nyuuzyou/wikis huggingface dataset:
Characterization of Closedness in terms of Nets

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $E \subseteq X$ be a subset. 
Then $E$ is closed for each $x \in X$ we have $x \in E$ :
:there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$.
== Proof ==
From Set is Closed iff Equals Topological Closure, we have that $E$ is closed :
:$x \in E$ $x \in \map \cl E$
From Point in Set Closure iff Limit of Net, for $x \in X$ we have $x \in \map \cl E$ :
:there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $E$ converging to $x$.
Category:Nets (Set Theory)
Characterization of Closure by Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB \subseteq \tau$ be a basis.
Let $A$ be a subset of $S$.
Let $x$ be a point of $S$.
Then $x \in A^-$ :
:$\forall U \in \BB: x \in U \implies A \cap U \ne \O$
where:
:$A^-$ denotes the closure of $A$
== Proof ==
=== Sufficient Condition ===
Let $x \in A^-$.
Let $U \in \BB$.
By definition of basis, $U$ is an open set of $T$.
Thus from Condition for Point being in Closure:
:if $x \in U$ then $A \cap U \ne \O$.
=== Necessary Condition ===
Let $x$ be such that for every $U \in \BB$:
:if $x \in U$
:then $A \cap U \ne \O$.
By Condition for Point being in Closure, to prove that $x \in \operatorname{Fr} A$ it is enough to prove
that:
:for every open set $U$ of $T$:
::if $x \in U$ then $A \cap U \ne \O$.
Let $U$ be an open set of $T$.
Let $x \in U$.
By definition of (analytic) basis, there exists $V \in \BB$ such that:
:$x \in V \subseteq U$
By assumption:
:$A \cap V \ne \O$
From the corollary to Set Intersection Preserves Subsets:
:$A \cap V \subseteq A \cap U$
So:
:$A \cap U \ne \O$
and hence the result.
== Sources ==
* 
* 
Category:Set Closures
Characterization of Closure by Open Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Let $x$ be a point of $T$.
Let $A^-$ denote the closure of $A$.
Then $x \in A^-$ :
:for every open set $U$ of $T$:
::$x \in U \implies A \cap U \ne \O$
== Proof ==
=== Sufficient Condition ===
Let $x \in A^-$.
 there exists an open set $U$ of $T$ such that:
:$x \in U$ and $A \cap U = \O$
We have that $U$ is open in $T$.
So by definition of closed set, $\relcomp S U$ is closed in $T$.
Then:
But we have:
:$x \in A^-$
and also:
:$x \in U$
and thus by definition of set intersection:
:$x \in A^- \cap U$
This contradicts $A^- \cap U = \O$
Hence by Proof by Contradiction the assumption that there exists an open set $U$ of $T$ such that $x \in U$ and $A \cap U = \O$ was false.
So for every open set $U$ of $T$:
:$x \in U \implies A \cap U \ne \O$
 
=== Necessary Condition ===
Let $x$ be such that for every open set $U$ of $T$:
:$x \in U \implies A \cap U \ne \O$
 $x \notin A^-$.
Then:
:$x \in \relcomp S {A^-}$
Then by assumption:
:$A \cap \relcomp S {A^-} \ne \O$
as $\relcomp S {A^-}$ is open.
By definition of complement:
:$A \cap \relcomp S {A^-} = \O$
So by Empty Intersection iff Subset of Complement:
:$A \nsubseteq A^-$
From Set is Subset of its Topological Closure:
:$A \subseteq A^-$
But from Set Complement inverts Subsets:
:$\relcomp S {A^-} \subseteq \relcomp S A$
from which by Empty Intersection iff Subset of Complement:
:$A \cap \relcomp S {A^-} = \O$
Hence by Proof by Contradiction the assumption that $x \notin A^-$ was false.
So $x \in A^-$.
== Sources ==
* 
* 
Category:Set Closures
Characterization of Compact Element in Complete Lattice

