nyuuzyou/wikis huggingface dataset:
Exhausting Sequence of Sets on the Strictly Positive Real Numbers

== Theorem ==
 
For each $k \in \N$, let $S_k = \openint {\dfrac 1 k} k$.
Then $\sequence {S_k}_k$ is an exhausting sequence of sets on $\R_{>0}$. 
== Proof ==
To prove that $\sequence {S_k}_k$ is exhausting $\R_{>0}$, it is sufficient to show:
:$(1): \quad \forall k \in \N: S_k \subseteq S_{k + 1}$
:$(2): \quad \ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$
=== $\sequence {S_k}_k$ is increasing ===
Let $k \in \N$.
Let $k = 1$.
Then:
:$S_k = \openint {\dfrac 1 k} k = \O$
Thus, by Empty Set is Subset of All Sets:
:$\openint {\dfrac 1 k} k \subseteq \openint {\dfrac 1 {k + 1} } {k + 1}$
If $k \ge 2$:
It follows that $\sequence {S_k}_k$ is increasing.
=== $\ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$ ===
Let $x \in \R_{>0}$.
==== Case 1: $1 < x$ ====
Let $1 < x$.
Let $k = \ceiling x$.
From Real Number is between Ceiling Functions:
:$x < k$
From Ordering of Reciprocals:
:$1 < k \implies \dfrac 1 k < 1$
So:
:$\dfrac 1 k < x < k$
and so $x \in S_k$.
==== Case 2: $x = 1$ ====
Let $x = 1$.
Let $k = 2$.
Then:
:$\dfrac 1 2 < 1 < 2$
and hence $x \in S_2$.
==== Case 3: $0 < x < 1$ ====
Let $0 < x < 1$.
Let $k = \ceiling {\dfrac 1 x}$.
From Ordering of Reciprocals:
:$0 < x < 1 \implies 1 < \dfrac 1 x$
From Real Number is between Ceiling Functions:
:$1 < \dfrac 1 x < k$
From Ordering of Reciprocals:
:$\dfrac 1 k < x < 1$
So:
:$\dfrac 1 k < x < k$
and so $x \in S_k$.
Hence:
:$\forall x \in \R_{>0} : \exists k \in \N : x \in S_k$
The result follows from the definition of exhausting sequence of sets.
Category:Real Numbers
Exist Term in Arithmetic Sequence Divisible by Number

== Theorem ==
Let $\sequence {a_k}$ be an $n$-term arithmetic sequence in $\Z$ defined by:
:$a_k = a_0 + k d$ for $k = 0, 1, 2, \dots, n - 1$
Let $b$ be a (strictly) positive integer such that $b$ and $d$ are coprime and $b \le n$.
Then there exists a term in $\sequence {a_k}$ that is divisible by $b$.
== Proof ==
We claim that at least one of the first $b$ terms:
:$a_0, a_0 + d, a_0 + 2 d, \dots, a_0 + \paren {b - 1} d$
is divisible by $b$.
Consider the remainders of each term after division by $b$.
They can takes on values of $0 \le r < b$.
If one of them gives $r = 0$ then we are done.
 not.
Since there are $b$ terms but only $b - 1$ possible remainders, by Pigeonhole Principle at least two terms must share a remainder.
That is:
:$a_0 + i d \equiv a_0 + j d \pmod b$
for some $i, j$ where and $0 \le i < j \le b - 1$.
But then:
:$\paren {j - i} d \equiv 0 \pmod b$
so $b \divides \paren {j - i} d$.
Since $b \perp d$, by Euclid's Lemma we have $b \divides \paren {j - i}$.
Since $0 < j - i < b$ we must have $b \nmid \paren {j - i}$ by Absolute Value of Integer is not less than Divisors.
This is a contradiction.
Therefore there is at least one term that is divisible by $b$.
Category:Arithmetic Sequences
Category:Divisibility
Existence-Uniqueness Theorem for First-Order Differential Equation

