is compatible with $\circ$.

Proof:
We have that an equivalence relation is a (symmetric) preordering.
Thus the result Preordering of Products under Operation Compatible with Preordering can be applied directly.
{{qed}}
Theorem: Outer Measure of Limit of Increasing Sequence of Sets

Let $\mu^*$ be an outer measure on a set $X$.
Let $\sequence {S_n}$ be an increasing sequence of $\mu^*$-measurable sets, and let $S_n \uparrow S$ (as $n \to \infty$).
Then for any subset $A \subseteq X$:
:$\ds \map {\mu^*} {A \cap S} = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$

Proof:
By the monotonicity of $\mu^*$, it suffices to prove that:
:$\ds \map {\mu^*} {A \cap S} \le \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
Assume that $\map {\mu^*} {A \cap S_n}$ is finite for all $n \in \N$, otherwise the statement is trivial by the monotonicity of $\mu^*$.
Let $S_0 = \O$.
Then $x \in S$ {{iff}} there exists an integer $n \ge 0$ such that $x \in S_{n + 1}$. 
Taking the least possible $n$, it follows that $x \notin S_n$, and so:
:$x \in S_{n + 1} \setminus S_n$
Therefore:
:$\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n}$
From Intersection Distributes over Union:
:$\ds A \cap S = A \cap \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n} = \bigcup_{n \mathop = 0}^\infty \paren {A \cap \paren {S_{n + 1} \setminus S_n} }$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {\mu^*} {A \cap S}
      | o = \le
      | r = \sum_{n \mathop = 0}^\infty \map {\mu^*} {A \cap \paren {S_{n + 1} \setminus S_n} }
      | c = {{Defof|Countably Subadditive Function}}
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {A \cap S_{n + 1} } - \map {\mu^*} {A \cap S_{n + 1} \cap S_n} }
      | c = {{Defof|Measurable Set of Arbitrary Outer Measure}}
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\map {\mu^*} {A \cap S_{n + 1} } - \map {\mu^*} {A \cap S_n} }
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n} - \map {\mu^*} {A \cap \O}
      | c = Telescoping Series
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}
      | c = Intersection with Empty Set and {{Defof|Outer Measure}}
}}
{{end-eqn}}
{{qed}}
Theorem: Half-Open Rectangles form Semiring of Sets

The half-open $n$-rectangles $\JJ_{ho}^n$ form a semiring of sets.

Proof:
By definition, $\O$ is considered to be a half-open $n$-rectangle.
That $\JJ_{ho}^n$ is $\cap$-stable follows from Half-Open Rectangles Closed under Intersection.
Thus, it remains to show condition $(3')$ for a semiring of sets:
:$(3'):\quad$ If $A, B \in \JJ_{ho}^n$, then there exists a finite sequence of pairwise disjoint half-open $n$-rectangles $A_1, A_2, \ldots, A_m \in \JJ_{ho}^n$ such that $\ds A \setminus B = \bigcup_{k \mathop = 1}^m A_k$.
To prove this, proceed by induction on $n$:
Theorem: Lebesgue Pre-Measure is Pre-Measure

The Lebesgue pre-measure $\lambda^n$ on the half-open $n$-rectangles $\JJ_{ho}^n$ is a pre-measure.

