avewright/proofwiki-math huggingface dataset:
Theorem: Fixed Points of Idempotent Operator

Let $X$ be a vector space.
Let $T : X \to X$ be an idempotent operator.
Then the set of fixed points of $T$ is precisely the range $\Rng T$.

Proof:
If $x$ is a fixed point of $T$, then:
:$T x = x$
and we clearly have $x \in \Rng T$. 
Conversely, if $y \in \Rng T$ then there exists $x \in X$ such that $y = T x$. 
Then we have:
{{begin-eqn}}
{{eqn	| l = T y
		| r = T^2 x
}}
{{eqn	| r = T x
		| c = since $T^2 = T$
}}
{{eqn	| r = y
}}
{{end-eqn}}
So $y$ is a fixed point of $T$.
{{qed}}
Category:Idempotent Operators
c93z0nv3cz77933pb3g3s9ex9xmi6ni
Theorem: Identity Element in Unital *-Algebra is Hermitian

Let $\struct {A, \ast}$ be a unital $\ast$-algebra.
Let ${\mathbf 1}_A$ be the identity element of $A$. 
Then we have:
:${\mathbf 1}_A^\ast = {\mathbf 1}_A$

Proof:
From Product of Element in *-Star Algebra with its Star is Hermitian, we have:
:$\paren { {\mathbf 1}_A {\mathbf 1}_A^\ast}^\ast = {\mathbf 1}_A {\mathbf 1}_A^\ast$
We have, since $\ast$ is an involution: 
:$\paren { {\mathbf 1}_A {\mathbf 1}_A^\ast} = {\mathbf 1}_A^{\ast \ast} = {\mathbf 1}_A$
and:
:${\mathbf 1}_A {\mathbf 1}_A^\ast = {\mathbf 1}_A^\ast$
Hence we conclude:
:${\mathbf 1}_A^\ast = {\mathbf 1}_A$
{{qed}}
Theorem: Inverse of Star of Element in Unital *-Algebra

Let $\struct {A, \ast}$ be a unital $\ast$-algebra.
Let $a \in A$ be invertible.
Then $a^\ast$ is invertible and:
:$\paren {a^\ast}^{-1} = \paren {a^{-1} }^\ast$

Proof:
We have: 
:$a a^{-1} = a^{-1} a = {\mathbf 1}_A$
From $(C^\ast 3)$ in the definition of an involution, we have:
:$\paren {a^{-1} }^\ast a^\ast = a^\ast \paren {a^{-1} }^\ast = {\mathbf 1}_A^\ast$ 
From Identity Element in Unital *-Algebra is Hermitian, we therefore have:
:$\paren {a^{-1} }^\ast a^\ast = a^\ast \paren {a^{-1} }^\ast = {\mathbf 1}_A$
Hence we have:
:$\paren {a^\ast}^{-1} = \paren {a^{-1} }^\ast$
{{qed}}
Theorem: Spectrum of Star of Element in *-Algebra

Let $\struct {A, \ast}$ be a unital $\ast$-algebra over $\C$.
Let $a \in A$. 
Let $\sigma_A$ denote the spectrum.
Then:
:$\map {\sigma_A} {a^\ast} = \set {\overline \lambda : \lambda \in \map {\sigma_A} a}$

Proof:
From the definition of an involution, we have:
:$\paren {\lambda {\mathbf 1}_A - a}^\ast = \overline \lambda {\mathbf 1}_A^\ast - a^\ast$ 
From Identity Element in Unital *-Algebra is Hermitian, we therefore have:
:$\paren {\lambda {\mathbf 1}_A - a}^\ast = \overline \lambda {\mathbf 1}_A - a^\ast$ 
From Inverse of Star of Element in Unital *-Algebra: Corollary, we deduce that:
:$\lambda {\mathbf 1}_A - a$ is invertible {{iff}} $\overline \lambda {\mathbf 1}_A - a^\ast$ is invertible.
That is to say:
:$\lambda \in \map {\sigma_A} a$ {{iff}} $\overline \lambda \in \map {\sigma_A} {a^\ast}$.
We conclude:
:$\map {\sigma_A} {a^\ast} = \set {\overline \lambda : \lambda \in \map {\sigma_A} a}$
{{qed}}
Theorem: Resolvent Set of Element in Subalgebra