== Theorem ==
Let $P$ and $Q$ be continuous real functions on some open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
There exists a unique function $\map f x = y$ on $I$ that satisfies the linear first order ordinary differential equation: 
:$(1): \quad y' + \map P x y = \map Q x$
along with the initial condition:
:$\map f a = b$
This function is:
:$\ds \map f x = be^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t$
where $\ds \map A x = \int_a^x \map P t \rd t$
== Proof ==
=== Existence ===
Because $P$ and $Q$ are continuous, they are integrable.
Hence we may use the Fundamental Theorem of Calculus.
Therefore:
:$\map {f'} x + \map P x \map f x = \map Q x$
For the initial condition:
=== Uniqueness ===
Let $f$ be a solution to $(1)$ which satisfies the initial condition.
Let $\map g x = \map f x e^{\map A x}$.
Moreover:
:$\map g a = b$
By the Fundamental Theorem of Calculus:
:$\ds \map g x = \int_a^x \map Q t e^{\map A t} \rd t + b$
Furthermore:
:$\map f x = \map g x e^{-\map A x}$
Therefore, we can conclude that:
:$\ds \map f x = b e^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t$
== Sources ==
* : $\S 8.3$
Category:Calculus Category:Differential Equations
Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation

== Theorem ==
Let $\map P x$ be a continuous function on an open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
Let $\map f x = y$ be a function satisfying the differential equation:
:$y' + \map P x y = 0$
and the initial condition:
:$\map f a = b$
Then there exists a unique function satisfying these initial conditions on the interval $I$.
That function takes the form:
:$\map f x = b e^{-\map A x}$
where:
:$\ds \map A x = -\int_a^x \map P t \rd t$
== Proof ==
=== Existence ===
Differentiating $\map f x = b e^{-\map A x}$ $x$:
So thedifferential equation becomes:
:$\map {f'} x + \map P x \map f x = -\map P x \map f x + \map P x \map f x = 0$
For the initial condition:
Thus such a function exists satisfying the conditions.
=== Uniqueness ===
Suppose that $f$ is a function satisfying the initial conditions.
Let $\map g x = \map f x e^{\map A x}$.
By Product Rule for Derivatives:
So $\map g x$ must be constant.
Therefore:
:$\map g x = \map g a = \map f a e^{\map A a} = \map f a = b$
From this, we conclude that:
:$\map f x = \map g x e^{-\map A x} = b e^{-\map A x}$
== Sources ==
* : $\S 8.2$
Category:Calculus
Category:Differential Equations
Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra

== Theorem ==
 
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra of $\Sigma$.
Then there exists an integrable random variable $Z$ on $\struct {\Omega, \GG, \Pr}$ such that: 
:$\ds \int_G Z \rd \Pr = \int_G X \rd \Pr$ for each $G \in \GG$.
Further, if $Z$ and $Z'$ are two integrable random variables satisfying this condition, we have: 
:$Z = Z'$ almost everywhere.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $9.2$
Category:Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra
Category:Conditional Expectation
Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra/Proof 1

