nyuuzyou/wikis huggingface dataset:
Cover of Doubletons of Infinite Particular Point Space has no Locally Finite Refinement

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Let $\CC$ be the open cover of $T$ defined as:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
Then $\CC$ has no open refinement which is locally finite.
== Proof ==
Suppose $T$ is an infinite particular point space.
As $S$ is infinite, $\CC$ is also infinite.
Let $x \in S, x \ne p$.
Then any neighborhood of $x$ must contain $p$, by the nature of the particular point topology.
But $p$ is contained in all elements of $\CC$.
That is:
:$\forall C \in \CC: p \in C$
So any neighborhood of $x$ intersects with all elements of the open cover $\CC$.
As $\CC$ is infinite, it therefore has no open refinement which is locally finite.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $16$
Category:Particular Point Topology
Category:Covers
Cover of Interval By Closed Intervals is not Pairwise Disjoint

== Theorem ==
Let $\closedint a b$ be a closed interval in $\R$.
Let $\JJ$ be a set of two or more closed intervals contained in $\closedint a b$ such that $\ds \bigcup \JJ = \closedint a b$.
Then the intervals in $\JJ$ are not pairwise disjoint.
== Proof ==
 that the intervals of $\JJ$ are pairwise disjoint.
Let $I = \closedint p q$ be the unique interval of $\JJ$ such that $a \in I$.
Let $J = \closedint r s$ be the unique interval of $\JJ$ containing the least real number not in $I$.
These choices are possible since $\JJ$ has at least two elements, and they are supposed disjoint.
If $q = r$ then $I \cap J \ne \O$, a contradiction.
If $r < q$ then $I \cap J \ne \O$, a contradiction.
If $r > q$ then there is some real number with $q < \alpha < r$.
Therefore $\alpha \notin I$.
Since $J$ contains the least real number not in $I$ it follows that there is $\beta \in J$ with $\beta < \alpha$.
But we also have that $\alpha < r \le \beta$ for all $\beta \in J$, a contradiction.
This exhausts all the possibilities, and we conclude that the intervals of $\JJ$ are not pairwise disjoint.
Category:Measure Theory
Cover of Set/Examples/Arbitrary Example 1

== Example of Cover of Set ==
Let $S = \set {1, 2, 3, 4, 5, 6}$.
Let $H \subseteq S = \set {1, 3}$.
Let $\CC = \set {\set {1, 2}, \set {3, 4} }$.
Then $\CC$ is a cover of $H$.
== Sources ==
* 
* 
Category:Examples of Covers of Sets
Cowen's Theorem

== Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
Let $M$ be the class of all $x$ such that $x \in M_x$.
Then $M$ is minimally superinductive under $g$.
== Proof ==
== Also presented as ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem
Category:Cowen's Theorem
Category:Progressing Mappings
Category:Superinductive Classes
Cowen's Theorem/Also presented as

== Cowen's Theorem: Also presented as ==
Let $g$ be a progressing mapping.
There exists a class $M$ whose elements are exactly those $x$ which are elements of all classes that are superinductive under $g$.
Also:
:$x \in M$ $x$ is an element of every subset of $\powerset x$ that:
::contains $\O$
::is closed under $g$ relative to $x$
::is closed under chain unions.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Theorem $7.13$
Category:Cowen's Theorem
Cowen's Theorem/Lemma 1

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
We have that:
:$\powerset x$ is $x$-special $g$.
== Proof ==
By definition of $x$-special:
:$S$ is special for $x$ ( $g$)
:
In this context:
:$S = \powerset x$
We have from Empty Set is Element of Power Set:
:$\O \in \powerset x$
Then by definition of closed under $g$ relative to $x$:
:$y$ is closed under $g$ relative to $x$
:
:$\forall z \in y \cap \powerset x: \map g z \in y$
In this context we have:
:$\forall z \in \powerset x: \map g z \in \powerset x$
This is trivially true, as $\Img g \subseteq \powerset x$ by definition.
Then by definition of closed under chain unions:
:$A$ is closed under chain unions
:
:for every chain $C$ of elements of $A$, $\ds \bigcup C$ is also in $A$
where $\ds \bigcup C$ denotes the union of $C$
Replacing $A$ with $\powerset x$, this is also trivially true.
Hence the result.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Lemma $7.4$
Category:Cowen's Theorem
Category:Power Set
Cowen's Theorem/Lemma 2

