avewright/proofwiki-math huggingface dataset:
Theorem: Stone-Weierstrass Theorem/Locally Compact Hausdorff Space/Complex-Valued

Let $X$ be a locally compact Hausdorff space.
Let $\struct {\map {\CC_0} {X, \C}, \norm {\, \cdot \,} }$ be the Banach algebra of complex-valued continuous functions vanishing at infinity on $X$.
Let $\AA$ be a $\ast$-subalgebra of $\map {\CC_0} {X, \C}$ such that:
:$(1) \quad$ for each $x, y \in X$ with $x \ne y$ there exists $h_{x y} \in \AA$ such that $\map {h_{x y} } x \ne \map {h_{x y} } y$
:$(2) \quad$ for each $x \in X$ there exists $f_x \in \AA$ such that $\map {f_x} x \ne 0$. 
Then $\AA$ is everywhere dense in $\struct {\map {\CC_0} {X, \C}, \norm {\, \cdot \,} }$.

Proof:
This proof is essentially identical to Stone-Weierstrass Theorem: Compact Space: Complex-Valued.
We define:
:$\BB = \set {\map \Re f : f \in \AA} = \set {\map \Im f : f \in \AA}$
Replicating the proof in Stone-Weierstrass Theorem: Compact Space: Complex-Valued, it can be shown:
:$\BB = \set {\map \Re f : f \in \AA} = \set {\map \Im f : f \in \AA}$
It is then shown that $\BB$ separates points.
For $(2)$, for each $x \in X$ there exists $f_x \in \AA$ with $\map {f_x} x \ne 0$.
Then either $\map \Re {\map {f_x} x} \ne 0$ or $\map \Im {\map {f_x} x} \ne 0$. 
We have $\map \Re {f_x}, \map \Im {f_x} \in \BB$, so $(2)$ is satisfied.
Hence we can apply Stone-Weierstrass Theorem: Locally Compact Space to obtain:
:$\BB^- = \map {\CC_0} {X, \R}$
Now let $f \in \map {\CC_0} {X, \C}$. 
Write $f = \map \Re f + i \map \Im f$ with $\map \Re f, \map \Im f \in \map \CC {X, \R}$.
From Real and Imaginary Parts of Complex-Valued Function Vanishing at Infinity Vanish at Infinity, we have $\map \Re f, \map \Im f \in \map {\CC_0} {X, \R}$. 
Let $\epsilon > 0$. 
Since $\BB^- = \map {\CC_0} {X, \R}$, there exists $\phi, \psi \in \BB \subseteq \AA$ such that:
:$\norm {\phi - \map \Re f}_\infty < \dfrac \epsilon 2$
and:
:$\norm {\psi - \map \Im f}_\infty < \dfrac \epsilon 2$
We then have: 
{{begin-eqn}}
{{eqn	| l = \norm {f - \paren {\phi + i \psi} }_\infty 
		| r = \norm {\paren {\map \Re f - \phi} + i \paren {\map \Im f - \psi} }_\infty 
}}
{{eqn	| o = \le
		| r = \norm {\map \Re f - \phi}_\infty + \cmod i \norm {\map \Im f - \psi}_\infty 
		| c = {{NormAxiomVector|3}}, {{NormAxiomVector|1}}
}}
{{eqn	| o = <
		| r = \frac \epsilon 2 + \frac \epsilon 2
}}
{{eqn	| r = \epsilon 
}}
{{end-eqn}}
Since $\phi, \psi \in \AA$ and $\AA$ is a subalgebra, we have $\phi + i\psi \in \AA$.
Hence $\AA^- = \map {\CC_0} {X, \C}$.
{{qed}}
Category:Stone-Weierstrass Theorem
59enrn1ibfldecoh95k6dt7vu9yoiv5
Theorem: Real and Imaginary Parts of Complex-Valued Function Vanishing at Infinity Vanish at Infinity

Let $X$ be a locally compact Hausdorff space.
Let $f : X \to \C$ be a complex-valued function vanishing at infinity.
Then $\map \Re f$ and $\map \Im f$ vanish at infinity.