== Theorem ==
== Proof ==
 
First take $X \ge 0$. 
Define a function $\mu : \GG \to \R$ by: 
:$\ds \map \mu A = \int_A X \rd \Pr$ 
for each $A \in \GG$.
From Integral of Integrable Function over Measurable Set is Well-Defined, this is well-defined. 
From Measure with Density is Measure, $\mu$ is a measure.
Note that if $\map \Pr A = 0$ for $A \in \GG$, we have $\map \mu A = 0$ from Integral of Integrable Function over Null Set.
So $\mu$ is absolutely continuous with respect to $\Pr \restriction_\GG$.
So by the Radon-Nikodym Theorem, there exists a $\Pr \restriction_\GG$-integrable function $Z \ge 0$ that is $\GG$-measurable such that: 
:$\ds \map \mu A = \int_A Z \rd \Pr \restriction_\GG$
for each $A \in \GG$.
Then from Integral of Positive Measurable Function with respect to Restricted Measure, we have: 
:$\ds \map \mu A = \int_A Z \rd \Pr$ 
for each $A \in \GG$.
That is: 
:$\ds \int_A X \rd \Pr = \int_A Z \rd \Pr$
for each $A \in \GG$.
Finally, we have that: 
:$\ds \int Z \rd \Pr = \int X \rd \Pr < \infty$
so $Z$ is integrable.
So to conclude, $Z$ is a $\GG$-measurable integrable random variable with: 
:$\ds \int_A X \rd \Pr = \int_A Z \rd \Pr$
for all $A \in \GG$. 
Now take $X$ a general integrable random variable. 
Then from Positive Part of Real-Valued Random Variable is Real-Valued Random Variable, Negative Part of Real-Valued Random Variable is Real-Valued Random Variable and the definition of an integrable random variable:
:$X^+$ and $X^-$ are integrable random variables.
Since $X^+ \ge 0$ and $X^- \ge 0$, by our previous work there exists integrable random variables $Z_1$, $Z_2$ with: 
:$\ds \int_A X^+ \rd \Pr = \int_A Z_1 \rd \Pr$
and:
:$\ds \int_A X^- \rd \Pr = \int_A Z_2 \rd \Pr$
for all $A \in \GG$, with $Z_1$ and $Z_2$ $\GG$-measurable. 
Then from Integral of Integrable Function is Additive: Corollary 2 and Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable, we have: 
:$Z_1 - Z_2$ is an integrable random variable that is $\GG$-measurable.
Then, we have: 
So setting $Z = Z_1 - Z_2$, we have: 
:$\ds \int_A X \rd \Pr = \int_A Z \rd \Pr$
for all $A \in \GG$.
Now let $Z'$ be another integrable random variable that is $\GG$-measurable with: 
:$\ds \int_A X \rd \Pr = \int_A Z' \rd \Pr$
for all $A \in \GG$.
Then: 
:$\ds \int_A Z \rd \Pr = \int_A Z' \rd \Pr$
for all $A \in \GG$.
Then, from Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal, we have that: 
:$Z = Z'$ almost everywhere
completing the proof.
== Sources ==
* : $9.2$
Category:Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra
Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra/Proof 2

== Theorem ==
== Proof ==
Observe that:
:$\map {L^2} {\Omega, \GG, \Pr} \subseteq \map {L^2} {\Omega, \Sigma, \Pr}$
is a closed linear space.
Let:
:$P : \map {L^2} {\Omega, \Sigma, \Pr} \to \map {L^2} {\Omega, \GG, \Pr}$
be the orthogonal projection.
Observe that for all $f \in \map {L^2} {\Omega, \Sigma, \Pr}$ and $g \in \map {L^2} {\Omega, \GG, \Pr}$:
Let $f \in \map {L^2} {\Omega, \Sigma, \Pr}$.
Let:
:$\ds g := \chi_{\set {P f \ge 0} } - \chi_{\set {P f < 0} }$
so that:
:$\size {\map P f} = \map P f g$
Since $g \in \map {L^2} {\Omega, \GG, \Pr}$, we have:
On the other hand, by Cauchy inequality and the density of simple functions:
:$\map {L^2} {\Omega, \Sigma, \Pr} \subseteq \map {L^1} {\Omega, \Sigma, \Pr}$
is a dense subspace.
Therefore, we can extend:
:$P : \map {L^1} {\Omega, \Sigma, \Pr} \to \map {L^1} {\Omega, \GG, \Pr}$
so that:
:$\ds \forall f \in \map {L^1} {\Omega, \Sigma, \Pr} : \norm {P f}_{\map {L^1} {\Omega, \Sigma, \Pr} } \le \norm f_{\map {L^1} {\Omega, \GG, \Pr} }$
Let $X \in \map {L^1} {\Omega, \Sigma, \Pr}$.
Let $\sequence {X_n} \subseteq \map {L^2} {\Omega, \Sigma, \Pr}$ such that:
:$\ds \lim_{n \mathop \to \infty} \norm {X_n - X}_{\map {L^1} {\Omega, \Sigma, \Pr} }$
Then for each $G \in \GG$:
Hence we can choose $Z = \map P X$.
Now let $Z'$ be another integrable random variable that is $\GG$-measurable with: 
:$\ds \int_A X \rd \Pr = \int_A Z' \rd \Pr$
for all $A \in \GG$.
Then:
By Measurable Function Zero A.E. iff Absolute Value has Zero Integral:
:$Z = Z'$ almost everywhere
Category:Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra
Existence and Uniqueness Theorem for 1st Order IVPs

