avewright/proofwiki-math huggingface dataset:
Theorem: Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $I$ be a closed ideal.
Let $x \in I$ be normal.
Let $\map {\sigma_A} x \subseteq \R$ be the spectrum of $x$ in $A$.
Let $K$ be a compact set such that $\map {\sigma_A} x \cup \set 0 \subseteq K$.  
Let $f : K \to \R$ be a continuous function with $\map f 0 = 0$.
Let $\map f x$ be obtained from the continuous functional calculus, restricting $f$ to $\map {\sigma_A} x$.
Then $\map f x \in I$.

Proof:
From Closed Ideal of C*-Algebra is Self-Adjoint, we have $x^\ast \in I$. 
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map \CC K$.
From Subalgebra Generated by Inclusion and Conjugate is Everywhere Dense in Space of Continuous Functions on Compact Subset of Complex Numbers, the set:
:$\ds \AA = \set {z \mapsto \sum_{j \mathop = 1}^N \alpha_j z^{\lambda_j} \overline z^{\mu_j} : n \in \N, \, \alpha_j \in \C, \, \lambda_j, \mu_j \ge 0} \subseteq \map \CC K$
is everywhere dense in $\map \CC K$.
Take $\sequence {p_n}_{n \mathop \in \N}$ in $\AA$ so that:
:$\norm {p_n - f}_\infty \to 0$ as $n \to \infty$.
We therefore have $\map {p_n} 0 \to \map f 0 = 0$. 
Set $q_n = p_n - \map {p_n} 0$. 
Hence $\sequence {q_n}_{n \mathop \in \N}$ is such that:
:$\norm {q_n - f}_\infty \to 0$
and there exists $N_n$, $\alpha_{j, n}$, $\lambda_{j, n}$ and $\mu_{j, n}$ such that:
:$\ds \map {q_n} z = \sum_{j \mathop = 1}^{N_n} \alpha_{j, n} z^{\lambda_{j, n} } \overline z^{\mu_{j, n} }$
where for each $j$ and $n$ we have $\lambda_{j, n} \ge 1$ or $\mu_{j, n} \ge 1$. 
We have:
:$\ds \map {q_n} x = \sum_{j \mathop = 1}^{N_n} \alpha_{j, n} x^{\lambda_{j, n} } \paren {x^\ast}^{\mu_{j, n} }$
Hence $\map {q_n} x \in I$ for each $n \in \N$ since $I$ is an ideal. 
Since the continuous functional calculus is an linear isometry, we have:
:$\norm {\map {q_n} x - \map f x}_\infty \to 0$ as $n \to \infty$.
Hence since $I$ is closed, we have $\map f x \in I$.
{{qed}}
Category:Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal
Category:Continuous Functional Calculus
jkt40jmdy94yocr3l9wsi6vtjvyz8ef
Theorem: Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal/Corollary

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\struct {A_+, \ast, \norm {\, \cdot \,}_\ast}$ be the unitization of $\struct {A, \ast, \norm {\, \cdot \,} }$.
Let $A_0 = \set {\tuple {a, 0} : a \in A}$. 
Let $x \in I$ be normal.
Let $\map {\sigma_{A_+} } x \subseteq \R$ be the spectrum of $\tuple {x, 0}$ in $A_+$.
Let $f : \map {\sigma_{A_+} } x \to \R$ be a continuous function with $\map f 0 = 0$.
Let $\map f {\tuple {x, 0} }$ be obtained from the continuous functional calculus.
Then $\map f {\tuple {x, 0} } \in A_0$.
Hence there exists $u \in A$ such that $\map f {\tuple {x, 0} } = \tuple {u, 0}$.

Proof:
From Normed Algebra Embeds into Unitization as Closed Ideal, $A_0$ is a closed ideal of $A$.
We have that $\tuple {x, 0}$ is normal in $A_+$.
The result follows from Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal.
{{qed}}
Category:Continuous Function Vanishing at Zero applied to Normal Element of Closed Ideal of Unital C*-Algebra is contained in Ideal
qn6yesfynshn0ruf8e1ks9u0s3edqws
Theorem: Inverse Image Mapping of Bijection is Inverse of Direct Image Mapping

Let $S$ and $T$ be sets.
Let $f :S \to T$ be a bijection.
Let $f^\to$ be the direct image mapping of $f$.
Let $f^\gets$ be the inverse image mapping of $f$.
Then:
:$\paren {f^\to}^{-1} = f^\gets$
where $\paren {f^\to}^{-1}$ denotes the inverse of $f^\to$.

