nyuuzyou/wikis huggingface dataset:
Completely Hausdorff Property is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Then:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
That is, for any two distinct elements $x, y \in S$ there exist open sets $U, V \in \tau$ containing $x$ and $y$ respectively whose closures are disjoint.
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U, V \in \tau: x \in U, y \in V, U^- \cap V^- = \O$
As $x, y \in H$ we have that:
:$x \in U \cap H, y \in V \cap H: \paren {U^- \cap H} \cap \paren {V^- \cap H} = \O$
From corollary $1$ to Closure of Subset in Subspace we have that:
:$\paren {U \cap H}^- \subseteq U^- \cap H$
:$\paren {V \cap H}^- \subseteq V^- \cap H$
where $\paren {U \cap H}^-$ and $\paren {V \cap H}^-$ are the closures in $H$ of the respective sets.
So:
:$x \in U \cap H, y \in V \cap H: \paren {U \cap H}^- \cap \paren {V \cap H}^- = \O$
and so the $T_{2 \frac 1 2}$ (completely Hausdorff) axiom is satisfied in $H$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Completely Hausdorff Spaces
Category:Topological Subspaces
Completely Hausdorff Space is Hausdorff Space

== Theorem ==
Let $\struct {S, \tau}$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a completely Hausdorff space.
From the definition:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
We have that Set is Subset of its Topological Closure and so $U \subseteq U^-$ and $V \subseteq V^-$.
This leads to:
:$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
which is precisely the definition of a Hausdorff ($T_2$) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Hausdorff Spaces
Category:Completely Hausdorff Spaces
Category:Hausdorff Spaces
Completely Hausdorff Space is Preserved under Closed Bijection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space, then so is $T_B$.
== Proof ==
Let $T_A$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Then:
:$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A^- \cap V_A^- = \O$
where $U_A^-$ signifies the closure of $U_A$.
Suppose that $T_B$ is not completely Hausdorff.
Then:
:$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B^- \cap V_B^- \ne \O$
That is, there exists at least one pair of points $a$ and $b$ for which the closures of all the open sets containing $a$ and $b$ are not disjoint.
From Topological Closure is Closed, both $U_B^-$ and $V_B^-$ are closed.
From Topology Defined by Closed Sets, it follows that $W_B = U_B^- \cap V_B^-$ is also a closed set.
From Preimage of Intersection under Mapping, we have:
:$\phi^{-1} \sqbrk {U_B^- \cap V_B^-} = \phi^{-1} \sqbrk {U_B^-} \cap \phi^{-1} \sqbrk {V_B^-}$
Let $H_A = \phi^{-1} \sqbrk {U_B^-}, K_A = \phi^{-1} \sqbrk {V_B^-}, W_A = \phi^{-1} \sqbrk {W_B}$.
Then we have:
:$H_A \cap K_A = W_A \ne \O$
By Bijection is Open iff Closed, $\phi$ is an open bijection.
By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.
From Continuity Defined from Closed Sets, we have that $H_A$, $K_A$ and $W_A$ are all closed.
Also, let $U_A = \phi^{-1} \sqbrk {U_B}$ and $V_A = \phi^{-1} \sqbrk {V_B}$.
From Continuity Defined by Closure, we have that:
:$\phi^{-1} \sqbrk {U_B^-} \subseteq \paren {\phi^{-1} \sqbrk {U_B} }^-$
:$\phi^{-1} \sqbrk {V_B^-} \subseteq \paren {\phi^{-1} \sqbrk {V_B} }^-$
that is:
:$H_A \subseteq U_A^-$
:$K_A \subseteq V_A^-$
As $H_A \cap K_A \ne \O$ it follows that $U_A^- \cap V_A^- \ne \O$.
From the bijective nature of $\phi$, we have that:
:$a \in U_B \implies \map {\phi^{-1} } a \in U_A$
:$b \in V_B \implies \map {\phi^{-1} } b \in V_A$
Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$.
Thus we have that:
:$\exists x, y \in X_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A^- \cap V_A^- \ne \O$
contradicting the fact that $T_A$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
Hence $T_B$ must after all be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Completely Hausdorff Spaces
Completely Hausdorff Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space, then so is $T_B$.
== Proof ==
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Closed Bijection.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Completely Hausdorff Spaces
Category:Homeomorphisms (Topological Spaces)
Completely Irreducible Element equals Infimum of Subset implies Element Belongs to Subset

