nyuuzyou/wikis huggingface dataset:
External Direct Product of Congruence Relations

== Theorem ==
Let $\struct {A, \odot}$ and $\struct {B, \circledast}$ be algebraic structures.
Let $\RR$ and $\SS$ be congruence relations on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.
Let $\struct {A / \RR, \odot_\RR}$ and $\struct {B / \SS, \circledast_\SS}$ denote the quotient structures defined by $\RR$ and $\SS$ respectively.
Let $\struct {A \times B, \otimes}$ be the external direct product of $\struct {A, \odot}$ and $\struct {B, \circledast}$.
Let $\TT$ be the relation on $\struct {A \times B, \otimes}$ defined as:
:$\forall \tuple {u, v}, \tuple {x, y} \in A \times B: \tuple {u, v} \mathrel \TT \tuple {x, y} \iff u \mathrel \RR x \land v \mathrel \SS y$
Then:
:$\TT$ is a congruence relation on $\struct {A \times B, \otimes}$
and the mapping $h: \struct {A / \RR, \odot_\RR} \times \struct {B / \SS, \circledast_\SS} \to \struct {\paren {A \times B} / \TT, \otimes_\TT}$ defined as:
:$\forall \tuple {\eqclass x \RR, \eqclass y \SS} \in \struct {A / \RR, \odot_\RR} \times \struct {B / \SS, \circledast_\SS}: \map h {\eqclass x \RR, \eqclass y \SS} = \eqclass {\tuple {x, y} } \TT$
is an isomorphism.
== Proof ==
We have that $\RR$ and $\SS$ are congruence relations on $\struct {A, \odot}$ and $\struct {B, \circledast}$ respectively.
Hence $\RR$ and $\SS$ are equivalence relations on $A$ and $B$ respectively.
Thus from Cartesian Product of Equivalence Relations we have that $\TT$ is an equivalence relation.
Let $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {c_1, d_1}, \tuple {c_2, d_2} \in A \times B$ be such that:
Then:
Hence by definition $\TT$ is a congruence relation on $\struct {A \times B, \otimes}$.
It remains to be shown that $h$ as defined is an isomorphism.
From Cartesian Product of Equivalence Relations, the equivalence class of $\tuple {a, b} \in A \times B$ is:
:$\eqclass {\tuple {a, b} } \TT = \set {\tuple {x, y}: x \in \eqclass a \RR, y \in \eqclass b \SS}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.14 \ \text{(b)}$
Category:Congruence Relations
Category:External Direct Products
Category:Quotient Mappings
External Direct Product of Groups is Group

== Theorem ==
Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups whose identity elements are $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the external direct product of $G_1$ and $G_2$.
Then $\struct {G_1 \times G_2, \circ}$ is a group whose identity element is $\tuple {e_1, e_2}$.
=== Finite Product ===
== Proof ==
Taking the group axioms in turn:
=== ===
From External Direct Product Closure it follows that $\struct {G_1 \times G_2, \circ}$ is closed.
=== ===
From External Direct Product Associativity it follows that $\circ$ is associative on $G_1 \times G_2$.
=== ===
From External Direct Product Identity it follows that $\tuple {e_1, e_2}$ is the identity element of $\struct {G_1 \times G_2, \circ}$.
=== ===
From External Direct Product Inverses it follows that $\tuple {g_1^{-1}, g_2^{-1} }$ is the inverse element of $\tuple {g_1, g_2}$ in $\struct {G_1 \times G_2, \circ}$.
All group axioms are fulfilled, and so $\struct {G_1 \times G_2, \circ}$ is a group whose identity element is $\tuple {e_1, e_2}$.
== Also see ==
* External Direct Product of Abelian Groups is Abelian Group
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.1$: Corollary
* : Introduction: Notation
* : Chapter $2$: The Definition of Group Structure: $\S 26 \zeta$
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
* : Chapter $6$: An Introduction to Groups: Exercise $1$
* : Chapter $1$: Definitions and Examples: Example $1.10$
Category:External Direct Products
Category:Group Direct Products
External Direct Product of Groups is Group/Finite Product