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
Let $M$ be the class of all $x$ such that $x \in M_x$.
We have that:
:$\O \in M$
== Proof ==
By Lemma $1$:
:$\powerset \O$ is $\O$-special $g$.
By Power Set of Empty Set:
:$\powerset \O = \set \O$
Hence:
:$\set \O$ is $\O$-special $g$.
The only other subset of $\set \O$ is $\O$, which is by definition empty.
So $\set \O$ is the only $\O$-special subset of $\powerset \O$ $g$.
Hence:
:$M_\O = \set \O$
Because $\O \in \set \O$, we have:
:$\O \in M_\O$
and so:
:$\O \in M$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Proposition $7.5$
Category:Cowen's Theorem
Category:Empty Set
Cowen's Theorem/Lemma 3

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
We have that:
:$x \subseteq y \implies M_x \subseteq M_y$
== Proof ==
Let us recall the definition of $x$-special $g$.
:$S$ is $x$-special ( $g$)
:
Let $x \subseteq y$.
First we show that $M_y \cap \powerset x$ is $x$-special $g$.
We take the criteria one by one:
;$(1): \quad \O \in M_y \cap \powerset x$
From Lemma $1$ we have that $\powerset x$ is $x$-special $g$.
Thus $\O \in \powerset x$.
We have by definition that $\O \in S$ for all $y$-special $S$ $g$.
Hence $\O$ is an element of the intersection of all $y$-special sets $g$.
That is:
:$\O \in M_y$
Hence:
:$\O \in M_y \cap \powerset x$
;$(2): \quad M_y \cap \powerset x$ is closed under $g$ relative to $x$
Let $z \in M_y \cap \powerset x$ be arbitrary.
Suppose $\map g z \subseteq x$.
Because $x \subseteq y$ we have by Subset Relation is Transitive that $\map g z \subseteq y$.
Because $z \in M_y$ and $\map g z \subseteq y$, we have that $\map g z \in M_y$.
Thus:
:$\map g z \in \powerset x$
and:
:$\map g z \in M_y$
Hence:
:$\map g z \in M_y \cap \powerset x$
and so $M_y \cap \powerset x$ is closed under $g$ relative to $x$.
;$(3): \quad M_y \cap \powerset x$ is closed under chain unions
We have by definition that $M_y$ and $\powerset x$ are both closed under chain unions.
From Closure under Chain Unions is Preserved by Intersection:
:$M_y \cap \powerset x$ is closed under chain unions.
All the criteria $(1)$, $(2)$ and $(3)$ are fulfilled by $M_y \cap \powerset x$.
Hence $M_y \cap \powerset x$ is $x$-special $g$.
By Intersection is Subset:
:$M_y \cap \powerset x \subseteq \powerset x$
and so:
:$M_x \subseteq M_y \cap \powerset x$
That is:
:$M_x \subseteq M_y$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Lemma $7.6$
Category:Cowen's Theorem
Category:Subset Relation
Cowen's Theorem/Lemma 4

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
Let $M$ be the class of all $x$ such that $x \in M_x$.
We have that:
:$M$ is closed under chain unions.
== Proof ==
Let $C$ be a chain of elements of $M$.
For each $x \in C$, we have $x \subseteq \ds \bigcup C$.
Hence by Lemma $3$:
:$M_x \subseteq M_{\mathop \cup C}$
Also, because $x \in M_x$, we have:
:$x \in M_x$
Hence:
:$x \in M_{\mathop \cup C}$
Thus:
:$C \subseteq M_{\mathop \cup C}$
Because $M_{\mathop \cup C}$ is closed under chain unions:
:$\ds \bigcup C \in M_{\mathop \cup C}$
Hence:
:$\ds \bigcup C \in M$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Proposition $7.7$
Category:Cowen's Theorem
Category:Closure under Chain Unions
Cowen's Theorem/Lemma 6