Proof:
Let $\epsilon > 0$. 
Since $f$ vanishes at infinity, there exists a compact $K \subseteq X$ such that:
:$\cmod {\map f x}^2 < \epsilon^2$ for each $x \in X \setminus K$. 
Then we have:
:$\paren {\map \Re {\map f x} }^2 + \paren {\map \Im {\map f x} }^2 < \epsilon^2$ 
So that:
:$\paren {\map \Re {\map f x} }^2 < \epsilon^2$
and:
:$\paren {\map \Im {\map f x} }^2 < \epsilon^2$
for each $x \in X \setminus K$. 
Then:
:$\cmod {\map \Re {\map f x} } < \epsilon$
and:
:$\cmod {\map \Im {\map f x} } < \epsilon$
for each $x \in X \setminus K$. 
Hence both $\map \Re f$ and $\map \Im f$ vanish at infinity.
{{qed}}
Category:Complex-Valued Functions Vanishing at Infinity
bahxhttoj3s8pldv1tgignubifqqhlj
Theorem: Newton's Three-Eighths Rule

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.
Let $P = \set {a = x_0, x_1, x_2, \ldots, x_{3 n - 1}, x_{3 n} = b}$ form a normal subdivision of $\closedint a b$:
:$\forall r \in \set {1, 2, \ldots, 3 n}: x_r - x_{r - 1} = \dfrac {b - a} {3 n}$
Then the definite integral of $f$ {{WRT|Integration}} $x$ from $a$ to $b$ can be approximated as:
{{begin-eqn}}
{{eqn | l = \int_a^b \map f x \rd x
      | o = \approx
      | r = \dfrac {3 h} 8 \paren {\map f {x_0} + 3 \sum_{r \mathop = 1}^n \paren {\map f {x_{3 i - 2} } + \map f {x_{3 i - 1} } } + 2 \sum_{r \mathop = 1}^{n - 1} \map f {x_{3 i} } + \map f {x_{3 n} } }
      | c = 
}}
{{eqn | r = \dfrac {3 h} 8 \paren {\map f {x_0} + 3 \map f {x_1} + 3 \map f {x_2} + 2 \map f {x_3} + 3 \map f {x_4} + 3 \map f {x_5} + 2 \map f {x_6} + \cdots + \map f {x_{3 n} } }
      | c = 
}}
{{end-eqn}}
where:
:$h = \dfrac {b - a} {3 n}$

Proof:
{{ProofWanted|Graphical approach}}
Theorem: Spectrum of Element of Unital Commutative Banach Algebra/Corollary 1

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$ that is not unital as an algebra.

Proof:
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the unitization of $\struct {A, \norm {\, \cdot \,} }$.
By the definition of the spectrum, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
Let $\Phi_{A_+}$ be the spectrum of $A_+$.
From Unitization of Commutative Algebra over Field is Commutative, $A_+$ is commutative.
Hence by Spectrum of Element of Unital Commutative Banach Algebra we have:
:$\map {\sigma_A} x = \set {\map \phi {x, 0} : \phi \in \Phi_{A_+} }$
From Characterization of Character on Banach Algebra, we have:
:$\Phi_{A_+} = \set {\widetilde \phi : \phi \in \Phi_A} \cup \set \psi$
where:
:$\map {\widetilde \phi} {x, \lambda} = \map \phi x + \lambda$
and:
:$\map \psi {x, \lambda} = \lambda$
for each $\tuple {x, \lambda} \in A_+$. 
Hence we have:
:$\map {\sigma_A} x = \set {\map \phi x : \phi \in \Phi_A} \cup \set {\map \psi 0} = \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$.
{{qed}}
Category:Spectrum of Element of Unital Commutative Banach Algebra
pvfnisnm3487lqvjw2fg4t3rmf3qi2j
Theorem: Existence of Maximal Commutative Subalgebra/Unital

Let $K$ be a field. 
Let $A$ be an unital algebra over $K$.
Let $S \subseteq A$ be a non-empty set such that:
:for all $x, y \in S$ we have $x y = y x$. 
Then there exists a commutative unital subalgebra of $A$ containing $S$ that is maximal (among all subalgebras) {{WRT}} set inclusion.