Proof:
From Mapping is Bijection iff Direct Image Mapping is Bijection:
:$f^\to$ is a bijection
Let $\paren {f^\to}^{-1}$ denote the inverse of $f^\to$.
We have:
{{begin-eqn}}
{{eqn | l = f^\gets \circ f^\to
      | r = \operatorname{id}_{\powerset S} \circ \paren{f^\gets \circ f^\to}
      | c = Identity Mapping is Left Identity
}}
{{eqn | r = \paren{\paren {f^\to}^{-1} \circ f^\to} \circ \paren{f^\gets \circ f^\to}
      | c = {{Defof|Inverse Mapping}} 
}}
{{eqn | r = \paren {f^\to}^{-1} \circ \paren{f^\to \circ \paren{f^\gets \circ f^\to} }
      | c = Composition of Mappings is Associative
}}
{{eqn | r = \paren {f^\to}^{-1} \circ f^\to
      | c = Direct Image of Inverse Image of Direct Image equals Direct Image Mapping
}}
{{eqn | r = \operatorname{id}_{\powerset S}
      | c = {{Defof|Inverse Mapping}} 
}}
{{end-eqn}}
Similarly, we have:
{{begin-eqn}}
{{eqn | l = f^\to \circ f^\gets
      | r = \paren{f^\gets \circ f^\to} \circ \operatorname{id}_{\powerset T}
      | c = Identity Mapping is Right Identity
}}
{{eqn | r = \paren{f^\gets \circ f^\to} \circ \paren{f^\to \circ \paren {f^\to}^{-1} }
      | c = {{Defof|Inverse Mapping}} 
}}
{{eqn | r = \paren{\paren{f^\to \circ f^\gets} \circ f^\to} \circ \paren {f^\to}^{-1}
      | c = Composition of Mappings is Associative
}}
{{eqn | r = f^\to \circ \paren {f^\to}^{-1}
      | c = Direct Image of Inverse Image of Direct Image equals Direct Image Mapping
}}
{{eqn | r = \operatorname{id}_{\powerset T}
      | c = {{Defof|Inverse Mapping}}
}}
{{end-eqn}}
It follows that $f^\gets$ is the inverse of $f^\to$ by definition.
From Inverse Mapping is Unique:
:$\paren {f^\to}^{-1} = f^\gets$
{{qed}}
Category:Direct Image Mappings
Category:Inverse Image Mappings
jih432jzug1b5tbexp81715zls2mxav
Theorem: Generalized Polar Decomposition in C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $x \in A$.
Let $\cmod x$ be the modulus of $x$. 
Let $0 < \beta < 1$ be a real number.
Then there exists $u \in A$ such that:
:$x = u \cmod x^\beta$
where $\cmod x^\beta$ is obtained from the continuous functional calculus.

Proof:
Let $\le_A$ be the canonical preordering of $A$.
From Product of Element of C*-Algebra with its Star is Positive, we have ${\mathbf 0}_A \le_A x^\ast x$.
Hence since $x^\ast x \le_A x^\ast x$ and $0 < \beta/2 < 1/2$, there exists $u \in A$ such that:
:$x = u \paren {x^\ast x}^{\beta/2}$
from Factorization Theorem for C*-Algebra in terms of Bound on Modulus.
From Power of Power of Positive Element of Unital C*-Algebra, we have:
:$u \paren {x^\ast x}^{\beta/2} = u \paren {\paren {x^\ast x}^{1/2} }^\beta = u \cmod x^\beta$
{{qed}}
Theorem: Hilbert Space Projections with Zero Product are Pointwise Orthogonal

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $P, Q : \HH \to \HH$ be projections (in the Hilbert space sense) such that:
:$P Q = 0$ 
Then:
:$\innerprod {Q x} {P y} = 0$ for each $x, y \in \HH$.