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X \subseteq S$, $p \in S$ such that
:$p$ is completely irreducible and $p = \inf X$
Then $p \in X$
== Proof ==
:$p \notin X$
By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element:
:$\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$
where $p^\succeq$ denotes the upper closure of $p$.
We will prove that
:$q$ is lower bound for $X$
Let $x \in X$.
By definitions of infimum and lower bound
:$p \preceq x$
By assumption:
:$x \ne p$
By definition of strictly precede:
:$p \prec x$
Thus by step condition:
:$q \preceq x$
By definition of strictly precede:
:$p \preceq q$
We will prove that
:$\forall b \in S: b$ is lower bound for $X \implies b \preceq q$
Let $b \in S$ such that
:$b$ is lower bound for $X$.
By definition of infimum:
:$b \preceq p$
Thus by definition of transitivity:
:$b \preceq q$
By definition of infimum:
:$q = \inf X = p$
Thus this contradicts $p \ne q$ by definition of strictly precede.
== Sources ==
* 
* 
Category:Meet Irreducible Elements
Completely Irreducible and Subset Admits Infimum Equals Element implies Element Belongs to Subset

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $x \in S$ such that
:$x$ is completely irreducible.
Let $X \subseteq S$ such that
:$X$ admits an infimum and $x = \inf X$
Then $x \in X$
== Proof ==
:$x \notin X$
By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element
:$\exists q \in S: x \prec q \land \left({\forall s \in S: x \prec s \implies q \preceq s}\right) \land x^\succeq = \left\{ {x}\right\} \cup q^\succeq$
where $x^\succeq$ denotes the upper closure of $x$.
We will prove that
:$X \subseteq q^\succeq$
Let $y \in X$.
By definitions of infimum and lower bound:
:$x \preceq y$
By definition of strictly precede:
:$x \prec y$
By existence of $q$:
:$q \preceq y$
Thus by definition of upper closure of element:
:$y \in q^\succeq$
By Infimum of Upper Closure of Element:
:$\inf\left({q^\succeq}\right) = q$
By Infimum of Subset:
:$q \preceq x$
Thus this contrsdicts $x \prec q$
== Sources ==
* 
* 
Category:Meet Irreducible Elements
Completely Irreducible implies Infimum differs from Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $p \in S$ such that
:$p$ is completely irreducible.
Then $\map \inf {p^\succeq \setminus \set p} \ne p$
where $p^\succeq$ donotes the upper closure of $p$.
== Proof ==
By definition of completely irreducible:
:$p^\succeq \setminus \set p$ admits a minimum.
Then:
:$p^\succeq \setminus \set p$ admits a infimum and $\map \inf {p^\succeq \setminus \set p} \in p^\succeq \setminus \set p$
By definition of difference:
:$\map \inf {p^\succeq \setminus \set p} \notin \set p$
Thus by definition of singleton:
:$\map \inf {p^\succeq \setminus \set p} \ne p$
== Sources ==
* 
Category:Meet Irreducible Elements
Completely Irreducible implies Meet Irreducible

