nyuuzyou/wikis huggingface dataset:
Adjoint of Identity Transformation

== Theorem ==
Let $\tuple {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space.
Let $I_\HH$ be the identity transformation on $\HH$.
Then: 
:${I_\HH}^* = I_\HH$
where ${I_\HH}^*$ denotes the adjoint of $I_\HH$. 
== Proof ==
From Identity Mapping on Normed Vector Space is Bounded Linear Operator:
:$I_\HH$ is a bounded linear transformation.
So, from the existence part of Existence and Uniqueness of Adjoint:
:$I_\HH$ has an adjoint ${I_\HH}^*$.
That is:
:$\innerprod {I_\HH h} g_\HH = \innerprod h { {I_\HH}^* g}_\HH$
for all $h, g \in \HH$.
Note that by the definition of the identity transformation, we also have: 
:$\innerprod {I_\HH h} g_\HH = \innerprod h {I_\HH g}_\HH$
for all $h, g \in \HH$.
So, from the uniqueness part of Existence and Uniqueness of Adjoint:
:$I_\HH = {I_\HH}^*$ 
Category:Adjoints
Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\struct {\map D T, T}$ be a symmetric densely-defined linear operator.
Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $T$.
Then $\map D T \subseteq \map D {T^\ast}$ and: 
:$T x = T^\ast x$ for each $x \in \map D T$. 
== Proof ==
For each $y \in \HH$, define the linear functional $f_x : \map D T \to \Bbb F$ by:
:$\map {f_y} x = \innerprod {T x} y$ for each $x \in \map D T$.
We show that for $y \in \map D T$, $f_y$ is bounded.
Let $y \in \map D T$, then: 
:$\innerprod {T x} y = \innerprod x {T y}$ for each $x \in \map D T$.
Then we have: 
So $f_y$ is bounded and $\map D T \subseteq \map D {T^\ast}$. 
Now, for $x, y \in \map D T$, we have: 
:$\innerprod {T x} y = \innerprod x {T y}$ 
since $\struct {\map D T, T}$ is symmetric and: 
:$\innerprod {T x} y = \innerprod x {T^\ast y}$ 
from the definition of the adjoint.
So we have: 
:$\innerprod x {T y - T^\ast y} = 0$
for each $x \in \map D T$. 
Taking a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ such that $x_n \to T y - T^\ast y$. 
Since: 
:$\innerprod {x_n} {T y - T^\ast y} = 0$
for each $n \in \N$, we have: 
:$\innerprod {T y - T^\ast y} {T y - T^\ast y} = \norm {T y - T^\ast y}^2 = 0$ 
from Inner Product is Continuous.
So $T y = T^\ast y$ for each $y \in \map D T$.
Category:Symmetric Densely-Defined Linear Operators
Category:Adjoints (Densely-Defined Linear Operators)
Adjugate Matrix/Examples

== Examples of Adjugate Matrices ==
=== $2 \times 2$ Square Matrix ===
=== $3 \times 3$ Square Matrix ===
=== Arbitrary Matrix $1$ ===
=== Arbitrary Matrix $2$ ===
=== Arbitrary Matrix $3$ ===
=== Arbitrary Matrix $4$ ===
Category:Examples of Adjugate Matrices
Adjugate Matrix/Examples/2 x 2

== Example of Adjugate Matrix ==
Let $\mathbf A$ be the square matrix of order $2$:
:$\mathbf A = \begin {pmatrix} a & b \\ c & d \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} d & -b \\ -c & a \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Adjugate Matrices
Adjugate Matrix/Examples/3 x 3

== Example of Adjugate Matrix ==
Let $\mathbf A$ be the square matrix of order $3$:
:$\mathbf A = \begin {pmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end {pmatrix}$
Let $A_{i j}$ denote the cofactor of element $a_{ij}$.
Then the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} A_{1 1} & A_{2 1} & A_{3 1} \\ A_{1 2} & A_{2 2} & A_{3 2} \\ A_{1 3} & A_{2 3} & A_{3 3} \end {pmatrix}$
== Sources ==
* 
* 
Category:Examples of Adjugate Matrices
Adjugate Matrix/Examples/Arbitrary Matrix 1

