nyuuzyou/wikis huggingface dataset:
Ordinals are Well-Ordered/Proof 1

== Theorem ==
== Proof ==
Recall that the Ordinals are Totally Ordered.
Let $A$ be a non-empty set of ordinals.
Let $\alpha \in A$.
Let $B = \alpha^+ \cap A$, where $\alpha^+$ denotes the successor set of $\alpha$.
Recall that $\alpha^+$ is an ordinal.
Note that $\alpha \in B$, so $B$ is non-empty.
By Intersection is Subset, $B \subseteq \alpha^+$.
It follows that there exists a smallest element $\kappa$ of $B$.
We claim that $\kappa$ is the smallest element of $A$.
So let $\beta \in A$. We need to show that $\kappa = \beta$ or $\kappa \in \beta$.
By Ordinal Membership is Trichotomy, either $\beta \in \alpha^+$, $\alpha^+ = \beta$, or $\alpha^+ \in \beta$.
If $\beta \in \alpha^+$, then $\beta \in B$; it follows by the definition of $\kappa$ that $\kappa = \beta$ or $\kappa \in \beta$.
If $\alpha^+ = \beta$ or $\alpha^+ \in \beta$, then it follows by the transitivity of $\beta$ that $\alpha^+ \subseteq \beta$.
Since $\kappa \in \alpha^+$, it follows that $\kappa \in \beta$.
Category:Ordinals are Well-Ordered
Ordinals are Well-Ordered/Proof 2

== Theorem ==
== Proof ==
By Ordinals are Totally Ordered, the ordinals are totally ordered by $\subseteq$:
Thus:
:the strict ordering $\subsetneqq$ on ordinals
and
:the strict ordering $\in$ on ordinals
are the same.
 the ordinals were not well-ordered by $\subsetneqq$.
Then we could find a sequence $\sequence {X_n}_{n \mathop = 0}^\infty$ of ordinals such that:
:$X_0 \supsetneqq X_1 \supsetneqq X_2 \cdots$
So for all $n > 0$:
:$X_n \subsetneqq X_0$
so:
:$X_n \in X_0$
Thus $\sequence {X_{n + 1} }_{n \mathop = 0}^\infty$ is a decreasing sequence under $\subsetneqq$ of elements of $\sequence {X_n}$.
But since $X_0$ is an ordinal it is well-ordered by $\subsetneqq$.
From Infinite Sequence Property of Well-Founded Relation, this demonstrates a contradiction.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals
Category:Ordinals are Well-Ordered
Ordinals have No Zero Divisors

== Theorem ==
Let $x$ and $y$ be ordinals.
Then:
:$\paren {x \cdot y} = 0 \iff \paren {x = 0 \lor y = 0}$
== Proof ==
=== Necessary Condition ===
Suppose that $\paren {x \cdot y} = 0$ and that $x \ne 0$.
By Ordinal Multiplication by Zero:
: $\paren {x \cdot 0} = 0$
Therefore:
: $\paren {x \cdot y} = \paren {x \cdot 0}$
Thus, by Ordinal Multiplication is Left Cancellable, we have that $y = 0$.
=== Sufficient Condition ===
If $x = 0$, then by Leibniz's Rule:
:$\paren {x \cdot y} = \paren {0 \cdot y}$
If $y = 0$, then also by Leibniz's Rule:
:$\paren {x \cdot y} = \paren {x \cdot 0}$
By Ordinal Multiplication by Zero, both $\paren {0 \cdot y} = 0$ and $\paren {x \cdot 0} = 0$.
== Sources ==
* : $\S 8.22$
Category:Ordinal Arithmetic
Ordinals under Addition form Ordered Monoid

== Theorem ==
$\struct {\On, +, \le}$ forms an ordered monoid, where:
:$\On$ denotes the class of all ordinals
:$+$ denotes ordinal addition.
== Proof ==
The result follows from Ordinals under Addition form Monoid and Ordinals under Addition form Ordered Semigroup.
Category:Ordinal Arithmetic
Ordinals under Addition form Ordered Semigroup