== Theorem ==
Let $L = \struct{S, \preceq}$ be a complete lattice.
Let $a \in S$.
:
:$(1)\quad a$ is a compact element
:$(2)\quad \forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
:$(3)\quad \forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$ 
== Proof ==
Recall, $a$ is a compact element :
:for every directed subset $D$ of $S$ such that $a \preceq \sup D$
::$\exists d \in D: a \preceq d$
=== Statement (1) implies Statement (3) ===
Let $a$ be a compact element.
=== Statement (3) implies Statement (2) ===
Let $a$ satisfy:
:$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$ 
=== Statement (2) implies Statement (1) ===
Let $a$ satisfy:
:$\forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
== Sources ==
* : Chapter $\text {II}$: Introduction to Locales, $\S3.1$ Lemma
Category:Complete Lattices
Category:Characterization of Compact Element in Complete Lattice
Characterization of Compact Element in Complete Lattice/Statement 1 implies Statement 3

== Theorem ==
Let $L = \struct{S, \preceq}$ be a complete lattice.
Let $a \in S$ be a compact element
Then:
:$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$ 
== Proof ==
Recall, $a$ is a compact element :
:for every directed subset $D$ of $S$ such that $a \preceq \sup D$
::$\exists d \in D: a \preceq d$
Let $A \subseteq S : a \preceq \sup A$.
Let $D = \leftset{b \in A : \exists F \subseteq A : F}$ is finite $\rightset{: b = \sup F}$
==== $D$ is a Directed Subset ====
We show that $D$ is a directed subset.
Let $x, y \in D$.
By definition of $D$:
:$\exists F, G \subseteq A: F, G$ are finite $: x = \sup F, y = \sup G$
Let $H = F \cup G$.
From Union of Finite Sets is Finite:
:$H$ is finite of $A$
Hence:
:$z = \sup H \in D$.
Furthermore:
:$z = \sup \set{x, y}$
By definition of supremum:
:$x, y \preceq z$
Since $x,y$ were arbitrary, then $D$ is a directed subset by definition.
==== Supremum of $D$ Equals Supremum of $A$ ====
We show that $\sup D = \sup A$.
Let $x \in D$.
By definition of $D$:
:$\exists F \subseteq A: F$ is finite $: x = \sup F$
From Supremum of Subset:
:$\sup F \preceq \sup A$
Since $x$ was arbitrary, it follows that:
:$\sup A$ is an upper bound for $D$
By definition of supremum:
:$\sup D \preceq \sup A$
Note that for all $b \in A$:
:$b = \sup \set b \in D$
Hence:
:$\sup D$ is an upper bound for $A$
By definition of supremum:
:$\sup A \preceq \sup D$
From :
:$\sup A = \sup D$
By definition of compact element:
:$\exists d \in D : a \preceq d$
By definition of $D$:
:$\exists F \subseteq A : \sup F = d$
Hence:
:$\exists F \subseteq A : F$ is finite $: a \preceq \sup F$
Since $A$ was arbitrary:
:$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$ 
Category:Characterization of Compact Element in Complete Lattice
Characterization of Compact Element in Complete Lattice/Statement 2 implies Statement 1

