nyuuzyou/wikis huggingface dataset:
Characterization of Lipschitz Continuity on Shift of Finite Type by Variations

== Theorem ==
Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $f : X_\mathbf A \to \C$ be a continuous mapping.
Let $\theta \in \openint 0 1$.
Let $C > 0$.
Then:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
:
:$\forall n \in \N : \map {\mathrm {var}_n} f \le C \theta ^n$
where:
:$d_\theta$ denotes the metric
:$\map {\mathrm {var}_n} f$ denotes the $n$th variation of $f$.
That is:
:$\ds \size f_\theta = \sup \set { \dfrac {\cmod {\map f x - \map f y} }{\map {d_\theta} {x,y} } : x,y \in X_\mathbf A, x \ne y}$
where $\size \cdot_\theta $ is the Lipschitz seminorm.
== Proof ==
=== Sufficient Condition ===
Suppose:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
Let $n\in\N$.
By definition of the metric:
:$\forall x, y \in X_\mathbf A:$
::$\forall i \in \openint {-n} n : x_i = y_i \implies \map {d _\theta} {x, y} \le \theta ^n$
Thus, by the assumption:
:$\forall x, y \in X_\mathbf A:$
::$\forall i \in \openint {-n} n :x_i = y_i \implies \size {\map f x - \map f y} \le C \map {d_\theta} {x, y} \le C \theta ^n$
Therefore:
:$\map {\mathrm {var}_n} f \le C \theta ^n$
=== Necessary Condition ===
Suppose:
:$\forall n \in \N : \map {\mathrm {var}_n} f \le C \theta ^n$
Let $x, y \in X_\mathbf A$.
If $x = y$, then obviously:
:$\size {\map f x - \map f y} = 0 \le C \map {d_\theta} {x, y}$
So, let $x \ne y$.
Then:
:$\exists N \in \N : d_\theta (x,y) = \theta ^N$
In particular:
:$\forall i \in \openint {-N} N : x_i = y_i$
Therefore:
:$\size {\map f x - \map f y}\le \map {\mathrm {var}_N} f\le C \theta ^N = C \map {d_\theta} {x, y}$
Category:Lipschitz Continuity
Category:Metric Spaces
Characterization of Locale

== Theorem ==
Let $L = \struct{S, \preceq}$ be an ordered set.
:$(1)\quad L$ is a locale
:$(2)\quad L$ is a frame
:$(3)\quad L$ is a complete lattice satisfying the infinite join distributive law
:$(4)\quad L$ is a complete Heyting algebra
:$(5)\quad L$ is a complete Brouwerian lattice
== Proof ==
=== Statement $(1)$ Statement $(2)$ ===
The equivalence of Statement $(1)$ and Statement $(2)$ follows immediately from the definition of locale
=== Statement $(2)$ Statement $(3)$ ===
The equivalence of Statement $(2)$ and Statement $(3)$ follows immediately from the definition of frame
=== Statement $(3)$ implies Statement $(4)$ ===
Let $L$ be a complete lattice satisfying the infinite join distributive law.
=== Statement $(4)$ implies Statement $(5)$ ===
Let $L$ be a complete Heyting algebra.
That $L$ is a complete Brouwerian lattice follows immediately from the definition of Heyting algebra.
=== Statement $(5)$ implies Statement $(3)$ ===
Let $L$ be a complete Brouwerian lattice.
Category:Locales
Category:Frames
Category:Brouwerian Lattices
Category:Heyting Algebras
Category:Complete Lattices
Category:Characterization of Locale
Characterization of Locale/Statement 3 Implies Statement 4

== Theorem ==
Let $L = \struct{S, \preceq}$ be an complete lattice satisfying the infinite join distributive law.
Then:
:$L$ is a Heyting algebra
== Proof ==
Let $a, b \in S$.
Let $a \to b = \sup \set{c \in S : a \wedge c \preceq b}$
We have:
Hence:
:$a \to b$ is the greatest element $c$ such that:
::$c \wedge a \preceq b$
It follows that $a \to b$ is a relative psuedocomplement by definition.
Since $a, b$ were arbitrary, then:
:$\forall a, b \in S : \exists a \to b \in S : a \to b$ is the greatest element $c \in L$ such that $a \wedge c \preceq b$
Hence $L$ is a Heyting algebra by definition.
Category:Characterization of Locale
Characterization of Locale/Statement 5 Implies Statement 3