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $p \in S$.
Then if $p$ is completely irreducible, then $p$ is meet irreducible.
== Proof ==
Assume that
:$p$ is completely irreducible.
By Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element
:$\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$
where $p^\succeq$ denotes the upper closure of $p$.
Let $a, b \in S$ such that
:$p = a \wedge b$
:$a \ne p$ and $b \ne p$
By Meet Precedes Operands:
:$p \preceq b$
By definition of strictly precede:
:$p \prec b$
Then
:$q \preceq b$
By definition of strictly precede:
:$p \preceq q$
By Meet Precedes Operands:
:$p \preceq a$
By definition of strictly precede:
:$p \prec a$
Then
:$q \preceq a$
By Meet Semilattice is Ordered Structure:
:$q \preceq p$
By definition of antisymmetry:
:$p = q$
Thus this contradicts $p \prec q$ by definition of strictly precede.
== Sources ==
* 
Category:Meet Irreducible Elements
Completely Metrizable Topology is Metrizable Topology

== Theorem ==
Let $\struct {S, \tau}$ be a completely metrizable topological space.
Then, $\struct {S, \tau}$ is a metrizable topological space.
== Proof ==
Since $\struct {S, \tau}$ is completely metrizable, $\tau$ is induced by a (complete) metric $d$.
In particular, $\struct {S, d}$ is a metric space.
As $\struct {S, \tau}$ is homeomorphic to itself by the identity mapping, it is metrizable.
Category:Metrizable Topologies
Category:Completely Metrizable Topologies
Completely Multiplicative Function is Multiplicative

== Theorem ==
Let $f: \Z \to \Z$ be a function on the integers $\Z$.
Let $f$ be completely multiplicative.
Then $f$ is multiplicative.
== Proof ==
By definition of complete multiplicativity:
:$\forall m, n \in \Z: \map f {m n} = \map f m \map f n$
Hence by True Statement is implied by Every Statement:
:$\forall m, n \in \Z: m \perp n \implies \map f {m n} = \map f m \map f n$
So $f$ is multiplicative.
Category:Number Theory
Category:Completely Multiplicative Functions
Category:Multiplicative Functions
Completely Multiplicative Function of Quotient

== Theorem ==
Let $f: \R \to \R$ be a completely multiplicative function.
Then:
:$\forall x, y \in \R, y \ne 0: \map f {\dfrac x y} = \dfrac {\map f x} {\map f y}$
whenever $\map f y \ne 0$.
== Proof ==
Let $z = \dfrac x y$.
Then:
Category:Multiplicative Functions
Completely Normal Space is Normal Space

== Theorem ==
Let $\struct {S, \tau}$ be a completely normal space.
Then $\struct {S, \tau}$ is also a normal space.
== Proof ==
Let $\struct {S, \tau}$ be a completely normal space.
From the definition, $\struct {S, \tau}$ is a completely normal space :
:$\struct {S, \tau}$ is a $T_5$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
We have that a $T_5$ space is a $T_4$ space.
So:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
which is precisely the definition of a normal space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Regular and Normal Spaces
Category:Completely Normal Spaces
Category:Normal Spaces
Completely Normal Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a completely normal space, then so is $T_B$.
== Proof ==
We have that $\struct {S, \tau}$ is a completely normal space :
:$\struct {S, \tau}$ is a $T_5$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
From $T_5$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_5$ space, then so is $T_B$.
From $T_1$ Space is Preserved under Homeomorphism:
:If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Homeomorphisms (Topological Spaces)
Category:Completely Normal Spaces
Completely Normal iff Every Subspace is Normal

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then $T$ is a completely normal space every subspace of $T$ is normal.
== Proof ==
From the definitions, we have that:
:$T$ is a completely normal space :
::$\struct {S, \tau}$ is a $T_5$ space
::$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
:$T$ is a normal space :
::$\struct {S, \tau}$ is a $T_4$ space
::$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
From $T_1$ Property is Hereditary, any subspace of a $T_1$ space is also a $T_1$ space.
Then we have that a space is $T_5$ iff every subspace is $T_4$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Completely Normal Spaces
Category:Normal Spaces
Category:Topological Subspaces
Completely Prime Filter Induced by Frame Homomorphism Induced by Completely Prime Filter

