nyuuzyou/wikis huggingface dataset:
Spacing Limit Theorem

== Theorem ==
Let $X_{\paren i}$ be the $i$th ordered statistic of $N$ samples from a continuous random distribution with density function $\map {f_X} x$.
Then the spacing between the ordered statistics given $X_{\paren i}$ converges in distribution to exponential for sufficiently large sampling according to:
:$N \paren {X_{\paren {i + 1} } - X_{\paren i} } \xrightarrow D \map \exp {\dfrac 1 {\map f { X_{\paren i} } } }$
as $N \to \infty$ for $i = 1, 2, 3, \dotsc, N - 1$.
== Proof ==
Given $i$ and $N$, the ordered statistic $X_{\paren i}$ has the probability density function:
:$\map {f_{X_{\paren i} } } {x \mid i, N} = \dfrac {N!} {\paren {i - 1}! \paren {N - i}!} \map {F_X} x^{i - 1} \paren {1 - \map {F_X} x}^{N - i} \map {f_X} x$
where $\map {F_X} x$ is the cumulative distribution function of $X$.
Let $Y_i = N \paren {X_{\paren {i + 1} } - X_{\paren i} }$ be an independent spacing variable valid for $i = 1, 2, 3, \dotsc, N - 1$ which is always positive.
The joint density function of both $X_{\paren i}$ and $Y_{i}$ is then:
:$\map {f_{X_{\paren i}, Y_i} } {x, y \mid i, N} = \dfrac {\paren {N - 1}!} {\paren {i - 1}! \paren {N - i - 1}!} \map {F_X} x^{i - 1} \paren {1 - \map {F_X} {x + \dfrac y N} }^{N - i - 1} \map {f_X} x \map {f_X} {x + \dfrac y N}$
The conditional density function of $Y_i$ given $X_{\paren i}$ is:
:$f_{Y_i} = \dfrac {f_{X_{\paren i}, Y_i} } {f_{X_{\paren i} } }$
which turns into:
:$\map {f_{Y_i} } {y \mid x = X_{\paren i}, i, N} = \dfrac {N - i} N \dfrac {\paren {1 - \map {F_X} {x + \dfrac y N} }^{N - i - 1} } {\paren {1 - \map {F_X} x}^{N - i} } \map {f_X} {x + \dfrac y N}$
The conditional cumulative function of $Y_i$ given $X_{\paren i}$ is:
:$\map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 - \paren {\dfrac {1 - \map {F_X} {x + \dfrac y N} } {1 - \map {F_X} x} }^{N - i}$
The following Taylor expansion in $y$ is an approximation of $\map {F_X} {x + \dfrac y N}$:
:$\map {F_X} {x + \dfrac y N} = \map {F_X} x + \map {f_X} x \dfrac y N + \map \OO {N^{-2} }$
Inserting this produces:
:$\map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 - \paren {1 - \dfrac {\map {f_X} x y} {N \paren {1 - \map {F_X} x} } + \map \OO {N^{-2} } }^{N - i}$
The limit as $N$ gets large is the exponential function:
:$\map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 -e^{- \map {f_X} x y \dfrac {1 - \dfrac i N} {1 - \map {F_X} x} } + \map \OO {N^{-1} }$
The distribution of $F_X$ is uniform by definition of a random pick:
:$\map {F_X} x \sim \map U {0, 1}$
Therefore $i$ is uniformly distributed as well:
:$\map {F_X} {X_{\paren i} } \approx \dfrac i N$
The limit of $F_{Y_i}$ is then:
:$\ds \lim_{N \mathop \to \infty} \map {F_{Y_i} } {y \mid x = X_{\paren i}, i, N} = 1 - e^{- \map {f_X} x y}$
Category:Probability Theory
Category:Named Theorems
Span of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x is preserved under Differentiation wrt x