== Theorem ==
Let $L = \struct{S, \preceq}$ be a complete Brouwerian lattice.
Then:
:$L$ satisfies the infinite join distributive law
== Proof ==
Let $A \subseteq S$.
Let $a \in S$.
==== Lemma 1 ====
:$\sup \set{a \wedge b : b \in A} \preceq a \wedge \sup A$
==== Proof of Lemma 1 ====
By definition of meet::
:$\forall b \in A : a \wedge b \preceq a, a \wedge b \preceq b$
By definition of upper bound:
:$a$ is an upper bound for $\set{a \wedge b : b \in A}$
By definition of supremum:
:$\sup \set{a \wedge b : b \in A} \preceq a$
From Finer Supremum Precedes Supremum:
:$\sup \set{a \wedge b : b \in A} \preceq \sup A$
By definition of meet:
:$\sup \set{a \wedge b : b \in A} \preceq a \wedge \sup A$
==== Lemma 2 ====
:$a \wedge \sup A \preceq \sup \set{a \wedge b : b \in A}$
==== Proof of Lemma 2 ====
By definition of supremum:
:$\forall b \in A : a \wedge b \preceq \sup \set{a \wedge b : b \in A}$
From Relative Pseudocomplement Preserves Order:
:$\forall b \in A : a \to a \wedge b \preceq a \to \sup \set{a \wedge b : b \in A}$
By definition of supremum:
:$\sup \set{a \to a \wedge b : b \in A} \preceq a \to \sup \set{a \wedge b : b \in A}$
From Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice:
:$\forall b \in A : b \preceq a \to a \wedge b$
From Finer Supremum Precedes Supremum:
:$\sup A \preceq \sup \set{a \to a \wedge b : b \in A}$
By :
:$\sup A \preceq a \to \sup \set{a \wedge b : b \in A}$
From Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice:
:$a \wedge \sup A \preceq \sup \set{a \wedge b : b \in A}$
By :
:$a \wedge \sup A = \sup \set{a \wedge b : b \in A}$
Since $A$ and $a$ were arbitrary:
:$\forall A \subseteq S, a \in S : a \wedge \sup A = \sup \set{a \wedge b : b \in A}$
It follows that $L$ satisfies the infinite join distributive law by definition.
Category:Characterization of Locale
Characterization of Localic Mapping Induced by Continuous Mapping

== Theorem ==
Let $T_1 = \struct{S_1, \tau_1}, T_2 = \struct{S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to \T_2$ be a continuous mapping.
Let $\map \Omega {T_1}, \map \Omega {T_2}$ be the locales of $T_1$ and $T_2$ respectively.
Let $F : \map \Omega {T_1} \to \map \Omega {T_2}$ be the localic mapping induced by $f$.
Then:
:$\forall U \in \tau_1 : \map F U = S_2 \setminus f \sqbrk{S_1 \setminus U}^-$
where:
:$S_2 \setminus f \sqbrk{S_1 \setminus U}^-$ denotes the relative complement of $f \sqbrk{S_1 \setminus U}^-$ in $S_2$ 
:$f \sqbrk{S_1 \setminus U}^-$ denotes the closure of $f \sqbrk{S_1 \setminus U}$ in $T_2$
:$f \sqbrk{S_1 \setminus U}$ denotes the image of $S_1 \setminus U$ under $f$
:$S_1 \setminus U$ denotes the relative complement of $U$ in $S_1$
== Proof ==
Let $G : \map \Omega {T_2} \to \map \Omega {T_1}$ be the frame homomorphism of $f$.
By definition of localic mapping induced by $f$, $F$ is the upper adjoint to $G$.
Hence:
:$\tuple{F, G}$ is a Galois connection
We have:
Next, we have:
And we have:
From $\paren 2$, $\paren 3$ and set equality:
:$\forall U \in \tau_1 : \map F U = S_2 \setminus f \sqbrk {S_1 \setminus U}^-$
Category:Localic Mappings
Characterization of Lower Semicontinuity