Proof:
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Note that ${\mathbf 1}_A$ commutes with every element of $S$.
Hence $S' = S \cup \set { {\mathbf 1}_A}$ is such that:
:for all $x, y \in S'$ we have $x y = y x$. 
Hence applying Existence of Maximal Commutative Subalgebra to $S'$, there exists a commutative subalgebra of $A$ containing $S'$ that is maximal (among all subalgebras) {{WRT}} set inclusion.
In particular:
:there exists a commutative unital subalgebra of $A$ containing $S$ that is maximal (among all subalgebras) {{WRT}} set inclusion.
{{qed}}
Category:Existence of Maximal Commutative Subalgebra
mpcec00m64kj345kp7dg4d1t2a3x3sh
Theorem: Gelfand Representation Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative Banach algebra over $\C$ such that:
:the spectrum $\Phi_A$ of $A$ is non-empty. 
Let $\hat a$ be the Gelfand transform of $a$.
Let $\struct {\map {\CC_0} {\Phi_A}, \norm {\, \cdot \,}_\infty}$ be the set of complex-valued functions vanishing at infinity with the supremum norm.
Define $G : A \to \map {\CC_0} {\Phi_A}$ by:
:$\map G a = \hat a$
for each $a \in A$.
Let $\map {\sigma_A} a$ be the spectrum of $a$ in $A$.
Let $r_A$ be the spectral radius of $a$ in $A$.
Then $G$ is an algebra homomorphism with:
:$(1) \quad \norm {\map G a}_\infty \le \norm a$ for each $a \in A$,
:$(2) \quad \map {r_A} a = \norm {\hat a}_\infty$ for each $a \in A$,
:$(3) \quad \map {\sigma_A} a = \hat a \sqbrk {\Phi_A}$ if $A$ is unital and $\map {\sigma_A} a = \hat a \sqbrk {\Phi_A} \cup \set 0$ otherwise.

Proof:
=== Proof that $G$ is an algebra homomorphism ===
Let $a, b \in A$, $\lambda \in \C$ and $\phi \in \Phi_A$.
Then we have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {\map G {a + \lambda b} } } \phi 
		| r = \map {\widehat {a + \lambda b} } \phi 
}}
{{eqn	| r = \map \phi {a + \lambda b}
}}
{{eqn	| r = \map \phi a + \lambda \map \phi b
		| c = since $\phi$ is linear
}}
{{eqn	| r = \map {\hat a} \phi + \lambda \map {\hat a} \phi 
}}
{{eqn	| r = \map {\paren {\map G a} } \phi + \lambda \map {\paren {\map G b} } \phi 
}}
{{end-eqn}}
So we obtain:
:$\map G {a + \lambda b} = \map G a + \lambda \map G b$
So $G$ is linear.
Now let $a, b \in A$. 
We have:
{{begin-eqn}}
{{eqn	| l = \map {\paren {\map G {a b} } } \phi 
		| r = \map {\hat {a b} } \phi 
}}
{{eqn	| r = \map \phi {a b}
}}
{{eqn	| r = \map \phi a \map \phi b
		| c = {{Defof|Character (Banach Algebra)}}
}}
{{eqn	| r = \map {\paren {\map G a} } \phi \map {\paren {\map G b} } \phi 
}}
{{end-eqn}}
So:
:$\map G {a b} = \map G a \map G b$
so $G$ is an algebra homomorphism.
{{qed|lemma}}
Theorem: Spectrum of Element of Unital Commutative Banach Algebra/Corollary 2

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $\Phi_A$ be the spectrum of $A$.
Then $\Phi_A \ne \O$.

Proof:
From Spectrum of Element of Banach Algebra is Non-Empty, we have $\map {\sigma_A} x \ne \O$.
That is, there exists $\lambda \in \map {\sigma_A} x$.
From Spectrum of Element of Unital Commutative Banach Algebra, there exists a character $\phi$ with $\map \phi x = \lambda$.
In particular, $\phi \in \Phi_A$.
So $\Phi_A \ne \O$.
{{qed}}
Category:Spectrum of Element of Unital Commutative Banach Algebra
1f0mhdljnx29ewqmn8accmq270806zg
Theorem: Image of Algebra under Algebra Homomorphism is Subalgebra

Let $R$ be a ring.
Let $A, B$ be $R$-algebras.
Let $\phi : A \to B$ be an $R$-algebra homomorphism.
Then $\phi \sqbrk A$ is a subalgebra of $B$.

