nyuuzyou/wikis huggingface dataset:
Local Basis Generated from Neighborhood Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be an element of $S$.
Let $\BB$ be a neighborhood basis of $x$.
For any subset $A \subseteq S$, let $A^\circ$ denote the interior of $A$.
Then the set:
:$\BB' = \set {H^\circ: H \in \BB}$
is a local basis of $x$. 
== Proof ==
First it must be shown that $\BB'$ is a set of open neighborhoods of $x$.
From the definition of the interior of a subset, $\BB'$ is a set of open sets.
Let $H \in \BB$. 
By assumption, $H$ is a neighborhood of $x$.
From the definition of a neighborhood:
:$\exists U \in \tau : x \in U \subseteq H$
From the definition of the interior of a subset:
:$U \subseteq H^\circ$
Hence $x \in H^\circ$.
It follows that $\BB'$ is a set of open neighborhoods of $x$.
It remains to show that $\BB'$ is a neighborhood basis.
Let $N$ be a neighborhood of $x$.
Since $\BB$ is a neighborhood basis of $x$ then:
:$\exists H \in \BB : H \subseteq N$
By definition of the interior, $H^\circ \subseteq H$.
From Subset Relation is Transitive $H^\circ \subseteq N$.
Hence $\BB'$ is a neighborhood basis.
The result follows.
Category:Local Bases
Category:Neighborhood Bases
Local Basis Test

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $\BB$ be a local basis for $x$ in $\struct {S, \tau}$.
Let $\CC$ be a set of open neighborhoods of $x$.
Then:
:$\CC$ is a local basis :
::$\forall B \in \BB \implies \exists C \in \CC: C \subseteq B$
== Proof ==
=== Necessary Condition ===
Let $\CC$ be a local basis.
Let $B \in \BB$.
Since $\BB$ is a local basis, by the definition of a local basis then $B$ is open.
Since $\CC$ is a local basis, by the definition of a local basis then:
:$\exists C \in \CC : C\subseteq B$.
=== Sufficient Condition ===
Let $\CC$ satisfy:
:$\forall B \in \BB \implies \exists C \in \CC: C\subseteq B$
Let $U \in \tau$ and $x \in U$.
By the definition of a local basis then:
:$\exists B \in \BB : B\subseteq U$
By the assumption then:
:$\exists C \in \CC: C\subseteq B \subseteq U$
By the definition of a local basis then $\CC$ is a local basis.
Category:Local Bases
Local Basis Test for Adherent Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in S$.
Let $\BB_x$ be a local basis of $x$.
Then $x \in S$ is an adherent point of $H$ :
:$\forall U \in \BB_x : H \cap U \ne \O$
== Proof ==
=== Necessary Condition ===
Let $x \in S$ be an adherent point of $H$.
By definition of an adherent point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a local basis of $T$:
:$\BB_x \subseteq \tau$
The result follows.
=== Sufficient Condition ===
Let $x$ satisfy:
:$\forall U \in \BB_x : H \cap U \ne \O$
Let $V$ be any open neighborhood of $x$.
By definition of a local basis of $T$:
:$\exists U \in \BB : x \in U \subseteq V$
Then:
:$H \cap U \ne \O$
From the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
:$H \cap V \ne \O$
Thus $x$ is an adherent point of $H$ by definition.
Category:Adherent Points
Local Basis Test for Isolated Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in H$.
Let $\BB_x$ be a local basis of $x$.
Then $x$ is an isolated point of $H$ :
:$\exists U \in \BB_x : U \cap H = \set x$
== Proof ==
=== Necessary Condition ===
Let $x \in H$ be an isolated point of $H$.
By definition of an isolated point:
:$\exists U \in \tau: U \cap H = \set x$
By definition of a local basis of $T$:
:$\exists V \in \BB_x : x \in V \subseteq U$
From Set Intersection Preserves Subsets:
:$V \cap H \subseteq U \cap H = \set x$
From Singleton of Element is Subset:
:$\set x \subseteq V \cap H$
From set equality:
:$V \cap H = \set x$
=== Sufficient Condition ===
Let $U \in \BB_x : U \cap H = \set x$.
By definition of local basis of $T$:
:$U \in \tau$
Then $x$ is an isolated point of $H$ by definition.
Category:Isolated Points
Local Basis Test for Limit Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in S$.
Let $\BB_x$ be a local basis of $x$.
Then $x \in S$ is a limit point of $H$ :
:$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$
== Proof ==
=== Necessary Condition ===
Let $x \in S$ be a limit point of $H$.
By definition of a limit point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
By definition of a local basis of $T$:
:$\BB_x \subseteq \tau$
The result follows.
=== Sufficient Condition ===
Let $x$ satisfy:
:$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$
Let $V$ be any open neighborhood of $x$.
By definition of a local basis of $T$:
:$\exists U \in \BB : x \in U \subseteq V$
Then:
:$H \cap U \setminus \set x \ne \O$
From the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
:$H \cap V \setminus \set x \ne \O$
Thus $x$ is a limit point of $H$ by definition.
Category:Limit Points
Local Basis of Open Subspace iff Local Basis