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
Let $M$ be the class of all $x$ such that $x \in M_x$.
We have that:
:$\forall z: M_z \subseteq M$
== Proof ==
Let $x \in M_z$.
Let $y = x \cup z$.
From Set is Subset of Union:
:$z \subseteq y$
Hence by Lemma $3$:
:$M_z \subseteq M_y$
Hence:
:$x \in M_y$
Also, we have:
:$x \subseteq y$
and so from Lemma $5$:
:$M_y \subseteq M_x \cup \paren {\powerset y \setminus \powerset x}$
Hence:
:$x \in M_x \cup \paren {\powerset y \setminus \powerset x}$
But we have that:
:$x \notin \powerset y \setminus \powerset x$
Hence:
:$x \in M_x$
Thus:
:$x \in M$
and so:
:$M_z \subseteq M$
and the result is shown to hold.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Proposition $7.9$
Category:Cowen's Theorem
Cowen's Theorem/Lemma 7

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
Let $M$ be the class of all $x$ such that $x \in M_x$.
We have that:
:$M$ is closed under $g$ relative to $x$.
== Proof ==
Let $x \in M$.
Then $x \in M_x$.
Because $x \subseteq x \cup \map g x$, we have from Lemma $3$:
:$M_x \subseteq M_{x \cup \map g x}$
Hence:
:$x \in M_{x \cup \map g x}$
Also:
:$\map g x \subseteq x \cup \map g x$
and so:
:$\map g x \in M_{x \cup \map g x}$
Also, from Lemma $6$:
:$M_{x \cup \map g x} \subseteq M$
Hence:
:$\map g x \in M$
and the result follows.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Proposition $7.10$
Category:Cowen's Theorem
Cowen's Theorem/Lemma 8

== Lemma for Cowen's Theorem ==
Let $g$ be a progressing mapping.
Let $x$ be a set.
Let $\powerset x$ denote the power set of $x$.
Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ $g$.
Let $A$ be a class which is superinductive under $g$.
Then:
:$M_x \subseteq A$ 
== Proof ==
Let $A$ be a class which is superinductive under $g$.
We show that $A \cap \powerset x$ is $x$-special $g$.
Let us recall the definition of $x$-special $g$.
:$S$ is $x$-special ( $g$)
:
We take the criteria one by one:
;$(1): \quad \O \in A \cap \powerset x$
We have by definition that $\O \in A$ and $\O \in \powerset x$.
Hence:
:$\O \in A \cap \powerset x$
;$(2): \quad A \cap \powerset x$ is closed under $g$ relative to $x$
Let $y \in A \cap \powerset x$ and $\map g y \in \powerset x$.
Because $y \in A$ we have:
:$\map g y \in A$
Hence:
:$\map g y \in A \cap \powerset x$
That is, $A \cap \powerset x$ is closed under $g$ relative to $x$.
;$(3): \quad A \cap \powerset x$ is closed under chain unions
Because $A$ and $\powerset x$ are both closed under chain unions, then so is $A \cap \powerset x$.
We have demonstrated that $A \cap \powerset x$ is $x$-special $g$.
We have that:
:$A \cap \powerset x \subseteq \powerset x$
Hence:
:$M_a \subseteq A \cap \powerset x$
Hence it follows that:
:$M_y \subseteq A$
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Lemma $7.11$
Category:Cowen's Theorem
Cowen's Theorem/Proof

== Theorem ==
== Proof ==
=== Lemma $1$ ===
=== Lemma $2$ ===
=== Lemma $3$ ===
=== Lemma $4$ ===
=== Lemma $5$ ===
=== Lemma $6$ ===
=== Lemma $7$ ===
Recall:
:From Lemma $2$:
::$\O \in M$
:From Lemma $7$:
::$M$ is closed under $g$ relative to $x$
:From Lemma $4$:
::$M$ is closed under chain unions.
Hence by definition:
:$M$ is superinductive under $g$.
=== Lemma $8$ ===
We have that $M$ is superinductive under $g$.
Let $A$ be superinductive under $g$.
Let $x \in M$.
Then:
:$x \in M_x$
and from Lemma $8$:
:$M_x \subseteq A$
Hence:
:$x \in A$
As $x$ is arbitrary, it follows by definition of subclass that:
:$M \subseteq A$.
Hence by definition $M$ is minimally superinductive under $g$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 7$ Cowen's theorem: Proposition $7.12$
Category:Cowen's Theorem
Cowen-Engeler Lemma