Proof:
From Image of Submodule under Linear Transformation is Submodule, $\phi \sqbrk A$ is a submodule of $B$. 
Let $a, b \in A$. 
We then have: 
:$\map \phi a \map \phi b = \map \phi {a b} \in \phi \sqbrk A$
since $\phi$ is a $R$-algebra homomorphism.
Hence $\phi \sqbrk A$ is a subalgebra of $B$.
{{qed}}
Category:Algebras
3jsvt37zrf660moailwo691ffv778f8
Theorem: Image of *-Algebra under *-Algebra Homomorphism is *-Subalgebra

Let $\struct {A, \ast}$ and $\struct {B, \square}$ be a $\ast$-algebras over $\C$.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Then $\phi \sqbrk A$ is a $\ast$-subalgebra of $B$.

Proof:
From Image of Algebra under Algebra Homomorphism is Subalgebra, $\phi \sqbrk A$ is a subalgebra of $B$. 
Further, for $a \in A$ we have:
:$\paren {\map \phi a}^\square = \map \phi {a^\ast} \in \phi \sqbrk A$
So:
:$x^\ast \in \phi \sqbrk A$ for $x \in \phi \sqbrk A$.
Hence $\phi \sqbrk A$ is a $\ast$-subalgebra of $B$.
{{qed}}
Category:*-Algebras
jzcfo1smrp3rc33gdx09xkljtuuwn3g
Theorem: Image of Closed Set under Linear Isometry from Banach Space to Normed Vector Space is Closed

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a Banach space over $\GF$. 
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a normed vector space over $\GF$. 
Let $T : X \to Y$ be a linear isometry.
Let $A$ be a closed subset of $X$.
Then $T \sqbrk A$ is closed in $X$.

Proof:
Let $\sequence {y_n}_{n \mathop \in \N}$ be a convergent sequence in $T \sqbrk A$ with:
:$y_n \to y$ as $n \to \infty$. 
From Convergent Sequence is Cauchy Sequence, $\sequence {y_n}_{n \mathop \in \N}$ is a Cauchy sequence. 
Then there exists $x_n \in X$ with $y_n = T x_n$.
Let $\epsilon > 0$.
Let $N \in \N$ be such that:
:$\norm {y_n - y_m}_Y = \norm {\map T {x_n - x_m} } < \epsilon$ for $n, m \ge N$.
Since $T$ is a linear isometry, we therefore have:
:$\norm {x_n - x_m} < \epsilon$ for $n, m \ge N$.
So $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $X$.
Hence there exists $x \in X$ such that $x_n \to x$.
Since $A$ is closed in $X$, we further have $x \in A$ from Subset of Metric Space contains Limits of Sequences iff Closed.
Since $T$ is continuous, we have $T x_n \to T x$. 
From Limit in Normed Vector Space is Unique, we have $y = T x$ with $x \in A$.
Hence $y \in T \sqbrk A$. 
Hence every convergent sequence in $T \sqbrk A$ has its limit in $T \sqbrk A$.
Hence $T \sqbrk A$ is closed.
{{qed}}
Category:Linear Isometries
Category:Banach Spaces
2n5r561grvxjfxoedg4wfca5zz7ewrb
Theorem: Locale of Open Sets Functor is Covariant

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:
{{DefineFunctor
|ob = $\map \Omega T := $ the locale of topological space $T$
|mor = $\map \Omega f := $ the continuous map induced by continuous mapping $f$
}}
Theorem: Pedal Circle of Incenter is Incircle

Let $\triangle ABC$ be a triangle whose incenter is $H$.
Then the pedal circle of $H$ is the incircle of $\triangle ABC$.