== Theorem ==
Let $x' = \map f {t, x}$, $\map x {t_0} = x_0$ be an explicit ODE of dimension $n$.
Let there exist an open ball $V = \sqbrk {t_0 - \ell_0, t_0 + \ell_0} \times \map {\overline B} {x_0, \epsilon}$ of $\tuple {t_0, x_0}$ in phase space $\R \times \R^n$ such that $f$ is Lipschitz continuous on $V$.
Then there exists $\ell < \ell_0$ such that there exists a unique solution $\map x t$ defined for $t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0}$.
== Proof ==
For $0 < \ell < \ell_0$, let $\XX = \map \CC {\closedint {t_0 - \ell_0} {t_0 + \ell_0}; \R^n}$ endowed with the sup norm be the Banach Space of Continuous Functions on Compact Space $\closedint {t_0 - \ell_0} {t_0 + \ell_0} \to \R^n$.
By Fixed Point Formulation of Explicit ODE it is sufficient to find a fixed point of the map $T: \XX \to \XX$ defined by:
:$\ds \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$
We also have Closed Subset of Complete Metric Space is Complete.
Therefore the Banach Fixed-Point Theorem it is sufficient to find a non-empty subset $\YY \subseteq \XX$ such that:
:$\YY$ is closed in $\XX$
:$T \YY \subseteq \YY$
:$T$ is a contraction on $\YY$
First note that $V$ is closed and bounded, hence compact by the Heine-Borel Theorem.
Therefore since $f$ is continuous, by the extreme value theorem, the maximum $\ds m = \sup_{\tuple {t, x} \mathop \in V} \size {\map f {t, x} }$ exists and is finite.
Let $\kappa$ be the Lipschitz constant of $f$.
Let:
:$\YY = \set {y \in \XX: \norm {\map y t - x_0} \le m \size {t - t_0}, t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0} }$
be the cone in $\XX$ centred at $\tuple {t_0, x_0}$.
Clearly $\YY$ is closed in $\XX$.
Also for $y \in \YY$ we have:
Therefore $T \YY \subseteq \YY$.
Finally we must show that $T$ is a contraction on $\YY$ (we will find that this restricts our choice of $\ell$).
Let $y_1, y_2 \in \YY$.
We have:
Taking the supremum over $t$ we have:
:$\norm {T y_1 - T y_2}_\sup \le \kappa \ell \norm {y_1 - y_2}_\sup$
for all $y_1, y_2 \in \YY$.
Therefore choosing $\ell < \kappa^{-1}$, $T$ is a contraction on $\YY$ as required.
This completes the proof.
Category:Differential Equations
Existence and Uniqueness of Adjoint

== Theorem ==
Let $\mathbb F \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\mathbb F$.
Let $A: \HH \to \KK$ be a bounded linear transformation. 
Then there exists a unique bounded linear transformation $B: \KK \to \HH$ such that: 
:$\innerprod {\map A x} y_\KK = \innerprod x {\map B y}_\HH$
for all $x \in \HH$ and $y \in \KK$. 
That is:
:each bounded linear transformation between Hilbert spaces has a unique adjoint.
== Proof ==
We first show that such a unique mapping $B$ exists, without first insisting on a bounded linear transformation.
=== Lemma 1 ===
We now show that $B$ is a linear transformation.
=== Lemma 2 ===
Finally, we show that $B$ is bounded.
=== Lemma 3 ===
So $B$ is the unique bounded linear transformation such that:
:$\innerprod {\map A x} y_\KK = \innerprod x {\map B y}_\HH$
for all $x \in \HH$ and $y \in \KK$. 
== Sources ==
* $13.1$: Existence of Hilbert Adjoint
Category:Adjoints
Category:Existence and Uniqueness of Adjoint
Existence and Uniqueness of Adjoint/Lemma 1