Let $A$ be a Banach algebra.
Let $B$ be a closed unital subalgebra of $A$.
Let $x \in B$.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively. 
Then:
:$\map {\rho_B} x \subseteq \map {\rho_A} x$

Proof:
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively. 
From Spectrum of Element in Unital Subalgebra, we have:
:$\map {\sigma_A} x \subseteq \map {\sigma_B} x$
From Set Complement inverts Subsets, we have:
:$\C \setminus \map {\sigma_B} x \subseteq \C \setminus \map {\sigma_A} x$
From the definition of the resolvent set, we have:
:$\map {\rho_B} x \subseteq \map {\rho_A} x$
{{qed}}
Category:Resolvent Sets
onshpxpkwcwfak2a5fvida1zr26w3lk
Theorem: Resolvent Set of Element in Closed Unital Subalgebra of Banach Algebra is Clopen in Resolvent Set

Let $A$ be a Banach algebra.
Let $B$ be a closed unital subalgebra of $A$.
Let $x \in B$.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively. 
Then $\map {\rho_B} x$ is clopen in $\map {\rho_A} x$.

Proof:
Let $\map G A$ and $\map G B$ be the group of units of $A$ and $B$ respectively. 
First, from Resolvent Set of Element in Subalgebra we have:
:$\map {\rho_B} x \subseteq \map {\rho_A} x$ 
Define $R : \map {\rho_A} x \to A$ by: 
:$\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$ 
for each $\lambda \in \map {\rho_A} x$. 
From Resolvent Mapping is Continuous: Banach Algebra, $R$ is continuous.
Hence from the definition of a continuous mapping and since $B$ is closed, we have that:
:$R^{-1} \sqbrk B$ is closed in $\map {\rho_A} x$.
Note that $\lambda {\mathbf 1}_A - x \in \map G B$ {{iff}} $\lambda {\mathbf 1}_A - x \in \map G A$ and $\paren {\lambda {\mathbf 1}_A - x}^{-1} \in B$.
That is, $R^{-1} \sqbrk B = \map {\rho_B} x$.
So $\map {\rho_B} x$ is closed in $\map {\rho_A} x$.
From Resolvent Set of Element of Banach Algebra is Open, $\map {\rho_B} x$ is open in $A$.
Hence $\map {\rho_B} x$ is open in $\map {\rho_A} x$.
We conclude that $\map {\rho_B} x$ is clopen in $\map {\rho_A} x$.
{{qed}}
Category:Resolvent Sets
mzb7rnsc4huebypwxpttnx6aa0nsau1
Theorem: Not All Matroids are Base-Orderable

Not all matroids are base-orderable.

Proof:
{{ProofWanted}}
Category:Matroid Theory
hnkkybqnkcy878q8ufhn3f8qvipag2o
Theorem: Open Balls form Local Basis of Metric Space

Let $\struct {M, d}$ be a metric space. 
Let $\tau$ be the topology on $M$ induced by $d$. 
Let $x \in X$. 
For each $\epsilon > 0$, let $\map {B_\epsilon} x$ be the open $\epsilon$-ball around $x$.
Let $\BB_x = \set {\map {B_\epsilon} x : \epsilon > 0}$.
Then $\BB_x$ is a local basis at $x$.

Proof:
Let $U$ be an open neighborhood of $x$. 
From the definition of an open set in $\struct {M, d}$:
:there exists $\epsilon > 0$ such that $\map {B_\epsilon} x \subseteq U$. 
Hence:
:there exists $V \in \BB_x$ such that $V \subseteq U$. 
So $\BB_x$ is a local basis at $x$.
{{qed}}
Category:Local Bases
Category:Metric Spaces
omun205oinnir1jq0y06ss86t5p9d6o
Theorem: Normed Vector Space is Locally Connected

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$. 
Then $\struct {X, \norm {\, \cdot \,} }$ is locally connected.