Proof:
We have, for each $x \in \HH$:
:$\map {\paren {P Q} } x = {\mathbf 0}_\HH$
Hence:
:$\map P {Q x} = {\mathbf 0}_\HH$
So we have $Q x \in \ker P$.
Since $x \in \HH$ was arbitrary we have $\Rng Q \subseteq \ker P$.
From the definition of a projections (in the Hilbert space sense), we have:
:$\ker P = \paren {\Rng Q}^\bot$
Hence:
:$\Rng Q \subseteq \paren {\Rng P}^\bot$
Hence for each $x \in \HH$, we have $Q x \in \paren {\Rng Q}^\bot$.
Hence for each $y \in \HH$ we have $Q x \perp P y$ from the definition of orthocomplement.
That is, $\innerprod {Q x} {P y} = 0$. 
{{qed}}
Category:Orthogonal Projections
4gpfpihsxu5glbn1la6rrjrfrpptvit
Theorem: Orthogonality of Resolution of the Identity evaluated at Disjoint Sets

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $A, B \in \map \BB X$ be such that $A \cap B = \O$.
Then:
:$\innerprod {\map \EE A x} {\map \EE B y} = 0$ 
for each $x, y \in \HH$.

Proof:
From $(2)$ in the definition of a resolution of the identity, we have:
:$\map \EE A \map \EE B = \map \EE {A \cap B} = \map \EE \O$
From $(1)$ in the definition of a resolution of the identity, we then have:
:$\map \EE A \map \EE B = 0$
From Hilbert Space Projections with Zero Product are Pointwise Orthogonal, we therefore havE:
:$\innerprod {\map \EE A x} {\map \EE B y} = 0$ 
for each $x, y \in \HH$.
{{qed}}
Category:Resolutions of the Identity
l6krzx2lbuos8eulroc5gjzzd6kqt5b
Theorem: Scalar-Valued Measure associated with Resolution of Identity is Positive Measure if Equal Vectors

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
If $x = y$, then $\EE_{x, y} := \EE_x$ is a measure with total variation $\norm x^2$.

Proof:
Let $A \in \map \BB X$. 
We then have:
{{begin-eqn}}
{{eqn	| l = \map {\EE_x} A
		| r = \innerprod {\map \EE A x} x
}}
{{eqn	| r = \innerprod {\map \EE A^2 x} x
		| c = Orthogonal Projection is Projection 
}}
{{eqn	| r = \innerprod {\map \EE A x} {\map \EE A x}
		| c = from Characterization of Projections, $\map \EE A$ is Hermitian
}}
{{eqn	| r = \norm {\map \EE A x}^2
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| o = \ge 
		| r = 0
}}
{{end-eqn}}
Hence $\EE_x$ is a measure.
Let $\norm {\EE_x}$ be the total variation of $\EE_x$.
Then:
{{begin-eqn}}
{{eqn	| l = \norm {\EE_x}
		| r = \map {\EE_x} X
		| c = {{Defof|Total Variation|subdef = Measure Theory}}
}}
{{eqn	| r = \innerprod {\map \EE X x} x
}}
{{eqn	| r = \innerprod x x
		| c = $(1)$ in {{Defof|Resolution of the Identity}}
}}
{{eqn	| r = \norm x^2 
		| c = {{Defof|Inner Product Norm}}
}}
{{end-eqn}}
{{qed}}
Category:Resolutions of the Identity
oyw7bcwzr3c4zfgaq9vkud0rtilh43d
Theorem: Null Sets Closed under Countable Union/Resolution of the Identity

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $\set {A_j : j \in \N} \subseteq \map \BB X$ such that:
:$\map \EE {A_j} = 0$ for each $j \in \N$.
Let:
:$\ds A = \bigcup_{j \mathop = 1}^\infty A_j$
Then $\map \EE A = 0$.