Proof:
We employ Characterization of Pre-Measures.
It is known that $\map {\lambda^n} \O = 0$ by definition of Lebesgue pre-measure.
The only possibility for two disjoint half-open $n$-rectangles to constitute a single, large half-open $n$-rectangle is when they are of the form:
:$\horectr {\mathbf a} {\mathbf b} \quad \horectr {\mathbf a'} {\mathbf b'}$
such that we have for some $i$ with $1 \le i \le n$:
:$i \ne j \implies a_j = a'_j$
:$i \ne j \implies b_j = b'_j$
:$i = j \implies a'_j = b_j$
which intuitively can be visualised as two cubes that together form one large bar, namely $\horectr {\mathbf a} {\mathbf b'}$.
In this situation, we have:
{{begin-eqn}}
{{eqn | l = \map {\lambda^n} {\horectr {\mathbf a} {\mathbf b} } + \map {\lambda^n} {\horectr {\mathbf a'} {\mathbf b'} }
      | r = \prod_{j \mathop = 1}^n \paren {b_j - a_j} + \prod_{j \mathop = 1}^n \paren {b'_j - a'_j}
      | c = {{Defof|Lebesgue Pre-Measure}}
}}
{{eqn | r = \paren {b_i - a_i + b'_i - a'_i} \prod_{\substack{j \mathop = 1 \\ j \mathop \ne i} } \paren {b_j - a_j}
      | c = By the noted properties of $a_j, b_j, a'_j, b'_j$
}}
{{eqn | r = \map {\lambda^n} {\horectr {\mathbf a} {\mathbf b'} }
      | c = {{Defof|Lebesgue Pre-Measure}}
}}
{{end-eqn}}
Thus it is verified that $\lambda^n$ is finitely additive.
Finally, suppose that $\horectr {\mathbf a_m} {\mathbf b_m} \downarrow \O$ is a decreasing sequence of sets, with limit $\O$.
Then there exists at least one $j$ with $1 \le j \le n$ such that:
:$\ds \lim_{m \mathop \to \infty} a_{m, j} = \lim_{m \mathop \to \infty} b_{m, j}$
which by Combination Theorem for Sequences is equivalent to:
:$\ds \lim_{m \mathop \to \infty} b_{m, j} - a_{m, j} = 0$
The fact that the sequence is decreasing means that, from Cartesian Product of Subsets, for all $m \in \N$, for all $1 \le i \le n$:
:$\hointr {a_{m, i} } {b_{m, i} } \subseteq \hointr {a_{1, i} } {b_{1, i} }$
and whence $b_{m, i} - a_{m, i} \le b_{1, i} - a_{m, 1}$.
Hence we have:
{{begin-eqn}}
{{eqn | l = \lim_{m \mathop \to \infty} \map {\lambda^n} {\horectr {\mathbf a_m} {\mathbf b_m} }
      | r = \lim_{m \mathop \to \infty} \prod_{i \mathop = 1}^n \paren {b_{m, i} - a_{m, i} }
      | c = {{Defof|Lebesgue Pre-Measure}}
}}
{{eqn | o = \le
      | r = \lim_{m \mathop \to \infty} \paren {b_{m, j} - a_{m, j} } \prod_{\substack{i \mathop = 1 \\ i \mathop \ne j} } \paren {b_{1, i} - a_{1, i} }
      | c = By the above discussion
}}
{{eqn | r = 0
      | c = Combination Theorem for Sequences
}}
{{end-eqn}}
This verifies the last condition for Characterization of Pre-Measures, since $\lambda^n$ only takes finite values.
Hence $\lambda^n$ is a pre-measure.
{{qed}}
Theorem: Existence and Uniqueness of Lebesgue Measure

Let $\lambda^n$ be the Lebesgue pre-measure on the half-open $n$-rectangles $\JJ_{ho}^n$.
Then Lebesgue measure, the extension of $\lambda^n$ to the Borel $\sigma$-algebra $\BB \left({\R^n}\right)$, exists and is unique.

Proof:
From Lebesgue Pre-Measure is Pre-Measure, $\lambda^n$ is a pre-measure on $\JJ_{ho}^n$.
By Half-Open Rectangles form Semiring of Sets, $\JJ_{ho}^n$ is a semiring of sets.
Also, from Characterization of Euclidean Borel Sigma-Algebra, $\map \BB {\R^n} = \map \sigma {\JJ_{ho}^n}$.
Now observe that the half-open $n$-rectangles $\horectr {-n} n$ form an increasing sequence of sets with limit $\R^n$.
Also, by definition of $\lambda^n$, have:
:$\map {\lambda^n} {\horectr {-n} n} = \ds \prod_{i \mathop = 1}^n n - \paren {-n} = \paren {2 n}^n < +\infty$
Hence, Carathéodory's Theorem and its corollary apply.
These yield existence and uniqueness of Lebesgue measure, the extension of $\lambda^n$ to $\map \BB {\R^n}$.
{{qed}}
Theorem: Lebesgue Measure is Diffuse

Let $\lambda^n$ be Lebesgue measure on $\R^n$.
Then $\lambda^n$ is a diffuse measure.