== Example of Adjugate Matrix ==
Let $\mathbf A$ be the square matrix:
:$\mathbf A = \begin {pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} 1 & -1 & 1 & -1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \end {pmatrix}$
== Proof ==
For a square matrix $\mathbf A = a_{i j}$ of order $4$, the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} A_{11} & A_{21} & A_{31} & A_{41} \\ A_{12} & A_{22} & A_{32} & A_{42} \\ A_{13} & A_{23} & A_{33} & A_{43} \\ A_{14} & A_{24} & A_{34} & A_{44} \end {pmatrix}$
For each $a_{i j}$ in $\mathbf A$, we calculate the cofactors $A_{i j}$:
Hence:
:$\adj {\mathbf A} = \begin {pmatrix} 1 & -1 & 1 & -1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \end {pmatrix}$
We check this result, using Product of Matrix with Adjugate equals Determinant by Unit Matrix.
First we note that $\map \det {\mathbf A} = 1$, by Determinant of Upper Triangular Matrix.
and all is well.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.13 \ \text {(a)}$
Category:Examples of Adjugate Matrices
Adjugate Matrix/Examples/Arbitrary Matrix 2

== Example of Adjugate Matrix ==
Let $\mathbf A$ be the square matrix:
:$\mathbf A = \begin {pmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \\ \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end {pmatrix}$
== Proof ==
For a square matrix $\mathbf A = a_{i j}$ of order $3$, the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end {pmatrix}$
For each $a_{i j}$ in $\mathbf A$, we calculate the cofactors $A_{i j}$:
Hence:
:$\adj {\mathbf A} = \begin {pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end {pmatrix}$
We check this result, using Product of Matrix with Adjugate equals Determinant by Unit Matrix.
First we note that $\map \det {\mathbf A} = -1$, by Expansion Theorem for Determinants, expanding using column $3$.
and all is well.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.13 \ \text {(b)}$
Category:Examples of Adjugate Matrices
Adjugate Matrix/Examples/Arbitrary Matrix 3

== Example of Adjugate Matrix ==
Let $\mathbf A$ be the square matrix:
:$\mathbf A = \begin {pmatrix} 1 & 2 & 0 \\ 0 & -1 & 2 \\ -1 & 2 & 0 \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} -4 & 0 & 4 \\ -2 & 0 & -2 \\ -1 & -4 & -1 \end {pmatrix}$
== Proof ==
For a square matrix $\mathbf A = a_{i j}$ of order $3$, the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end {pmatrix}$
For each $a_{i j}$ in $\mathbf A$, we calculate the cofactors $A_{i j}$:
Hence:
:$\adj {\mathbf A} = \begin {pmatrix} -4 & 0 & 4 \\ -2 & 0 & -2 \\ -1 & -4 & -1 \end {pmatrix}$
We check this result, using Product of Matrix with Adjugate equals Determinant by Unit Matrix.
First we note that $\map \det {\mathbf A} = -8$, by Expansion Theorem for Determinants, for example expanding using column $1$.
and all is well.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.13 \ \text {(c)}$
Category:Examples of Adjugate Matrices
Adjugate Matrix/Examples/Arbitrary Matrix 4

== Example of Adjugate Matrix ==
Let $\mathbf A$ be the square matrix:
:$\mathbf A = \begin {pmatrix} -1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & -3 & 3 \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} 12 & 6 & -2 \\ -6 & -3 & 1 \\ 6 & 3 & -1 \end {pmatrix}$
== Proof ==
For a square matrix $\mathbf A = a_{i j}$ of order $3$, the adjugate matrix of $\mathbf A$ is:
:$\adj {\mathbf A} = \begin {pmatrix} A_{1 1} & A_{2 1} & A_{3 1} \\ A_{1 2} & A_{2 2} & A_{3 2} \\ A_{1 3} & A_{2 3} & A_{3 3} \end {pmatrix}$
For each $a_{i j}$ in $\mathbf A$, we calculate the cofactors $A_{i j}$:
Hence:
:$\adj {\mathbf A} = \begin {pmatrix} 12 & 6 & -2 \\ -6 & -3 & 1 \\ 6 & 3 & -1 \end {pmatrix}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.14 \ \text {(a)}$
Category:Examples of Adjugate Matrices
Admissible Curve in Riemannian Manifold has Unique Forward Reparametrization by Arc Length

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold with or without boundary.
Let $I \subseteq \R$ be a real interval.
Let $\gamma : I \to M$ be an admissible Curve.
Then $\gamma$ has a unique forward reparametrization by arc length.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Definitions
Category:Riemannian Geometry
Admissible Family of Curves Symmetry Lemma