Proof:
Let $x \in \HH$. 
Define $\EE_{x, x} : \map \BB X \to \C$ by:
:$\map {\EE_{x, x} } A = \innerprod {\map \EE A x} x$
for each $A \in \map \BB X$.
Since $\map \EE {A_j} = 0$ for each $j \in \N$, we have $\map {\EE_{x, x} } {A_j} = 0$ for each $j \in \N$. 
Since $\EE_{x, x}$ is a complex measure, we have:
:$\map {\EE_{x, x} } A = 0$ 
from countable additivity.
From Scalar-Valued Measure associated with Resolution of Identity is Positive Measure if Equal Vectors, we have:
:$\map {\EE_{x, x} } A = \norm {\map \EE A x}^2 = 0$ 
From {{NormAxiomVector|1}}, we have $\map \EE A x = 0$.
Further, $x \in \HH$ is arbitrary so we have $\map \EE A = 0$. 
{{qed}}
Theorem: Poincaré Duality Theorem

Let $M$ be an $n$-manifold.
Let its $n$th homology group $\map {H_n} M$ be infinite and cyclic (that is, that $M$ is an orientable manifold).
Then $\map {H_r} M$ is homeomorphic to the $\paren {n - r}$th cohomology group $\map {H^{n - r} } M$ for all $r$.

Proof:
{{ProofWanted}}
{{Namedfor|Jules Henri Poincaré|cat = Poincaré}}
Theorem: Null Sets Closed under Subset/Resolution of the Identity

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $E, E' \in \map \BB X$ be such that:
:$E \subseteq E'$ and $\map \EE {E'} = {\mathbf 0}_{\map B \HH}$
Then $\map \EE E = {\mathbf 0}_{\map B \HH}$.

Proof:
For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
We then have:
:$\map {\EE_{x, y} } {E'} = 0$
for each $x, y \in \HH$.
From Null Sets Closed under Subset, we have $\map {\EE_{x, y} } E = 0$ for each $x, y \in \HH$.
That is, $\innerprod {\map \EE E x} y = 0$ for all $x, y \in \HH$.
From Linear Subspace Dense iff Zero Orthocomplement, we have $\HH^\bot = \set { {\mathbf 0}_\HH}$ and hence $\map \EE E x = {\mathbf 0}_\HH$ for each $x \in \HH$.
Hence $\map \EE E = {\mathbf 0}_{\map B \HH}$.
{{qed}}
Theorem: Null Sets Closed under Subset/Resolution of the Identity

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $E, E' \in \map \BB X$ be such that:
:$E \subseteq E'$ and $\map \EE {E'} = {\mathbf 0}_{\map B \HH}$
Then $\map \EE E = {\mathbf 0}_{\map B \HH}$.

Proof:
Let $\le_{\map B \HH}$ be the canonical preordering on $\map B \HH$.
From Measure is Monotone: Resolution of the Identity, we have:
:$\map \EE E \le_{\map B \HH} \map \EE {E'} = {\mathbf 0}_{\map B \HH}$
On the other hand from Bounds on Projection in Unital C*-Algebra, we have:
:${\mathbf 0}_{\map B \HH} \le_{\map B \HH} \map \EE E$
From Canonical Preordering of C*-Algebra is Antisymmetric, we then have:
:$\map \EE E = {\mathbf 0}_{\map B \HH}$
{{qed}}
Theorem: Canonical Preordering of C*-Algebra is Antisymmetric

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering of $A$.
Let $a, b \in A$ such that:
:$a \le_A b$
and:
:$b \le_A a$
Then $a = b$.
That is, $\le_A$ is antisymmetric and hence $\le_A$ is a partial ordering.