== Theorem ==
Let Let $T = \struct{S, \tau}$ be a topological space.
Let $U \subseteq S$ be an open subset
Let $\tau_U$ denote the subspace topology on $U$.
Let $s \in U$.
Let $\BB \subseteq \powerset U$.
Then:
:$\BB$ is a local basis of $s$ in $\struct{U, \tau_U}$ $\BB$ is a local basis of $s$ in $\struct{S, \tau}$
== Proof ==
Let $\map \BB s$ denote the set of open sets containing $s$ in $\struct{S, \tau}$
Let $\map \CC s$ denote the set of open sets containing $s$ in $\struct{U, \tau_U}$
From Open Set in Open Subspace:
:$\BB \subseteq \map \BB s$ $\BB \subseteq \map \CC s$
=== Necessary Condition ===
Let $\BB$ be a local basis of $s$ in $\struct{U, \tau_U}$.
Let $W \in \map \BB s$.
By definition of subspace topology:
:$W' = W \cap U \in \tau_U$
By definition of local basis:
:$\exists V \in \BB : x \in V \subseteq W'$
From Subset Relation is Transitive:
:$V \subseteq W$
We have:
:$\exists V \in \BB : x \in V \subseteq W$
It follows that:
:$\BB$ is a local basis of $s$ in $\struct{S, \tau}$
=== Sufficient Condition ===
Let $\BB$ be a local basis of $s$ in $\struct{S, \tau}$.
Let $W \in \map \CC s$.
From Open Set in Open Subspace:
:$W \in \map \BB s$.
Hence:
:$\exists V \in \BB : x \in V \subseteq W$
It follows that:
:$\BB$ is a local basis of $s$ in $\struct{U, \tau_U}$
Category:Topological Subspaces
Category:Local Bases
Local Basis of P-adic Number

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Then the set of open balls $\set {\map {B_{p^{-n} } } a : n \in Z}$ is a local basis of $a$ consisting of clopen sets.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $\BB_a$ be the set of all open balls of $Q_p$ centered on $a$.
That is:
:$\BB_a = \set{\map {B_\epsilon} a : \epsilon \in \R_{>0}}$
From Open Balls Centered on P-adic Number is Countable:
:$\BB_a = \set {\map {B_{p^{-n} } } a : n \in Z}$
From Open Balls form Local Basis for Point of Metric Space, $\BB_a$ is a local basis of $a$.
From P-adic Numbers form Non-Archimedean Valued Field:
:the $p$-adic numbers is a non-Archimedean division ring.
From Open Balls are Clopen In Non-Archimedean Division Ring:
:the set $\BB_a = \set {\map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of clopen sets.
Category:Topology of P-adic Numbers
Local Basis of P-adic Number/Closed Balls

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Then the set of closed balls $\set {\map {B^-_{p^{-n} } } a: n \in Z}$ is a local basis of $a$ consisting of clopen sets.
== Proof ==
From Local Basis of P-adic Number the set $\set {\map {B_{p^{-n } } } a: n \in \Z}$ is a local basis of clopen sets.
From Open Ball in P-adic Numbers is Closed Ball:
:$\set {\map {B_{p^{-n} } } a: n \in \Z} = \set {\map {B^-_{p^{-\paren{n + 1} } } } a: n \in \Z} = \set {\map {B^-_{p^{-n} } } a : n \in \Z}$
The result follows.
Category:Topology of P-adic Numbers
Local Basis of P-adic Number/Cosets