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $I = \closedint a b$ is a closed real interval.
Let $J$ is an open real interval.
Let $\Gamma : J \times I \to M$ be an admissible family of curves.
Let $\tuple {a_0, a_1, a_2, \ldots, a_{n - 1}, a_n}$ be a finite subdivision of $I$.
Then:
:$\forall i \in \N_{> 0} : i \le n : \forall s \in J : \forall t \in \closedint {a_{i - 1} }{a_i} : D_s \partial_t \Gamma = D_t \partial_s \Gamma$
where $D_t$ denotes the covariant derivative along the main curve, and $D_s$ denotes the covariant derivative along the transverse curve.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics and Minimizing Curves
Category:Riemannian Manifolds
Category:Curves
Affine Coordinates are Well-Defined

== Theorem ==
Let $\EE$ be an affine space with difference space $V$ over a field $k$.
Let $\RR = \left({p_0, e_1, \ldots, e_n}\right)$ be an affine frame in $\EE$.
Define a mapping $\Theta_\RR : k^n \to \EE$ by:
:$\ds \map {\Theta_\RR} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$
Then $\Theta_\RR$ is a bijection.
== Proof ==
=== Proof of Surjection ===
Let $p \in \EE$.
Let $v = p - p_0 \in V$.
Let $\tuple {\lambda_1, \ldots, \lambda_n}$ be coordinates of $v$ in the basis $\tuple {e_1, \ldots, e_n}$.
Then:
thus demonstrating that $\Theta_\RR$ is a surjection.
=== Proof of Injection ===
Let:
:$\map {\Theta_\RR} {\lambda_1, \ldots, \lambda_n} = \map {\Theta_\RR} {\mu_1, \ldots, \mu_n}$
That is:
:$\ds p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i = p_0 + \sum_{i \mathop = 1}^n \mu_i e_i$
Then by $(3)$ of Properties of Affine Spaces:
:$\ds \sum_{i \mathop = 1}^n \lambda_i e_i = \sum_{i \mathop = 1}^n \mu_i e_i$
By Expression of Vector as Linear Combination from Basis is Unique:
:$\lambda_i = \mu_i$
for $i = 1, \ldots, n$.
Hence the result.
Category:Affine Geometry
Affine Group of One Dimension as Semidirect Product

== Theorem ==
Let $\map {\operatorname{Af}_1} \R$ be the $1$-dimensional affine group on $\R$.
Let $\R^+$ be the additive group of real numbers.
Let $\R^\times$ be the multiplicative group of real numbers.
Let $\phi: \R^\times \to \Aut {\R^+}$ be defined as:
:$\forall b \in \R^\times: \map \phi b = \paren {a \mapsto a b}$
Let $\R^+ \rtimes_\phi \R^\times$ be the corresponding semidirect product.
Then:
:$\map {\operatorname {Af}_1} \R \cong \R^+ \rtimes_\phi \R^\times$
where $\cong$ denotes (group) isomorphism.
== Proof ==
By definition, a (group) isomorphism is a (group) homomorphism which is a bijection.
Recall the definition of underlying set of $1$-dimensional affine group on $\R$:
:$S = \set {f_{a b}: x \mapsto a x + b : a \in \R_{\ne 0}, b \in \R}$
So the bijection $\psi: \map {\operatorname {Af}_1} \R \to \R^+ \rtimes_\phi \R^\times$ defined by $\map \psi {f_{a b} } = \paren {b, a}$ arises naturally.
It remains to show that $\psi$ is a (group) homomorphism:
Let $f_{a b}, f_{c d} \in \map {\operatorname {Af}_1} \R$.
Then:
Let $\tuple {b, a}, \tuple {d, c} \in \R^+ \rtimes_\phi \R^\times$.
Then:
So:
:$\map \psi {f_{a b} } \, \map \psi {f_{c d} } = \map \psi {f_{a b} \circ f_{c d} }$
So the bijection $\psi$ is a (group) homomorphism, and thus a (group) isomorphism.
== Also see ==
* Affine Group as Semidirect Product
Category:Affine Groups
Affine Group of One Dimension is Group