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a frame.
Let $\struct{\mathbf 2, \vee, \wedge, \preceq}$ denote the (Boolean Lattice) $\mathbf 2$.
For every completely prime filter $p$ of $L$, let:
:$\phi_p : L \to \mathbf 2$ be the frame homomorphism defined by:
::$\forall a \in L : \map {\phi_p} a = \begin{cases}
\top & : a \in p\\
\bot & : a \notin p 
\end{cases}$
For every frame homomorphism $\phi : L \to \mathbf 2$, let:
:$p_\phi = \map {\phi^{-1}} \top$ 
where 
:$\map {\phi^{-1}} \top$ denotes the preimage of $\top \in \mathbf 2$ under the mapping $\phi$
which is the completely prime filter of $L$ induced by $\phi$ 
Let $p$ be a completely prime filter of $L$.
Then:
:$p_{\phi_p} = p$
== Proof ==
From Frame Homomorphism Onto Two Induced by Completely Prime Filter:
:For every completely prime filter $p$ of $L$, $\phi_p : L \to \mathbf 2$ is a frame homomorphism
From Completely Prime Filter Induced by Frame Homomorphism Onto Two:
:For every frame homomorphism $\phi : L \to \mathbf 2$, $p_\phi = \map {\phi^{-1}} \top$ is a completely prime filter of $L$
We have:
Category:Completely Prime Filters
Category:Two (Boolean Lattice)
Category:Frame Homomorphisms
Completely Prime Filter Induced by Frame Homomorphism Onto Two

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a frame.
Let $\struct{\mathbf 2, \vee, \wedge, \preceq}$ denote the (Boolean Lattice) $\mathbf 2$.
Let $\phi : L \to \mathbf 2$ be a frame homomorphism.
Then:
:$\map {\phi^{-1}} \top$ is a completely prime filter
where $\map {\phi^{-1}} \top$ denotes the preimage of $\top \in \mathbf 2$ under the mapping $\phi$
== Proof ==
From Singleton of Greatest Element in Two is Completely Prime Filter:
:$\set \top$ is a completely prime filter
From Inverse Image Mapping of Frame Homomorphism Preserves Completely Prime Filter:
:$\map {\phi^\gets}{\set \top}$ is a completely prime filter
We have:
The result follows.
Category:Completely Prime Filters
Category:Two (Boolean Lattice)
Category:Frame Homomorphisms
Completely Prime Filter Induced by Meet Irreducible Element

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a frame.
Let $b$ be a meet-irreducible element of $L$.
Let:
:$p = \set{a \in L : a \npreceq b}$
Then:
:$p$ is a completely prime filter of $L$.
== Proof ==
From meet irreducible element, it is sufficient to show:
:$(1)\quad\forall A \subseteq L : \bigvee A \in p \iff \paren{\exists a \in A : a \in p}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in p \iff \paren{\forall a \in A : a \in p}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$
By definition of $p$, it is sufficient to show:
:$(1')\quad\forall A \subseteq L : \bigvee A \npreceq b \iff \paren{\exists a \in A : a \npreceq b}$
:$(2')\quad\forall $ finite $A \subseteq L : \bigwedge A \npreceq b \iff \paren{\forall a \in A : a \npreceq b}$
=== Statement $(1')$ ===
Statement $(1')$ follows from the contrapositive statement of Characterization of Supremum Precedes Element.
=== Statement $(2')$ ===
Let $A \subseteq L$ be finite.
Suppose $\bigwedge A \npreceq b$.
By definition of infimum:
:$\bigwedge A$ is a lower bound for $A$.
By definition of lower bound:
:$\forall a \in A : \bigwedge A \preceq a$
By :
:$\forall a \in A : a \npreceq b$
Suppose $\bigwedge A \preceq b$.
From Prime Element iff Meet Irreducible in Distributive Lattice:
:$b$ is a prime element
From Characterization of Prime Element in Meet Semilattice:
:$\exists a \in A : a \preceq b$
It follows that Statement $(2')$ holds.
The result follows.
Category:Meet Irreducible Elements
Category:Completely Prime Filters
Completely Prime Filter Induced by Meet Irreducible Induced by Completely Prime Filter