== Theorem ==
$\struct {\On, +, \le}$ forms an ordered semigroup, where:
:$\On$ denotes the class of all ordinals
:$+$ denotes ordinal addition.
== Proof ==
The result follows from:
:Ordinals under Addition form Semigroup
:Subset is Compatible with Ordinal Addition.
Category:Ordinal Arithmetic
Ordinals under Multiplication form Ordered Monoid

== Theorem ==
$\struct {\On, \times, \le}$ forms an ordered monoid, where:
:$\On$ denotes the class of all ordinals
:$\times$ denotes ordinal multiplication.
== Proof ==
The result follows from Ordinals under Multiplication form Monoid and Ordinals under Multiplication form Ordered Semigroup.
Category:Ordinal Arithmetic
Ordinals under Multiplication form Ordered Semigroup

== Theorem ==
$\struct {\On, \times, \le}$ forms an ordered semigroup, where:
:$\On$ denotes the class of all ordinals
:$\times$ denotes ordinal multiplication.
== Proof ==
The result follows from Ordinals under Multiplication form Semigroup and Subset is Compatible with Ordinal Multiplication.
Category:Ordinal Arithmetic
Ore's Theorem

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph of order $n \ge 3$.
Let $G$ be an Ore graph, that is:
:For each pair of non-adjacent vertices $u, v \in V$:
::$(1): \quad \deg u + \deg v \ge n$
Then $G$ is Hamiltonian.
== Proof ==
From Ore Graph is Connected it is not necessary to demonstrate that $G$ is connected.
 it were possible to construct a graph that fulfils condition $(1)$ which is not Hamiltonian.
For a given $n \ge 3$, let $G$ be the graph with the most possible edges such that $G$ is non-Hamiltonian which satisfies $(1)$.
Although it does not contain a Hamilton cycle, $G$ has to contain a Hamiltonian path $\tuple {v_1, v_2, \ldots, v_n}$.
Otherwise it would be possible to add further edges to $G$ without making $G$ Hamiltonian.
Since $G$ is not Hamiltonian, $v_1$ is not adjacent to $v_n$, otherwise $\tuple {v_1, v_2, \ldots, v_n, v_1}$ would be a Hamilton cycle.
By $(1)$, we have that:
:$\deg v_1 + \deg v_n \ge n$
By the Pigeonhole Principle, for some $i$ such that $2 \le i \le n - 1$, $v_i$ is adjacent to $v_1$, and $v_{i - 1}$ is adjacent to $v_n$.
But the cycle $\tuple {v_1, v_2, \ldots, v_{i - 1}, v_n, v_{n - 1}, \ldots, v_i, v_1}$ is then a Hamilton cycle.
So $G$ is Hamiltonian after all.
== Also see ==
* Hamiltonian Graph is not necessarily Ore Graph, demonstrating that the converse of Ore's Theorem does not hold.
* Dirac's Theorem
== Historical Note ==
Category:Hamiltonian Graphs
Category:Ore Graphs
Ore Graph is Connected

== Theorem ==
Let $G = \struct {V, E}$ be an Ore graph.
Then $G$ is connected.
== Proof ==
Let $G$ be an Ore graph of order $n$.
 $G$ is not connected.
Then it has at least two components.
Call these components $C_1$ and $C_2$.
Thus, there exist non-adjacent vertices $u$ and $v$ such that $u$ is in $C_1$ and $v$ is in $C_2$.
Let $m_1$ and $m_2$ be the number of vertices in $C_1$ and $C_2$ respectively.
It is clear that:
:$m_1 + m_2 \le n$
By definition of Ore graph, $G$ is simple.
Thus it follows that:
:$\map {\deg_G} u \le m_1 - 1$ and $\map {\deg_G} v \le m_2 - 1$
Thus:
That is:
:$\map {\deg_G} u + \map {\deg_G} v < n$
But by definition of Ore graph:
:$\map {\deg_G} u + \map {\deg_G} v \ge n$
By Proof by Contradiction, it follows that our assumption that $G$ is not connected was false.
Hence the result.
== Also see ==
* Ore's Theorem
Category:Ore Graphs
Ore Number/Examples/140

== Example of Ore Number ==
:$\map H {140} = 5$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:140
Ore Number/Examples/1638

== Example of Ore Number ==
:$\map H {1638} = 9$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:1638
Ore Number/Examples/270

== Example of Ore Number ==
:$\map H {270} = 6$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:270
Ore Number/Examples/28

== Example of Ore Number ==
:$\map H {28} = 3$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:28
Ore Number/Examples/496

== Example of Ore Number ==
:$\map H {496} = 5$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:496
Ore Number/Examples/6

== Example of Ore Number ==
:$\map H 6 = 2$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:6
Ore Number/Examples/672

== Example of Ore Number ==
:$\map H {672} = 8$
where $\map H n$ denotes the harmonic mean of the divisors of $n$.
== Proof ==
From Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum:
:$\map H n = \dfrac {n \, \map {\sigma_0} n} {\map {\sigma_1} n}$
where:
:$\map {\sigma_0} n$ denotes the divisor count function: the number of divisors of $n$
:$\map {\sigma_1} n$ denotes the divisor sum function: the sum of the divisors of $n$.
Category:Ore Numbers
Category:672
Orientation of Contour is Well-Defined

== Theorem ==
Let $C$ be a contour in the complex plane $\C$.
Let $D \subseteq \C$ be a connected domain.
Let $\Img C \subseteq \partial D$, where $\Img C$ denotes the image of $C$, and $\partial D$ denotes the boundary of $D$.
The orientation of $C$ with respect to $D$, if it exists, does not depend on the choice of parameterization of $C$.
== Proof ==
By definition of contour, $C$ is a concatenation of a finite sequence of directed smooth curves $C_1, \ldots, C_n$.
Let each directed smooth curve $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k}{b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Let $\sigma_k: \closedint {c_k}{d_k}$ be a reparameterization of $C_k$.
By definition of directed smooth curve, there exists a bijective differentiable strictly increasing real function:
:$\phi_k: \closedint c d \to \closedint a b$
such that $\sigma_k = \gamma_k \circ \phi_k$.
Let $\sigma: \closedint c d \to \C$ be the parameterization of $C$ defined by $\sigma_1, \ldots, \sigma_n$.
Set $K_\sigma := \set { t \in \closedint c d : \textrm{ $\sigma$ is not differentiable at $t$ } }$.
If $t \in \openint c d \setminus K_\sigma$, then there exists $k \in \set {1, \ldots, n}$ such that $t \in \openint {c_k}{d_k}$, and:
Suppose that $C$ is positively oriented with respect to $D$.
It follows that there exists $r \in \R_{>0}$ such that:
:$\forall \epsilon \in \openint 0 r : \map \gamma t + \epsilon i \map {\gamma_k'}{ \map {\phi_k} t } \in D, \map \gamma t - \epsilon i \map {\gamma_k'}{ \map {\phi_k} t } \notin D$
When we replace the expression for $\map {\gamma_k'}{ \map {\phi_k} t }$, we have:
:$\map \gamma t + \dfrac {\epsilon}{ \map {\phi_k'} t } i \map {\sigma_k'} t \in D,\map \gamma t -\dfrac {\epsilon}{ \map {\phi_k'} t } i \map {\sigma_k'} t \notin D$
where $\dfrac {\epsilon}{ \map {\phi_k'} t } \in \openint 0 {\dfrac r {\map {\phi_k'} t} }$.
This shows that $C$ is positively oriented with respect to $D$, when we use the parameterization $\sigma$ of $C$.
When $C$ is negatively oriented with respect to $D$, the proof is similar.
Category:Orientation of Complex Contour
Orientation of Simple Closed Contour is with Respect to Interior

== Theorem ==
Let $C$ be a simple closed contour in the complex plane $\C$.
Let $\Int C$ denote the interior of $C$.
$C$ is positively oriented, $C$ is positively oriented with respect to $\Int C$.
$C$ is negatively oriented, $C$ is negatively oriented with respect to $\Int C$.
== Proof ==
=== Sufficient condition ===
Suppose $C$ is positively oriented.
Let $\gamma : \closedint a b$ be a parameterization of $C$.
Set $K := \set { t \in \closedint a b : \textrm{ $\gamma$ is not differentiable at $t$ } }$.
By definition of positive orientation, it follows that for all $t \in \openint a b \setminus K$, there exists $r_0 \in \R_{>0}$ such that:
:for all $\epsilon \in \openint 0 {r_0}$ : $\map \gamma t + \epsilon i \map {\gamma'} t \in \Int C$
Complex Plane is Homeomorphic to Real Plane shows that we can identify the complex plane $\C$ with the real plane $\R^2$ by the homeomorphism $\map \phi {x, y} = x + i y$.
Interior of Simple Closed Contour is Well-Defined shows that the image $\Img C$ can be identified with the image of a Jordan curve $g: \R^2 \to \R^2$.
From the same theorem, it follows that $\Int C$ can be identified with the interior of $g$.
From the Jordan Curve Theorem, it follows that $\Int C$ is an open connected set with $\Img C$ as its boundary.
It follows that $\Int C$ is a connected domain with $\Img C \subseteq \partial \Int C$, which is the first requirement for $C$ to be positively oriented with respect to $\Int C$.
The next requirement is that for all $t \in \openint a b \setminus K$, we need to find $r \in \R_{>0}$ such that:
:for all $\epsilon \in \openint 0 r$ : $\map \gamma t + \epsilon i \map {\gamma'} t \in \Int C$, and $\map \gamma t - \epsilon i \map {\gamma'} t \notin \Int C$
From All Normal Vectors of Simple Closed Contour Cannot Point into Interior, it follows that there exists $r_1 \in \R_{>0}$ such that $\map \gamma t - \epsilon i \map {\gamma'} t \notin \Int C$ for all $\epsilon \in \openint 0 {r_1}$.
Then $r := \map \min {r_0 , r_1}$ fulfils the criteria.
When $C$ is negatively oriented, the proof is similar.
=== Necessary condition ===
This follows directly from the definitions of orientation of contour.
Category:Orientation of Complex Contour
Orthic Triangle of Obtuse Triangle

== Theorem ==
Let $\triangle ABC$ be an obtuse triangle such that $A$ is the obtuse angle.
Let $H$ be the orthocenter of $\triangle ABC$.
Let $\triangle DEF$ be the orthic triangle of $\triangle ABC$.
Then $\triangle DEF$ is also the orthic triangle of $\triangle HBC$, which is an acute triangle.
== Proof ==
:
By construction:
:$CE \perp BH$
:$BF \perp CH$
:$HD \perp BC$
Thus by definition $CE$, $BF$ and $HD$ are the altitudes of $\triangle HBC$.
Also by construction, $A$ lies on $CE$, $BF$ and $HD$.
Hence $\triangle DEF$ is the orthic triangle of $\triangle HBC$.
Category:Orthic Triangles
Category:Obtuse Triangles
Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $O$ be the circumcenter of $\triangle ABC$.
Let $G$ be the centroid of $\triangle ABC$.
Let $H$ be the orthocenter of $\triangle ABC$.
Then $O$, $G$ and $H$ are the same points $\triangle ABC$ is equilateral.
If $\triangle ABC$ is not equilateral, then $O$, $G$ and $H$ are all distinct.
== Proof ==
=== Necessary Condition ===
Let $\triangle ABC$ be an equilateral triangle.
By definition, each side of $\triangle ABC$ is the base of an isosceles triangle.
:
Let $AE$, $BF$ and $CD$ be the altitudes of $\triangle ABC$ through $A$, $B$ and $C$ respectively.
From Altitudes of Triangle Meet at Point, let them intersect at $G$.
From Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles:
:$AE$, $BF$ and $CD$ are the medians of $\triangle ABC$
and:
:$AE$, $BF$ and $CD$ are the perpendicular bisectors of $\triangle ABC$.
They all meet at a single point, that is $G$.
Hence by definition, the circumcenter, centroid and orthocenter of $\triangle ABC$ are the same point.
=== Converse Condition ===
Let $\triangle ABC$ not be an equilateral triangle.
Thus at least one pair of sides of $\triangle ABC$ is not equal.
, suppose $AC \ne BC$.
Then $AB$ is not the base of an isosceles triangle.
:
Thus from Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles, the altitude from $AB$, the midpoint of $AB$ and the perpendicular bisector of $AB$ are all different.
So the circumcenter, centroid and orthocenter of $\triangle ABC$ likewise cannot be the same point.
Category:Equilateral Triangles
Orthocenter and Incenter Coincide if Triangle is Equilateral

== Theorem ==
Let $\triangle ABC$ be an equilateral triangle.
Let $G$ be the orthocenter of $\triangle ABC$.
Then:
$G$ is the incenter of $\triangle ABC$.
== Proof ==
Draw the circumcircle of $\triangle ABC$ through points $A, B,$ and $C$.
Given: $G$ is the orthocenter of $\triangle ABC$.
By Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral:
:$G$ is also the centroid and circumcenter of $\triangle ABC$.
By mutatis mutandis:
all the smaller triangles are congruent.
So
:$DG = EG = FG$
By the definition of incircle:
:$DG$ is a radius of the incircle of $\triangle ABC$.
By definition of center: 
:$G$ is the center of the incircle of $\triangle ABC$
Category:Equilateral Triangles
Category:Orthocenters of Triangles
Category:Incircles of Triangles
Orthocenter of Self-Conjugate Triangle

== Theorem ==
Let $\CC$ be a circle.
Let $\triangle PQR$ be a self-conjugate triangle with respect to $\CC$.
Then the orthocenter of $\triangle PQR$ is the center of $\CC$.
== Proof ==
By definition of self-conjugate triangle:
:$PR$ is the polar of $Q$
:$QR$ is the polar of $P$
and from Self-Conjugate Triangle needs Two Sides to be Specified:
:$PQ$ is the polar of $R$
all with respect to $\CC$.
Let $O$ be the center of $\CC$.
Then from Polar of Point is Perpendicular to Line through Center:
:$PQ \perp OR$
:$QR \perp OP$
:$PR \perp OQ$
That is: $OP$, $OQ$ and $OR$ are lines perpendicular to the sides of $\triangle PQR$ which all pass through $O$.
Hence, by definition, $O$ is the orthocenter of $\triangle PQR$.
== Sources ==
* : Chapter $\text {III}$. The Circle: $9$. Conjugate triangles
Category:Conjugate Triangles
Category:Orthocenters of Triangles
Orthocomplement Reverses Subset

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space space.
Let $A, B$ be subsets of $V$ with $A \subseteq B$.
Then: 
:$B^\perp \subseteq A^\perp$
where $\perp$ denotes orthocomplementation.
== Proof ==
Let: 
:$h \in B^\perp$
Then, from the definition of orthocomplement, we have:
:$h \perp b$ for each $b \in B$.
Since $A \subseteq B$, we in particular have: 
:$h \perp a$ for each $a \in A$.
So, from the definition of orthocomplement:
:$h \in A^\perp$
So:
:$h \in B^\perp$ implies $h \in A^\perp$.
So, from the definition of subset:
:$B^\perp \subseteq A^\perp$
Category:Inner Product Spaces
Category:Orthocomplements
Orthocomplement equal to Orthocomplement of Linear Span

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space over $\GF$. 
Let $S \subseteq V$ be non-empty.
Then:
:$S^\bot = \paren {\map \span S}^\bot$
where $\bot$ denotes orthocomplement.
== Proof ==
From Orthocomplement Reverses Subset, we have:
:$\paren {\map \span S}^\bot \subseteq S^\bot$ 
Conversely let $y \in S^\bot$. 
Let $x \in \map \span S$.
From the definition of the linear span there exists $x_1, \ldots, x_n \in S$ and $\alpha_1, \ldots, \alpha_n \in \GF$ such that:
:$\ds x = \sum_{j \mathop = 1}^n \alpha_j x_j$
We then have:
So $\innerprod x y = 0$ for all $x \in \map \span S$.
So $y \in \paren {\map \span S}^\bot$.
Hence we have $S^\bot \subseteq \paren {\map \span S}^\bot$.
We conclude:
:$S^\bot = \paren {\map \span S}^\bot$
Category:Orthocomplements
Orthocomplement is Closed Linear Subspace

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $A \subseteq V$ be a subset of $V$.
Then the orthocomplement $A^\perp$ of $A$ is a closed linear subspace of $H$.
== Proof ==
Let $\sequence {x_n} \subset A^\perp$ be a convergent sequence.
Let $x$ be the limit of $\sequence {x_n}$.
Then, by the definition of orthocomplement:
:$\forall n \in \N, y \in A: \innerprod {x_n} y = 0$
Passing to the limit, from Inner Product is Continuous we have:
:$\forall y \in A: \innerprod x y = 0$
So:
:$x \in A^\perp$
This shows that $A^\perp$ is closed.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
Category:Inner Product Spaces
Category:Orthocomplements
Orthocomplement of Closure

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $S \subseteq V$ be non-empty.
Then:
:$S^\bot = \map \cl S^\bot$
where $\bot$ denotes orthocomplementation.
== Proof ==
From Orthocomplement Reverses Subset, we have:
:$\map \cl S^\bot \subseteq S^\bot$
Conversely, let $y \in S^\bot$.
Let $x \in \map \cl S$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that $x_n \to x$.
Since $y \in S^\bot$, we have $\innerprod {x_n} y = 0$ for each $n \in \N$.
From Inner Product is Continuous, we have $\innerprod x y = 0$ taking $n \to \infty$.
Since $x \in \map \cl S$ was arbitrary we have $y \in \map \cl S^\bot$.
Category:Orthocomplements
Orthocomplement of Subset of Orthocomplement is Superset

== Theorem ==
Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $A, B \subseteq V$ be subsets of $V$ such that $B \subseteq A^\perp$, where $A^\perp$ is the orthocomplement of $A$.
Then: 
:$A \subseteq B^\perp$.
== Proof ==
Let $B \subseteq A^\perp$.
Then by Orthocomplement Reverses Subset:
:$A^{\perp\perp} \subseteq B^\perp$
By Double Orthocomplement is Closed Linear Span and the definition of closed linear span:
:$A \subseteq A^{\perp\perp}$
Hence, by Subset Relation is Transitive:
:$A \subseteq B^\perp$
Category:Inner Product Spaces
Category:Orthocomplements
Orthogonal Difference is Closed Linear Subspace

== Theorem ==
Let $H$ be a Hilbert space.
Let $M, N$ be closed linear subspaces of $H$.
Then the orthogonal difference $M \ominus N$ is also a closed linear subspace of $H$.
== Proof ==
By definition, $M \ominus N = M \cap N^\perp$.
By Orthocomplement is Closed Linear Subspace, $N^\perp$ is a closed linear subspace of $H$.
Hence the result, by Closed Linear Subspaces Closed under Intersection.
Category:Hilbert Spaces
Orthogonal Group is Group

== Theorem ==
Let $k$ be a field.
The $n$th orthogonal group on $k$ is a group.
== Proof ==
A direct corollary of Orthogonal Group is Subgroup of General Linear Group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(iv) (b)}$
Category:Orthogonal Groups
Orthogonal Group is Subgroup of General Linear Group

== Theorem ==
Let $k$ be a field.
Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.
Let $\GL {n, k}$ be the $n$th general linear group on $k$.
Then $\map {\operatorname O} {n, k}$ is a subgroup of $\GL {n, k}$.
== Proof ==
From Unit Matrix is Orthogonal, the unit matrix $\mathbf I_n$ is orthogonal.
Let $\mathbf A, \mathbf B \in \map {\operatorname O} {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are orthogonal.
Then by Inverse of Orthogonal Matrix is Orthogonal:
:$\mathbf B^{-1}$ is a orthogonal matrix.
By Product of Orthogonal Matrices is Orthogonal Matrix:
:$\mathbf A \mathbf B^{-1}$ is a orthogonal matrix.
Thus by definition of orthogonal group:
:$\mathbf A \mathbf B^{-1} \in \map {\operatorname O} {n, k}$
Hence the result by One-Step Subgroup Test.
Category:Orthogonal Groups
Category:General Linear Group