Proof:
Let $x \in X$.
Let $\epsilon > 0$. 
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball around $x$ in $\struct {X, \norm {\, \cdot \,} }$. 
From Open Balls form Local Basis of Metric Space, the set $\BB_x = \set {\map {B_\epsilon} x : \epsilon > 0}$ is a local basis at $x$.
From Open Ball in Normed Vector Space is Connected, $\map {B_\epsilon} x$ is connected.
Hence $x$ has a local basis of connected sets.
Since $x$ was arbitrary, $\struct {X, \norm {\, \cdot \,} }$ is locally connected.
{{qed}}
Category:Locally Connected Spaces
Category:Normed Vector Spaces
eiyk67bubamftu87zwd6gfhe20j4ak9
Theorem: Locally Connected Separable Topological Space has Countably Many Components

Let $\struct {X, \tau}$ be a locally connected separable topological space. 
Let $\CC$ be the set of components of $\struct {X, \tau}$.
Then $\CC$ is countable.

Proof:
From Component of Locally Connected Space is Open:
:$\CC \subseteq \tau$
From Equivalence Classes are Disjoint, we have:
:the sets in $\CC$ are pairwise disjoint.
From Collection of Pairwise Disjoint Open Sets in Separable Topological Space is Countable, it follows that:
:$\CC$ is countable.
{{qed}}
Category:Locally Connected Spaces
Category:Components (Topology)
q1b4bnudygjd90vt6juroqj0dezc3rb
Theorem: Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $B$ be a closed unital subalgebra of $A$. 
Let $x \in B$.
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively.
Then:
:$\partial \map {\sigma_B} x \subseteq \partial \map {\sigma_A} x$
where $\partial$ denotes topological boundary.

Proof:
Let:
:$\lambda \in \partial \map {\sigma_B} x$. 
From Boundary is Intersection of Closure with Closure of Complement and Spectrum of Element of Banach Algebra is Closed, we have:
:$\partial \map {\sigma_B} x = \map {\sigma_B} x \cap \map \cl {\map {\rho_B} x}$
where $\map {\rho_B} x$ is the resolvent set of $x$ in $B$. 
Likewise let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$. 
{{AimForCont}} that $\lambda \in \map {\rho_A} x$.
Then:
:$\lambda \in \map \cl {\map {\rho_B} x} \cap \map {\rho_A} x$
That is, from Closure of Subset in Subspace:
:$\lambda \in \map {\cl_{\map {\rho_A} x} } {\map {\rho_B} x}$
where $\cl_{\map {\rho_A} x}$ is the closure taken in the subspace topology of $\map {\rho_A} x$.
From Resolvent Set of Element in Closed Unital Subalgebra of Banach Algebra is Clopen in Resolvent Set, we have:
:$\map {\rho_B} x$ is clopen in $\map {\rho_A} x$. 
From Set is Closed iff Equals Topological Closure, we therefore have:
:$\map {\cl_{\map {\rho_A} x} } {\map {\rho_B} x} = \map {\rho_B} x$
Hence $\lambda \in \map {\sigma_B} x \cap \map {\rho_B} x$. 
Since $\map {\rho_B} x = \C \setminus \map {\sigma_B} x$, this contradicts the definition of set complement. 
Hence $\lambda \in \C \setminus \map {\rho_A} x = \map {\sigma_A} x$.
From Resolvent Set of Element in Subalgebra, we have $\map {\rho_B} x \subseteq \map {\rho_A} x$. 
Hence from Set Closure Preserves Set Inclusion, we have $\map \cl {\map {\rho_B} x} \subseteq \map \cl {\map {\rho_A} x}$. 
We therefore have $\lambda \in \map {\sigma_A} x \cap \map \cl {\map {\rho_A} x} = \partial \map {\sigma_A} x$.
{{qed}}
Theorem: Resolvent Set of Element in Banach Algebra has same Unique Unbounded Component as Resolvent Set of Element in Closed Subalgebra

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $B$ be a closed unital subalgebra of $A$. 
Let $x \in B$.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively.
Then $\map {\rho_A} x$ and $\map {\rho_B} x$ both have unique unbounded component, and further these components coincide.