== Theorem ==
Let $\map {\operatorname {Af}_1} \R$ be the $1$-dimensional affine group on $\R$.
Then $\map {\operatorname {Af}_1} \R$ is a group.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 3.2$: Groups; the axioms: Exercise $(2)$
Category:Affine Groups
Affine Group of One Dimension is Group/Proof 1

== Theorem ==
== Proof ==
Taking the group axioms in turn:
=== ===
Let :
:$a, c \in \R_{\ne 0} \land b, d \in \R$
Let:
:$f_{ab}, f_{cd} \in \map {\operatorname {Af}_1} \R$
Then:
By the field axioms:
:$a c \in \R_{\ne 0} \land a d + b \in \R$
Thus $f_{ab} \circ f_{cd} \in \map {\operatorname {Af}_1} \R$ and so $\map {\operatorname {Af}_1} \R$ is closed.
=== ===
From Composition of Mappings is Associative, it follows directly that $\circ$ is associative on $\map {\operatorname {Af}_1} \R$.
=== ===
By Identity of Affine Group of One Dimension, $\map {\operatorname {Af}_1} \R$ has $f_{1, 0}$ as an identity element.
=== ===
By Inverse in Affine Group of One Dimension, every element $f_{a b}$ of $\map {\operatorname {Af}_1} \R$ has an inverse $f_{c d}$ where $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
All the group axioms are thus seen to be fulfilled, and so $\map {\operatorname {Af}_1} \R$ is a group.
Category:Affine Groups
Affine Subspace of 3 Dimensional Euclidean Space

== Theorem ==
Let $\R^3$ denote the (real) Euclidean $3$-space.
Let $\SS$ be an affine subspace of $\R^3$.
Then $\SS$ is one of the following:
:a point
:a straight line
:a plane
:the whole space $\R^3$.
== Proof ==
== Sources ==
* 
Category:Examples of Affine Subspaces
Category:Euclidean Space
Affirming the Consequent

== Fallacy ==
Let $p \implies q$ be a conditional statement.
Let its consequent $q$ be true.
Then it is a fallacy to assert that the antecedent $p$ is also necessarily true.
That is:
This fallacy is called affirming the consequent.
== Proof ==
We apply the Method of Truth Tables.
:$\begin{array}{|ccc|c||c|} \hline
p & \implies & q & q & p \\
\hline
\F & \T & \F & \F & \F \\
\F & \T & \T & \T & \F \\
\T & \F & \F & \F & \T \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, when $q$ is true, and so is $p \implies q$, then it is not always the case that $p$ is also true.
== Also see ==
* Denying the Antecedent
* Conditional and Converse are not Equivalent
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
* : $2$: Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables
* 
* 
Category:Affirming the Consequent
Category:Formal Fallacies
Category:Fallacies
Category:Conditional
Al-Khwarizmi/The Algebra/Inheritance Problem

== Problem ==
:A woman dies, leaving her husband, a son and three daughters.
:She also leaves $\frac 1 8 + \frac 1 7$ of her estate to a stranger.
:According to law:
::the husband receives one quarter of the estate
::the son receives double the share of a daughter
:but this division is made only after the legacy to the stranger has been paid.
How should the inheritance be divided?
== Solution ==
The stranger gets $300$ of $1120$ parts of the legacy.
The husband gets $205$ of $1120$ parts of the legacy.
The son gets $246$ of $1120$ parts of the legacy.
The daughters each get $123$ of $1120$ parts of the legacy.
== Proof ==
Let $x$ be the total legacy.
The stranger gets $x \paren {\dfrac 1 7 + \dfrac 1 8} = \dfrac {15 x} {56}$.
This leaves $x \paren {1 - \dfrac {15} {56} } = \dfrac {41 x} {56}$ to be divided between the family.
The husband gets $\dfrac 1 4 \times \dfrac {41 x} {56}$
The son gets twice what the daughters get, so the remaining $\dfrac 3 4 \times \dfrac {41 x} {56}$ gets split $5$ ways:
:$\dfrac 1 5 \times \dfrac 3 4 \times \dfrac {41 x} {56}$ to each of the $3$ daughters
:$\dfrac 2 5 \times \dfrac 3 4 \times \dfrac {41 x} {56}$ to the son.
To expressing each share in the same denominator, we make sure each fraction is a multiple of $\dfrac 1 {20 \times 56} = \dfrac 1 {1120}$.
Hence:
:the stranger's portion: $\dfrac {20 \times 15 x} {1120} = \dfrac {300 x} {1120}$
:the husband's portion: $\dfrac {5 \times 41 x} {1120} = \dfrac {205 x} {1120}$
:the son's portion: $\dfrac {2 \times 3 \times 41 x} {1120} = \dfrac {246 x} {1120}$
:each daughter's portion: $\dfrac {1 \times 3 \times 41 x} {1120} = \dfrac {123 x} {1120}$
== Sources ==
* 
* : Al-Khwarizmi (c. $825$ ): $37$
Category:Examples of Algebra Problems
Aleph-Null