Proof:
A singleton $\set {\mathbf x} \subseteq \R^n$ is seen to be closed by combining:
:Euclidean Space is Complete Metric Space
:Metric Space is Hausdorff
:{{Corollary|Compact Subspace of Hausdorff Space is Closed}}
Hence by Closed Set Measurable in Borel Sigma-Algebra:
:$\set {\mathbf x} \in \map \BB {\R^n}$
where $\map \BB {\R^n}$ is the Borel $\sigma$-algebra on $\R^n$.
Write $\mathbf x + \epsilon = \tuple {x_1 + \epsilon, \ldots, x_n + \epsilon}$ for $\epsilon > 0$.
Then:
:$\ds \set {\mathbf x} = \bigcap_{m \mathop \in \N} \horectr {\mathbf x} {\mathbf x + \frac 1 m}$
where $\horectr {\mathbf x} {\mathbf x + \dfrac 1 m}$ is a half-open $n$-rectangle.
{{handwaving|justify equality}}
By definition of Lebesgue measure, we have (for all $m \in \N$):
:$\ds \map {\lambda^n} {\horectr {\mathbf x} {\mathbf x + \frac 1 m} } = \prod_{i \mathop = 1}^n \frac 1 m = m^{-n}$
From Characterization of Measures, it follows that:
:$\ds \map {\lambda^n} {\set {\mathbf x} } = \lim_{m \mathop \to \infty} m^{-n}$
which equals $0$ from Sequence of Powers of Reciprocals is Null Sequence.
Therefore, for each $\mathbf x \in \R^n$:
:$\map {\lambda^n} {\set {\mathbf x} } = 0$
that is, $\lambda^n$ is a diffuse measure.
{{qed}}
Theorem: Diffuse Measure of Countable Set

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Suppose that for all $x \in X$, the singleton $\set x$ is in $\Sigma$.
Suppose further that $\mu$ is a diffuse measure.
Let $E \in \Sigma$ be a countable measurable set.
Then $\map \mu E = 0$.

Proof:
It holds trivially that:
:$\ds E = \bigcup_{e \mathop \in E} \set e$
and in particular, this union is countable.
Also, $\map \mu {\set e} = 0$ for all $e \in E$ as $\mu$ is diffuse.
Hence Null Sets Closed under Countable Union applies to yield:
:$\map \mu E = 0$
{{qed}}
Theorem: Half-Open Rectangles Closed under Intersection

Let $\horectr {\mathbf a} {\mathbf b}$ and $\horectr {\mathbf c} {\mathbf d}$ be half-open $n$-rectangles.
Then $\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d}$ is also a half-open $n$-rectangle.

Proof:
As $\O$ is trivially a half-open $n$-rectangle, let us assume that:
:$\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} \ne \O$
By Cartesian Product of Intersections: General Case, it follows that:
:$\ds \horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d} = \prod_{i \mathop = 1}^n \hointr {a_i} {b_i} \cap \hointr {c_i} {d_i}$
which leaves only to prove that the $\hointr {a_i} {b_i} \cap \hointr {c_i} {d_i}$ are half-open intervals.
Now let $x \in \hointr {a_i} {b_i} \cap \hointr {c_i} {d_i}$.
Then $x$ is subject to:
* $x \ge a_i$ and $x \ge c_i$, i.e., $x \ge \max \set {a_i, c_i}$
* $x < b_i$ and $x < d_i$, i.e., $x < \min \set {b_i, d_i}$
and we see that these conditions are satisfied {{iff}}:
:$x \in \hointr {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
Thus, we conclude:
:$\hointr {a_i} {b_i} \cap \hointr {c_i} {d_i} = \hointr {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
showing that indeed the intersection is a half-open interval.
Combining this with the above reasoning, it follows that indeed:
:$\horectr {\mathbf a} {\mathbf b} \cap \horectr {\mathbf c} {\mathbf d}$
is again a half-open $n$-rectangle.
{{qed}}
Theorem: Homomorphism to Group Preserves Inverses

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, *}$ be a group.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.

Proof:
By hypothesis, $\struct {T, *}$ is a group.
By {{Group-axiom|2}}, $\struct {T, *}$ has an identity.
Thus Homomorphism with Identity Preserves Inverses can be applied.
{{Qed}}
Theorem: Equivalence Induced by Epimorphism is Congruence Relation

Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Then the induced equivalence $\RR_\phi$ is a congruence relation for $\oplus$.