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a frame.
Let $\top$ denote the greatest element of $L$.
For every completely prime filter $p$ of $L$, let:
:$b_p = \bigvee \set{a \in L : a \notin p}$
where:
:$\bigvee \set{a \in L : a \notin p}$ denotes the supremum of the set $\set{a \in L : a \notin p}$
which is a meet irreducible element not equal to $\top$
For every meet-irreducible element $b$ of $L$ not equal to $\top$, let:
:$p_b = \set{a \in L : a \npreceq b}$
where
:$p_b$ is a completely prime filter of $L$.
Let $p$ be a completely prime filter $p$ of $L$.
Then:
:$p_{b_p} = p$
== Proof ==
We have:
Category:Completely Prime Filters
Category:Meet Irreducible Elements
Completely Prime Filter Induces Meet-Irreducible Open Set

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\FF$ be a completely prime filter in the complete lattice $\struct{\tau, \subseteq}$.
Let $W = \bigcup \set{U \in \tau : U \notin \FF}$.
Then:
:$W$ is a meet-irreducible open set
== Proof ==
By definition of completely prime filter:
:$\FF$ is a proper subset of $\tau$
Hence:
:$\set{U \in \tau : U \notin \FF} \ne \O$
By :
:$W \in \tau$
From Filter Contains Greatest Element:
:$S \in \FF$
By definition of completely prime filter:
:$W \notin \FF$
and so:
:$W \ne S$
Let $U_1, U_2 \in \tau$:
:$U_1 \cap U_2 \subseteq W$
By the contrapositive statement of :
:$U_1 \cap U_2 \notin \FF$
as $W \notin \FF$
By the contrapositive statement of :
:either $U_1 \notin \FF$ or $U_2 \notin \FF$
Hence:
:either $U_1 \subseteq W$ or $U_2 \subseteq W$
From Characterization of Meet-Irreducible Open Set, $W$ is a meet-irreducible open set.
== Also see ==
* Topology forms Complete Lattice
Category:Meet-Irreducible Open Sets
Category:Completely Prime Filters
Completely Prime Ideal is Dual of Completely Prime Filter

== Theorem ==
Let $\struct{L, \preceq}$ be a complete lattice.
Let $K \subseteq L$.
The following are dual statements:
:$K$ is a completely prime filter of $\struct {L, \preceq}$
:$K$ is a completely prime ideal of $\struct {L, \preceq}$
== Proof ==
By definition of completely prime filter:
:$K$ is a completely prime filter of $\struct {L, \preceq}$
:
:$(1)\quad K$ is a proper filter
:$(2)\quad \forall A \subseteq L: \paren{\wedge A \in K \implies A \cap K \ne \O}$
where $\wedge A$ denotes the infimum of $A$
By the duality principle, the dual of this statement is:
:$(1')\quad K$ is a proper ideal
:$(2')\quad \forall A \subseteq L: \paren{\vee A \in K \implies A \cap K \ne \O}$
where $\vee A$ denotes the supremum of $A$
By definition of completely prime ideal:
:$K$ is a completely prime ideal of $\struct {L, \preceq}$
Category:Dual Pairs (Order Theory)
Category:Completely Prime Filters
Category:Completely Prime Ideals
Completeness Criterion (Metric Spaces)