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\Z_p$ be the $p$-adic integers.
Then the set $\set {a + p^n \Z_p: n \in Z}$ is a local basis of $a$ consisting of clopen sets.
== Proof ==
From Local Basis of P-adic Number the set $\set { \map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of clopen sets.
From Open Balls of P-adic Number:
:$\set {\map {B_{p^{-n} } } a : n \in \Z} = \set {a + p^{n + 1} \Z_p : n \in \Z} = \set {a + p^n \Z_p : n \in \Z}$
The result follows.
== Also see ==
* Closed Balls of P-adic Number
* Open Balls of P-adic Number
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.5$
Category:Topology of P-adic Numbers
Local Basis of Topological Vector Space

== Theorem ==
Let $\struct {\XX, \tau}$ be a topological vector space.
Let $0_\XX$ denote the zero vector of $\XX$.
Then there exists a local basis $\BB$ of $0_\XX$ with the following properties:
:$(1): \quad \forall W \in \BB: \exists V \in \BB$ such that $V + V \subseteq W$ (where the addition $V + V$ is meant in the sense of the Minkowski sum)
:$(2): \quad$ Every $W \in \BB$ is star-shaped (balanced)
:$(3): \quad$ Every $W \in \BB$ is absorbent.
:$(4): \quad \ds \bigcap_\BB W = \set {0_\XX}$.
== Proof ==
The proof will be carried out in various steps.
We will construct a collection of star-shaped neighborhoods of $0_\XX$.
Then we will show that it is indeed a local basis with the required properties.
Firstly we define the following set:
:$\BB_0 := \set {W \in \tau: 0 \in W, W \text{ is star-shaped} }$
=== $\BB_0$ is a local basis for $0_\XX$ ===
Let $U \ni 0_\XX$ be an open set.
Notice that:
:$0 \cdot 0_\XX = 0_\XX$
This way we have proved that $\BB_0$ is a local basis for $0_\XX$.
Since $\cdot: K \times \XX \to \XX$ is a continuous mapping, there is a neighborhood of $\tuple {0, 0_\XX} \in K \times \XX$ in the form:
:$\openint {-\epsilon} \epsilon \times W$
such that:
:$\map \cdot {\openint {-\epsilon} \epsilon \times W} \subseteq U$
which means that
:$\openint {-\epsilon} \epsilon \cdot W \subseteq U$
Let:
:$\ds G := \bigcup_{\lambda \mathop \in \openint {-\epsilon} \epsilon, \lambda \mathop \ne 0} \lambda W \subseteq U$
We have that:
:$(1): \quad$ For every $\lambda \ne 0$, the set $\lambda \cdot W$ is open.
Hence $G$ is open as union of open sets.
:$(2): \quad 0_\XX \in G$ since $0 \in W$.
:$(3): \quad$ If $x \in G$ then $-x \in G$.
Therefore, $G$ is open, star-shaped, contains $0_\XX$ and $G \subseteq U$.
=== Condition $(1)$ is satisfied by $\BB_0$ ===
Let $V \in \BB_0$.
We have that $+\tuple {0_\XX, 0_\XX} = 0_\XX \in W$.
Since $+$ is a continuous mapping and $\BB_0$ is a local basis of $0_\XX$, there exists a $V \in \BB_0$ such that:
:$+\tuple {V, V} = V + V \subseteq W$
=== $\BB_0$ consists of Absorbent Sets ===
We notice that:
:$\tuple {\dfrac 1 n, x} \overset n \to \tuple {0, x}$
and:
:$\map \cdot {\dfrac 1 n, x} \overset n \to \map \cdot {0, x} = 0_\XX$
Thus, there is a $n \in \N$ such that:
:$\frac 1 n x \in V$
or what amounts to the same thing:
:$x \in n V$
=== Condition $(4)$ is satisfied by $\BB_0$ ===
It suffices that we find a $V \in \BB_0$ such that $x \notin V$.
This is possible since $\tau$ is a Hausdorff topology.
The proof is now complete.
Local Compactness is Preserved under Open Continuous Surjection

== Theorem ==
Let $T_A = \left({S_A, \tau_A}\right)$ and $T_B = \left({S_B, \tau_B}\right)$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous mapping which is also an open mapping and a surjection.
If $T_A$ is locally compact, then $T_B$ is also locally compact.
== Proof ==
== Also see ==
* Weak Local Compactness is Preserved under Open Continuous Surjection
== Sources ==
* : Chapter $6$: Compactness: $\S18$: Locally Compact Spaces: Theorem $18.5$
Category:Locally Compact Spaces
Category:Open Mappings
Local Connectedness is not Preserved under Continuous Mapping