Proof:
From the definition of $\le_A$ we have:
:$b - a$ is positive.
Hence $b - a$ is Hermitian and:
:$\map {\sigma_A} {b - a} \subseteq \hointr 0 \infty$
From the Spectral Mapping Theorem for Polynomials, we also have:
:$\map {\sigma_A} {a - b} = \set {-t : t \in \map {\sigma_A} {a - b} } \subseteq \hointl {-\infty} 0$ 
Again from the definition of $\le_A$ we have:
:$a - b$ is positive.
Hence:
:$\map {\sigma_A} {a - b} \subseteq \hointr 0 \infty$
Hence we have:
:$\map {\sigma_A} {a - b} \subseteq \hointr 0 \infty \cap \hointl {-\infty} 0 = \set 0$
From Spectrum of Element of Banach Algebra is Non-Empty, we have $\map {\sigma_A} {a - b} \ne \O$.
Hence $\map {\sigma_A} {a - b} = \set 0$.
So $\map {r_A} {a - b} = 0$ where $r_A$ denotes spectral radius.
Hence from Spectral Radius of Normal Element of C*-Algebra Equal to Norm: Corollary we have $a - b = {\mathbf 0}_A$.
Hence $a = b$.
{{qed}}
Category:Canonical Preorderings of C*-Algebras
0yntviqfti43lvv853hd6hr931lpp7g
Theorem: Characterization of Injective Linear Transformations with Closed Image

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be Banach spaces. 
Let $T : X \to Y$ be a bounded linear transformation.
Then $T$ is injective and $\Img T$ is closed {{iff}}:
:there exists $c > 0$ such that $\norm {T x}_Y \ge c \norm x_X$ for each $x \in X$.

Proof:
=== Sufficient Condition ===
Suppose that:
:there exists $c > 0$ such that $\norm {T x}_Y \ge c \norm x_X$ for each $x \in X$.
First, if $T x = {\mathbf 0}_Y$ for some $x \in X$, then we have:
:$c \norm x_X \le 0$
Since $c > 0$ and $\norm x_X \ge 0$, we have $\norm x_X = 0$.
Hence by {{NormAxiomVector|1}}, we have $x = {\mathbf 0}_X$.
So $T$ is injective.
Now let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $\Img T$ such that $y_n \to y$ with $y \in Y$. 
We show that $y \in \Img T$. 
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$ such that $y_n = T x_n$ for each $n \in \N$.
Then:
:$\norm {x_n - x_m}_X \le c^{-1} \norm {T x_n - T x_m}_Y = c^{-1} \norm {y_n - y_m}_Y$ 
from the linearity of $T$.
Since $\sequence {y_n}_{n \mathop \in \N}$ is convergent, it is Cauchy by Convergent Sequence is Cauchy Sequence.
Let $\epsilon > 0$.
Since $\sequence {y_n}_{n \mathop \in \N}$ is Cauchy, there exists $N \in \N$ such that:
:$\norm {y_n - y_m} < c \epsilon$ for $n, m \ge N$.
Then for $n, m \ge N$, we have $\norm {x_n - x_m}_X < \epsilon$.
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Since $X$ is Banach, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$.
Hence since $T$ is bounded, we have $T x_n \to T x$ from Continuous Mapping is Sequentially Continuous.
That is, $y_n \to T x$ with $T x \in \Img T$. 
So:
:whenever $\sequence {y_n}_{n \mathop \in \N}$ is a sequence in $\Img T$ converging to $y \in Y$, we have $y \in \Img T$.
Hence $\Img T$ is closed.
{{qed|lemma}}
Theorem: Characterization of Spectrum of Bounded Linear Operator in Hilbert Space in terms of Approximate Eigenvalues

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\lambda \in \map \sigma T$ {{iff}} $\lambda$ is an approximate eigenvalue of $T$ or $\overline \lambda$ is an eigenvalue of $T^\ast$.