Proof:
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively. 
From Spectrum of Element of Banach Algebra is Bounded, we have:
:$\map {\sigma_A} x$ and $\map {\sigma_B} x$ are bounded.
Hence, from Complement of Bounded Set in Complex Plane has at most One Unbounded Component:
:$\map {\rho_A} x$ and $\map {\rho_B} x$ both have a unique unbounded component.
Let $U$ be the unbounded component of $\map {\rho_A} x$.
From Complement of Closed Disk in Complex Plane is Path-Connected, we have that:
:$\set {z \in \C : \cmod z > \norm x}$ is path-connected.
From Path-Connected Space is Connected:
:$\set {z \in \C : \cmod z > \norm x}$ is connected.
From Spectrum of Element of Banach Algebra is Bounded, we have:
:$\map {\rho_A} x \supseteq \set {z \in \C : \cmod z > \norm x}$
and:
:$\map {\rho_B} x \supseteq \set {z \in \C : \cmod z > \norm x}$
and hence:
:$U \cap \map {\rho_B} x \supseteq \set {z \in \C : \cmod z > \norm x} \ne \O$
From Component of Resolvent Set of Element in Unital Banach Algebra is Disjoint from or Component of Resolvent Set in Closed Subalgebra, we therefore have:
:$U$ is a component of $\map {\rho_B} x$.
Since $U$ is unbounded, we conclude that $U$ is also the unique unbounded component of $\map {\rho_B} x$. 
{{qed}}
Category:Resolvent Sets
Category:Components (Topology)
Category:Unital Banach Algebras
hsb0zhgeteyin373b5aybpkdo2hd4ew
Theorem: Spectral Permanence Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.
Let $B$ be a closed unital subalgebra of $A$. 
Let $x \in B$.
Let $\map {\sigma_A} x$ and $\map {\sigma_B} x$ be the spectra of $x$ in $A$ and $B$ respectively.
Let $\map {\rho_A} x$ and $\map {\rho_B} x$ be the resolvent sets of $x$ in $A$ and $B$ respectively. 
Then there exists a indexed family $\family {U_i}_{i \mathop \in I}$ of bounded components of $\map {\rho_A} x$ such that:
:$\ds \map {\sigma_B} x = \map {\sigma_A} x \cup \bigcup_{i \mathop \in I} U_i$ 
Further, $I$ can be made countable.

Proof:
Let $\family {V_\alpha}_{\alpha \mathop \in J}$ be the set of bounded components of $\map {\rho_A} x$.
From Component of Resolvent Set of Element in Unital Banach Algebra is Disjoint from or Component of Resolvent Set in Closed Subalgebra, for each $\alpha \in J$ we either have:
:$V_\alpha \subseteq \map {\sigma_B} x$
or:
:$V_\alpha \cap \map {\sigma_B} x = \O$ 
Let $\family {U_i}_{i \mathop \in I}$ be the subset of $\family {V_\alpha}_{\alpha \mathop \in J}$ consisting of those $V_\alpha$ such that $V_\alpha \subseteq \map {\sigma_B} x$. 
From Spectrum of Element in Unital Subalgebra, we have $\map {\sigma_A} x \subseteq \map {\sigma_B} x$ and hence:
:$\ds \map {\sigma_A} x \cup \bigcup_{i \mathop \in I} U_i \subseteq \map {\sigma_B} x$
From Resolvent Set of Element in Banach Algebra has same Unique Unbounded Component as Resolvent Set of Element in Closed Subalgebra:
:$\map {\rho_A} x$ and $\map {\rho_B} x$ both have a unique unbounded component $S$.
Specifically:
:if $z \in \map {\sigma_B} x \cap \map {\rho_A} x$, $z$ is contained in a bounded component of $\map {\rho_A} x$.
That is, for $z \in \map {\sigma_B} x$ we either have $z \in \map {\sigma_A} x$ or there exists $i \in I$ such that $z \in U_i$. 
This is precisely the statement that: 
:$\ds \map {\sigma_B} x \subseteq \map {\sigma_A} x \cup \bigcup_{i \mathop \in I} U_i$ 
Hence we conclude:
:$\ds \map {\sigma_B} x = \map {\sigma_A} x \cup \bigcup_{i \mathop \in I} U_i$ 
It remains to show that $I$ can be made countable.
From Complex Plane is Separable:
:$\C$ is separable.
From Normed Vector Space is Locally Connected:
:$\C$ is locally connected.
From Open Subset of Locally Connected Space is Locally Connected, Subspace of Separable Metric Space is Separable and Resolvent Set of Element of Banach Algebra is Open:
:$\map {\rho_A} x$ is locally connected.
From Locally Connected Separable Topological Space has Countably Many Components:
:there are countably many distinct components of $\map {\rho_A} x$. 
Hence among the $\family {U_i}_{i \mathop \in I}$, only countably many are distinct. 
By deleting duplicates, we can therefore make $I$ countable.
{{qed}}
Category:Spectra (Spectral Theory)
Category:Unital Banach Algebras
Category:Named Theorems
tkz2mdzptckrv32gohesi6n5c8cvsx7
Theorem: Finite Set Formed by Substitution has Larger Intersection