== Theorem ==
Let $f: S \to \overline \R$ be an extended real valued function.
Let $S$ be endowed with a topology $\tau$.
The following are equivalent:
:$(1): \quad$ $f$ is lower semicontinuous (LSC) on $S$.
:$(2): \quad$ The epigraph $\map {\operatorname{epi}} f$ of $f$ is a closed set in $S \times \R$ with the product topology.
:$(3): \quad$ All lower level sets of $f$ are closed in $S$.
== Proof ==
The proof is carried out in the following three steps:
=== LSC implies Closed Epigraph ===
Let $f: S \to \overline \R$ be LSC on $S$.
Let $\map {\operatorname {epi} } f$ denote the epigraph of $f$.
Take a sequence $\sequence {\tuple {x_n, a_n} }_{n \mathop \in \N} \in \map {\operatorname {epi} } f$ such that:
:$\tuple {x_n, a_n} \to \tuple {\bar x, \bar a} \in S \times \R$
as $n \to \infty$.
This implies that $x_n \to \bar x$ and $a_n \to \bar a$ by definition of the product topology on $S \times \R$ and because a Continuous Mapping is Sequentially Continuous.
Consequently:
Thus:
:$\tuple{\bar x, \bar a} \in \map {\operatorname {epi} } f$
=== Closed Epigraph implies Closed Level Sets ===
Let $\map {\operatorname{epi} } f$ be a closed set in $S \times \R$.
Let $\alpha \in \R$.
Let $\operatorname {lev} \limits_{\mathop \le \alpha} f$ denote the $\alpha$-lower level set of $f$.
Then:
:$\ds \operatorname {lev} \limits_{\mathop \le \alpha} f = \map {\operatorname {epi} } f \cap S \times \set \alpha$
is closed in $S$ because:
:closedness is preserved under Intersection|intersection
and:
:$S \times \set \alpha$ is a closed set with respect with the product topology of $S \times \R$.
=== Closed Level Sets implies LSC ===
Let all lower level sets of $f$ be closed.
Let $x \in S$.
Let $\ds a: = \liminf_{t \mathop \to x} \map f t$.
In order to prove that $f$ is LSC it suffices to show that $a = \map f x$. 
We know already that $a \le \map f x$.
Thus it suffices to show that $\map f x \le a$.
Let $a = \infty$.
As $\infty \le \infty$ the proof is complete.
Let $a < \infty$.
Then we can find a sequence $\sequence {x_n}_{n \mathop \in \N}$ such that:
:$x_n \to x$ and $\map f {x_n} \to a$
as $n \to \infty$.
For any $b > a$:
:$\map f {x_n} \le b$
or equivalently:
:$x_n \in \ds \operatorname {lev} \limits_{\mathop \le b} f$
which is closed.
Therefore $\ds \operatorname {lev} \limits_{\mathop \le b} f$ contains the limit $x$.
We have that for all $b > a$:
:$\map f x \le b$
Therefore:
:$\map f x \le a$
Hence the result.
So:
:$(1) \implies (2) \implies (3)$
and the proof is complete.
== Sources ==
* : $\S 2$: Proposition $1.3$
* : Proposition $1.7$
Category:Limits of Mappings
Characterization of Matroid Independent Sets in Terms of Bases

== Theorem == 
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ be the set of bases of the matroid $M$.
Then:
:$\mathscr I = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
== Proof ==
Let $\mathscr I^\prime = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
From Independent Subset is Contained in Base:
:$\mathscr I \subseteq \mathscr I^\prime$
By definition of matroid base:
:$\mathscr B \subseteq \mathscr I$
By matroid axiom $\text I 2$:
:$\mathscr I^\prime \subseteq \mathscr I$
By set equality:
:$\mathscr I = \mathscr I^\prime$
Category:Matroid Bases
Category:Matroid Independent Subsets
Characterization of Maximal Ideal of Unital Commutative Banach Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $M \subseteq A$.
Then $M$ is maximal there exists a character $\phi$ with $M = \ker \phi$.
== Proof ==
=== Necessary Condition ===
This is Maximal Ideal of Unital Commutative Banach Algebra is Kernel of Character.
=== Sufficient Condition ===
This is Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal.
Category:Maximal Ideals of Algebras
Category:Commutative Banach Algebras
Characterization of Measurable Functions

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \overline \R$ be an extended real-valued function.
Then the following are all equivalent:
== Proof ==
Each of $(2)$ up to $(5')$ is equivalent to $(1)$ by combining Mapping Measurable iff Measurable on Generator and Generators for Extended Real Sigma-Algebra.
== Sources ==
* : $8.1$
Category:Measurable Functions
Characterization of Measures