== Theorem ==
Let $\omega$ denote the minimally inductive set.
:$\omega = \aleph_0$
where $\aleph$ denotes the aleph mapping.
== Proof ==
For all $n \in \omega$, $n \notin \NN'$ by the definition of the class of infinite cardinals.
Therefore, $\omega \le \aleph_0$.
Moreover, $\omega \in \NN'$ by Minimally Inductive Set is Infinite Cardinal.
Therefore, $\aleph_x = \omega$ for some ordinal $x$.
It follows that $\aleph_0 \le \aleph_x$ since $0 \le x$ and the definition of the aleph mapping, so $\aleph_0 \le \omega$.
Thus, $\aleph_0 = \omega$.
Category:Aleph Mapping
Aleph Product is Aleph

== Theorem ==
Let $x$ be an ordinal.
Then:
:$\left|{\aleph_x \times \aleph_x}\right| = \aleph_x$
where $\aleph$ denotes the aleph mapping.
== Proof ==
== Sources ==
* : $\S 10.46 \ (2)$
Category:Aleph Mapping
Aleph Zero is less than Cardinality of Continuum

== Theorem ==
$\aleph_0 < \mathfrak c$
where
:$\aleph$ denotes the aleph mapping,
:$\mathfrak c$ denotes the cardinality of the continuum.
== Proof ==
By Power Set of Natural Numbers has Cardinality of Continuum:
:$\mathfrak c = \card {\powerset \N}$
where:
:$\powerset \N$ denotes the power set of $\N$
:$\card {\powerset \N}$ denotes the cardinality of $\powerset \N$.
By Cardinality of Set less than Cardinality of Power Set:
:$\card \N < \card {\powerset \N}$ 
Thus by Aleph Zero equals Cardinality of Naturals:
:$\aleph_0 < \mathfrak c$
== Sources ==
* 
Category:Infinite Sets
Category:Aleph Mapping
Category:Cardinality of Continuum
Aleph is Infinite

== Theorem ==
Let $x$ be an ordinal.
:$\aleph_x \ge \omega$
where:
:$\aleph$ denotes the aleph mapping
:$\omega$ denotes the minimally inductive set.
== Proof ==
Since $0 \le x$, it follows that $\aleph_0 \le \aleph_x$ by definition of the aleph mapping.
But $\aleph_0 = \omega$ by Aleph-Null.
Therefore, $\omega \le \aleph_x$.
Category:Aleph Mapping
Aleph is Infinite Cardinal

== Theorem ==
Let $x$ be an ordinal.
Then $\aleph_x$ is an infinite cardinal where $\aleph$ denotes the aleph mapping.
== Proof ==
Let $\On$ denote the class of all ordinals.
By definition of the aleph mapping:
:$\aleph: \On \to \NN'$
where $\NN'$ denotes the class of infinite cardinals.
The theorem statement is an immediate consequence of this fact.
Category:Aleph Mapping
Alexander's Compactness Theorem