== Lemma ==
Let $\mathbb F \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\HH$.
Let $\KK$ be a Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\KK$.
Let $A : \HH \to \KK$ be a bounded linear transformation. 
Then:
:There exists a unique mapping $B : \KK \to \HH$ such that: 
:$\innerprod {A x} y_\KK = \innerprod x {B y}_\HH$
for all $x \in \HH$ and $y \in \KK$. 
</onlyinclude>
== Proof ==
Let $\norm \cdot_\HH$ be the inner product norm of $\HH$.
Let $\norm \cdot_\KK$ be the inner product norm of $\KK$.
For each $y \in \KK$, define the linear functional $f_y : \HH \to \mathbb F$ by: 
:$\map {f_y} x = \innerprod {\map A x} y_\KK$
Let $\norm A$ denote the norm on $A$.
We have that $A$ is a bounded linear transformation.
From Norm on Bounded Linear Transformation is Finite:
:$\norm A$ is finite.
We therefore have: 
Taking $M = \norm A \norm y_\KK$, we have: 
:$\size {\map {f_y} x} \le M \norm x_\HH$
for each $x \in \HH$, with $M$ independent of $x$. 
So, $f_y$ is bounded.
So, by the Riesz Representation Theorem (Hilbert Spaces), there exists a unique $\map z y \in \HH$ such that: 
:$\map {f_y} x = \innerprod x {\map z y}_\HH$ 
for each $x \in \HH$. 
That is, for each $y \in \KK$ there exists precisely one $\map z y \in \HH$ such that: 
:$\innerprod {A x} y_\KK = \innerprod x {\map z y}_\HH$
for all $x \in \HH$.
Define the mapping $B : \KK \to \HH$ by: 
:$B y = \map z y$ 
for each $y \in \KK$.
This map has: 
:$\innerprod {A x} y_\KK = \innerprod x {B y}_\HH$ 
for each $x \in \HH$ and $y \in \KK$.
Since the choice of $\map z y$ was unique, the map $B$ must also be unique, so $B$ is the unique map with the required properties.
Category:Existence and Uniqueness of Adjoint
Existence and Uniqueness of Adjoint/Lemma 2

== Lemma ==
Let $\mathbb F \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\HH$.
Let $\KK$ be a Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\KK$.
Let $A : \HH \to \KK$ be a bounded linear transformation.
Let $B : \KK \to \HH$ be the unique mapping satisfying: 
:$\innerprod {A x} y_\KK = \innerprod x {B y}_\HH$
for each $x \in \HH$ and $y \in \KK$. 
Then Wwe have: 
:$\map B {\alpha x + \beta y} = \alpha B x + \beta B y$
for all $\alpha, \beta \in \mathbb F$ and $x, y \in \KK$.
That is, $B$ is a linear transformation.
== Proof ==
Let $\alpha, \beta \in \mathbb F$. 
Let $x, y \in \KK$. 
We have: 
Recall that there exists exactly one $z \in \HH$ such that: 
:$\innerprod {A x} {\alpha x + \beta y}_\HH = \innerprod x z_\KK$
So we must have: 
:$\map B {\alpha x + \beta y} = \alpha B x + \beta B y$
So: 
:$\map B {\alpha x + \beta y} = \alpha B x + \beta B y$
for each $\alpha, \beta \in \mathbb F$ and $x, y \in \KK$.
So:
:$B$ is a linear transformation.
Category:Existence and Uniqueness of Adjoint
Existence and Uniqueness of Adjoint/Lemma 3

== Lemma ==
Let $\mathbb F \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\HH$.
Let $\KK$ be a Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\KK$.
Let $A : \HH \to \KK$ be a bounded linear transformation.
Let $B : \KK \to \HH$ be the unique mapping satisfying: 
:$\innerprod {A x} y_\KK = \innerprod x {B y}_\HH$
for each $x \in \HH$ and $y \in \KK$. 
Then $B$ is a bounded linear transformation.
== Proof ==
Let $\norm \cdot_\HH$ be the inner product norm of $\HH$.
Let $\norm \cdot_\KK$ be the inner product norm of $\KK$.
From Existence and Uniqueness of Adjoint: Lemma 2, we have that: 
:$B$ is a linear transformation.
It remains to show that $B$ is bounded.
Let $\norm A$ denote the norm on $A$.
We have that $A$ is a bounded linear transformation.
From Norm on Bounded Linear Transformation is Finite:
:$\norm A$ is finite.
Then: 
Note that for all $y \in \KK$ such that: 
:$B y \ne 0$ 
we have: 
:$\norm {B y}_\HH \le \norm A \norm y_\KK$
From the definition, a norm is positive definite.
Hence this inequality holds if $B y = 0$.
So:
:$\forall y \in \KK: \norm {B y}_\HH \le \norm A \norm y_\KK$
That is: 
:$B$ is a bounded linear transformation.
Category:Existence and Uniqueness of Adjoint
Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 1