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Denote $\overline \R_{\ge 0}$ for the set of positive extended real numbers.
A mapping $\mu: \Sigma \to \overline \R_{\ge 0}$ is a measure :
:$(1):\quad \map \mu \O = 0$
:$(2):\quad \mu$ is finitely additive
:$(3):\quad$ For every increasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$, if $E_n \uparrow E$, then:
::::$\map \mu E = \ds \lim_{n \mathop \to \infty} \map \mu {E_n}$
where $E_n \uparrow E$ denotes limit of increasing sequence of sets.
Alternatively, and equivalently, $(3)$ may be replaced by either of:
:$(3'):\quad$ For every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map \mu {E_1}$ is finite, if $E_n \downarrow E$, then:
::::$\map \mu E = \ds \lim_{n \mathop \to \infty} \map \mu {E_n}$
:$(3):\quad$ For every decreasing sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ for which $\map \mu {E_1}$ is finite, if $E_n \downarrow \O$, then:
::::$\ds \lim_{n \mathop \to \infty} \map \mu {E_n} = 0$
where $E_n \downarrow E$ denotes limit of decreasing sequence of sets.
== Proof ==
=== Necessary Condition ===
To show is that a measure $\mu$ has the properties $(1)$, $(2)$, $(3)$, $(3')$ and $(3)$.
Property $(1)$ is also part of the definition of measure, and hence is immediate.
Property $(2)$ is precisely the statement of Measure is Finitely Additive Function.
Property $(3)$ is precisely the statement of Measure of Limit of Increasing Sequence of Measurable Sets.
For $(3)$, note that it is a special case of $(3')$.
For property $(3')$, let $\sequence {E_n}_{n \mathop \in \N} \downarrow E$ be a decreasing sequence in $\Sigma$.
Suppose that $\map \mu {E_1} < +\infty$.
By Measure is Monotone, this implies:
:$\forall n \in \N: \map \mu {E_n} < +\infty$
and also:
:$\map \mu E < +\infty$
Now define:
:$\forall n \in \N: F_n := E_1 \setminus E_n$
Then:
:$F_n \uparrow E_1 \setminus E$
Hence, property $(3)$ can be applied as follows:
Here, all expressions involving subtraction are well-defined as $\mu$ takes finite values.
It follows that:
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$
as required.
=== Sufficient Condition ===
The mapping $\mu$ is already satisfying axiom $(1)$ for a measure by the imposition on its codomain.
Also, axiom $(3')$ is identical to assumption $(1)$.
It remains to check axiom $(2)$.
So let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
Define, for $n \in \N$:
:$F_n = \ds \bigcup_{k \mathop = 1}^n E_k$
Then:
:$\forall n \in \N: F_n \subseteq F_{n + 1}$
Also, by Additive Function is Strongly Additive:
:$\ds \forall n \in \N: \map \mu {F_n} = \map \mu {\bigcup_{k \mathop = 1}^n E_k} = \sum_{k \mathop = 1}^n \map \mu {E_k}$
Hence, using condition $(3)$ on the $F_n$, obtain:
This establishes that $\mu$ also satisfies axiom $(2)$ for a measure, and so it is a measure.
Now to show that $(3')$ and $(3)$ can validly replace $(3)$.
As $(3')$ clearly implies $(3)$ (which is a special case of the former), it will suffice to show that $(3)$ implies $(3)$.
== Sources ==
* : $4.4$
Category:Measure Theory
Characterization of Meet-Irreducible Open Set

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $W \in \tau$.
:
:$(1)\quad W$ is a meet-irreducible open set
:$(2)\quad \forall U, V \in \tau : \paren {U \cap V = W \implies U = W \text { or } V = W}$
:$(3)\quad \forall U, V \in \tau : \paren {U \cap V \subseteq W \implies U \subseteq W \text { or } V \subseteq W}$
== Proof ==
By definition of meet-irreducible open set:
:$W$ is meet-irreducible open set 
:
:$W$ is meet-irreducible in the frame $\struct {\tau, \subseteq}$
By definition of meet-irreducible:
:$W$ is meet-irreducible in the frame $\struct {\tau, \subseteq}$ 
:
:$\forall U, V \in \tau : \paren {U \cap V = W \implies U = W \text { or } V = W}$
From Prime Element iff Meet Irreducible in Distributive Lattice:
:$W$ is meet-irreducible in the frame $\struct {\tau, \subseteq}$ 
:
:$W$ is prime element in the frame $\struct {\tau, \subseteq}$
By definition of prime element:
:$W$ is prime element in the frame $\struct {\tau, \subseteq}$
:$\forall U, V \in \tau : \paren {U \cap V \subseteq W \implies U \subseteq W \text { or } V \subseteq W}$
The result follows.
Category:Meet-Irreducible Open Sets
Characterization of Meet Irreducible Element