== Theorem ==
Let $\struct {A, \tau_1}$ and $\struct {B, \tau_2}$ be topological spaces.
Let $f: A \to B$ be a continuous mapping.
Let $\struct {\Img f, \tau_3}$ be the image of $f$ with the subspace topology of $B$.
Let $\struct {A, \tau_1}$ be locally connected.
Then it is not necessarily the case that $\struct {\Img f, \tau_3}$ is also locally connected.
== Proof ==
Proof by Counterexample:
Let $\struct {A, \tau_1}$ be any countable discrete space.
Let $B \subseteq \R$ be the set of all points on $\R$ defined as:
:$B := \set 0 \cup \set {\dfrac 1 n: n \in \Z_{>0} }$
Let $\struct {B, \tau_2}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.
From Discrete Space is Locally Connected, $\struct {A, \tau_1}$ is locally connected.
Let $f: A \to B$ be a bijection.
From Mapping from Discrete Space is Continuous, $f: A \to B$ is a continuous mapping.
From Integer Reciprocal Space with Zero is not Locally Connected, $\struct {B, \tau_2} = \struct {\Img f, \tau_3}$ is not locally connected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $2 \ \text{(b)}$
Category:Locally Connected Spaces
Category:Continuous Mappings
Local Connectedness is not Preserved under Infinite Product

== Theorem ==
The property of local connectedness is not preserved under the operation of forming an infinite product space.
== Proof ==
Let $T = \struct {\CC, \tau_d}$ be the Cantor space.
Let $A_n = \struct {\set {0, 2}, \tau_n}$ be the discrete space of the two points $0$ and $2$.
Let $\ds A = \prod_{n \mathop = 1}^\infty A_n$.
Let $T' = \struct {A, \tau}$ be the product space where $\tau$ is the product topology on $A$.
From Cantor Space as Countably Infinite Product, $T'$ is homeomorphic to $T$.
From Totally Disconnected and Locally Connected Space is Discrete, we have that $A_n$ is locally connected.
But we also have that the Cantor Space is not Locally Connected.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $9$
Category:Cantor Space
Category:Locally Connected Spaces
Local Expression for Metric of Product Riemannian Manifold

== Theorem ==
Let $\struct {M_1, g_1}$ and $\struct {M_2, g_2}$ be Riemannian manifolds.
Let their smooth local coordinates be $\tuple {x^1, \dots, x^n}$ and $\tuple {x^{n + 1}, \dots, x^{n + m}}$ respectively.
Let $M_1 \times M_2$ be the product Riemannian manifold with smooth local coordinates $\tuple {x^1, \dots, x^{n + m}}$.
Then the product Riemannian metric $g = g_1 \oplus g_2$ has the local expression $g = g_{ij} dx^i dx^j$ where $\paren {g_{i j} }$ is the block diagonal matrix:
:$\paren {g_{i j} } = \begin{pmatrix}
 \paren {g_1}_{ab} & 0 \\
 0 & \paren {g_2}_{cd}
\end{pmatrix}$
and the indices $a,b$ run from $1$ to $n$, while $c,d$ run from $n + 1$ to $n + m$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Local Maxima of Number of Goldbach Decompositions

== Theorem ==
Let $\mathbb E$ be the set of even positive integers.
Let $G: \mathbb E \to \N$ be the mapping defined as:
:$\forall n \in \mathbb E: \map G n =$ the number of Goldbach decompositions of $n$.
Then $G$ has local maxima when $n$ is a multiple of $6$.
== Proof ==
== Historical Note ==
== Sources ==
* : $6$
Category:Number Theory
Local Membership of Equalizer

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $e: E \to C$ be the equalizer of $f, g : C \to D$.
Then a variable element $z: Z \to C$ is a local member of $e$ $f \circ z = g \circ z$:
:$z \in_C e \iff f \circ z = g \circ z$
== Proof ==
Firstly, note that by Equalizer is Monomorphism, local membership of $e$ is defined.
=== Necessary Condition ===
Suppose that $z \in_C e$.
By definition of local membership, there is an $h: Z \to E$ such that $z = e \circ h$.
Then since $e$ is the equalizer of $f$ and $g$:
:$f \circ e = g \circ e$
from which we deduce:
:$f \circ z = f \circ e \circ h = g \circ e \circ h = g \circ z$
by using metacategory axiom $(\text C 3)$, that is associativity of $\circ$.
=== Sufficient Condition ===
Suppose that $f \circ z = g \circ z$.
Since $e$ is the equalizer of $f$ and $g$, we find $h: Z \to E$ such that:
:$z = e \circ h$
Hence $z \in_C e$, by definition of local membership.
== Sources ==
* : $\S 5.1$: Example $5.3$
Category:Subobjects
Category:Category Theory
Local Minimum of Gamma Function on Positive Domain