== Theorem ==
The external direct product of a finite sequence of groups is itself a group.
== Proof ==
Let $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ be groups.
Let $\ds \struct {G, \circ} = \prod_{k \mathop = 1}^n G_k$ be the external direct product of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$.
Taking the group axioms in turn:
=== ===
From External Direct Product Closure: General Result it follows that $\struct {G, \circ}$ is closed.
=== ===
From External Direct Product Associativity: General Result it follows that $\struct {G, \circ}$ is associative.
=== ===
Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ respectively.
From External Direct Product Identity: General Result it follows that $\tuple {e_1, e_2, \ldots, e_n}$ is the identity element of $\struct {G, \circ}$.
=== ===
Let $g_1, g_2, \ldots, g_n$ be arbitrary elements of $G_1, G_2, \ldots, G_n$
Let $g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1}$ be the inverse elements of $g_1, g_2, \ldots, g_n$ in $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ respectively.
From External Direct Product Inverses: General Result it follows that $\tuple {g_1^{-1}, g_2^{-1}, \ldots, g_n^{-1} }$ is the inverse element of $\tuple {g_1, g_2, \ldots, g_n}$ in $\struct {G, \circ}$.
All group axioms are fulfilled, hence $\struct {G, \circ}$ is a group.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.10$: Corollary
Category:Group Theory
Category:Group Direct Products
External Direct Product of Projection with Canonical Injection

== Theorem ==
Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {S_1 \times S_2, \circ}$ be the external direct product of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
Let:
:$\pr_1: \struct {S_1 \times S_2, \circ} \to \struct {S_1, \circ_1}$ be the first projection from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_1, \circ_1}$
:$\pr_2: \struct {S_1 \times S_2, \circ} \to \struct {S_2, \circ_2}$ be the second projection from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_2, \circ_2}$.
Let:
:$\inj_1: \struct {S_1, \circ_1} \to \struct {S_1 \times S_2, \circ}$ be the canonical injection from $\struct {S_1, \circ_1}$ to $\struct {S_1 \times S_2, \circ}$
:$\inj_2: \struct {S_2, \circ_2} \to \struct {S_1 \times S_2, \circ}$ be the canonical injection from $\struct {S_2, \circ_2}$ to $\struct {S_1 \times S_2, \circ}$.
Then:
:$(1): \quad \pr_1 \circ \inj_1 = I_{S_1}$
:$(2): \quad \pr_2 \circ \inj_2 = I_{S_2}$
where $I_{S_1}$ and $I_{S_2}$ are the identity mappings on $S_1$ and $S_2$ respectively.
=== General Result ===
== Proof ==
Let $\tuple {s_1, s_2} \in S_1 \times S_2$.
So, $s_1 \in S_1$ and $s_2 \in S_2$.
From the definition of the canonical injection, we have:
:$\map {\inj_1} {s_1} = \tuple {s_1, e_2}$
:$\map {\inj_2} {s_2} = \tuple {e_1, s_2}$
From the definition of the first projection:
:$\map {\pr_1} {s_1, e_2} = s_1$
and similarly from the definition of the second projection:
:$\map {\pr_2} {e_1, s_2} = s_2$
Thus:
:$\map {\pr_1 \circ \inj_1} {s_1} = s_1$
:$\map {\pr_2 \circ \inj_2} {s_2} = s_2$
and the result follows from the definition of the identity mapping.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.3$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.3$
Category:Canonical Injections
Category:External Direct Products
Category:Projections
External Direct Product of Projection with Canonical Injection/General Result