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $A \subseteq S$ be a dense subset.
Suppose that every Cauchy sequence in $A$ converges in $M$.
Then $M$ is complete.
== Proof 1 ==
== Proof 2 ==
Category:Metric Spaces
Category:Complete Metric Spaces
Category:Completeness Criterion (Metric Spaces)
Completeness Criterion (Metric Spaces)/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $S$.
For each $n$ pick a Cauchy sequence $\sequence {y_{n, m} }_{m \mathop \in \N}$ in $A$ converging to $x_n$ like so:
:
Let $N \in \N$ be such that $\map d {x_{n_1}, x_{n_2} } < \epsilon / 3$ for all $n_1, n_2 > N$.
Let $M \in \N$ be such that $\map d {y_{n_i, m}, x_{n_i} } < \epsilon / 3$ for all $m > M$ and all $n_1, n_2 > N$. 
Let $m > M$.
Let $n_1, n_2 > N$.
We have:
Therefore $\sequence {y_{m, n} }_{n \mathop \in \N}$ is Cauchy in $A$ for $m > M$.
So $\sequence {y_{m, n} }_{n \mathop \in \N}$ converges to some limit $y_n \in S$.
Category:Completeness Criterion (Metric Spaces)
Completeness Criterion (Metric Spaces)/Proof 2

== Theorem ==
== Proof ==
Let $\left\langle{x_n}\right\rangle$ be a Cauchy sequence in $M$.
Since $A$ is dense, we can invoke the Axiom of Countable Choice to choose for each $n$ some $y_n \in A$ which is within $1 / n$ of $x_n$.
We will show that $\left\langle{y_n}\right\rangle$ is Cauchy.
Let $\epsilon > 0$.
Since $\left\langle{x_n}\right\rangle$ is Cauchy, there exists some $N'$ such that:
:$\forall n, m \ge N': d \left({x_n, x_m}\right) < \frac 1 3 \epsilon$
Let $N \in \N$ be greater than $N'$ and $3 \epsilon^{-1}$.
Note that $n, m \ge N$ implies that both:
: $(1): \quad n, m \ge N'$
: $(2): \quad \dfrac 1 n, \dfrac 1 m \le \dfrac 1 N < \dfrac 1 3 \epsilon$
Thus for $n, m \ge N$, by the Triangle Inequality and the above:
Hence $\left\langle{y_n}\right\rangle$ is a Cauchy sequence in $A$.
By assumption, it converges to some limit $y \in S$.
Now, we verify that $\left\langle{x_n}\right\rangle$ converges to $y$ as well.
Let $\epsilon > 0$.
Since $\left\langle{y_n}\right\rangle$ converges, there exists some $N'$ such that:
:$\forall n \ge N': d \left({y_n, y}\right) < \dfrac 1 2 \epsilon$
Let $N \in \N$ be greater than $N'$ and $2 \epsilon^{-1}$.
Note that $n \ge N$ implies that both:
: $(1): \quad n \ge N'$
: $(2): \quad \dfrac 1 n \le \dfrac 1 N < \dfrac 1 2 \epsilon$
Thus, for $n \ge N$, we have by the Triangle Inequality and above that:
Hence $\left\langle{x_n}\right\rangle$ converges to $y$.
Category:Completeness Criterion (Metric Spaces)
Completeness Theorem

=== Completeness Theorem for Semantic Tableaux ===
=== Completeness Theorem for Propositional Tableaux and Boolean Interpretations ===
==== Extended Completeness Theorem for Propositional Tableaux and Boolean Interpretations ====
=== Completeness Theorem for Hilbert Proof System Instance 2 and Boolean Interpretations ===
=== Gödel's Completeness Theorem ===
== Sources ==
* 
* 
Category:Completeness Theorem
Category:Mathematical Logic
Completeness Theorem for Hilbert Proof System Instance 2 and Boolean Interpretations

== Theorem ==
Instance 2 of the Hilbert proof systems is a complete proof system for boolean interpretations.
That is, for every WFF $\mathbf A$:
:$\models_{\mathrm{BI}} \mathbf A$ implies $\vdash_{\mathscr H_2} \mathbf A$
== Proof ==
== Also see ==
The Soundness Theorem for Hilbert Proof System Instance 2 and Boolean Interpretations in which it is proved that:
:If $\vdash \mathbf A$ then $\models \mathbf A$.
== Sources ==
* : $\S 4.8$: Completeness: Theorem $14$
Category:Completeness Theorem
Category:Hilbert Proof System Instance 2
Category:Named Theorems