== Theorem ==
Let $L = \struct{S, \preceq}$ be a complete lattice.
Let $a \in S$ satisfy:
:$\forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
Then:
:$a$ is a compact element
== Proof ==
Let $D$ be a directed subset of $S$:
:$a \preceq \sup D$
Let $I = {b \in S : \exists d \in D : b \preceq d}$.
==== $I$ is an Ideal of $S$ ====
We will show that $I$ satisfies the join semilattice ideal axioms.
===== $I$ is a Lower Section =====
Let $x \in I$.
Let $y \preceq x$.
By definition of $I$:
:$\exists d \in D : x \preceq d$
By :
:$y \preceq d$
Hence:
:$y \in I$
It follows that $I$ is a lower section of $S$.
===== $I$ is a Join Subsemilattice =====
Let $x, y \in I$.
By definition of $I$:
:$\exists d_1, d_2 \in D : x \preceq d, y \preceq e$
By definition of a directed subset:
:$\exists d \in D : d_1, d_2 \preceq d$
By :
:$x, y \preceq d$
By definition of join:
:$x \vee y \preceq d$
Hence:
:$x \vee y \in I$
This proves $I$ satisfies the join semilattice ideal axioms.
It follows that $I$ is an ideal by definition.
==== Supremum of $I$ Equals Supremum of $D$ ====
We show that $\sup I = \sup D$.
Let $x \in I$.
By definition of $I$:
:$\exists d \in D: x \preceq d$
By definition of supremum:
:$x \preceq \sup D$
Since $x$ was arbitrary, it follows that:
:$\sup D$ is an upper bound for $I$
By definition of supremum:
:$\sup I \preceq \sup D$
Note that for all $d \in D$:
:$d \in I$
Hence:
:$\sup I$ is an upper bound for $D$
By definition of supremum:
:$\sup D \preceq \sup I$
From :
:$\sup I = \sup D$
We have :
:$a \in I$
By definition of $I$:
:$\exists d \in D : a \preceq d$
Since $D$ was arbitrary:
:for every directed subset $D$ of $S$ such that $a \preceq \sup D$
::$\exists d \in D: a \preceq d$
Hence $a$ is a compact element by definition.
Category:Characterization of Compact Element in Complete Lattice
Characterization of Compact Element in Complete Lattice/Statement 3 implies Statement 2

== Theorem ==
Let $L = \struct{S, \preceq}$ be a complete lattice.
Let $a \in S$ satisfy:
:$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$ 
Then:
:$\forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
== Proof ==
Let:
:$I \subseteq S : I$ is an ideal $: a \preceq \sup I$
We have :
:$\exists F \subseteq I : F$ is finite $: a \preceq \sup F$
By :
:$\sup F \in I$
By :
:$a \in I$
Since $I$ was arbitrary:
:$\forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
Category:Characterization of Compact Element in Complete Lattice
Characterization of Compact Element in Frame or Locale

== Theorem ==
Let $L = \struct{S, \preceq}$ be a frame or locale.
Let $a \in S$.
:
:$(1)\quad a$ is a compact element
:$(2)\quad \forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
:$(3)\quad \forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$ 
:$(4)\quad \forall A \subseteq S : a = \sup A \implies \exists F \subseteq A : F$ is finite $: a = \sup F$ 
== Proof ==
Recall, a frame or locale is a complete lattice satisfying the infinite join distributive law:
From Characterization of Compact Element in Complete Lattice:
:Statements $(1)$, $(2)$ and $(3)$ are equivalent.
=== Statement $(3)$ implies Statement $(4)$ ===
Let $a$ satisfies:
:$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$
Let $A \subseteq S : a = \sup A$.
We have :
:$\exists F \subseteq A : F$ is finite $: a \preceq \sup F$
From Supremum of Subset:
:$\sup F \preceq \sup A = a$
By :
:$a = \sup F$
The result follows.
=== Statement $(4)$ implies Statement $(3)$ ===
Let $a$ satisfies:
:$\forall A \subseteq S : a = \sup A \implies \exists F \subseteq A : F$ is finite $: a = \sup F$
Let $A \subseteq S : a \preceq \sup A$.
From Predecessor is Infimum:
:$a = a \wedge \sup A$
By the infinite join distributive law:
:$a = \sup \set{a \wedge x : x \in A}$
We have :
:$\exists F \subseteq A : F$ is finite $: a = \sup \set{a \wedge x : x \in F}$
By the infinite join distributive law:
:$a = a \wedge \sup F$
From Predecessor is Infimum:
:$a \preceq \sup F$
The result follows.
== Sources ==
* : Chapter $\text {II}$: Introduction to Locales, $\S3.1$
Category:Frames
Category:Locales
Characterization of Completely Prime Filter in Complete Lattice