== Theorem ==
Let $\struct{S, \wedge, \preceq}$ be a meet semilattice.
Let $z \in S$.
Then:
:$z$ is meet irreducible
:$\forall x, y \in S : \leftparen{z \prec x}$ and $\rightparen{z \prec y} \implies \paren{ z \prec x \wedge y}$
where $z \prec x$ denotes that $z \preceq x$ and $z\neq x$. 
== Proof ==
=== Necessary Condition ===
Let $z$ be a meet irreducible element.
Let:
:$x, y \in S : \leftparen{z \prec x}$ and $\rightparen{z \prec y}$
We have:
:$z \ne x$
:$z \ne y$
By definition of meet irreducible element:
:$z \ne x \wedge y$
By definition of infimum:
:$z \preceq x \wedge y$
Hence:
:$z \prec x \wedge y$
The result follows.
=== Sufficient Condition ===
Let:
:$\forall x, y \in S : \leftparen{z \prec x}$ and $\rightparen{z \prec y} \implies \paren{ z \prec x \wedge y}$
Let:
:$x,y \in S : z = x \wedge y$
By definition of infimum:
:$z \preceq x$
:$z \preceq y$
By the contrapositive statement of the hypothesis:
:either $z = x$ or $z = y$
As $x$ and $y$ were arbitrary:
:$\forall x, y \in S : \paren{ z = x \wedge y} \implies \leftparen{z = x}$ or $\rightparen{z = y}$
It follows that $z$ is meet irreducible by definition.
Category:Meet Irreducible Elements
Characterization of Metacategory via Equations

== Theorem ==
Let $\mathbf C_0$ and $\mathbf C_1$ be collections of objects.
Let $\operatorname{Cdm}$ and $\operatorname{Dom}$ assign to every element of $\mathbf C_1$ an element of $\mathbf C_0$.
Let $\operatorname{id}$ assign to every element of $\mathbf C_0$ an element of $\mathbf C_1$.
Denote with $\mathbf C_2$ the collection of pairs $\tuple {f, g}$ of elements of $\mathbf C_1$ satisfying:
:$\Dom g = \Cdm f$
Let $\circ$ assign to every such pair an element of $\mathbf C_1$.
Then $\mathbf C_0, \mathbf C_1, \operatorname{Cdm}, \operatorname{Dom}, \operatorname{id}$ and $\circ$ together determine a metacategory $\mathbf C$ the following seven axioms are satisfied:
where $A$ and $f, g, h$ are arbitrary elements of $\mathbf C_0$ and $\mathbf C_1$, respectively.
Further, in the last two lines, it is presumed that all compositions are defined.
Hence it follows that:
* $\mathbf C_0$ and $\mathbf C_1$ represent the collections of objects and morphisms of $\mathbf C$
* $\operatorname{Dom}$ and $\operatorname{Cdm}$ represent the domain and codomain of a morphism of $\mathbf C$
* $\operatorname{id}$ represents the identity morphisms of $\mathbf C$
* $\mathbf C_2$ represents the collection of composable morphisms of $\mathbf C$
* $\circ$ represents the composition of morphisms in $\mathbf C$
== Proof ==
Let $\mathbf C_0$ and $\mathbf C_1$ be collections of objects.
Let $A$ in $C_O$ be arbitrary.
Let $f, g, h$ in $C_1$ be arbitrary.
By definition of identity morphism:
:$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$
:$f \circ \operatorname{id}_A = f$
:$\operatorname{id}_A \circ g = g$
whenever $A$ is the domain of $f$ or the codomain of $g$, respectively.
From definition of identity morphism, since: 
:$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$:
Therefore:
:$\Dom{\operatorname{id}_A} = A$
Similarly, from definition of identity morphism, since: 
:$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$:
Therefore:
:$\Cdm{\operatorname{id}_A} = A$
From definition of identity morphism again, since:
:$f \circ \operatorname{id}_A = f$
and:
:$\Dom{\operatorname{id}_A} = A$
Therefore:
:$f \circ \operatorname{id}_A = f \circ \operatorname{id}_{\Dom f} $
Hence:
:$f \circ \operatorname{id}_{\Dom f} = f$
From definition of identity morphism again, since:
:$\operatorname{id}_A \circ g = g$
and:
:$\Cdm{\operatorname{id}_A} = A$
Therefore:
:$\operatorname{id}_A \circ f = \operatorname{id}_{\Cdm f} \circ f $
Hence:
:$\operatorname{id}_{\Cdm f} \circ f = f$
== Sources ==
* : $\S 3.1$
Category:Category Theory
Characterization of Minimal Element