== Theorem ==
Let $a, b, x, \alpha_1, \alpha_2 \in \R$ be real numbers.
Let $f_1$ and $f_2$ be the real functions defined as:
Let $\map \CC \R$ be the space of continuous real-valued functions.
Let $\tuple {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be the vector space of continuously differentiable real-valued functions.
Let $S = \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.
Let $\map \span S$ be the span of $S$.
Then differentiation $x$ preserves $\map \span S$.
== Proof ==
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Real Analysis
Spanning Criterion of Normed Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.
Let $X^\ast$ be the vector space of bounded linear functionals on $X$.
Let $A \subseteq X$ be a subset.
Let $\vee A$ be the closed linear span of $A$, i.e. the closure of the linear span of $A$.
Then $z \in \vee A$ :
:$\forall \ell \in X^\ast : \ell \restriction_A = 0 \implies \map \ell z = 0$
where $\ell \restriction_A$ denotes the restriction of $\ell$ to $A$.
== Proof ==
=== Necessary condition ===
Let $z \in \vee A$.
Let $\ell \in X^\ast$ such that $\ell \restriction_A = 0$.
Recall, by , there is a $C > 0$ such that:
:$(1):\quad \forall x \in X : \size {\map \ell x} \le C \norm x$
On the other hand::
Let $\epsilon > 0$.
Then there exist:
:$n \in \Z_{>0}$
:$\alpha_1, \ldots, \alpha_n \in \GF$
:$a_1, \ldots, a_n \in A$
such that:
:$\ds (3):\quad \norm {z - \sum_{i \mathop = 1}^n \alpha_i a_i} < \epsilon$.
Therefore:
By $\epsilon \to 0$, we obtain:
:$\map \ell z = 0$
=== Sufficient condition ===
Let $z \not \in \vee A$.
As $\vee A = \map \cl {\map \span A}$, we have:
:$\ds d := \inf_{x \in \map \span A} \norm {z - x} > 0$
Consider the linear subspace:
:$V := \map \span A + \GF z$
Define the linear functional $ \ell_0 : V \to \GF$ by:
:$\map {\ell_0} {x + \alpha z} = \alpha$
where $x \in \map \span A$ and $\alpha \in \GF$.
$\ell_0$ is well-defined, since if:
:$x + \alpha z = x' + \alpha' z$
then:
:$\paren {\alpha - \alpha'} z = x' - x \in \map \span A \cap \GF z = \set 0$
Observe, if $\alpha \ne 0$:
If $\alpha = 0$, then:
:$\map {\ell_0} {x + \alpha z} = 0$
Thus:
:$\forall v \in V : \size { \map {\ell_0} v} \le d^{-1} \norm v$
By Hahn-Banach Theorem, there exists an $\ell \in X^\ast$ such that:
:$\forall x \in X : \size { \map \ell v} \le d^{-1} \norm x$
:$\ell \restriction_V = \ell_0$
In particular:
:$\ell \restriction_A = {\ell_0} \restriction_A = 0$
:$\map \ell z = \map {\ell_0} z = 1 \ne 0$
== Sources ==
* : $8.2$: Extension of Bounded Linear Functionals
Category:Normed Vector Spaces
Sparse Matrix/Examples/Arbitrary Example 1

== Example of Sparse Matrix ==
A tridiagonal matrix is an example of a sparse matrix
:$\begin {pmatrix} a & b & 0 & \cdots & 0 & 0 \\ c & d & e & \cdots & 0 & 0 \\ 0 & f & g & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & w & x \\ 0 & 0 & 0 & \cdots & y & z \end {pmatrix}$
particularly when its order is significantly greater than $3$.
== Sources ==
* 
* 
Category:Examples of Sparse Matrices
Spearman's Rank Correlation Coefficient/Examples/Examination Results

== Example of Use of Spearman's Rank Correlation Coefficient ==
Let $S$ be a class of students.
Let $\sequence {x_i}$ and $\sequence {y_i}$ be rankings of $S$ based on the examination results in French and Mathematics respectively.
Spearman's rank correlation coefficient may be appropriate to determine the correlation between $\sequence {x_i}$ and $\sequence {y_i}$.
== Sources ==
* 
* 
Category:Examples of Spearman's Rank Correlation Coefficient
Special Highly Composite Number/Examples/1

== Example of Special Highly Composite Number ==
$1$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.
== Proof ==
We have that $1$ is highly composite.
From One Divides all Integers, it follows trivially that $1$ is a divisor of all larger highly composite numbers.
Thus, by definition, $1$ is a special highly composite number.
== Sources ==
* 
Category:Special Highly Composite Numbers
Special Highly Composite Number/Examples/12

== Example of Special Highly Composite Number ==
$12$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.
== Proof ==
By inspection of the sequence of highly composite numbers, $12$ is highly composite.
 $n > 12$ is a highly composite number which is not divisible by $12$.
We have that $6$ is a special highly composite number.
Therefore $6$ is a divisor of $n$.
As $12$ is not a divisor of $n$, it follows that the multiplicity of $2$ in $n$ is $1$.
From Prime Decomposition of Highly Composite Number, that means:
: $n = 2 \times 3 \times 5 \times r$
where $r$ is a possibly vacuous square-free product of prime numbers strictly greater than $5$.
Then:
The result follows by Proof by Contradiction.
== Sources ==
* 
Category:Special Highly Composite Numbers
Category:12
Special Highly Composite Number/Examples/2