== Theorem ==
=== Definition by Open Covers ===
=== Definition Elements of Sub-Basis ===
== Proof ==
Let every open cover of $S$ have a finite subcover.
Let $\BB$ be a sub-basis of $\tau$. 
By definition of a compact space, from every cover of $S$ by elements of $\BB$, a finite subcover can be selected.
Let the space $T$ have a sub-basis $\BB$ such that every cover of $S$ by elements of $\BB$ has a finite subcover.
 $T$ is not such that every open cover of $S$ has a finite subcover.
Use Zorn's Lemma to find an open cover $\CC$ which has no finite subcover that is maximal among such open covers.
So if:
:$V$ is an open set
and:
:$V \notin \CC$
then $\CC \cup \set V$ has a finite subcover, necessarily of the form:
:$\CC_0 \cup \set V$
for some finite subset $\CC_0$ of $\CC$.
Consider $\CC \cap \BB$, that is, the sub-basic subset of $\CC$.
Suppose $\CC \cap \BB$ covers $S$.
Then, by hypothesis, $\CC \cap \BB$ would have a finite subcover.
But $\CC$ does not have a finite subcover.
So $\CC \cap \BB$ does not cover $S$.
Let $x \in S$ that is not covered by $\CC \cap \BB$.
We have that $\CC$ covers $S$, so:
:$\exists U \in \CC: x \in U$
We have that $\BB$ is a sub-basis.
So for some $B_1, \ldots, B_n \in \BB$, we have that:
:$x \in B_1 \cap \cdots \cap B_n \subseteq U$
Since $x$ is not covered, $B_i \notin \CC$.
As noted above, this means that for each $i$, $B_i$ along with a finite subset $\CC_i$ of $\CC$, covers $S$.
But then $U$ and all the $\CC_i$ cover $S$.
Hence $\CC$ has a finite subcover.
This contradicts our supposition that we can construct $\CC$ so as to have no finite subcover.
It follows that we cannot construct an open cover $\CC$ of $S$ which has no finite subcover.
== Notes ==
Although this proof makes use of Zorn's Lemma, the proof does not need the full strength of the Axiom of Choice.
Instead, it relies on the intermediate Ultrafilter Lemma.
== Also known as ==
Alexander's Compactness Theorem is also known as:
: Alexander's Sub-Basis Theorem
: Alexander's Sub-Base Theorem
which can also be seen in the form the Alexander Sub-Basis Theorem, and so on.
Sub-Base and Sub-Basis can also be seen here rendered as Subbase and Subbasis.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
* 
Category:Compact Spaces
Alexandroff Extension is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $T^*$ is a compact topological space.
== Proof ==
Let $\UU$ be an open cover of $T^*$.
At least one $V \in \UU$ contains $p$.
Because $p \notin S$, $V$ is not an open set of $T$.
Therefore, by definition of the Alexandroff extension, $V$ must be the complement relative to $S^*$ of a closed, compact subset $\relcomp {S^*} V$ of $T$.
Because $\relcomp {S^*} V$ is compact, it can be covered by a finite number of sets in $\UU$.
So:
:$T^*$ can be covered by $V$
and:
:that finite number of sets in $\UU$ are a cover for $\relcomp {S^*} V$.
That is, $\UU$ has a finite subcover.
Hence the result by definition of compact topological space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $34$. One Point Compactification Topology: $1$
Category:Alexandroff Extensions
Category:Compact Spaces
Alexandroff Extension of Rational Number Space is Biconnected

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a biconnected space.
== Proof ==
From Particular Point of Alexandroff Extension of Rational Number Space is Dispersion Point, $p$ is a dispersion point of $T^*$.
The result follows from Set with Dispersion Point is Biconnected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $5$
Category:Alexandroff Extensions
Category:Biconnected Sets
Category:Rational Number Space
Alexandroff Extension of Rational Number Space is Connected

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a connected space.
== Proof ==
By definition $T^*$ is a connected space of $T^*$ it admits no separation.
 $T^*$ does admit a separation.
That is, there exist open sets $A, B \in \tau^*$ such that $A, B \ne \O$, $A \cup B = \Q^*$ and $A \cap B = \O$.
That is, both $A$ and $B = \relcomp {\Q^*} A$ are open in $T^*$.
, Let $p \in A$.
Then $B$ is a closed and compact subset of $\Q$.
From Compact Set of Rational Numbers is Nowhere Dense, $B$ is nowhere dense.
From Set is Closed iff Equals Topological Closure, $B$ equals its closure.
Thus by definition of nowhere dense, the interior of $B$ is empty.
We have that $B$ is open in $T^*$ and $p \notin B$.
Thus it follows by definition of Alexandroff extension that $B$ is open in $\struct {\Q, \tau_d}$.
But from Interior of Open Set, $B$ equals its interior.
This contradicts the deduction above that the interior of $B$ is empty.
Hence by Proof by Contradiction $B$ cannot be open in $T^*$.
That is, $A$ and $B$ cannot form a separation of $T^*$.
Thus $T^*$ is a connected space by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $5$
Category:Alexandroff Extensions
Category:Connected Spaces
Category:Rational Number Space