== Theorem ==
Let $X$ and $Y$ be sets.
Let $M$ be a set of mappings from subsets of $X$ to $Y$.
Thus each element of $M$ is a partial mapping from $X$ to $Y$.
Define a mapping:
:$\Phi: X \to \powerset Y$ 
thus:
:$\map \Phi x = \set {\map f x: f \in M \text{ and } x \in \Dom f}$
That is:
:$\map \Phi x = \set {y \in Y: \exists f \in M: \tuple {x, y} \in f}$
Suppose that the following hold:
Then there exists an element of $M$ with domain $X$.
== Proof ==
Let $\map {\operatorname {Fin} } X$ be the set of finite subsets of $X$.
For each $S \in \map {\operatorname {Fin} } X$, let:
:$\Gamma_S = \set {f \in M: S \subseteq \Dom f}$
By $(2)$, $\Gamma_S$ is non-empty for each $S \in \map {\operatorname {Fin} } X$.
For each $S \in \map {\operatorname {Fin} } X$:
:$\Gamma_S \cap \Gamma_T = \Gamma_{S \cup T}$:
If:
:$f \in \Gamma_{S \cup T}$
then:
:$S \cup T \subseteq \Dom f$
Then:
:$S \subseteq \Dom f$ 
and:
:$T \subseteq \Dom f$
so:
:$f \in \Gamma_S \land f \in \Gamma_T$
Thus:
:$f \in \Gamma_S \cap \Gamma_T$
If:
:$f \in \Gamma_S \cap \Gamma_T$
then:
:$f \in \Gamma_S \land f \in \Gamma_T$
Thus:
:$S \subseteq \Dom f \land T \subseteq \Dom f$
Thus:
:$S \cup T \subseteq \Dom f$
So:
:$f \in \Gamma_{S \cup T}$.
Thus $\set {\Gamma_S: S \in \map {\operatorname {Fin} } X}$ has the finite intersection property.
Then by the corollary to the Ultrafilter Lemma, there is an ultrafilter $\UU$ on $M$ which includes $\set {\Gamma_S: S \in \map {\operatorname {Fin} } X}$ as a subset.
We will now construct a mapping:
:$\phi: X \to Y$
Fix $x \in X$.
Note that by the definitions of $\Phi$ and $\Gamma$:
:$\map \Phi x = \set {\map f x: f \in \Gamma_{\set x} }$
For each $y \in \map \Phi x$, let:
:$K_y = \set {f \in \Gamma_{\set x} : \map f x = y}$
Then:
:$\set {K_y : y \in \map \Phi x}$ 
is a partition of $\Gamma_{\set x}$.
Because $\set x \in \map {\operatorname {Fin} } X$:
:$\Gamma_{\set x} \in \UU$
By $(1)$, $\map \Phi x$ is finite.
Thus by Component of Finite Union in Ultrafilter, there is exactly one $y \in \map \Phi x$ such that:
:$K_y \in \UU$
We define $\map \phi x$ to be that $y$.
By the definition of $K_y$, for each $x \in X$ we have:
:$\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \UU$
We must show that:
:$\phi \in M$
Let $S$ be a finite subset of $X$.
Let:
:$\ds \Psi = \bigcap_{x \mathop \in S} \set {f \in \Gamma_{\set x}: \map f x = \map \phi x}$
Since:
:$\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \UU$ for each $x \in X$
:$S$ is finite
:$\UU$ is a filter
it follows that:
:$\Psi \in \UU$
Thus:
:$\Psi \ne \O$
Choose any $f \in \Psi$.
Then:
:$\map f x \in \Gamma_{\set x}$
for each $x \in S$.
So $f \in M$ and $S$ is a finite subset of the domain of $f$.
Since $M$ has finite character, $f \restriction S \in M$.
Furthermore, by the definition of $\Psi$:
:$\map f x = \map \phi x$
for each $x \in S$.
Thus:
:$\phi \restriction S = f \restriction S \in \phi$
So we see that:
:$\phi \restriction S \in M$ 
for each finite subset $S$ of $X$.
Since $M$ has finite character, $\phi \in M$.
== Sources ==
* : $\S 6.35$
Category:Set Theory
Cramer's Rule