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $F \subseteq L$.
Then:
:$F$ is a completely prime filter
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$
== Proof ==
=== Necessary Condition ===
Let $F$ be a completely prime filter.
=== Sufficient Condition ===
Let $F$ satisfy:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
Category:Complete Lattices
Category:Completely Prime Filters
Category:Characterization of Completely Prime Filter in Complete Lattice
Characterization of Completely Prime Filter in Complete Lattice/Necessary Condition

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $F \subseteq L$ be a completely prime filter.
Then:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$
== Proof ==
==== $F$ satisfies Statement $(1)$ ====
Let $A \subseteq L$.
Let $\bigvee A \in F$
By definition of completely prime filter:
:$A \cap F \ne \O$
Let $x \in A \cap F$
By definition of set intersection:
:$x \in A$ and $x \in F$
Let $x \in A : x \in F$
By definition of supremum:
:$x \preceq \bigvee A$
By definition of filter:
:$\bigvee A \in F$
Hence:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
==== $F$ satisfies Statement $(2)$ ====
Let $A \subseteq L$ be finite.
Let $\bigwedge A \in F$
By definition of infimum:
:$\forall x \in A : \bigwedge A \preceq x$
By definition of filter:
:$\forall x \in A : x \in F$
Let:
:$\forall x \in A : x \in F$
Let $A \ne \O$.
By definition of filter:
:$F$ is sublattice of $L$
From Existence of Non-Empty Finite Infima in Meet Semilattice:
:$\bigwedge A \in F$
Let $A = \O$.
From Infimum of Empty Set is Greatest Element:
:$\top = \bigwedge A$
From Filter Contains Greatest Element:
:$\top \in F$
Hence:
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
The result follows.
Category:Characterization of Completely Prime Filter in Complete Lattice
Characterization of Completely Prime Filter in Complete Lattice/Sufficient Condition

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $F \subseteq L$ satisfy:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$
Then:
:$F$ is a completely prime filter
== Proof ==
==== $F$ is a Proper Subset ====
From Supremum of Empty Set is Smallest Element:
:$\bot = \bigvee \O$
By definition of empty set:
:$\forall x \in F : x \notin \O$
By the contrapositive statement of $(1)$:
:$\bot = \bigvee \O \notin F$
It follows that $F \neq L$
==== $F$ is not Empty ====
From Infimum of Empty Set is Greatest Element:
:$\top = \bigwedge \O$
By definition of empty set, $\O$ vacuously satisfies:
:$\forall x \in \O : x \in F$
By $(2)$:
:$\top = \bigwedge \O \in F$
It follows that $F \neq \O$
Hence $F$ satisfies filter axiom (1).
==== $F$ is a Down-Directed Subset of $L$ ====
Let $x, y \in F$.
From $(2)$:
:$x \wedge y \in F$
From Meet Precedes Operands:
:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
Hence $F$ satisfies filter axiom (2).
==== $F$ is an Upper Section of $L$ ====
Let $x \in F$.
Let $x \preceq y$.
From Successor is Supremum:
:$x \vee y = y$
From $(1)$:
:$y \in F$
Hence $F$ satisfies filter axiom (3).
It follows that $F$ is a filter.
==== $F$ is Completely Prime ====
Let $A \subseteq L$.
Let $\bigvee A \in F$.
From $(1)$:
:$\exists a \in A : a \in F$
By definition of subset:
:$a \in A \cap F$
Hence $A \cap F \ne \O$.
Hence $F$ is a completely prime filter by definition.
Category:Characterization of Completely Prime Filter in Complete Lattice
Characterization of Completely Prime Ideal in Complete Lattice