== Theorem ==
The local minimum of the Gamma function on the positive real numbers occurs at the point:
:$\tuple {1 \cdotp 46163 21449 68362 34126 26595, 0 \cdotp 88560 31944 10888 70027 88159}$
== Proof ==
== Sources ==
* : $0,88560 31944 \ldots$
* : $\S 1.2.5$: Permutations and Factorials: Figure $7$
Category:Gamma Function
Local Normal Form for Immersions

== Theorem ==
Let $\Omega\subset\R^k$ be open.
Let $f: \Omega \to \R^n$ be an immersion.
Let $p \in \Omega$.
Then:
:$k \le n$
and there exists a local diffeomorphism $\phi$ around $\map f p$ such that:
:$\phi \circ \map f x = \tuple {x, 0}$
for all $x$ in a neighborhood of $p$.
== Proof ==
== Also see ==
* Local Normal Form for Submersions
Category:Immersions
Category:Implicit Functions
Local Normal Form for Submersions

== Theorem ==
Let $\Omega\subset\R^n$ be open.
Let $f : \Omega \to \R^k$ be an submersion.
Let $p\in\Omega$.
Then $n\geq k$, and there exists a local diffeomorphism $\phi$ around $f(p)$ such that
:$\phi\circ f (x, y) = x$ for all $(x, y)$ in a neighborhood of $p$.
== Proof ==
== Also see ==
* Local Normal Form for Immersions
Category:Submersions
Category:Implicit Functions
Local Orthonormal Coframe defines Riemannian Density

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $\tuple {\epsilon^i}$ be an local orthonormal coframe.
Let $\mu$ be a Riemannian density.
Then there exists $\mu$ such that:
:$\mu = \size {\epsilon^1 \wedge \ldots \wedge \epsilon^n}$
where $\wedge$ denotes the wedge product.
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Riemannian Geometry
Local Orthonormal Frame and Coframe related by Index Raising

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $\tuple {E_i}$ be the local frame of $M$.
Let $\tuple {\epsilon^i}$ the local coframe dual to $\tuple {E_i}$.
Let $\sharp$ be the sharp operator.
Then the following are equivalent:
:$\tuple {E_i}$ is orthonormal.
:$\tuple {\epsilon^i}$ is orthonormal.
:$\forall i \in \N_{1 \mathop \le i \mathop \le n} : \tuple {\epsilon^i}^\sharp = E_i$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Riemannian Geometry
Local Uniform Convergence Implies Compact Convergence

== Theorem ==
Let $X$ be a topological space.
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {f_n}$ be a sequence of mappings $f_n : X \to M$.
Let $f_n$ converge locally uniformly to $f : X \to M$.
Then $f_n$ converges compactly to $f$.
== Proof ==
Recall that a sequence $f_n$ converges compactly to $f$, if $f_n$ converges uniformly to $f$ over any compact subset $K \subset X$.
Consider a sequence $f_n$ which converges locally uniformly to $f : X \to M$.f
Let $K$ be an arbitrary subset $K \subset X$.
For $x \in K$ there exists an open neighborhood $U_x$ such that over $U_x$ the sequence $f_n$ converges uniformly to $f$.
Observe that $\set {U_x : x \in K}$ is an open cover of $K$.
Since $K$ is compact ( the induced topology), every open cover for $K$ has a finite subcover.
So there exists a finite collection of points $x_1, \ldots, x_\ell$ such that $\set {U_{x_1}, \ldots, U_{x_\ell} }$ is an open subcover of $K$. 
Fix $\epsilon > 0$.
Consider $i \in \set {1, \ldots, \ell}$.
Then there exists $N_i \in \N$ such that:
:$\forall n \ge N_i: \sup \set {\map d {\map {f_n} z, \map f z} : z \in U_{x_i} } < \epsilon$
Take $N = \max \set {N_1, \ldots, N_\ell}$. 
Then for $\ds z \in \bigcup_{i \mathop = 1}^\ell U_{x_i}$:
:$\sup \set {\map d {\map {f_n} z, \map f z} : z \in U_{x_i} } < \epsilon$
But since:
:$\ds K \subset \bigcup_{i \mathop = 1}^\ell U_{x_i}$
and $\epsilon$ is arbitrary, we conclude that $f_n$ converges uniformly over $K$.
The fact that $K$ is an arbitrary compact subset finishes the proof. 
== Also see ==
* Compact Convergence Implies Local Uniform Convergence if Weakly Locally Compact
* Equivalence of Local Uniform Convergence and Compact Convergence
Category:Uniform Convergence
Category:Compact Convergence
Locale of Open Sets Functor is Covariant