:$\card{R \cap S} = \card{T \cap S} + 1$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \card{R \cap S}
      | r = \card{\paren{\paren{T \setminus \set y} \cup \set x} \cap S}
}}
{{eqn | r = \card{\paren{\paren{T \setminus \set y} \cap S } \cup \paren{ \set x \cap S} }
      | c = Intersection Distributes over Union
}}
{{eqn | r = \card{\paren{\paren{T \setminus \set y} \cap S } \cup \set x}
      | c = As $x \in S$
}}
{{eqn | r = \card{\paren{\paren{T \cap S} \setminus \paren{\set y \cap S} } \cup \set x}
      | c = Set Intersection Distributes over Set Difference
}}
{{eqn | r = \card{\paren{\paren{T \cap S} \setminus \O } \cup \set x}
      | c = As $y \notin S$
}}
{{eqn | r = \card{\paren{T \cap S} \cup \set x}
      | c = Set Difference with Empty Set is Self
}}
{{eqn | r = \card{T \cap S} +  \card {\set x}
      | c = Cardinality of Pairwise Disjoint Set Union and $x \notin T$
}}
{{eqn | r = \card{T \cap S} +  1
      | c = Cardinality of Singleton
}}
{{end-eqn}}
{{qed}}
Category:Finite Sets
Category:Set Intersection
faovpcyynod23lf2bwhnkfj4no5l7y0
Theorem: Matrix Scalar Product Distributes over Matrix Entrywise Addition

Let $\mathbf A$ and $\mathbf B$ be matrices both of order $m \times n$.
Let $k$ be a scalar.
Then:
:$k \paren {\mathbf A + \mathbf B} = k \mathbf A + k \mathbf B$
where:
:$+$ denotes matrix entrywise addition
:$k \mathbf A$ etc. denotes matrix scalar product.

Proof:
<onlyinclude>
Let $a_{i j}$ and $b_{i j}$ denote the $\tuple {i, j}$th entry in $\mathbf A$ and $\mathbf B$ respectively.
{{begin-eqn}}
{{eqn | o = 
      | r = k \mathbf A + k \mathbf B
      | c = 
}}
{{eqn | q = \forall i \in \closedint 1 m, \forall j \in \closedint 1 n
      | r = k a_{i j} + k b_{i j}
      | c = {{Defof|Matrix Scalar Product}}
}}
{{eqn | r = k \paren {a_{i j} + b_{i j} }
      | c = Distributive Property
}}
{{eqn | r = k \paren {\mathbf A + \mathbf B}
      | c = {{Defof|Matrix Entrywise Addition}}
}}
{{end-eqn}}
{{qed}}
Theorem: Finite Set Formed by Substitution has Same Cardinality

:$\card R = \card T$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \card R
      | r = \card{\paren{T \setminus \set y} \cup \set x}
}}
{{eqn | r = \card{T \setminus \set y} + \card{\set x}
      | c = Cardinality of Pairwise Disjoint Set Union
}}
{{eqn | r = \paren{\card T - \card{\set y} } + \card{\set x}
      | c = Cardinality of Set Difference with Subset
}}
{{eqn | r = \paren{\card T - 1 } + 1
      | c = Cardinality of Singleton
}}
{{eqn | r = \card T
      | c = Canceling terms
}}
{{end-eqn}}
{{qed}}
Category:Finite Sets
Category:Cardinality
b8bc6rpn3394or1i1j9nqcm73d5dtxx
Theorem: Spectrum of Element of Banach Algebra is Non-Empty/Corollary

Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\C$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Then $\map {\sigma_A} x \ne \O$.