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $I \subseteq L$.
Then:
:$I$ is a completely prime ideal
:$(1)\quad\forall A \subseteq L : \bigwedge A \in I \iff \paren{\exists a \in A : a \in I}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigvee A \in I \iff \paren{\forall a \in A : a \in I}$
where:
:* $\bigwedge A$ denotes the infimum of $A$ in $L$
:* $\bigvee A$ denotes the supremum of $A$ in $L$
== Proof ==
This is the dual statement of Characterization of Completely Prime Filter in Complete Lattice by Dual Pairs (Order Theory).
The result follows from the Duality Principle.
Category:Complete Lattices
Category:Completely Prime Ideals
Characterization of Constant-Curvature Metrics

== Theorem ==
Let $M$ be a complete connected $n$-dimensional Riemannian manifold.
Let $M$ have constant sectional curvature.
Let $\tilde M$ be the Euclidean space, sphere, or the hyperbolic space with the constant sectional curvature.
Let $\Gamma$ be a discrete group of isometries of $\tilde M$ that acts freely on $\tilde M$.
Then, up to isometry, $M$ is the Riemannian quotient of the form $\tilde M / \Gamma$.
== Proof ==
== Sources ==
* : $\S 1$: What Is Curvature? Curvature in Higher Dimensions
Category:Riemannian Manifolds
Category:Curvature
Characterization of Continuity in terms of Nets

== Theorem ==
Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_Y}$ be topological spaces.
Let $f : \struct {X, \tau_X} \to \struct {Y, \tau_Y}$ be a function.
Let $x \in X$. 
Then $f$ is continuous at $x$ :
:for every directed set $\struct {\Lambda, \preceq}$ and net $\family {x_\lambda}_{\lambda \in \Lambda}$ such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, \tau_X}$, we have that: 
::the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$ in $\struct {Y, \tau_Y}$. 
== Proof ==
=== Necessary Condition ===
Suppose that $f$ is continuous at $x$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$.
Let $U$ be an open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$.
Then there exists an open neighborhood $V$ of $x$ in $\struct {X, \tau_X}$ such that: 
:$f \sqbrk V \subseteq U$
Since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda_0 \in \Lambda$ such that: 
:$x_n \in V$ for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$. 
Then: 
:$\map f {x_n} \in U$
for $\lambda_0 \preceq \lambda$.
Since $U$ was an arbitrary open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$, we have that $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$. 
=== Sufficient Condition ===
Suppose that $f$ is not continuous at $x$.
Let $\Lambda$ be the set of open neighborhoods of $\map f x$. 
Then $\struct {\Lambda, \supseteq}$ is a directed set from Open Neighborhoods of Point form Directed Ordering.
Since $f$ is not continuous at $x$, there exists an open neighborhood of $\map f x$ in $\struct {Y, \tau_Y}$ such that: 
:$f \sqbrk V \not \subseteq U$
for all open neighborhoods $V$ of $x$ in $\struct {X, \tau_X}$.
So for each open neighborhood $V$ of $x$, we can pick $x_V \in V$ such that: 
:$\map f {x_V} \not \in U$
Since $U$ is an open neighborhood of $\map f x$, we certainly have: 
:$\family {\map f {x_V} }_{V \in \Lambda}$ does not converge to $\map f x$.
We show that: 
:$\family {x_U}_{U \in \Lambda}$ converges to $x$.
If $V$ is an open neighborhood of $x$ and $V \supseteq U$, we have: 
:$x_U \in U$
so that: 
:$x_U \in V$
So $\family {x_U}_{U \in \Lambda}$ converges to $x$.
So:
:if $f$ is not continuous at $x$, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ such that $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, \tau_X}$ but $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ does not converge to $\map f x$ in $\struct {Y, \tau_Y}$.
By contraposition, we have the demand.
Category:Continuous Mappings
Category:Nets (Set Theory)
Characterization of Continuity of Linear Functional in Weak-* Topology