== Theorem ==
Let $n \in \N$.
Let $b_1, b_2, \dots, b_n$ be real numbers.
Let $\mathbf b = \tuple {b_1, b_2, \dots, b_n}^\intercal$.
Let $x_1, x_2, \dots, x_n$ be real numbers.
Let $\mathbf x = \tuple {x_1, x_2, \dots, x_n}^\intercal$.
Let $A$ be a nonsingular $n \times n$ matrix whose elements are in $\R$.
For each $i \in \set {1, \dots, n}$, let $A_i$ be the matrix obtained by replacing the $i$th column with $\mathbf b$.
Let:
:$A \mathbf x = \mathbf b$
Then:
:$\mathbf x_i = \dfrac {\map \det {A_i} } {\map \det A}$
for each $i \in \set {1, \dots, n}$.
== Proof ==
Let $C$ be the cofactor matrix of $A$.
By definition, $C^\intercal$ is the adjugate matrix of $A$.
Therefore by Matrix Product with Adjugate Matrix:
:$A \cdot C^\intercal = \det A \cdot I_n$
Because $A$ is nonsingular, $A^{-1}$ exists.
From Matrix is Nonsingular iff Determinant has Multiplicative Inverse, $1 / \det A$ also exists.
Therefore:
:$A^{-1} = \dfrac 1 {\det A} \cdot C^\intercal$
Since $A \mathbf x = \mathbf b$:
:$\mathbf x = A^{-1} \mathbf b$
Therefore:
:$\mathbf x = \paren {\dfrac 1 {\det A} C^\intercal} \mathbf b$
For each $r, s \in \set {1, 2, \ldots, n}$, let $A_{r s}$ denote the element of $C$ whose index is $\tuple {r, s}$.
Then by the definition of transpose:
:$C^\intercal = \begin {bmatrix} A_{11} & A_{21} & \dots & A_{n1} \\
A_{12} & A_{22} & \dots & A_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
A_{1n} & A_{2n} & \dots & A_{nn}
\end{bmatrix}$
By the definition of cofactor matrix, $A_{r s}$ is the cofactor of the element of $A$ whose index is $\tuple {r, s}$.
Let $i \in \set {1, 2, \ldots, n}$.
Then by definition of matrix product:
:$\ds x_i = \dfrac 1 {\det A} \paren {\sum_{j \mathop = 1}^n A_{ji} b_j}$
Recall that $A_i$ is the matrix obtained from $A$ by replacing the $i$th column with $\mathbf b$. 
Then if:
:$A = \begin {bmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,n} \\
a_{2,1} & a_{2,2} & \dots & a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n,1} & a_{n,2} & \dots & a_{n,n}
\end{bmatrix}$
then:
:$A_i = \begin{bmatrix}
a_{1,1} & a_{1,2} & \dots & a_{1, i - 1} & b_1 & a_{1, i + 1} & \dots & a_{1,n} \\
a_{2,1} & a_{2,2} & \dots & a_{2, i - 1} & b_2 & a_{2, i + 1} & \dots & a_{2,n} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
a_{n,1} & a_{n,2} & \dots & a_{n, i - 1} & b_n & a_{n, i + 1} & \dots & a_{n,n}
\end{bmatrix}$
For all $j \in \set {1, 2, \dots, n}$ the matrix obtained by deleting the $j$th row and the $i$th column of $A$ is equal to the matrix obtained by deleting the $j$th row and $i$th column of $A_i$.
Therefore, by definition, the cofactor of $\sqbrk {A_i}_{j,i}$ is equal to $A_{ji}$.
By the Expansion Theorem for Determinants, we now have:
:$\ds \det{A_i} = \sum_{j \mathop = 1}^n A_{ji} b_j$
Therefore:
:$x_i = \dfrac {\det {A_i} } {\det A}$
as desired.
== Historical Note ==
== Sources ==
* 
* 
* 
* 
Category:Linear Algebra
Creation of Orthogonal Vector from Independent Vectors