Proof:
=== Sufficient Condition ===
Suppose that $\lambda$ is an approximate eigenvalue of $T$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\HH$ such that:
:$\norm {x_n} = 1$ for each $n \in \N$
and:
:$\norm {T x_n - \lambda x_n} \to 0$ as $n \to \infty$.  
By Characterization of Injective Linear Transformations with Closed Range: Corollary, we either have:
:$T - \lambda I$ is not injective or $\Rng {T - \lambda I}$ is not closed.
If $\Rng {T - \lambda I}$ is not closed then $\Rng {T - \lambda I} \ne \HH$, so $T - \lambda I$ is not surjective in this case.
Either way, $\lambda \in \map \sigma T$. 
Now suppose that $\overline \lambda$ is an eigenvalue of $T^\ast$. 
That is:
:$\map \ker {T^\ast - \overline \lambda I} \ne \set { {\mathbf 0}_\HH}$
From Kernel of Linear Transformation is Orthocomplement of Image of Adjoint, we have:
:$\map \ker {T^\ast - \overline \lambda I} = \paren {\Rng {\paren {T^\ast - \overline \lambda I}^\ast} }^\bot = \Rng {T - \lambda I}^\bot$
from Adjoint is Involutive.
So:
:$\Rng {T - \lambda I}^\bot \ne \set { {\mathbf 0}_\HH}$
From Linear Subspace Dense iff Zero Orthocomplement, we therefore have:
:$\Rng {T - \lambda I}$ is not everywhere dense.
In particular, $\Rng {T - \lambda I} \ne \HH$, so $\lambda \in \map \sigma T$.
{{qed|lemma}}
Theorem: Eigenvalue of Bounded Linear Operator is Approximate Eigenvalue

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\lambda \in \C$ be an eigenvalue of $T$.
Then $\lambda$ is an approximate eigenvalue of $T$.

Proof:
Since $\lambda$ is an eigenvalue of $T$, there exists $y \in \HH$ with $y \ne {\mathbf 0}_\HH$ such that:
:$T y = \lambda y$
Setting:
:$\ds x_n = \frac y {\norm y}$
we have $\norm {x_n} = 1$ by {{NormAxiomVector|2}} and:
:$T x_n = \lambda x_n$ 
We then have:
:$\norm {T x_n - \lambda x_n} = 0$ for each $n \in \N$.
So:
:$\norm {T x_n - \lambda x_n} \to 0$
Hence $\lambda$ is an approximate eigenvalue of $T$.
{{qed}}
Category:Approximate Eigenvalues (Bounded Linear Operators)
8m9wrcus98dbswla01tzs0zc74zve7p
Theorem: Space of Bounded Linear Operators on Hilbert Space is Unital C*-Algebra

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B \HH$ be the space of bounded linear operators.
Let $\norm {\, \cdot \,}_{\map B \HH}$ be the operator norm on $\map B \HH$.
Let $\ast : \map B \HH \to \map B \HH$ denote the adjoint operation.
Then $\struct {\map B \HH, \ast, \norm {\, \cdot \,}_{\map B \HH} }$ is a unital $\text C^\ast$-algebra.

Proof:
From Space of Bounded Linear Transformations is Unital Banach Algebra, $\struct {\map B \HH, \norm {\, \cdot \,} }$ is a unital Banach algebra.
Theorem: Characterization of Homeomorphic Topological Spaces/Necessary Condition

Let $T_1 = \struct {S_1, \tau_1}$ be topological space.
Let $S_2$ be a set.
Let $\tau_2$ be a subset of the powerset $\powerset {S_2}$.
Let $\struct {S_2, \tau_2}$ be a topological space homeomorphic to $T_1$.
Then:
:there exists a mapping $f : S_1 \to S_2$:
::$(1): \quad f$ is a bijection
::$(2): \quad f^\to \restriction_{\tau_1}$ is a bijection from $\tau_1$ to $\tau_2$
:where
::$f^\to \restriction_{\tau_1}$ denotes the restriction of $f^\to$ to $\tau_1$
::$f^\to$ denotes the direct image mapping of $f$

Proof:
Let $f: S_1 \to S_2$ be a homeomorphism.
By definition of a homeomorphism:
:$f$ is a bijection
:$f$ is an open mapping
:$f$ is a continuous mapping
By definition of an open mapping:
:$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$
By definition of direct image mapping:
:$\forall U \in \tau_1 : \map {f^\to} U \in \tau_2$
By definition of continuous mapping:
:$\forall V \in \tau_2 : f^{-1} \sqbrk V \in \tau_1$
We have:
{{begin-eqn}}
{{eqn | q = \forall V \in \tau_2
      | l = \map {f^\to} {f^{-1} \sqbrk V}
      | r = f \sqbrk {f^{-1} \sqbrk V}
      | c = {{Defof|Direct Image Mapping}}
}}
{{eqn | r = \map {\paren{f \circ f^{-1} } } V
      | c = {{Defof|Composite Mapping}}
}}
{{eqn | r = V
      | c = Image of Preimage of Subset under Surjection equals Subset
}}
{{end-eqn}}
It follows that $f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$.
Theorem: Spectrum of Projection in *-Algebra

Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $p$ be a projection on $A$.
Let $\map {\sigma_A} p$ be the spectrum of $p$ in $A$.
Then $\map {\sigma_A} p \subseteq \set {0, 1}$.