== Lemma ==
On $\widehat G_\infty := \ds \coprod_{n \mathop \in \N} G_n$ the relation:
:$\tuple {x_n, n} \sim \tuple {y_m, m} \iff \exists k \ge n, m: \map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$
is an equivalence relation.
== Proof ==
=== Reflexivity ===
Since $g_{n, n} = \mathop {Id}_{G_n}$ we have:
:$\forall \tuple {x_n, n} \in \widehat G_\infty: \map {g_{n, n} } {x_n} = \map {g_{n, n} } {x_n}$
Hence:
:$\tuple {x_n, n} \sim \tuple {x_n, n}$
=== Symmetry ===
Let $\tuple {x_n, n} \sim \tuple {y_m, m}$.
Then there exists a $k \ge n, m$ such that:
:$\map {g_{n, k} } {x_n} = \map {g_{m, k} } {x_m}$
Hence also:
:$\map {g_{m, k} } {x_m} = \map {g_{n, k} } {x_n}$
That is:
:$\tuple {y_m, m} \sim \tuple {x_n, n}$
=== Transitivity ===
Let $\tuple {x_n, n} \sim \tuple {y_m, m}$ and $\tuple {y_m, m} \sim \tuple {z_r, r}$.
Then there exist $k \ge m, n$ and $l \ge n, r$ such that:
:$\map {g_{n, k} } {x_n} = \map {g_{m, k} } {y_m}$
:$\map {g_{m, l} } {y_m} = \map {g_{r, l} } {z_r}$
Let $q:= \max \set {k, l}$.
Then we have:
that is:
:$\tuple {x_n, n} \sim \tuple {z_r, r}$
Category:Existence and Uniqueness of Direct Limit of Sequence of Groups
Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 2

== Lemma ==
The following defines a group structure on $G_\infty$: 
Let $\struct {G_\infty, \cdot}$ be the algebraic structure defined as follows.
Let $\eqclass {\tuple {x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$ be arbitrary elements of $G_\infty$.
Let $l := \max \set {m, n}$.
Let the operation $\cdot$ on $G_\infty$ be defined as:
:$\tuple {\eqclass {\tuple {x_n, n} } {} \cdot \eqclass {\tuple {y_m, m} } {} } := \eqclass {\tuple {\map {g_{n l} } {x_n} \map {g_{m l} } {y_m}, l} } {}$
Then $\struct {G_\infty, \cdot}$ is a group.
== Proof ==
=== Well-Definedness ===
The definition depends on the choice $\tuple {x_n, n}$ and $\tuple {y_m, m}$ of representatives of $\eqclass {\tuple {x_n, n} } {}$ and $\eqclass {\tuple {y_m, m} } {}$. 
We have to show that the product element is independent of this choice.
Let $\tuple {x_{n'}, n'}$ and $\tuple {y_{m'}, m'}$ be different representatives of the chosen equivalence classes. 
Let $l' := \max \set {n', m'}$.
, suppose that $l' \ge l$.
We have that:
:$\tuple {x_n, n} \sim \tuple {x_{n'}, n'}$
and:
:$\tuple {y_m, m} \sim \tuple {y_{m'}, m'}$
		
and so:
:$\map {g_{n, l'} } {x_n} = \map {g_{n', l'} } {x_{n'} }$
and:
:$\map {g_{m, l'} } {y_m} = \map {g_{m', l'} } {y_{m'} }$
		