== Theorem ==
Let $\mathbf V$ be a vector space.
Let $\mathbf a$ and $\mathbf b$ be vectors of $\mathbf V$ such that $\mathbf a$ and $\mathbf b$ form a linearly independent set.
Let $\mathbf a$ and $\mathbf b$ be expressed as column vectors.
Let $\mathbf b'$ be calculated as:
:$\mathbf b' = \mathbf b - \dfrac {\mathbf a^\intercal \mathbf b} {\mathbf a^\intercal \mathbf a} \mathbf a$
where $\mathbf a^\intercal$ denotes the transpose of $\mathbf a$.
Then $\mathbf b'$ is orthogonal to $\mathbf a$.
== Proof ==
== Sources ==
* 
* 
Category:Orthogonality (Linear Algebra)
Category:Linear Independence
Criterion for Ring with Unity to be Topological Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $\tau$ be a topology over $R$.
Suppose that $+$ and $\circ$ are $\tau$-continuous mappings.
Then $\struct {R, +, \circ, \tau}$ is a topological ring.
== Proof ==
As we presume $\circ$ to be continuous, we need only prove that $\struct {R, +, \tau}$ is a topological group.
As we presume $+$ to be continuous, we need only show that negation is continuous.
As $\struct {R, \circ}$ is a semigroup and $\circ$ is continuous:
:$\struct{R, \circ, \tau}$ is a topological semigroup.
From Identity Mapping is Homeomorphism, the identity mapping $I_R : \struct {R, \tau} \to \struct {R, \tau}$ is continuous.
 
From Multiple Rule for Continuous Mappings to Topological Semigroup, the mapping $\paren{- 1_R} \circ I_R : R \to R$ defined by:
:$\forall b \in R : \map {\paren {\paren {-1_R} \circ I_R} } b = \paren {-1_R} \circ b$
is continuous.
From Product with Ring Negative, for each $b \in R : -b = \paren {-1_R} \circ b$.
Hence negation is continuous.
Critical Line Theorem

== Theorem ==
There exist an infinite number of nontrivial zeroes of the Riemann $\zeta$ function on the critical line.
== Proof ==
== Also see ==
* Riemann Hypothesis: If this is true, all nontrivial zeroes are on the critical line.
== Historical Note ==
== Sources ==
* 
* 
* : $0,5$
* 
* 
* 
* 
Category:Riemann Zeta Function
Category:Named Theorems
Critical Path Analysis/Examples/Aircraft Turnround

== Example of Critical Path Analysis ==
Consider the process of turning round an aeroplane after landing.
The precedence table for its activities is presented as follows:
:$\begin {array} {llll}
& \textit {Activity} & \textit {Duration (Minutes)} & \textit {Precedents} \\
\hline
\text A & \text {Disembark passengers} & 7 & \\
\text B & \text {Unload baggage} & 10 & \\
\text C & \text {Refuel aircraft} & 12 & \text A \\
\text D & \text {Clean cabin} & 15 & \text A \\
\text E & \text {Load catering requirements} & 6 & \text D \\
\text F & \text {Load baggage} & 12 & \text B \\
\text G & \text {Embark passengers} & 14 & \text C, \text E \\
\text H & \text {Final loading check} & 2 & \text F, \text G \\
\end{array}$
The following CPA network diagrams the above task breakdown.
:
The critical path is the path consisting of edges $\text A, \text D, \text E, \text G, \text H$. 
== Sources ==
* 
* 
Category:Examples of Critical Path Analysis
Cross-Relation Equivalence Classes on Natural Numbers are Cancellable for Addition

== Theorem ==
Let $\struct {\N, +}$ be the semigroup of natural numbers under addition.
Let $\struct {\N \times \N, \oplus}$ be the (external) direct product of $\struct {\N, +}$ with itself, where $\oplus$ is the operation on $\N \times \N$ induced by $+$ on $\N$.
Let $\boxtimes$ be the cross-relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
Let $\eqclass {x, y} {}$ denote the equivalence class of $\tuple {x, y}$ under $\boxtimes$.
The operation $\oplus$ on these equivalence classes is cancellable, in the sense that:
== Proof ==
Let $\eqclass {a_1, b_1} {}, \eqclass {a_2, b_2} {}, \eqclass {c_1, d_1} {}, \eqclass {c_2, d_2} {}$ be $\boxtimes$-equivalence classes such that $\eqclass {a_1, b_1} {} = \eqclass {a_2, b_2} {}$ and $\eqclass {c_1, d_1} {} = \eqclass {c_2, d_2} {}$.
Then:
Then we have:
== Sources ==
* : $\S 2.5$: Theorem $2.22$
Category:Natural Number Addition
Category:Cartesian Product
Category:Equivalence Relations
Cross-Relation is Congruence Relation

== Theorem ==
The cross-relation $\boxtimes$ is a congruence relation on $\struct {S_1 \times S_2, \oplus}$.
== Proof ==
From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation.
We now need to show that:
First we note that:
Then:
So $\boxtimes$ is a congruence relation on $\struct {S \times C, \oplus}$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Cross-Relations