Proof:
Suppose first that $\struct {A, \norm {\, \cdot \,} }$ is unital. 
From Completion Theorem (Normed Algebra), there exists a unital Banach algebra $\struct {\widetilde A, \widetilde {\norm {\, \cdot \,} } }$ and an isometric algebra homomorphism $\phi : A \to \widetilde A$ where:
:${\mathbf 1}_{\widetilde A} = \map \phi { {\mathbf 1}_A}$
From Spectrum of Element of Banach Algebra is Non-Empty, we have:
:$\map {\sigma_{\widetilde A} } {\map \phi x} \ne \O$
From Spectrum of Element in Unital Subalgebra, we have:
:$\map {\sigma_{\widetilde A} } {\map \phi x} \subseteq \map {\sigma_{\phi \sqbrk A} } {\map \phi x}$
so:
:$\map {\sigma_{\phi \sqbrk A} } {\map \phi x} \ne \O$
To finish we show that:
:$\map {\sigma_{\phi \sqbrk A} } {\map \phi x} = \map {\sigma_A} x$
We have $\lambda \in \map {\sigma_{\phi \sqbrk A} } {\map \phi x}$ {{iff}} $\lambda \map \phi { {\mathbf 1}_A} - \map \phi x \in \map G {\phi \sqbrk A}$ where $\map G {\phi \sqbrk A}$ is the group of units of $\phi \sqbrk A$.
Since $\phi$ is linear, we have $\lambda \in \map {\sigma_{\phi \sqbrk A} } {\map \phi x}$ {{iff}} $\map \phi {\lambda {\mathbf 1}_A - x} \in \map G {\phi \sqbrk A}$.
This is equivalent to the existence of $z \in A$ such that $\map \phi {\lambda {\mathbf 1}_A - x} \map \phi z = \map \phi z \map \phi {\lambda {\mathbf 1}_A - x} = \map \phi { {\mathbf 1}_A}$. 
Since $\phi$ is an algebra homomorphism, this is equivalent to:
:$\map \phi {\paren {\lambda {\mathbf 1}_A - x} z} = \map \phi {z \paren {\lambda {\mathbf 1}_A - x} } = \map \phi { {\mathbf 1}_A}$
Since $\phi$ is a bijection, this is equivalent to:
:$\paren {\lambda {\mathbf 1}_A - x} z = z \paren {\lambda {\mathbf 1}_A - x} = {\mathbf 1}_A$
for $z \in A$.
That is, equivalent to $\lambda {\mathbf 1}_A - x \in \map G A$.
Hence equivalent to $\lambda \in \map {\sigma_A} x$. 
So:
:$\map {\sigma_A} x = \map {\sigma_{\phi \sqbrk A} } {\map \phi x} \ne \O$
Suppose now that $\struct {A, \norm {\, \cdot \,} }$ is not unital.
Then, we have:
:$\map {\sigma_A} x = \map {\sigma_{A_+} } {\tuple {x, 0} }$
where $A_+$ is the normed unitization of $A$.
From the previous case, we have $\map {\sigma_{A_+} } {\tuple {x, 0} } \ne \O$ and hence $\map {\sigma_A} x \ne \O$.
{{qed}}
Category:Spectrum of Element of Banach Algebra is Non-Empty
d7ma52bfdqctwdfuv9fu6i3tt23rue3
Theorem: Gelfand-Mazur Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital normed algebra over $\C$ where:
:$\map G A = A \setminus \set { {\mathbf 0}_A}$ 
where $\map G A$ is the group of units of $A$.
Then $A$ is isometrically algebra isomorphic to $\C$.