Then we have, since all our maps are group homomorphisms:
 		
	
that is:
:$\map {g_{n, l} } {x_n} \map {g_{m,l} } {y_m} \sim \map {g_{n', l'} } {x_{n'} } \map {g_{m', l'} } {y_{m'} }$
This proves that our definition is independent of the choice of representative.
=== Group Axioms ===
, by the definition of the group operation, we may assume that the representatives are always in the same group $G_l \in \sequence {G_n}_{n \mathop \in \N}$.
To see this we note that we always consider a finite collection of group elements 
:$\set {\eqclass {\tuple {x_{n_1}, {n_1} } } {}, \ldots, \eqclass {\tuple {x_{n_k}, {n_k} } } {} } \subset G_\infty$
Define $l:= \max \set {n_1, \ldots, n_k}$.
Then:
:$\forall i \in \set {1, \ldots, k}: \map {g_{n_i, l} } {x_{n_i} } \in G_n \land \tuple {x_{n_1}, {n_1} } \sim \tuple {\map{ g_{n_i, l} } {x_{n_i} }, l}$
==== ====
Let $\eqclass {\tuple{x_n, n} } {}, \eqclass {\tuple {y_m, m} } {}, \eqclass {\tuple {y_n, n} } {}, \eqclass {\tuple {z_n, n} } {} \in G_\infty$. 
Then: 
==== ====
Let $\eqclass {\tuple {x_n, n} } {} \in G_\infty$ and let $1_n$ be the identity of $G_n$. 
Note that
:$\forall k, n \in \N : \paren {k \ge n \implies \map {g_{n k} } {1_n} = 1_k}$
because the maps $g_{n k}$ are group homomorphisms.
Then: 
Similarly we also find that $\eqclass {\tuple {1_n, n} } {} \cdot \eqclass {\tuple {x_n, n} } {} = \eqclass {\tuple {x_n, n} } {}$.
Thus $\eqclass {\tuple {1_n, n} } {}$ is the identity of $G_\infty$.
==== ====
Let $\eqclass {\tuple {x_n, n} } {} \in G_\infty$.
Then:
Similarly we also find that $\eqclass {\tuple {x^{-1}_n, n} } {} \cdot \eqclass {\tuple {x_n, n} } {} = \eqclass {\tuple {1_n, n} } {}$.
Thus $\eqclass {\tuple {x_n, n} } {}$ has an inverse, that is:
:$\eqclass {\tuple {x_n^{-1}, n} } {}$
in $G_\infty$.
Category:Existence and Uniqueness of Direct Limit of Sequence of Groups
Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 3

== Lemma ==
Let $h_\infty: G_\infty \to H$ be the mapping defined as:
:$\eqclass {\tuple {x_n, n} } {} \mapsto \map {h_n} {x_n}$
Then $h_\infty$ is a well-defined group homomorphism.
== Proof ==
=== Well-Definedness of $h_\infty$ ===
Let $\tuple {x_n, n}, \tuple{x_{n'}, n'} \in \eqclass {\tuple {x_n, n} } {}$.
, let $n' \ge n$. 
Then we have:
:$\map {g_{n, n'} } {x_n} = x_{n'}$
and:
This proves that $h_\infty$ is independent of the representative chosen.
That is, $h_\infty$ is well-defined.
=== Homomorphism Property ===
Let $\eqclass {\tuple{x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$.
, we may assume, by the definition of the group operation, that $n = m$.
See Lemma 2 for details.
It follows that:
Thus $h_\infty$ is a homomorphism.
Category:Existence and Uniqueness of Direct Limit of Sequence of Groups
Existence and Uniqueness of Domain of Relation

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Then the domain $\Dom \RR$ of $\RR$ exists and is unique.
== Proof ==
By the Axiom of Specification we can create the subclass of $V$:
:$\Dom \RR = \set {x \in V: \exists y \in V: \tuple {x, y} \in \RR}$
Hence $\Dom \RR$ exists.
Suppose $\QQ \subseteq V$ such that $\QQ$ and $\Dom \RR$ are both the domain of $\RR$.
Then:
:$\QQ = \set {x \in V: \exists y \in V: \tuple {x, y} \in \RR}$
Thus:
:$x \in \QQ \implies x \in \Dom \RR$
and:
:$x \in \Dom \RR \implies x \in \QQ$
Hence by the Axiom of Extension:
:$\QQ = \Dom \RR$
and uniqueness has been demonstrated.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 8$ Relations
Category:Relation Theory