== Theorem ==
Let $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$ be algebraic structures with identities $e_1, e_2, \dotsc, e_j, \dotsc, e_n$ respectively.
Let $\ds \struct {S, \circ} = \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$ be the external direct product of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \dotsc, \struct {S_j, \circ_j}, \dotsc, \struct {S_n, \circ_n}$.
Let $\pr_j: \struct {S, \circ} \to \struct {S_j, \circ_j}$ be the $j$th projection from $\struct {S, \circ}$ to $\struct {S_j, \circ_j}$.
Let $\inj_j: \struct {S_j, \circ_j} \to \struct {S, \circ}$ be the canonical injection from $\struct {S_j, \circ_j}$ to $\struct {S, \circ}$.
Then:
:$\pr_j \circ \inj_j = I_{S_j}$
where $I_{S_j}$ is the identity mapping from $S_j$ to $S_j$.
== Proof ==
Let $\ds \tuple {s_1, s_2, \dotsc, s_{j - 1}, s_j, s_{j + 1}, \dotsc, s_n} \in \prod_{i \mathop = 1}^n \struct {S_i, \circ_i}$.
So:
:$s_j \in S_j$
From the definition of the canonical injection:
:$\map {\inj_j} {s_j} = \tuple {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n}$
So from the definition of the $j$th projection:
:$\map {\pr_j} {e_1, e_2, \dotsc, e_{j - 1}, s_j, e_{j + 1}, \dotsc, e_n} = s_j$
Thus:
:$\map {\pr_j \circ \inj_j} {s_j} = s_j$
and the result follows from the definition of the identity mapping.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.12$
Category:External Direct Products
Category:Canonical Injections
External Direct Product of Ringoids is Ringoid

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be ringoids.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their external direct product.
Then $\struct {R, +, \circ}$ is a ringoid.
== Proof ==
By definition of ringoid, $\circ_k$ distributes over $+_k$ for all $k = 1, 2, \ldots, n$.
Let $x, y, z \in R$.
Then:
In the same way:
:$\paren {y + z} \circ x = \paren {y \circ x} + \paren {z \circ x}$
Thus $\circ$ is distributive over $+$.
By definition of ringoid, $\struct {R, +, \circ}$ is a ringoid.
Category:External Direct Products
Category:Distributive Operations
External Direct Product of Semigroups

== Theorem ==
The external direct product of two semigroups is itself a semigroup.
== Proof ==
* From External Direct Product Closure, the external direct product of two closed algebraic structures is closed.
* From External Direct Product Associativity, the external direct product of two associative algebraic structures is associative.
The result follows.
Category:External Direct Products
Category:Semigroups
External Direct Sum of Rings is Ring

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Then their (external) direct product:
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
is a ring.
== Proof ==
Consider the structures $\struct {R_1, +_1}, \struct {R_2, +_2}, \ldots, \struct {R_n, +_n}$.
By the definition of a ring, these are all groups.
From External Direct Product of Groups is Group we have that the their external direct product:
:$\ds \struct {R, +} = \prod_{k \mathop = 1}^n \struct {R_k, +_k}$
is a group.
Similarly, consider the structures $\struct {R_1, \circ_1}, \struct {R_2, \circ_2}, \ldots, \struct {R_n, \circ_n}$.
By the definition of a ring, these are all semigroups.
From External Direct Product of Semigroups we have that the their external direct product:
:$\ds \struct {R, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, \circ_k}$
is a semigroup.
Finally we note that from External Direct Product of Ringoids is Ringoid, $\circ$ as defined here is distributive over $+$.
Hence the result, by definition of ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.8$
* : $\S 3.1$: Direct sums
Category:Direct Sums of Rings
Category:Ring Theory
Extremal Length of Composition

== Theorem ==
Let $\Gamma_1$ and $\Gamma_2$ be families of (unions of) rectifiable curves on a Riemann surface $X$.
Let $\Gamma_1$ and $\Gamma_2$ be disjoint in the sense that there exist disjoint Borel sets $E_1 \subseteq X$ and $E_2 \subseteq X$ with $\bigcup \Gamma_1 \subset E_1$ and $\bigcup \Gamma_2 \subset E_2$.
The extremal length of the family:
:$\Gamma := \set {\gamma_1 \cup \gamma_2:\ \gamma_1 \in \Gamma_1 \text{ and }\gamma_2 \in \Gamma_2}$
satisfies:
:$\map \lambda \Gamma = \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
== Proof ==
By the Series Law for Extremal Length:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Hence it remains only to prove the opposite inequality.
Let $\rho$ be a metric as in the definition of extremal length, normalized such that $\map A \rho = 1$.
The claim to be proved is that:
:$\paren {\map L {\Gamma, \rho} }^2 \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
 