== Example of Special Highly Composite Number ==
$2$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.
== Proof ==
We have that $2$ is highly composite.
Let $n > 2$ be a highly composite number.
From Prime Decomposition of Highly Composite Number, the multiplicity of $2$ in $n$ is at least as high as the multiplicity of any other prime $p$ in $n$.
Thus if $p \divides n$ it follows that $2 \divides n$.
Thus, by definition, $2$ is a special highly composite number.
== Sources ==
* 
Category:Special Highly Composite Numbers
Category:2
Special Highly Composite Number/Examples/6

== Example of Special Highly Composite Number ==
$6$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.
== Proof ==
By inspection of the sequence of highly composite numbers, $6$ is highly composite.
 $n > 6$ is a highly composite number which is not divisible by $6$.
We have that $2$ is a special highly composite number.
Therefore $2$ is a divisor of $n$.
As $6$ is not a divisor of $n$, it follows that $3$ is also not a divisor of $n$.
By Prime Decomposition of Highly Composite Number, that means $n = 2^k$ for some $k \ge 3$.
Then:
But this contradicts our deduction that $n = 2^k$ where $k \ge 3$.
The result follows by Proof by Contradiction.
== Sources ==
* 
Category:Special Highly Composite Numbers
Category:6
Special Highly Composite Number/Examples/60

== Example of Special Highly Composite Number ==
$60$ is a special highly composite number, being a highly composite number which is a divisor of all larger highly composite numbers.
== Proof ==
By inspection of the sequence of highly composite numbers, $60$ is highly composite.
 $n > 60$ is a highly composite number which is not divisible by $60$.
We have that $12$ is a special highly composite number.
Therefore $12$ is a divisor of $n$.
As $60$ is not a divisor of $n$, it follows that while $3$ is a divisor of $n$, $5$ is not.
From Prime Decomposition of Highly Composite Number, no prime number greater than $5$ is a divisor of $n$.
Thus:
:$n = 2^a \times 3^b$
where $a \ge b \ge 1$.
This will be investigated on a case-by-case basis.
:$(1): \quad b = 1$
That is, $n = 2^a \times 3$.
We have that $n > 60$.
Therefore:
:$(1 \text a): \quad a \ge 5$
as $2^4 \times 3^1 = 48$.
Then:
It follows by Proof by Contradiction that $b \ne 1$.
:$(2): \quad b = 2$
That is, $n = 2^a \times 3^2$.
We have that $n > 60$.
Therefore:
:$(2 \text a): \quad a \ge 3$
as $2^2 \times 3^2 = 36$.
Then:
It follows by Proof by Contradiction that $b \ne 2$.
:$(3): \quad b \ge 3$
By Prime Decomposition of Highly Composite Number we have that $a \ge 3$.
Then:
By Proof by Cases it is seen that the existence of a highly composite $n$ not divisible by $60$ leads to a contradiction.
The result then follows by Proof by Contradiction.
== Sources ==
* 
Category:Special Highly Composite Numbers
Category:60
Special Linear Group is Normal Subgroup of General Linear Group

== Theorem ==
Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is a normal subgroup of the general linear group $\GL {n, K}$.
== Proof ==
From Special Linear Group is Subgroup of General Linear Group, we have that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
It remains to be shown that $\SL {n, K}$ is a normal subgroup of $\GL {n, K}$.
== Sources ==
* 
* 
Category:Special Linear Group
Category:General Linear Group
Category:Examples of Normal Subgroups
Special Linear Group is Subgroup of General Linear Group

== Theorem ==
Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is a subgroup of the general linear group $\GL {n, K}$.
== Proof ==
Because the determinants of the elements of $\SL {n, K}$ are not $0_K$, they are nonsingular.
So $\SL {n, K}$ is a subset of $\GL {n, K}$.
Now we need to show that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.
Let $\mathbf A$ and $\mathbf B$ be elements of $\SL {n, K}$.
As $\mathbf A$ is nonsingular we have that it has an inverse $\mathbf A^{-1} \in \GL {n, K}$.
From Determinant of Inverse Matrix:
:$\map \det {\mathbf A^{-1} } = \dfrac 1 {\map \det {\mathbf A} }$
and so:
:$\map \det {\mathbf A^{-1} } = 1$
So $\mathbf A^{-1} \in \SL {n, K}$.
Also, from Determinant of Matrix Product:
:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B} = 1$
Hence the result from the Two-Step Subgroup Test.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $6$
* : $\S 36$. Subgroups: Worked Example
* : Chapter $4$: Subgroups: Example $4.4$
Category:Matrix Algebra
Category:Special Linear Group
Category:General Linear Group
Special Linear Group is not Abelian