== Theorem ==
Let $C$ be a class.
Let $\prec$ be a relation on $C$.
Let $B$ be a subclass of $C$.
Let $x \in B$.
Let $S_x = \set {y \in C: y \prec x \text{ and } y \ne x}$ be the initial segment of $x$ in $C$.
Then $x$ is a minimal element of $B$ $B \cap S_x = \O$.
== Proof ==
=== Necessary Condition ===
Suppose $x$ is a minimal element of $B$.
Then for each $z \in B$ such that $z \ne x$, $z \nprec x$.
Thus $S_x \cap B = \O$.
=== Sufficient Condition ===
Suppose that $x$ is not a minimal element of $B$.
Then for some $z \in B$, $z \prec x$ and $z \ne x$.
Thus $z \in S_x$.
Since $z \in B$, $B \cap S_x \ne \O$.
Category:Class Theory
Characterization of N-Cube

== Theorem ==
Let $\struct {\R^n, d}$ be a Euclidean $n$-Space equipped with the usual metric $d$.
Let $x, y \in \R^n$, where:
:$x = \tuple {x_1, x_2, \ldots, x_n}$
:$y = \tuple {y_1, y_2, \ldots, y_n}$
Let $R > 0$ be fixed.
Let:
:$\ds Q = \set {x, y \in \R^n: \sup_{x, y} \max_i \size {y_i - x_i} \le R}$
Then $Q$ is an $n$-cube.
== Proof ==
For ease of presentation, denote:
:$y - x = r \in \R^n$
and:
:$y_j - x_j = r_j$
for $j = 1, 2, \ldots, n$.
== Sources ==
* : $\S \text P.6$
Category:Metric Spaces
Category:Real Analysis
Characterization of Neighborhood by Basis

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\BB$ be an analytic basis for $\tau$.
Let $N \subseteq S$.
Let $x \in N$.
Then $N$ is a neighborhood at $x$ :
:$\exists B \in \BB : x \in B : B \subseteq N$ 
== Proof ==
From Basis induces Local Basis:
:$\BB_x = \set {B \in \BB: x \in B}$ is a local basis at $x$
By definition of local basis, $N$ is a neighborhood at $x$ :
:$\exists B \in \BB_x : B \subseteq N$
By definition of $\BB_x$, $N$ is a neighborhood at $x$ :
:$\exists B \in \BB : x \in B : B \subseteq N$
Category:Neighborhoods
Characterization of Normal Operators

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\GF$.
Let $A$ be a bounded linear operator on $\HH$.
Then the following are equivalent:
:$(1): \quad A A^* = A^* A$, that is, $A$ is normal
:$(2): \quad \forall h \in H: \norm {A h}_\HH = \norm {A^*h}_\HH$
where:
:$A^*$ denotes the adjoint of $A$
:$\norm {\, \cdot\,}_\HH$ denotes the inner product norm of $\HH$
If $\GF = \C$, these are also equivalent to:
:$(3): \quad \map \Re A \map \Im A = \map \Im A \map \Re A$, that is, the real and imaginary parts of $A$ commute.
== Proof ==
=== $(3)$ equivalent to $(1)$ ===
Suppose $\GF = \C$. 
We have: 
and:
So $\map \Re A \map \Im A = \map \Im A \map \Re A$ 
:$A^\ast A - A A^\ast = -A^\ast A + A A^\ast$
This is equivalent to: 
:$A^\ast A = A A^\ast$
 
=== $(1)$ implies $(2)$ ===
We have: 
=== $(2)$ implies $(1)$ ===
As above, we have: 
:$\norm {A h}_\HH = \sqrt {\innerprod h {A^\ast A h}_\HH}$
So from Adjoint is Involutive, we have: 
:$\norm {A h}_\HH = \sqrt {\innerprod h {A A^\ast h}_\HH}$
So from Inner Product is Sesquilinear, we have: 
:$\innerprod h {\paren {A^\ast A - A A^\ast} h}_\HH = 0$ 
We have: 
So $A^\ast A - A A^\ast$ is Hermitian and: 
:$\innerprod {\paren {A^\ast A - A A^\ast} h} h_\HH = 0$
for each $h \in \HH$.
So by Norm of Hermitian Operator: Corollary, we have: 
:$A^\ast A = A A^\ast$
== Sources ==
* $II.2.16$
Category:Linear Transformations on Hilbert Spaces