Define $\alpha_j := \sqrt {\map A {\rho {\restriction_{E_j} } } }$ for $j = 1, 2$.
Suppose either $\map \lambda {\Gamma_1}$ or $\map \lambda {\Gamma_2}$ is infinite.
It trivially follows that both $\alpha_1$ and $\alpha_2$ are positive.
Assume both extremal lengths are finite.
Let $\alpha_j = 0$.
Then:
:$\map L {\Gamma_j, \rho} = 0$
Indeed, we can choose a metric $\tilde \rho$ that agrees with $\rho$ on $A_j$ and has total area $\epsilon > 0$.
Then:
:$\paren {\map L {\Gamma_j, \rho} }^2 = \paren {\map L {\Gamma_j, \tilde \rho} }^2 \le \epsilon \cdot \map \lambda {\Gamma_j}$
Because $\epsilon$ was arbitrary, the claim follows.
, suppose $\alpha_2 = 0$.
Then:
:$\paren {\map L {\Gamma_j, \rho} }^2 = \paren {\map L {\Gamma_1, \rho} + \map L {\Gamma_2, \rho} }^2 = \paren {\map L {\Gamma_j, \rho} }^2 \le \map \lambda {\Gamma_1} \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
Thus both $\alpha_1$ and $\alpha_2$ are positive.
Let:
:$\rho_j := \dfrac {\rho {\restriction_{E_j} } } {\alpha_j}$
Then $\map A \rho = 1$.
So:
In conclusion:
:$\map \lambda \Gamma = \sup_\rho \map L {\Gamma, \rho} \le \map \lambda {\Gamma_1} + \map \lambda {\Gamma_2}$
as claimed.
Category:Functional Analysis
Extremal Length of Union

== Theorem ==
Let $X$ be a Riemann surface.
Let $\Gamma_1$ and $\Gamma_2$ be families of rectifiable curves (or, more generally, families of unions of rectifiable curves) on $X$. 
Then the extremal length of their union satisfies:
:$\dfrac 1 {\map \lambda {\Gamma_1 \cup \Gamma_2} } \le \dfrac 1 {\map \lambda {\Gamma_1} } + \dfrac 1 {\map \lambda {\Gamma_2} }$
Suppose that additionally $\Gamma_1$ and $\Gamma_2$ are disjoint in the following sense: there exist disjoint Borel subsets:
:$A_1, A_2 \subseteq X$ such that $\ds \bigcup \Gamma_1 \subset A_1$ and $\ds \bigcup \Gamma_2 \subset A_2$
Then
:$\dfrac 1 {\map \lambda {\Gamma_1 \cup \Gamma_2} } = \dfrac 1 {\map \lambda {\Gamma_1} } + \dfrac 1 {\map \lambda {\Gamma_2} }$
== Proof ==
Set $\Gamma := \Gamma_1\cup \Gamma_2$.
Let $\rho_1$ and $\rho_2$ be conformal metrics as in the definition of extremal length, normalized such that:
:$\map L {\Gamma_1, \rho_1} = \map L {\Gamma_2, \rho_2} = 1$
We define a new metric by:
:$\rho := \map \max {\rho_1, \rho_2}$
Then:
:$\map L {\Gamma, \rho} \ge 1$
and:
:$\map A \rho \le \map A {\rho_1} + \map A {\rho_2}$
Hence:
Taking the infimum over all metrics $\rho_1$ and $\rho_2$, the claim follows.
Now suppose that the disjointness assumption holds, and let $\rho$ again be a Borel-measurable conformal metric, normalized such that $\map L {\Gamma, \rho} = 1$. 
We can define $\rho_1$ to be the restriction of $\rho$ to $A_1$, and likewise $\rho_2$ to be the restriction of $\rho$ to $A_2$.
By this we mean that, in local coordinates, $\rho_j$ is given by
:$\map {\rho_j} z \size {\d z} = \begin {cases}
\map \rho z \size {\d z} & : z \in A_j \\
0 \size {\d z} & : \text {otherwise}
\end {cases}$
Then:
:$\map A \rho = \map A {\rho_1} + \map A {\rho_2}$
and:
:$\map L {\Gamma_1, \rho_1}, \map L {\Gamma_2, \rho_2} \ge 1$
Hence:
Taking the infimum over all metrics $\rho$, we see that:
:$\dfrac 1 {\map \lambda {\Gamma_1 \cup \Gamma_2} } \ge \dfrac 1 {\map \lambda {\Gamma_1} } + \dfrac 1 {\map \lambda {\Gamma_2} }$
Together with the first part of the Proposition, this proves the claim.
Category:Geometric Function Theory
Extremal of Functional/Examples/Brachistochrone