== Theorem ==
Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.
Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.
Then $\SL {n, K}$ is not an abelian group.
== Proof ==
From Special Linear Group is Subgroup of General Linear Group we have that $\SL {n, K}$ is a group.
From Matrix Multiplication is not Commutative‎ it follows that $\SL {n, K}$ is not abelian.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 29 \alpha \ (5)$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP
Category:Special Linear Group
Special Orthogonal Group is Group

== Theorem ==
Let $k$ be a field.
The $n$th orthogonal group on $k$ is a group.
== Proof ==
A direct corollary of Special Orthogonal Group is Subgroup of Orthogonal Group.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(iv) (c)}$
Category:Orthogonal Groups
Special Orthogonal Group is Subgroup of Orthogonal Group

== Theorem ==
Let $k$ be a field.
Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.
Let $\map {\operatorname {SO} } {n, k}$ be the $n$th special orthogonal group on $k$.
Then $\map {\operatorname {SO} } {n, k}$ is a subgroup of $\map {\operatorname O} {n, k}$.
== Proof ==
We have that Unit Matrix is Proper Orthogonal, so $\map {\operatorname {SO} } {n, k}$ is not empty.
Let $\mathbf A, \mathbf B \in \map {\operatorname {SO} } {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are proper orthogonal.
Then by Inverse of Proper Orthogonal Matrix is Proper Orthogonal:
:$\mathbf B^{-1}$ is a proper orthogonal matrix.
By Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix:
:$\mathbf A \mathbf B^{-1}$ is a proper orthogonal matrix.
Thus by definition of special orthogonal group:
:$\mathbf A \mathbf B^{-1} \in \map {\operatorname {SO} } {n, k}$
Hence the result by One-Step Subgroup Test.
Category:Orthogonal Groups
Spectral Mapping Theorem

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $\map \CC {\map {\sigma_A} x}$ be the space of continuous functions on $\map {\sigma_A} x$.
Let $f \in \map \CC {\map {\sigma_A} x}$.
Let $\map {\Theta_x} f = \map f x$ be the continuous functional calculus of $x$ applied to $f$. 
Then:
:$\map {\sigma_A} {\map f x} = f \sqbrk {\map {\sigma_A} x}$
 
== Proof 1 ==
== Proof 2 ==
Category:Continuous Functional Calculus
Category:Spectral Mapping Theorem
Spectral Mapping Theorem/Proof 1

== Theorem ==
== Proof ==
Let $B \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, x^\ast}$.
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B$ is commutative.
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_A} {\map f x} = \map {\sigma_B} {\map f x}$
Let $\Phi_B$ be the spectrum of $B$. 
From Spectrum of Element of Unital Commutative Banach Algebra, we therefore have:
:$\map {\sigma_A} {\map f x} = \set {\map \phi {\map f x} : \phi \in \Phi_B}$
From Continuous Functional Calculus Commutes with Character on Generated C*-Subalgebra, we have $\map \phi {\map f x} = \map f {\map \phi x}$ for each $\phi \in \Phi_B$. 
Hence we have:
Hence we obtain the desired:
:$\map {\sigma_A} {\map f x} = f \sqbrk {\map {\sigma_A} x}$
Category:Spectral Mapping Theorem
Spectral Mapping Theorem/Proof 2

== Theorem ==
== Proof ==
Let $B \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, x^\ast}$.
In the proof of Existence and Uniqueness of Continuous Functional Calculus, it is shown that $\Theta_x : \map \CC {\map {\sigma_A} x} \to B$ is an isometric unital $\ast$-algebra isomorphism.
Hence by Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism, we have:
:$\map {\sigma_{\map \CC {\map {\sigma_A} x} } } f = \map {\sigma_B} {\map {\Theta_x} f}$
From Spectrum of Element of Space of Continuous Functions on Compact Hausdorff Space, we have:
:$\map {\sigma_{\map \CC {\map {\sigma_A} x} } } f = \set {\map f z : z \in \map {\sigma_A} x} = f \sqbrk {\map {\sigma_A} x}$
Hence:
:$\map {\sigma_B} {\map f x} = f \sqbrk {\map {\sigma_A} x}$
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra we have $\map {\sigma_A} x = \map {\sigma_B} x$ and hence:
:$\map {\sigma_A} {\map f x} = \map {\sigma_A} {\map {\Theta_x} f} = f \sqbrk {\map {\sigma_A} x}$
Category:Spectral Mapping Theorem
Spectral Mapping Theorem for Polynomials

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$. 
Let $p : \C \to \C$ be a polynomial with:
:$\ds \map p z = \sum_{j \mathop = 0}^n a_j z^j$
for all $z \in \C$, for some $a_0, \ldots, a_n \in \C$. 
Define:
:$\ds \map p y = a_0 {\mathbf 1}_A + \sum_{j \mathop = 1}^n a_j y^j$
for each $y \in A$. 
Let $x \in A$.
Let $\map {\sigma_A} x$ and $\map {\sigma_A} {\map p x}$ be the spectrum of $x$ and $\map p x$ respectively.
Then we have:
:$\map {\sigma_A} {\map p x} = p \sqbrk {\map {\sigma_A} x}$
== Proof ==
Let $\map G A$ be the group of units of $A$.
Suppose that $p$ is a constant polynomial, so that:
:$\map p z = \lambda$ 
for each $z \in \C$, and:
:$\map p x = \lambda {\mathbf 1}_A$
From Spectrum of Element of Banach Algebra is Non-Empty, we have:
:$\map {\sigma_A} x \ne \O$
so that:
:$p \sqbrk {\map {\sigma_A} x} = \set \lambda$
As to $\map {\sigma_A} {\map p x}$, we have:
Note that if $\mu \in \C$ is such that $\mu \ne \lambda$, we have:
:$\paren {\mu - \lambda} {\mathbf 1}_A \in \map G A$
with:
:$\ds \paren {\paren {\mu - \lambda} {\mathbf 1}_A}^{-1} = \frac 1 {\mu - \lambda} {\mathbf 1}_A$
If $\mu = \lambda$, we have:
:$\paren {\mu - \lambda} {\mathbf 1}_A = {\mathbf 0}_A$
Since ${\mathbf 0}_A x = {\mathbf 0}_A$ for each $x \in A$, we have:
:$\paren {\mu - \lambda} {\mathbf 1}_A \not \in \map G A$ for $\mu = \lambda$
Hence we conclude that:
:$\set {\mu \in \C : \paren {\mu - \lambda} {\mathbf 1}_A \not \in \map G A} = \set \lambda = p \sqbrk {\map {\sigma_A} x}$
in the case that $p$ is a constant polynomial.
Now suppose that $p$ is not a constant polynomial.
Let $\mu \in \C$. 
From Polynomial Factor Theorem: Corollary: Complex Numbers, there exists $\zeta_1, \ldots, \zeta_n \in \C$ and $\alpha \in \C \setminus \set 0$ such that:
:$\ds \mu - \map p z = \alpha \prod_{j \mathop = 1}^n \paren {\zeta_j - z}$
Then, we have:
:$\ds \mu - \map p x = \alpha \prod_{j \mathop = 1}^n \paren {\zeta_j {\mathbf 1}_A - x}$
From Product of Commuting Elements in Monoid is Unit iff Each Element is Unit, we have that:
:$\ds \mu - \map p x \in \map G A$ $\zeta_j {\mathbf 1}_A - x \in \map G A$ for all $1 \le j \le n$
so that:
:$\ds \mu - \map p x \not \in \map G A$ $\zeta_j {\mathbf 1}_A - x \not \in \map G A$ for some $1 \le j \le n$.
This is equivalent to:
:$\mu - \map p x \not \in \map G A$ $\zeta_j \in \map {\sigma_A} x$ for some $1 \le j \le n$.
Note that:
:$\set {z \in \C : \map p z = \mu} = \set {\zeta_1, \ldots, \zeta_n}$
So, we have:
:$\mu - \map p x \not \in \map G A$ there exists $z \in \map {\sigma_A} x$ such that $\map p z = \mu$.
That is:
:$\mu - \map p x \in \map G A$ $\mu \in p \sqbrk {\map {\sigma_A} x}$
So we obtain:
:$\map {\sigma_A} {\map p x} = p \sqbrk {\map {\sigma_A} x}$
Category:Spectra (Spectral Theory)
Spectral Permanence Theorem

== 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.
Category:Spectra (Spectral Theory)
Category:Unital Banach Algebras
Category:Named Theorems