Proof:
Define $\theta : \C \to A$ by: 
:$\map \theta \lambda = \lambda {\mathbf 1}_A$
for each $\lambda \in \C$.
For $z, w, \lambda \in \C$ we have:
{{begin-eqn}}
{{eqn	| l = \map \theta {z + \lambda w}
		| r = \paren {z + \lambda w} {\mathbf 1}_A
}}
{{eqn	| r = z {\mathbf 1}_A + \lambda \paren {w {\mathbf 1}_A}
}}
{{eqn	| r = \map \theta z + \lambda \map \theta w
}}
{{end-eqn}}
so $\theta$ is linear. 
Further, for $\lambda, \mu \in \C$ we have:
:$\map \theta {\lambda \mu} = \paren {\lambda \mu} {\mathbf 1}_A = \paren {\lambda {\mathbf 1}_A} \paren {\mu {\mathbf 1}_A} = \map \theta \lambda \map \theta \lambda$
Hence $\theta$ is an algebra homomorphism.
For $\lambda \in \C$, we have: 
{{begin-eqn}}
{{eqn	| l = \norm {\map \theta \lambda}
		| r = \norm {\lambda {\mathbf 1}_A}
}}
{{eqn	| r = \cmod \lambda \norm { {\mathbf 1}_A}
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| r = \cmod \lambda 
		| c = {{Defof|Unital Normed Algebra}}
}}
{{end-eqn}}
Hence $\theta$ is an isometry.
It remains to show that $\theta$ is surjective.
Let $x \in A$. 
From {{Corollary|Spectrum of Element of Banach Algebra is Non-Empty}}, we have:
:$\map {\sigma_A} x \ne \O$
where $\map {\sigma_A} x$ is the spectrum of $x$ in $A$. 
Let $\lambda \in \map {\sigma_A} x$. 
Then:
:$\lambda {\mathbf 1}_A - x \in A \setminus \map G A = \set { {\mathbf 0}_A}$
Hence we have:
:$x = \lambda {\mathbf 1}_A = \map \theta \lambda$
That is:
:$x \in \Img \theta$
So $\theta$ is surjective.
So $\theta$ is an isometric isomorphism onto $\C$.
Hence $A$ is isometrically isomorphic to $\C$.
{{qed}}
{{Namedfor|Stanisław Mieczysław Mazur|cat = Mazur, S|name2 = Israel Moiseevich Gelfand|cat2 = Gelfand, I}}
Category:Normed Algebras
Category:Algebra Isomorphisms
5qni1okgzkkdq8jdsa6d34zikeno32l
Theorem: Open Subset of Locally Compact Hausdorff Space is Locally Compact Hausdorff Space

Let $\struct {X, \tau_X}$ be a locally compact Hausdorff space.
Let $Y \subseteq X$ be open.
Let $\tau_Y$ be the subspace topology on $Y$ inherited from $\struct {X, \tau_X}$.
Then $\struct {Y, \tau_Y}$ is a locally compact Hausdorff space.

Proof:
From $T_2$ Property is Hereditary, $\struct {Y, \tau_Y}$ is Hausdorff.
Let $y \in Y$. 
Let $\CC$ be a neighborhood basis of $y$ in $\struct {X, \tau_X}$ consisting of compact sets. 
Let $U$ be a neighborhood of $y$ in $\struct {Y, \tau_Y}$.
Then there exists an open neighborhood $V$ of $y$ in $\struct {Y, \tau_Y}$ such that:
:$V \subseteq U$
From Open Set in Open Subspace:
:$V$ is open in $\struct {X, \tau_X}$.
Hence from the definition of a neighborhood basis, there exists $C \in \CC$ such that $y \in C \subseteq V$.
Again from Open Set in Open Subspace, we have that:
:$C$ is compact in $\struct {Y, \tau_Y}$. 
Hence $C$ is compact in $\struct {Y, \tau_Y}$ with $C \subseteq U$.
Since $C$ is compact and $U$ is arbitrary, we have that $\CC$ is a neighborhood basis for $y$ in $\struct {Y, \tau_Y}$.
{{qed}}
Category:Locally Compact Hausdorff Spaces
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Theorem: Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space/Corollary

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.

Proof:
Let $\phi \in \Phi_A$. 
From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have that:
:the condition that $\map \phi x \map \phi y = \map \phi {x y}$ for all $x, y \in A$ implies that $\map \phi { {\mathbf 1}_A} = 1$.
That is, $\map \phi x \map \phi y = \map \phi {x y}$ for all $x, y \in A$ implies that $\phi \ne {\mathbf 0}_{A^\ast}$. 
So, we have:
:$\Phi_A = \set {\phi \in A^\ast : \map \phi x \map \phi y = \map \phi {x y} \text { for all } x, y \in A}$
In the proof of Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space, it was established that $\struct {\set {\phi \in A^\ast : \map \phi x \map \phi y = \map \phi {x y} \text { for all } x, y \in A}, w^\ast}$ is compact and Hausdorff.
Hence $\struct {\Phi_A, w^\ast}$ is a compact Hausdorff space.
{{qed}}
Category:Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space
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