== Examples of Extremals of Functionals ==
The brachistochrone is the extremal for the functional:
:$\map \phi f = \ds \int_a^b \sqrt {\paren {\dfrac {1 + \paren {\map {f'} x}^2} {2 g \map f x} } } \rd x$
where $g$ denotes acceleration due to gravity.
== Proof ==
== Sources ==
* 
Category:Examples of Extremals of Functionals
Extremally Disconnected Metric Space is Discrete

== Theorem ==
Let $M = \struct {A, d}$ be a metric space which is extremally disconnected.
Then $M$ is the discrete topology.
== Proof ==
Let $M = \struct {A, d}$ be extremally disconnected.
Let $p \in A$.
As $M$ is a metric space, $\set p$ can be expressed as:
:$\set p = \ds \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$
where:
:$\map {B_{1 / n} } p$ denotes the open $1 / n$-ball of $p$
:$\paren {\map {B_{1 / n} } p}^-$ denotes the closure of $\map {B_{1 / n} } p$
That is, as the intersection of the closures of the open $1 / n$-ball of $p$ for all non-zero natural numbers.
Now let:
:$\ds U = \bigcup_{n \mathop \in \N_{>0} } \map {B_{1 / 2 n} } p \setminus \paren {\map {B_{1 / \paren {2 n + 1} } } p}^-$
Then either $U$ or the complementary set of annuli is an open set which has $p$ as a non-interior limit point provided $\set p$ is not open.
So if $M$ is not the discrete topology, it cannot be extremally disconnected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Extremally Disconnected Spaces
Category:Discrete Topology
Extremally Disconnected Space is Totally Separated

== Theorem ==
Let $T = \struct {S, \tau}$ be an extremally disconnected topological space.
Then $T$ is totally separated.
== Proof ==
Let $T = \struct {S, \tau}$ be an extremally disconnected topological space.
Then by definition $T$ is a $T_2$ (Hausdorff) space such that the closure of every open set of $T$ is open.
Let $x, y \in S: x \ne y$.
As $T$ is a $T_2$ (Hausdorff) space, there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
Let $U^-$ denote the closure of $U$.
From Topological Closure is Closed, $U^-$ is closed.
But $U$ is open, so from the definition of extremally disconnected space, then $U^-$ is also open.
So $U^-$ is clopen.
From Open Set Disjoint from Set is Disjoint from Closure, $U^- \cap V = \O$, so $y \notin U^-$.
Thus $\set {U^- \mid S \setminus U^-}$ is a partition where $x \in U^-$ and $y \in X \setminus U^-$.
Thus, by definition, $T$ is totally separated.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Totally Separated Spaces
Category:Extremally Disconnected Spaces
Extremally Disconnected by Disjoint Open Sets

== Theorem ==
=== Definition using Closures of Open Sets ===
=== Definition using Interiors of Closed Sets ===
=== Definition using Disjoint Open Sets ===
== Proof ==
=== Definition $1$ Definition $2$ ===
Follows directly from Complement of Interior equals Closure of Complement.
=== Definition $1$ implies Definition $3$ ===
Let $A, B \subseteq S$ be disjoint open sets.
Then:
As $A, B$ are arbitrary:
:the closures of every pair of open sets which are disjoint are also disjoint.
=== Definition $3$ implies Definition $1$ ===
Let $A \subseteq S$ be an open set.
By Topological Closure is Closed, $A^-$ is closed.
Hence $\relcomp S {A^-}$ is open.
We have:
By Set is Closed iff Equals Topological Closure, $\relcomp S {A^-}$ is closed.
Thus $\relcomp S {\relcomp S {A^-} } = A^-$ is open.
As $A$ is arbitrary:
:the closure of every open set is open.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Equivalence of Definitions of Extremally Disconnected Space
Extremally Disconnected by Interior of Closed Sets