Proof:
From the definition of a projection, we have $p^2 = p$.
Hence $p^2 - p = {\mathbf 0}_A$.
Hence from Spectrum of Zero Vector in Algebra, we have $\map {\sigma_A} {p^2 - p} = \map {\sigma_A} { {\mathbf 0}_A} = \set 0$.
From the Spectral Mapping Theorem for Polynomials, we have:
:$\set 0 = \set {z^2 - z : z \in \map {\sigma_A} p}$
In other words:
:if $z \in \map {\sigma_A} p$, then $z^2 - z = 0$.
That is:
:$z = 0$ or $z = 1$.
So we obtain $\map {\sigma_A} p \subseteq \set {0, 1}$.
{{qed}}
Category:Projections (*-Algebras)
Category:Spectra (Spectral Theory)
Category:Spectrum of Projection in *-Algebra
cmwxczezdpme5p6gcvinx1qgj2zo099
Theorem: Spectrum of Zero Vector in Algebra

Let $A$ be an algebra over $\C$. 
Let ${\mathbf 0}_A$ be the zero vector of $A$.
Let $\map {\sigma_A} { {\mathbf 0}_A}$ be the spectrum of ${\mathbf 0}_A$ in $A$.
Then:
:$\map {\sigma_A} { {\mathbf 0}_A} = \set 0$ if $A \ne \set { {\mathbf 0}_A}$
:$\map {\sigma_A} { {\mathbf 0}_A} = \O$ if $A = \set { {\mathbf 0}_A}$.

Proof:
First suppose that $A$ is unital with identity element ${\mathbf 1}_A$.
For $\lambda \in \C \setminus \set 0$, we have:
:$\lambda^{-1} {\mathbf 1}_A \paren {\lambda {\mathbf 1}_A - {\mathbf 0}_A} = {\mathbf 1}_A = \paren {\lambda {\mathbf 1}_A - {\mathbf 0}_A} \lambda^{-1} {\mathbf 1}_A$
Hence $\lambda \not \in \map {\sigma_A} { {\mathbf 0}_A}$.
If $A = \set { {\mathbf 0}_A}$, then we have:
:${\mathbf 0}_A^2 = {\mathbf 0}_A = {\mathbf 1}_A$
so $0 \not \in \map {\sigma_A} { {\mathbf 0}_A}$, and we get $\map {\sigma_A} { {\mathbf 0}_A} = \O$ in this case. 
Now suppose that $A \ne \set { {\mathbf 0}_A}$, then ${\mathbf 1}_A \ne {\mathbf 0}_A$.
We then have $a {\mathbf 0}_A = {\mathbf 0}_A a = {\mathbf 0}_A \ne {\mathbf 1}_A$ for all $a \in A$.
Hence $0 \in \map {\sigma_A} { {\mathbf 0}_A}$.
Hence we get $\map {\sigma_A} { {\mathbf 0}_A} = \set 0$ in this case.
Now suppose that $A$ is not unital.
Let $A_+$ be the unitization of $A$.
Then we have:
:$\map {\sigma_A} { {\mathbf 0}_A} = \map {\sigma_{A_+} } {\tuple { {\mathbf 0}_A, 0} } \cup \set 0$
From the unital case we have:
:$\map {\sigma_{A_+} } {\tuple { {\mathbf 0}_A, 0} } = \set 0$
and hence:
:$\map {\sigma_A} { {\mathbf 0}_A} = \set 0$
{{qed}}
Category:Spectra (Spectral Theory)
3itw5sgblcozt0a6gtz07zy04j9klx0