Proof:
Let $\CC$ denote the incircle of $\triangle ABC$.
{{AimForCont}} $\CC$ is not the pedal circle of $H$.
By definition, the sides of $\triangle ABC$ are tangent to $\CC$.
Hence from {{Porism|Line at Right Angles to Diameter of Circle}}, the line through $H$ to those point of tangency to $\CC$ are perpendicular to the sides.
Hence $\CC$ is the pedal circle of $H$ by definition.
Hence by Proof by Contradiction, the pedal circle of $H$ is the incircle.
{{qed}}
Theorem: Pedal Circle of Excenter is Excircle

Let $\triangle ABC$ be a triangle.
Let $H$ be an excenter of $\triangle ABC$.
Then the pedal circle of $H$ is the excircle of $\triangle ABC$ whose center is $H$.

Proof:
Let $\EE$ denote the excircle of $\triangle ABC$ whose center is $H$.
{{AimForCont}} $\EE$ is not the pedal circle of $H$.
By definition, the sides of $\triangle ABC$ are tangent to $\EE$.
Hence from {{Porism|Line at Right Angles to Diameter of Circle}}, the line through $H$ to those point of tangency to $\EE$ are perpendicular to the sides.
Hence $\EE$ is the pedal circle of $H$ by definition.
Hence by Proof by Contradiction, the pedal circle of $H$ is the excircle of $\triangle ABC$ whose center is $H$.
{{qed}}
Theorem: Non-Zero C*-Algebra contains Non-Zero Hermitian Element

Let $\tuple {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$ algebra such that:
:$A \ne \set { {\mathbf 0}_A}$
where ${\mathbf 0}_A$ is the zero vector of $A$. 
Then there exists a Hermitian element $x \in A \setminus \set { {\mathbf 0}_A}$.

Proof:
Let $a \in A \setminus \set { {\mathbf 0}_A}$.
From Product of Element in *-Star Algebra with its Star is Hermitian, $a a^\ast$ is Hermitian.
We just need to show that $a a^\ast \ne {\mathbf 0}_A$.
By the $\text C^\ast$ identity, we have:
:$\norm {a a^\ast} = \norm a^2 \ne 0$ 
from {{NormAxiomVector|1}}, since $a \ne {\mathbf 0}_A$.
Hence applying {{NormAxiomVector|1}} again, we have $a a^\ast \ne {\mathbf 0}_A$.
{{qed}}
Theorem: Non-Zero C*-Algebra contains Non-Zero Hermitian Element

Let $\tuple {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$ algebra such that:
:$A \ne \set { {\mathbf 0}_A}$
where ${\mathbf 0}_A$ is the zero vector of $A$. 
Then there exists a Hermitian element $x \in A \setminus \set { {\mathbf 0}_A}$.

Proof:
{{AimForCont}} that ${\mathbf 0}_A$ is the only Hermitian element of $A$. 
Let $a \in A$. 
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, there exists Hermitian elements $b, c$ such that:
:$a = b + ic$ 
Since ${\mathbf 0}_A$ is the only Hermitian element of $A$, we have $b = {\mathbf 0}_A$ and $c = {\mathbf 0}_A$.
We would then have $a = {\mathbf 0}_A$, contradicting that $A \ne \set { {\mathbf 0}_A}$. 
Hence there are Hermitian elements of $A$ other than ${\mathbf 0}_A$.
{{qed}}
Theorem: Spectrum of Commutative C*-Algebra is Non-Empty

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a commutative $\text C^\ast$-algebra such that:
:$A \ne \set { {\mathbf 0}_A}$
where ${\mathbf 0}_A$ is the zero vector of $A$.
Let $\Phi_A$ be the spectrum of $A$.
Then $\Phi_A \ne \O$.

Proof:
Let $\map {\sigma_A} a$ be the spectrum of $a$ in $A$.
Let $r_A$ denotes spectral radius in $A$.
From Non-Zero C*-Algebra contains Non-Zero Hermitian Element:
:there exists a Hermitian element $x \in A \setminus \set { {\mathbf 0}_A}$.
Hence from {{NormAxiomVector|1}}, we have $\norm a \ne 0$. 
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have:
:$\norm a = \map {r_A} a = \sup \set {\cmod \lambda : \lambda \in \map {\sigma_A} a} \ne 0$
Hence there exists $\lambda \in \map {\sigma_A} a \setminus \set 0$. 
From Spectrum of Element of Unital Commutative Banach Algebra and {{Corollary|Spectrum of Element of Unital Commutative Banach Algebra|1}}, we have:
:$\map {\sigma_A} a \subseteq \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$
Then:
:$\lambda \in \set {\map \phi x : \phi \in \Phi_A} \cup \set 0$
while $\lambda \ne 0$.
So there exists $\phi \in \Phi_A$ such that $\map \phi x = \lambda$.
In particular $\Phi_A \ne \O$.
{{qed}}
Category:C*-Algebras
ahk301i2j1huolkb6g65yhs0c1mtlvv
Theorem: Gelfand-Naimark Theorem/Commutative Case/Unital