Proof:
Let $x, x', y, y' \in S$ such that:
:$x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$
By definition of induced equivalence:
{{begin-eqn}}
{{eqn | l = x \mathrel {\RR_\phi} x'
      | o = \leadsto
      | r = \map \phi x = \map \phi {x'}
      | c = 
}}
{{eqn | l = y \mathrel {\RR_\phi} y'
      | o = \leadsto
      | r = \map \phi y = \map \phi {y'}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {x \oplus y}
      | r = \map \phi x * \map \phi y
      | c = {{Defof|Epimorphism (Abstract Algebra)}}
}}
{{eqn | r = \map \phi {x'} * \map \phi {y'}
      | c = equality shown above
}}
{{eqn | r = \map \phi {x' \oplus y'}
      | c = {{Defof|Epimorphism (Abstract Algebra)}}
}}
{{end-eqn}}
Thus $\paren {x \oplus y} \mathrel {\RR_\phi} \paren {x' \oplus y'}$ by definition of induced equivalence.
So $\RR_\phi$ is a congruence relation for $\oplus$.
Theorem: Unique Isomorphism from Quotient Mapping to Epimorphism Domain

Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.
Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induced by $\RR_\phi$.
Let $\struct {S / \RR_\phi, \oplus_{\RR_\phi} }$ be the quotient structure defined by $\RR_\phi$.
Then there is one and only one isomorphism:
:$\psi: \struct {S / \RR_\phi, \oplus_{\RR_\phi} } \to \struct {T, *}$
which satisfies:
:$\psi \circ q_{\RR_\phi} = \phi$
where $\circ$ denotes composition of mappings.

Proof:
From the Quotient Theorem for Surjections, there is a unique bijection from $S / \RR_\phi$ onto $T$ satisfying $\psi \circ q_{\RR_\phi} = \phi$.
Also:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
      | l = \map \psi {\eqclass x {\RR_\phi} \oplus_{\RR_\phi} \eqclass y {\RR_\phi} }
      | r = \map \psi {\eqclass {x \oplus y} {\RR_\phi} }
      | c = {{Defof|Quotient Structure}}
}}
{{eqn | r = \map \phi {x \oplus y}
      | c = {{Defof|Epimorphism (Abstract Algebra)}}
}}
{{eqn | r = \map \phi x * \map \phi y
      | c = {{Defof|Epimorphism (Abstract Algebra)}}
}}
{{eqn | r = \map \psi {\eqclass x {\RR_\phi} } * \map \psi {\eqclass y {\RR_\phi} }
      | c = {{Defof|Quotient Mapping}}
}}
{{end-eqn}}
Therefore $\psi$ is an isomorphism.
{{Qed}}
Theorem: Identity is in Kernel of Group Homomorphism

Let $G$ and $H$ be groups.
Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.
Let $\phi: G \to H$ be a (group) homomorphism from $G$ to $H$.
Then:
:$e_G \in \map \ker \phi$
where $\map \ker \phi$ is the kernel of $\phi$.

Proof:
From the definition of kernel:
:$\map \ker \phi = \set {x \in G: \map \phi x = e_H}$
From Group Homomorphism Preserves Identity we have that:
:$\map \phi {e_G} = e_H$
Hence the result.
{{Qed}}
Theorem: Union is Dominated by Disjoint Union

Let $I$ be an indexing set.
For all $i \in I$, let $S_i$ be a set.
Then:
:$\ds \bigcup_{i \mathop \in I} S_i \preccurlyeq \bigsqcup_{i \mathop \in I} S_i$
where $\preccurlyeq$ denotes domination, $\bigcup$ denotes union, and $\bigsqcup$ denotes disjoint union.