== Theorem ==
=== Definition using Closures of Open Sets ===
=== Definition using Interiors of Closed Sets ===
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) topological space such that the closure of every open set of $T$ is open.
Let $V \subseteq S$ be closed in $T$.
Then $S \setminus V$ is open by definition.
Then its closure $\paren {S \setminus V}^-$ is open by hypothesis.
By Complement of Interior equals Closure of Complement we have that:
:$\paren {S \setminus V}^- = S \setminus V^\circ$
where $V^\circ$ is the interior of $V$.
As $S \setminus V^\circ$ is open in $T$, it follows that $V^-$ is closed.
So the interior of every closed set of $T$ is closed.
By a similar argument we see that if the interior of every closed set of $T$ is closed in $T$, then the closure of every open set of $T$ is open.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Equivalence of Definitions of Extremally Disconnected Space
Extreme Set in Compact Convex Set contains Extreme Point

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$. 
Let $K$ be a non-empty compact convex subset of $X$.
Let $E$ be an extreme set of $K$. 
Then $E$ contains an extreme point of $K$.
== Proof ==
Let $P$ be the set of extreme sets in $K$ that are contained in $E$. 
Since $E \in P$, $P$ is certainly non-empty.
Define a relation $\preceq$ on $P$ by $A \preceq B$ $B \subseteq A$. 
From Subset Relation is Ordering and Dual Ordering is Ordering, we have:
:$\struct {P, \preceq}$ is an ordered set.
We show that every non-empty chain in $\struct {P, \preceq}$ has an upper bound. 
We will then invoke Zorn's Lemma.
=== Lemma ===
From Zorn's Lemma, we have that $P$ has a maximal element.
Let $M$ be a maximal element of $P$. 
Then $M$ is an extreme set of $K$ such that: 
:whenever $M_\ast \in P$ has $M \preceq M_\ast$ we have $M_\ast = M$
and:
:$M \subseteq E$ 
That is: 
:whenever a extreme set $M_\ast \subseteq E$ of $K$ has $M_\ast \subseteq M$, we have $M_\ast = M$. 
So $M$ does not properly contain any other extreme set of $K$.
With view to apply Point in Convex Set is Extreme Point iff Singleton is Extreme Set, we show that $M$ is a singleton. 
 suppose that $a, b \in M$ with $a \ne b$. 
Then from Normed Dual Space Separates Points, there exists an $f \in X^\ast$ with: 
:$\map f a \ne \map f b$
where $X^\ast$ is the normed dual of $X$. 
So either: 
:$\map f a > \map f b$
or:
:$\map f a < \map f b$ 
 suppose that $\map f a < \map f b$.
Otherwise, swap $a$ and $b$. 
Define: 
:$\ds M^f = \set {x \in M : \map f x = \max_{y \in M} \map f y}$
From Preimage of Maximum of Bounded Linear Functional on Extreme Set in Convex Compact Set is Extreme Set, we have that: 
:$M^f$ is an extreme set in $K$. 
We also have: 
:$M^f \subseteq M \subseteq E$ 
However, from the definition of maximum, we have: 
:$\ds \map f b \le \max_{y \in M} \map f y$ 
So we have: 
:$\ds \map f a < \max_{y \in M} \map f y$ 
So:
:$a \not \in M^f$
So:
:$M^f$ is a proper subset of $M$
and:
:$M^f$ is an extreme set of $K$ contained in $E$.
This contradicts the maximality of $M$. 
So $M$ is a singleton, and:
:$M = \set x$
for some $x \in X$.
Since $M \subseteq E$, in particular we have $x \in E$.
From Point in Convex Set is Extreme Point iff Singleton is Extreme Set, we have: 
:$x$ is an extreme point in $K$ that is contained in $E$.
== Sources ==
* $21.6$: The Krein-Milman Theorem
Category:Convex Sets (Vector Spaces)
Category:Extreme Sets
Category:Extreme Set in Compact Convex Set contains Extreme Point
Category:Extreme Points of Convex Sets