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a commutative unital $\text C^\ast$-algebra.
Let $\Phi_A$ be the spectrum of $A$.
Let $\struct {\map \CC {\Phi_A}, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of continuous complex-valued functions vanishing at infinity.
Define $G : A \to \map \CC {\Phi_A}$ by:
:$\map G a = \hat a$
where $\hat a$ is the Gelfand transform of $a$.
Then $G$ is a isometric unital $\ast$-algebra isomorphism.
That is, the commutative unital  $\text C^\ast$-algebras are precisely $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ for compact Hausdorff spaces $K$, up to isometric $\ast$-algebra isomorphism.

Proof:
From Complex-Valued Function on Compact Hausdorff Space Vanishes at Infinity, we have:
:$\map {\CC_0} K = \map \CC K$
for every compact Hausdorff space $K$.
Hence from Space of Continuous Functions Vanishing at Infinity is C*-Algebra, $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ is indeed a $\text C^\ast$-algebra for each compact Hausdorff space $K$.
Finally, recall that $\Phi_A$ is a compact Hausdorff space from {{Corollary|Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space}}.
Now from Gelfand-Naimark Theorem: Commutative Case, $G$ is a isometric $\ast$-algebra isomorphism.
Further, defining $\mathbf 1_{\map \CC K} : K \to \C$ by:
:$\map {\mathbf 1_{\map \CC K} } x = 1$
for each $x \in K$, we have:
:$\mathbf 1_{\map \CC K} \in \map \CC K$
and:
:$f \mathbf 1_{\map \CC K} = \mathbf 1_{\map \CC K} f = f$
for each $f \in \map \CC K$.
So $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ is unital.
We now have:
{{begin-eqn}}
{{eqn	| l = \map G { {\mathbf 1}_A}
		| r = \map {\hat { {\mathbf 1}_A} } \phi
}}
{{eqn	| r = \map \phi { {\mathbf 1}_A}
}}
{{eqn	| r = 1
		| c = Character on Unital Banach Algebra is Unital Algebra Homomorphism
}}
{{eqn	| r = \map {\mathbf 1_{\map \CC {\Phi_A} } } \phi
}}
{{end-eqn}}
for each $\phi \in \Phi_A$.
So $G$ is a isometric unital $\ast$-algebra isomorphism.
Hence every commutative unital $\text C^\ast$-algebra $A$ is isometrically $\ast$-algebra isomorphic to $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ for some compact Hausdorff space $K$: in particular $K = \Phi_A$. 
Conversely, we have established that $\struct {\map \CC K, \overline \cdot, \norm {\, \cdot \,} }$ is a unital $\text C^\ast$-algebra for each compact Hausdorff space $K$.
{{qed}}
Category:Gelfand-Naimark Theorem
ry5401vaqxerxkxrkngfy2jeh2efhji
Theorem: Proof using Axiom of Choice is Nonconstructive

A proof which depends upon the {{Axiom-link|Choice}} is a nonconstructive proof.

Proof:
The {{Axiom-link|Choice}} allows that an infinite number of selections be made without the need to specify the choice function.
Hence such a proof is nonconstructive.
{{qed}}
Theorem: Star of Product of Elements in *-Algebra

Let $\tuple {A, \ast}$ be a $\ast$-algebra.
Let $n \ge 2$. 
Let $x_1, \ldots, x_n \in A$.
Then:
:$\ds \paren {\prod_{j \mathop = 1}^n x_j}^\ast = \prod_{j \mathop = 1}^n x_{n - j + 1}^\ast$

Proof:
We proceed by induction on $n$. 
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \paren {\prod_{j \mathop = 1}^n x_j}^\ast = \prod_{j \mathop = 1}^n x_{n - j + 1}^\ast$ for all $x_1, \ldots, x_n \in A$.
Theorem: Image is Subset Iff Subset of Preimage

Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A \subseteq S, B \subseteq T : f \sqbrk A \subseteq B$ {{iff}} $A \subseteq f^{-1} \sqbrk B$

Proof:
=== Necessary Condition ===
We have:
{{begin-eqn}}
{{eqn | q = \forall A \subseteq S, B \subseteq T
      | l = f \sqbrk A
      | o = \subseteq
      | r = B
}}
{{eqn | ll = \leadsto
      | l = f^{-1} \sqbrk {f \sqbrk A} 
      | o = \subseteq
      | r = f^{-1} \sqbrk B
      | c = Preimage of Subset is Subset of Preimage
}}
{{eqn | ll = \leadsto
      | l = \paren{f^{-1} \circ f} \sqbrk A
      | o = \subseteq
      | r = f^{-1} \sqbrk B
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | ll = \leadsto
      | l = A
      | o = \subseteq
      | r = f^{-1} \sqbrk B
      | c = Subset of Domain is Subset of Preimage of Image
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Inverse Image Mapping is Upper Adjoint to Direct Image Mapping

Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Let $f^\to : \powerset S \to \powerset T$ denote the direct image mapping.
Let $f^\gets : \powerset T \to \powerset S$ denote the inverse image mapping.
Then:
:$f^\gets : \powerset T \to \powerset S$ is the upper adjoint to $f^\to : \powerset S \to \powerset T$
That is:
:$\forall A \subseteq S, B \subseteq T : A \subseteq \map {f^\gets} B$ {{iff}} $\map {f^\to} A \subseteq B$

Proof:
From Image is Subset Iff Subset of Preimage:
:$\forall A \subseteq S, B \subseteq T : f \sqbrk A \subseteq B$ {{iff}} $A \subseteq f^{-1} \sqbrk B$  
By definition of direct image mapping:
:$\forall A \subseteq S : \map {f^\to} A = f \sqbrk A$
By definition of inverse image mapping:
:$\forall B \subseteq T : \map {f^\gets} B = f^{-1} \sqbrk B$
Hence:
:$\forall A \subseteq S, B \subseteq T : A \subseteq \map {f^\gets} B$ {{iff}} $\map {f^\to} A \subseteq B$
By definition of Galois connection:
:$\tuple{f^\gets, f^\to}$ is a Galois connection
It follows that $f^\gets : \powerset T \to \powerset S$ is the upper adjoint to $f^\to : \powerset S \to \powerset T$
{{qed}}
Category:Direct Image Mappings
Category:Inverse Image Mappings
Category:Galois Connections
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Theorem: Preimage is Subset Iff Subset of Complement of Image of Complement

Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A \subseteq S, B \subseteq T : f^{-1} \sqbrk B \subseteq A$ {{iff}} $B \subseteq T \setminus f \sqbrk {S \setminus A}$

Proof:
We have:
{{begin-eqn}}
{{eqn | q = \forall A \subseteq S, B \subseteq T
      | l = f^{-1} \sqbrk B
      | o = \subseteq
      | r = A
}}
{{eqn | ll= \leadstoandfrom
      | l = S \setminus A
      | o = \subseteq
      | r = S \setminus {f^{-1} \sqbrk B}
      | c = Relative Complement inverts Subsets
}}
{{eqn | r = f^{-1} \sqbrk {T \setminus B}
      | c = Complement of Preimage equals Preimage of Complement
}}
{{eqn | ll= \leadstoandfrom
      | l = f \sqbrk {S \setminus A} 
      | o = \subseteq
      | r = T \setminus B
      | c = Image is Subset Iff Subset of Preimage
}}
{{eqn | ll= \leadstoandfrom
      | l = T \setminus \paren{T \setminus B}
      | o = \subseteq
      | r = T \setminus f \sqbrk {S \setminus A}
      | c = Relative Complement inverts Subsets
}}
{{eqn | ll= \leadstoandfrom
      | l = B
      | o = \subseteq
      | r = T \setminus f \sqbrk {S \setminus A}
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}
Category:Relative Complement
Category:Images
Category:Preimages
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