Proof:
For all $\ds x \in \bigcup_{i \mathop \in I} S_i$, there exists a $\map i x \in I$ such that $x \in S_{\map i x}$.
Thus the mapping $\ds \iota : \bigcup_{i \mathop \in I} S_i \to \bigsqcup_{i \mathop \in I} S_i$ defined by:
:$\map \iota x = \tuple {x, \map i x}$
is an injection.
{{handwaving}}
{{qed}}
Category:Set Theory
l6tatqqscsjgkyfmx7fq1vcp1x1al31
Theorem: Disjoint Union Preserves Domination

Let $I$ be an indexing set.
For all $i \in I$, let $A_i$ and $B_i$ be sets such that $A_i \preccurlyeq B_i$.
Here, $\preccurlyeq$ denotes domination.
Then:
:$\ds \bigsqcup_{i \mathop \in I} A_i \preccurlyeq \bigsqcup_{i \mathop \in I} B_i$
where $\bigsqcup$ denotes disjoint union.

Proof:
By definition of domination, for all $i \in I$, there exists an injection $\iota_i: A_i \to B_i$.
Thus the mapping $\ds \iota : \bigsqcup_{i \mathop \in I} A_i \to \bigsqcup_{i \mathop \in I} B_i$ defined by:
:$\map \iota {x, i} = \tuple {\map {\iota_i} x, i}$
is an injection.
{{handwaving|That statement needs proof. Probably best in a separate page.}}
{{qed}}
Category:Set Theory
aguwpsc9rqxomkvaj3a253s9hsc2nfu
Theorem: Epimorphism from Real Numbers to Circle Group

Let $\struct {K, \times}$ be the circle group, that is:
:$K = \set {z \in \C: \cmod z = 1}$
under complex multiplication.
Let $f: \R \to K$ be the mapping from the real numbers to $K$ defined as:
:$\forall x \in \R: \map f x = \cos x + i \sin x$
Then $f: \struct {\R, +} \to \struct {K, \times}$ is a group epimorphism.
Its kernel is:
:$\map \ker f = \set {2 \pi n: n \in \Z}$

Proof:
$f$ is a surjection from ...
{{link wanted|Needs a link to a result specifying that $f$ is surjective (may already exist).}}
{{qed|lemma}}
Then:
{{begin-eqn}}
{{eqn | l = \map f x \times \map f y
      | r = \paren {\cos x + i \sin x} \paren {\cos y + i \sin y}
      | c =
}}
{{eqn | r = \cos x \cos y + i \sin x \cos y + \cos x i \sin y + i \sin x i \sin y
      | c =
}}
{{eqn | r = \paren {\cos x \cos y - \sin x \sin y} + i \paren {\sin x \cos y + \cos x \sin y}
      | c = as $i^2 = -1$
}}
{{eqn | r = \map \cos {x + y} + i \map \sin {x + y}
      | c = Cosine of Sum and Sine of Sum
}}
{{eqn | r = \map f {x + y}
      | c =
}}
{{end-eqn}}
So $f$ is a (group) homomorphism.
{{qed|lemma}}
Thus $f$ is seen to be a surjective homomorphism.
Hence, by definition, it is a (group) epimorphism.
{{qed|lemma}}
From Cosine of Multiple of Pi:
:$\forall n \in \Z: \cos n \pi = \paren {-1}^n$
and from Sine of Multiple of Pi:
:$\forall n \in \Z: \sin n \pi = 0$
From Sine and Cosine are Periodic on Reals, it follows that these are the only values of $\Z$ for which this holds.
For $\cos x + i \sin x = 1 + 0 i$ it is necessary that:
:$\cos x = 1$
:$\sin x = 0$
and it can be seen that the only values of $x$ for this to happen is:
:$x \in \set {2 \pi n: n \in \Z}$
Hence, by definition of kernel:
:$\map \ker f = \set {2 \pi n: n \in \Z}$
{{qed}}
Theorem: Increasing Sequence of Sets induces Partition on Limit

Let $\sequence {S_n}_{n \mathop \in \N} \uparrow S$ be an increasing sequence of sets with limit $S$.
Define $T_1 = S_1$, and, for $n \in \N$, $T_{n + 1} = S_{n + 1} \setminus S_n$, where $\setminus$ denotes set difference.
Then $\sequence {T_n}_{n \mathop \in \N}$ is a countable partition of $S$.

Proof:
That $\sequence {T_n}_{n \mathop \in \N}$ partitions $S$, means precisely that:
:$(1): \quad$ The $T_n$ are pairwise disjoint
:$(2): \quad \ds \bigcup_{n \mathop \in \N} T_n = S$
It is more convenient to prove $(1)$ and $(2)$ separately: