nyuuzyou/wikis huggingface dataset:
Classical Runge-Kutta Method

== Proof Technique ==
Consider the first order ODE:
:$(1): \quad y' = \map f {x, y}$ subject to the initial condition $\map y {x_0} = y_0$
where $\map f {x, y}$ is continuous.
Let $\map y x$ be the particular solution of $(1)$.
For all $n \in \N_{>0}$, we define:
:$x_n = x_{n - 1} + h$
where $h \in \R_{>0}$.
Let the following numbers be calculated:
Then for all $n \in \N_{>0}$ such that $x_n$ is in the domain of $y$:
:$y_{n + 1} = y_n + \dfrac 1 6 \paren {m_1 + 2 m_2 + 2 m_3 + m_4}$
is an approximation to $\map y {x_{n + 1} }$.
== Proof ==
== Also known as ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods
* 
* 
Category:Runge-Kutta Methods
Category:Differential Equations
Category:Numerical Analysis
Classical Runge-Kutta Method/Also known as

== Classical Runge-Kutta Method: Also known as == 
The classical Runge-Kutta method method is also referred to as:
:the four-stage Runge-Kutta method
:RK4
but usually it is simply known as the Runge-Kutta method.
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods
* 
* 
Category:Runge-Kutta Methods
Classification of Bounded Sesquilinear Forms

== Theorem ==
Let $\mathbb F \in \set {\R, \C}$. 
Let $\HH$ a be Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\HH$.
Let $\KK$ a be Hilbert space over $\mathbb F$ with inner product ${\innerprod \cdot \cdot}_\KK$.
Let $u: \HH \times \KK \to \Bbb F$ be a bounded sesquilinear form with bound $M$.
Then there exist unique bounded linear transformations $A : \HH \to \KK$ and $B : \KK \to \HH$ such that:
:$\forall h \in \HH, k \in \KK: \map u {h, k} = \innerprod {A h} k_\KK = \innerprod h {B k}_\HH$
Furthermore:
:$\norm A \le M$
and:
:$\norm B \le M$
where $\norm \cdot$ denotes the norm of a bounded linear transformation. 
== Proof ==
== Also see ==
* Bounded Linear Transformation Induces Bounded Sesquilinear Form, which shows that considered inner products are in fact bounded sesquilinear forms.
* The notion of the adjoint of a bounded linear transformation, which relies on this theorem for being well-defined.
== Sources ==
* $\text {II}.2.2$
Category:Hilbert Spaces
Category:Linear Transformations on Hilbert Spaces
Classification of Compact One-Manifolds

== Theorem ==
Every compact connected one-dimensional manifold is diffeomorphic to either a circle or a closed interval.
=== Corollary ===
== Proof ==
=== Lemma 1 ===
Let $f$ be a Morse function on a one-manifold $X$.
Let $S$ be the union of the critical points of $f$ and $\partial X$.
As $S$ is finite, $X \setminus S$ consists of a finite number of one-manifolds, $L_1, L_2, \cdots, L_n$.
=== Lemma 2 ===
=== Lemma 3 ===
Category:Manifolds
Category:Classification of Compact One-Manifolds
Classification of Compact One-Manifolds/Corollary

== Corollary to Classification of Compact One-Manifolds ==
A compact one-manifold has an even number of points in its boundary.
== Proof ==
Let $M$ be a one-manifold.
That is, $M$ is a topological manifold of dimension $1$.
By Classification of Compact One-Manifolds:
Every compact connected one-dimensional manifold is diffeomorphic to either a circle or a closed interval.
It remains to be shown that a circle has an even number of points in its boundary.
It remains to be shown that a closed interval has an even number of points in its boundary.
By definition, the boundary of $M$ consists of all the points in the closure of $M$ which are not in the interior of $M$.
Thus, the boundary of $M$ is defined as:
:$\partial M := M^- \setminus M^\circ$
where $M^-$ denotes the closure and $M^\circ$ the interior of $M$.
Hence a circle has an even number of points in its boundary.
For the case of the real interval,let:
:$M = \closedint a b$
such that $a \ne b$, since it is a one-manifold.
By Closure of Real Interval is Closed Real Interval, it follows that the closure of $M$, $M^-$, is:
:$\closedint a b$.
By Interior of Closed Real Interval is Open Real Interval, it follows that the interior of $M$, $M^\circ$ is the open interval:
:$\openint a b$.
It follows that $a$, $b \in \partial M$.
If:
:$a \ne b$
then the number of points in its boundary is 2, which is an even number.
Hence a closed interval has an even number of points in its boundary.
Hence, a compact one-manifold has an even number of points in its boundary.
Category:Classification of Compact One-Manifolds
Classification of Compact One-Manifolds/Lemma 1

== Lemma for Classification of Compact One-Manifolds ==
Let $f$ be a function on $\closedint a b$ that is smooth and has a positive derivative everywhere except one interior point, $c$.
Then there exists a globally smooth function $g$ that agrees with $f$ near $a$ and $b$ and has a positive derivative everywhere.
== Proof ==
Let $r$ be a smooth nonnegative function that vanishes outside a compact subset of $\openint a b$, which equals $1$ near $c$, which satisfies:
:$\ds \int_a^b r = 1$
Define:
:$\ds \map g x = \map f a + \int_a^x \paren {k \map r s + \map {f'} s \paren {1 - \map r s} } \rd s$
where the constant:
:$\ds k = \map f b = - \map f a - \int_a^b \map {f'} s \paren {1 - \map r s} \d s$
Category:Classification of Compact One-Manifolds
Classification of Compact One-Manifolds/Lemma 2

== Lemma for Classification of Compact One-Manifolds ==
Let $f$ be a Morse function on a one-manifold $X$.
Let $S$ be the union of the critical points of $f$ and $\partial X$.
As $S$ is finite, $X - S$ consists of a finite number of one-manifolds, $L_1, L_2, \cdots, L_n$.
$f$ maps each $L_i$ diffeomorphically onto an open interval in $\R$.
== Proof ==
Let $L$ be any of the $L_i$.
Because $f$ is a local diffeomorphism and $L$ is connected, $f \sqbrk L$ is open and connected in $\R$.
We also have $f \sqbrk L \in f \sqbrk X$, the latter of which is compact.
Hence there are numbers $c$ and $d$ such that $f \sqbrk L = \openint c d$.
It suffices to show $f$ is one to one on $L$, because then $f^{-1}: \openint c d \to L$ is defined and locally smooth.
Let $p$ be any point of $L$.
Set $q = \map f p$.
It suffices to show that every other point $z \in L$ can be joined to $p$ by a curve $\gamma: \closedint q y \to L$ such that $f \circ \gamma$ is the identity and $\map \gamma y = z$.
Since $\map f z = y \ne q = \map f p$, this result shows $f$ is one to one.
So let $Q$ be the set of points $x$ that can be so joined.
Since $f$ is a local diffeomorphism, $Q$ is both open and Definition:Closed Set (Topology).
Hence $Q = L$.
Category:Classification of Compact One-Manifolds
Classification of Compact One-Manifolds/Lemma 3

== Lemma for Classification of Compact One-Manifolds ==
Let $L$ be a subset of $X$ diffeomorphic to an open interval in $\R$, where $\dim X = 1$.
Then the closure $\map \cl L$ contains at most two points not in $L$.
== Proof ==
Let $g$ be a diffeomorphism:
:$g: \openint a b \to L$
Let $p \in \map \cl L \setminus L$.
Let $J$ be a closed subset of $X$ diffeomorphic to $\closedint 0 1$ such that:
:$1$ corresponds to $p$
:$0$ corresponds to some $\map g t$ in $L$.
Consider the set $\set {s \in \openint a t: \map g s \in J}$.
This set is both open and closed in $\openint a b$.
Hence $J$ contains either $g \sqbrk {\openint a t}$ or $g \sqbrk {\openint t b}$.
Category:Classification of Compact One-Manifolds
Classification of Compact Three-Manifolds Supporting Zero-Curvature Geometry

== Theorem ==
Every closed, orientable, path connected $3$-dimensional Riemannian manifold which supports a geometry of zero curvature is homeomorphic to one of the following:
* Torus $\mathbb T^3$
* Half-Twist Cube
* Quarter-Twist Cube
* Hantschze-Wendt Manifold
* $\frac 1 6$-Twist Hexagonal Prism
* $\frac 1 3$-Twist Hexagonal Prism
The $3$-torus is described on the torus page.
The other manifolds can be described using quotient spaces on familiar prisms, with the equivalence relations described below.
File:Halftwistcube.JPG|The Half-Twist Cube
File:Hantschzewendt.JPG|The Hantschze-Wendt Manifold
File:Quartertwistcube.JPG|The Quarter-Twist Cube
File:Sixthtwisthexagon.JPG|The $\frac 1 6$-Twist Hexagonal Prism
File:Thirdtwisthexagon.JPG|The $\frac 1 3$-Twist Hexagonal Prism
== Proof ==
Category:Manifolds
Classification of Compact Two-Manifolds/Lemma

== Lemma for Classification of Compact Two-Manifolds ==
A compact, boundaryless $2$-manifold $S$ is diffeomorphic to a polyhedral disk $P$ with edges identified pairwise.
That is, for any closed, connected $2$-manifold, there exists a polyhedral disk $P$ and an equivalence relation $\sim$ such that:
:$S \cong P \setminus \sim$
== Proof ==
Category:Classification of Compact Two-Manifolds
Classification of Conic Sections by Coefficients of General Equation

== Theorem ==
Let $K$ be a conic section embedded in a Cartesian plane with the general equation:
:$a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0$
where $a, b, c, f, g, h \in \R$.
Let $\Delta$ denote the discriminant of $K$:
:$\Delta = \begin {vmatrix} a & h & g \\ h & b & f \\ g & f & c \end {vmatrix}$
If $\Delta \ne 0$, then $K$ is an ellipse, a parabola or a hyperbola, according to:
If $\Delta = 0$, then $K$ is a degenerate conic:
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:Conic Sections
Classification of Conic Sections by Coefficients of General Equation/Historical Note

== Historical Note on Classification of Conic Sections by Coefficients of General Equation ==
In , introduced the idea that:
:pairs of intersecting straight lines
:ellipses
:parabolas
:hyperbolas
are all in fact instances of the various types of conic section, depending on the relative positions of the foci.
In particular, he considered the parabola as a conic section with one focus at infinity.
== Sources ==
* 
* 
Category:Historical Notes
Classification of Convex Polyhedra whose Faces are Regular Polygons

== Theorem ==
The convex polyhedra whose faces are all regular polygons are as follows:
:The $5$ Platonic solids
:The uniform prisms and uniform antiprisms, countably infinite in number
:The $13$ Archimedean polyhedra
:The $92$ Johnson polyhedra.
== Proof ==
The Platonic solids are the convex polyhedra all of whose faces are congruent and whose vertices are regular.
From Five Platonic Solids, there are $5$ of these.
The uniform prisms are made from two regular polygons of an arbitrary number of sides forming the bases, separated by as many squares as there are sides of the two bases, forming the lateral faces.
From Uniform Prisms are Countably Infinite, they form a countably infinite set.
Note that when the bases are themselves squares, the uniform prism is then a cube, and so has already been counted amongst the Platonic solids.
The uniform antiprisms are made from two regular polygons of an arbitrary number of sides forming the bases, separated by twice as many equilateral triangles as there are sides of the two bases, forming the lateral faces.
From Uniform Antiprisms are Countably Infinite, they form a countably infinite set.
Note that when the bases are themselves equilateral triangles, the uniform antiprism is then a regular octahedron, and so has already been counted amongst the Platonic solids.
The Archimedean polyhedra consist of all isogonal convex polyhedra whose faces are all regular polygons, excluding:
:the Platonic solids
:the uniform prisms
:the uniform antiprisms.
From Thirteen Archimedean Polyhedra, there are $13$ of these.
The Johnson polyhedra are all the other convex polyhedra whose faces are all regular polygons.
From 92 Johnson Polyhedra, there are $92$ of these.
== Historical Note ==
== Sources ==
* 
* : $13$
* : $13$
Category:Convex Polyhedra
Classification of Convex Polyhedra whose Faces are Regular Polygons/Historical Note

== Historical Note on Classification of Convex Polyhedra whose Faces are Regular Polygons ==
The final step of this result was demonstrated by , who first enumerated what are now known as the Johnson polyhedra in $1966$.
He conjectured, but did not prove, that there were no more than $92$ of them.
In $1969$ proved that 's list was complete.
== Sources ==
* 
* : $13$
* : $13$
Category:Historical Notes
Classification of Finite Simple Groups

== Theorem ==
The finite simple groups can be classified as follows:
=== Abelian Simple Groups ===
The Abelian simple groups are the cyclic groups of prime order.
=== Alternating Groups ===
The alternating groups $\map A n$ for $n \ge 5$.
=== Groups of Lie Type ===
The various families of groups of Lie type.
=== Sporadic Groups ===
== Historical Note ==
== Sources ==
* 
* 
Category:Finite Groups
Category:Simple Groups
Classification of Groups of Order up to 15

== Theorem ==
Up to isomorphism, every group of order $\order G \le 15$ is one of the below:
Order	Abelian 	Non-Abelian
1 	$\Z_1$ 	
2 	$\Z_2$ 	
3 	$\Z_3$ 	
4 	$\Z_4, \Z_2 \oplus \Z_2$ 	
5 	$\Z_5$ 	
6 	$\Z_6$ 	$D_3 \cong S_3$
7 	$\Z_7$ 	
8 	$\Z_8, \Z_4 \oplus \Z_2, \Z_2 \oplus \Z_2 \oplus \Z_2$	$D_4, \Dic 2$
9 	$\Z_9, \Z_3 \oplus \Z_3$ 	
10 	$\Z_{10}$ 	$D_5$
11 	$\Z_{11}$ 	
12 	$\Z_{12}, \Z_6 \oplus \Z_2$ 	$D_6, A_4, \Dic 3$
13 	$\Z_{13}$ 	
14 	$\Z_{14}$ 	$D_7$
15 	$\Z_{15}$ 	
where:
: $D_n$ is the dihedral group of order $2 n$
: $S_n$ is the $n$th symmetric group
: $A_n$ is the alternating group on $n$ points
: $\Dic n$ is the dicyclic group of order $4 n$.
== Proof ==
The Abelian cases are the direct result of the Fundamental Theorem of Finite Abelian Groups.
The non-Abelian cases follow from seven separate theorems:
:$(1): \quad$ Trivial Group is Cyclic Group - determines theorem for order $1$
:$(2): \quad$ Prime Group is Cyclic - determines theorem for orders $2$, $3$, $5$, $7$, $11$, and $13$
:$(3): \quad$ Group of Order Prime Squared is Abelian - determines theorem for orders $4$ and $9$
:$(4): \quad$ Group of Order p q is Cyclic - determines theorem for order $15$
:$(5): \quad$ Groups of Order Twice a Prime - determines theorem for orders $6$, $10$, $14$
:$(6): \quad$ Groups of Order 8 - determines theorem for order $8$
:$(7): \quad$ Groups of Order 12 - determines theorem for order $12$
== Sources ==
* : Appendix $17$: Groups of order $15$ or less
Category:Order of Groups
Classification of Irrational Straight Lines derived from Apotome

== Theorem ==
== Proof ==
From :
:the square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.
From :
:the square on an apotome applied to a rational straight line produces as breadth a first apotome.
From :
:the square on a first apotome of a medial straight line applied to a rational straight line produces as breadth a second apotome.
From :
:the square on a second apotome of a medial straight line applied to a rational straight line produces as breadth a third apotome.
From :
:the square on a minor straight line applied to a rational straight line produces as breadth a fourth apotome.
From :
:the square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.
From :
:the square on the straight line which produces with a medial area a medial whole, if applied to a rational straight line, produces as breadth a sixth apotome.
All of these breadths so produced differ from the first and from each other:
:from the first because it is rational
and:
:from each other because they are different in order.
Thus it follows that the irrational straight lines themselves are different from one another.
From :
:an apotome is not the same as a binomial straight line.
Combining the above analysis with Classification of Irrational Straight Lines derived from Binomial Straight Line, it can be seen that the following $13$ types of irrational straight line are all different in nature:
:Medial
:Binomial
:First bimedial
:Second bimedial
:Major
:"Side" of a rational plus medial area
:"Side" of the sum of two medial areas
:Apotome
:First apotome of a medial straight line
:Second apotome of a medial straight line
:Minor
:Straight line which produces with a rational area a medial whole
:Straight line which produces with a medial area a medial whole.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Classification of Irrational Straight Lines derived from Binomial Straight Line

== Theorem ==
== Proof ==
From :
:the square on a medial, applied to a rational straight line, produces as breadth a rational straight line which is incommensurable in length with that to which it is applied.
From :
:the square on a binomial, applied to a rational straight line, produces as breadth the first binomial.
From :
:the square on a first bimedial, applied to a rational straight line, produces as breadth the second binomial.
From :
:the square on a second bimedial, applied to a rational straight line, produces as breadth the third binomial.
From :
:the square on the major, applied to a rational straight line, produces as breadth the fourth binomial.
From :
:the square on the side of a rational plus medial area, applied to a rational straight line, produces as breadth the fifth binomial.
From :
:the square on the side of the sum of two medial areas, applied to a rational straight line, produces as breadth the sixth binomial.
All of these breadths so produced differ from the first and from each other:
:from the first because it is rational
and:
:from each other because they are different in order.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Classification of Open Neighborhoods in Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $U \subseteq X$.
Let $x \in X$. 
Then $U$ is an open neighborhood of $x$ $U - x$ is an open neighborhood of ${\mathbf 0}_X$.
Equivalently, the open neighborhoods of $x$ are precisely the sets of the form $V + x$ where $V$ is an open neighborhood of ${\mathbf 0}_X$. 
== Proof ==
=== Necessary Condition ===
Suppose that $U$ is an open neighborhood of $x$.
Then ${\mathbf 0}_X \in U - x$ and from Translation of Open Set in Topological Vector Space is Open, $U - x$ is open.
So $U - x$ is an open neighborhood of ${\mathbf 0}_X$.
=== Sufficient Condition ===
Suppose that $U - x$ is an open neighborhood of ${\mathbf 0}_X$.
Note that $U = \paren {U - x} + x$.
So, we have $x \in U$ and from Translation of Open Set in Topological Vector Space is Open, $U$ is open.
So $U$ is an open neighborhood of $x$.
Category:Topological Vector Spaces
Classification of Stationary Points/Function of One Variable

== Theorem ==
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.
Let $P$ be a stationary point of $f$.
Then $P$ is either:
:a local maximum
:a local minimum
:a point of inflection.
== Proof ==
== Sources ==
* 
* 
Category:Classification of Stationary Points
Classification of Stationary Points/Function of Two Variables

== Theorem ==
Let $\SS$ be a surface defined by the Cartesian equation $z = \map f {x, y}$.
Let $P$ be a stationary point on $\SS$.
Then $P$ is either:
:a local maximum
:a local minimum
:a saddle point.
== Proof ==
== Sources ==
* 
* 
Category:Classification of Stationary Points
Clavius's Law

== Theorem ==
If, from the negation of a proposition $p$ we can derive $p$, we may conclude $p$:
=== Formulation 1 ===
=== Formulation 2 ===
== Also known as ==
== Also see ==
* Reductio ad Absurdum, of which this is a particular case.
Category:Clavius's Law
Category:Conditional
Category:Reductio ad Absurdum
Clavius's Law/Formulation 1

== Theorem ==
:$\neg p \implies p \vdash p$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Exercise $1 \ \text{(j)}$
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(a)}$
Category:Clavius's Law
Clavius's Law/Formulation 2

== Theorem ==
:$\vdash \left({\neg p \implies p}\right) \implies p$
== Proof 1==
== Proof 2==
== Proof by Truth Table==
== Sources ==
* : $\S \text{II}$: Exercise $11 \ \text{(a)}$
* : Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Exercise $1 \ \text{(e)}$
* 
Category:Clavius's Law
Clavius's Law/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations.
:$\begin{array}{|cccc|c|c|} \hline
(\neg & p & \implies & p) & \implies & p \\
\hline
\T & \F & \F & \F & \T & \F \\
\F & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S \text{II}$: Exercise $13$
Category:Clavius's Law
Category:Truth Table Proofs
Clear Registers Program

== URM Program ==
Let $a, b \in \N$ be natural numbers such that $0 < a$.
Then we define the URM program $\map Z {a, b}$ to be:
Line 		Command
$1$ 		$\map Z a$
$2$ 		$\map Z {a + 1}$
$3$ 		$\map Z {a + 2}$
$\vdots$ 		$\vdots$
$b - a + 1$		$\map Z b$
where $\map Z a$ denotes the zero instruction:
:$r_a \gets 0$
Hence this URM program zeroizes all the registers from $R_a$ through to $R_b$.
If $a > b$ then $\map Z {a, b}$ is the null URM program.
The length of $\map Z {a, b}$ is:
:$\map \lambda {\map Z {a, b} } = \begin {cases}
0 & : a > b \\
b - a + 1 & : a \le b
\end {cases}$
== Proof ==
Self-evident.
Category:URM Programs
Clopen Points in Arens-Fort Space

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Let $q \in S: q \ne \tuple {0, 0}$.
Then $\set q$ is both open and closed in $T$.
$\set {\tuple {0, 0} }$ itself is closed, but not open.
== Proof ==
We have that $\set q$ is finite so $\relcomp S {\set q}$ is cofinite.
So $\relcomp S {\set q}$ is open and so $\set q$ is closed.
Then we have that $\tuple {0, 0} \notin \set q$ so $\set q$ is open.
However, $\tuple {0, 0} \notin \relcomp S {\set {\tuple {0, 0} } }$ and $\relcomp S {\set {\tuple {0, 0} } }$ is clearly not open in $T$.
But as $\set {\tuple {0, 0} }$ is finite so $\relcomp S {\set {\tuple {0, 0} } }$ is cofinite.
Hence, as for $\set q$, we have that $\set {\tuple {0, 0} }$ is closed.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $7$
Category:Clopen Sets
Category:Arens-Fort Space
Clopen Points in Fort Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an infinite set $S$.
Let $q \in S: q \ne p$.
Then $\set q$ is both open and closed in $T$.
$\set p$ itself, on the other hand, is closed but not open.
== Proof ==
We have that $\set q$ is finite so $\relcomp S {\set q}$ is cofinite.
So $\relcomp S {\set q}$ is open and so $\set q$ is closed.
Then we have that $p \notin \set q$ so $\set q$ is open.
However, $p \notin \relcomp S {\set p}$ and $\relcomp S {\set p}$ is infinite.
So $\set p$ is not open in $S$.
But $\set p$ is finite, so $\relcomp S {\set p}$ is cofinite.
Hence, as for $\set q$, we have that $\set p$ is closed.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $7$
Category:Clopen Sets
Category:Fort Spaces
Clopen Points in Modified Fort Space

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then all points in $S \setminus \set {a, b}$ are both open and closed in $T$.
$a$ and $b$ themselves are not open in $T$, but they are closed in $T$.
== Proof ==
Let $p \in S: p \notin \set {a, b}$.
From the definition of modified Fort space, any subset of $S \setminus \set {a, b}$ is open in $T$.
It follows directly that as $p \in S \setminus \set {a, b}$ we have that $\set p \subseteq S \setminus \set {a, b}$.
Hence $p$ is open in $T$.
As for $a$ and $b$, we have that $S \setminus \set a$ and $S \setminus \set b$ are not finite and so $a$ and $b$ are not open in $T$.
For all points $p \in S$ (including $a$ and $b$), we have that $\set p$ is (trivially) finite.
It follows that while $S \setminus \set p$ contains either $a$ or $b$, it is cofinite.
Thus $S \setminus \set p$ is open in $T$.
It follows by definition that $p$ is closed in $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $2$
Category:Modified Fort Space