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Let $X^\ast$ be the normed dual space of $X$.
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$. 
Let $X^{\ast \ast}$ be the second normed dual of $X$.
Then a linear functional $\phi : \struct {X^\ast, w^\ast} \to \GF$ is continuous there exists $x \in X$ such that: 
:$\phi = x^\wedge$ 
where $x^\wedge$ is the evaluation linear transformation evaluated at $x$.
That is: 
:$\struct {X^\ast, w^\ast}^\ast = \iota X$ 
== Proof ==
Note that the weak-$\ast$ topology $w^\ast$ is generated as an initial topology by $\set {x^\wedge : x \in X}$. 
This result is then given by Continuity of Linear Functionals in Initial Topology on Vector Space Generated by Linear Functionals, taking $F = \set {x^\wedge : x \in X}$. 
== Sources ==
* : Theorem $3.16$
Category:Weak-* Topologies
Characterization of Continuity of Linear Functional in Weak Topology

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$ with weak topology $w$. 
Let $X^\ast$ be the topological dual space of $X$.
Let $f : X \to \GF$ be a linear functional.
Then $f$ is $w$-continuous $f \in X^\ast$. 
That is: 
:$\struct {X, w}^\ast = X^\ast$
== Proof ==
This is precisely Continuity of Linear Functionals in Initial Topology on Vector Space Generated by Linear Functionals, taking $F = X^\ast$. 
Category:Weak Topologies on Topological Vector Spaces
Characterization of Convergence in Locally Convex Space

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $\struct {X, \mathcal P}$ be a locally convex space over $\Bbb F$ with standard topology $\tau$.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$. 
Then: 
:$x_n \to x$ in $\struct {X, \tau}$
:
:$\map p {x_n - x} \to 0$ as $n \to \infty$ as a real sequence, for each $p \in \mathcal P$.
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$x_n \to x$ in $\struct {X, \tau}$
Let $p \in \mathcal P$. 
Let $\epsilon > 0$. 
We aim to show that there exists $N \in \N$ such that: 
:$\map p {x_n - x} < \epsilon$ for $n \ge N$
showing that: 
:$\map p {x_n - x} \to 0$ as $n \to \infty$. 
From the definition of a convergent sequence in a topological space, we have: 
:for each $U \in \tau$ such that $x \in U$, there exists $N \in \N$ such that $x_n \in U$ for $n \ge N$.
From the definition of the standard topology $\tau$, we have: 
:$U = \set {y \in X : \map p {y - x} < \epsilon}$ is open.
So there exists $N \in \N$ such that $x_n \in U$ for $n \ge N$. 
That is: 
:$\map p {x_n - x} < \epsilon$ for $n \ge N$. 
Since $\epsilon$ was arbitrary we have: 
:$\map p {x_n - x} \to 0$ as $n \to \infty$. 
=== Sufficient Condition ===
Suppose that:
:$\map p {x_n - x} \to 0$ as $n \to \infty$ as a real sequence, for each $p \in \mathcal P$.
Let $U \in \tau$ be such that $x \in U$.
We aim to show that there exists $N \in \N$ such that $x_n \in U$ for $n \ge N$.
From Open Sets in Standard Topology of Locally Convex Space, there exists $\epsilon > 0$ and $p_1, p_2, \ldots, p_m \in \mathcal P$ such that: 
:$\set {y \in X : \map {p_k} {y - x} < \epsilon \text { for each } k \in \set {1, 2, \ldots, m} } \subseteq U$
Since: 
:$\map {p_k} {x_n - x} \to 0$ as $n \to \infty$ 
for each $k \in \set {1, 2, \ldots, m}$, there exists $N_k \in \N$ such that: 
:$\map {p_k} {x_n - x} < \epsilon$ for $n \ge N_k$. 
Set: 
:$N = \max \set {N_1, N_2, \ldots, N_k}$
Then, for $n \ge N$, we have: 
:$\map {p_k} {x_n - x} < \epsilon$ for all $k \in \set {1, 2, \ldots, m}$.
Then: 
:$x_n \in U$ for $n \ge N$. 
Since $U$ was an arbitrary open set containing $x$, we have: 
:$x_n \to x$ in $\struct {X, \tau}$. 
Category:Locally Convex Spaces
Category:Convergent Sequences