== Theorem ==
Let $\mathbf{Top}$ denote the category of topological spaces.
Let $\mathbf{Loc}$ denote the category of locales.
Then:
:the open sets functor $\mathbf \Omega : \mathbf{Top} \to \mathbf{Loc}$ is a covariant functor
== Proof ==
Recall the open sets functor $\mathbf \Omega : \mathbf{Top} \to \mathbf{Loc}$ is defined by:
=== Object Functor is Well-Defined ===
From Locale of Topological Space is Locale:
:given a topological space $T$:
::$\map \Omega T$ is a locale.
Hence the object functor is well-defined.
=== Arrow Functor is Well-Defined ===
By definition of continuous map induced by continuous mapping:
:$\map \Omega f = F^{\operatorname{op}}$
where $F : \map \Omega {T_2} \to \map \Omega {T_1}$ is the frame homomorphism of $f$
From Frame Homomorphism of Continuous Mapping is Frame Homomorphism:
:the frame homomorphism of $f$ is a frame homomorphism.
Hence the arrow functor is well-defined.
=== Arrow Functor Preserves Composites ===
Let $T_1 = \struct{S_1, \tau_1}, T_2 = \struct{S_2, \tau_2}, T_3 = \struct{S_3, \tau_3}$ be topological spaces.
Let $f : T_1 \to T_2$ and $g : T_2 \to T_3$ be continuous mappings.
Let:
:$\map \Omega f = F^{\operatorname{op}}$
where $F : \map \Omega {T_2} \to \map \Omega {T_1}$ is the frame homomorphism of $f$
Let:
:$\map \Omega g = G^{\operatorname{op}}$
where $G : \map \Omega {T_3} \to \map \Omega {T_2}$ is the frame homomorphism of $g$
Let:
:$\map \Omega {g \circ f} = H^{\operatorname{op}}$
where $H : \map \Omega {T_3} \to \map \Omega {T_1}$ is the frame homomorphism of $g \circ f$
From Frame of Open Sets Functor is Contravariant:
:$H = F \circ G$
We have
It follows that the arrow functor of $\Omega$ preserves composites. 
=== Arrow Functor Preserves Identities ===
Let $T = \struct {S, \tau}$ be a topological space.
Let:
:$\map \Omega {\operatorname{id}_S} = F^{\operatorname{op}}$
where $I : \map \Omega T \to \map \Omega T$ is the frame homomorphism of $\operatorname{id}_S$
From Frame of Open Sets Functor is Contravariant:
:$I = \operatorname{id}_{\map \Omega T}$
By definition of :
:$\paren{\operatorname{id}_{\map \Omega T}}^{\operatorname{op}} = \operatorname{id}_{\map \Omega T}$
Hence:
:$\map \Omega {\operatorname{id}_S} = \operatorname{id}_{\map \Omega T}$
It follows that the arrow functor of $\Omega$ preserves identities. 
It follows that $\Omega$ is a covariant functor by definition.
== Sources ==
* : Chapter II: Frames and Locales. Spectra, $\S 1.3$
Category:Functors
Locale of Topological Space is Locale

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\map \Omega T$ denote the locale of $T$. 
Then:
:$\map \Omega T$ is a locale
== Proof ==
By definition of frame of topological space:
:the frame of $T$ is $\map \Omega T$
From Frame of Topological Space is Frame:
:$\map \Omega T$ is a frame
By definition of locale:
:$\map \Omega T$ is a locale
== Also see ==
* Continuous Mapping Induces Continuous Morphism
== Sources ==
* : Chapter II: Introduction to Locales, $\S1.1$
Category:Topological Spaces
Category:Locales