nyuuzyou/wikis huggingface dataset:
Bound on Total Variation of Scalar-Valued Measure associated with Resolution of the Identity

== Theorem ==
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 $A \in \map \BB X$.
Let $\norm {\EE_{x, y} }$ denote the total variation of $\EE_{x, y}$. 
Then:
:$\norm {\EE_{x, y} } \le \norm x \norm y$
== Proof ==
Let $\map P X$ be the set of finite partitions of $X$ into $\map \BB X$-measurable sets.
We have:
Let $\set {S_1, S_2, \ldots, S_n} \in \map P X$.
We have:
:$\ds \sum_{j \mathop = 1}^n \cmod {\map {\EE_{x, y} } {S_j} } = \sum_{j \mathop = 1}^n \cmod {\innerprod {\map \EE {S_j} x} y}$
Let $\alpha_j \in \C$ be such that $\cmod {\alpha_j} = 1$ and:
:$\cmod {\innerprod {\map \EE {S_j} x} y} = \alpha_j \innerprod {\map \EE {S_j} x} y$
We then have:
Since $\set {S_1, S_2, \ldots, S_n}$ is a partition, we have $S_i \cap S_j = \O$ for $i \ne j$. 
Hence from Orthogonality of Resolution of the Identity evaluated at Disjoint Sets, we have:
:$\innerprod {\map \EE {S_i} x} {\map \EE {S_j} x} = 0$ for $i \ne j$.
We then have:
and hence:
:$\ds \norm {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x} = \norm x$ 
Hence we have:
:$\ds \norm {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x} \norm y = \norm x \norm y$
Hence we have:
:$\ds \sum_{j \mathop = 1}^n \cmod {\map {\EE_{x, y} } {S_j} } \le \norm x \norm y$
Taking the supremum over $\set {S_1, \ldots, S_n} \in \map P X$ we obtain:
:$\norm {\EE_{x, y} } \le \norm x \norm y$
Category:Resolutions of the Identity
Boundary (Topology)/Examples/Half-Open Real Interval

== Examples of Boundaries in the context of Topology ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $\hointl a b$ be a half-open interval of $\R$.
Then the boundary of $\hointl a b$ is the set of its endpoints $\set {a, b}$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.32$
Category:Examples of Set Boundaries
Boundary (Topology)/Examples/Integers in Real Numbers

== Examples of Boundaries in the context of Topology ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $\Z$ be the set of integers.
Then the boundary of $\Z$ in $\struct {\R, \tau_d}$ is $\Z$ itself.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 31$
Category:Examples of Set Boundaries
Boundary (Topology)/Examples/Open Unit Interval

== Examples of Boundaries in the context of Topology ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $\openint 0 1$ be the open unit interval in $\R$.
Then the boundary of $\openint 0 1$ is the set of its endpoints $\set {0, 1}$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 31$
Category:Examples of Set Boundaries
Boundary (Topology)/Examples/Rationals in Closed Unit Interval

== Examples of Boundaries in the context of Topology ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $S$ be the set defined as:
:$S = \Q \cap \closedint 0 1$
where:
:$\Q$ denotes the set of rational numbers
:$\closedint 0 1$ denotes the closed unit interval.
Then the boundary of $S$ in $\struct {\R, \tau_d}$ is $\closedint 0 1$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 31$
Category:Examples of Set Boundaries
Boundary (Topology)/Examples/Reciprocals in Real Numbers

== Examples of Boundaries in the context of Topology ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $S$ be the set defined as:
:$S = \set {\dfrac 1 n: n \in \Z_{>0} }$
Then the boundary of $S$ in $\struct {\R, \tau_d}$ is $S \cup \set 0$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 31$
Category:Examples of Set Boundaries
Boundary Condition/Examples/Arbitrary Example 1

== Example of Boundary Condition ==
Consider the differential equation:
:$\dfrac {\d^2 y} {\d x^2} + 4 \dfrac {\d y} {\d x} = 0$
for $x \ge 0$.
From Solution to $y' ' + 4 y' = 0$ this has a general solution:
:$y = A + B e^{-4 x}$
Let the boundary conditions be:
Then substituting $x = 0$ into the general solution and its first derivative yields:
and so the particular solution:
:$4 y = 1 - e^{-4 x}$
== Sources ==
* 
* 
Category:Examples of Boundary Conditions
Boundary is Intersection of Closure with Closure of Complement

== Theorem ==
Let $T$ be a topological space.
Let $X \subseteq T$.
Let $\partial X$ denote the boundary of $X$, defined as:
:$\partial X = X^- \setminus X^\circ$
Let $\overline X = T \setminus X$ denote the complement of $X$ in $T$.
Let $X^-$ denote the closure of $X$.
Then:
:$\partial X = X^- \cap \paren {\overline X}^-$
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Closures
Category:Set Boundaries
Boundary of Boundary is Contained in Boundary

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Then:
:$\map \partial {\partial H} \subseteq \partial H$
where $\partial H$ is the boundary of $H$.
That is, the boundary of the boundary of $H$ is contained in the boundary of $H$.
== Proof ==
Let $B = \partial H$.
From Boundary of Set is Closed we have that $B$ is closed in $T$.
Let $B^-$ denote the closure of $B$.
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial B = B^- \cap \paren {T \setminus B}^-$
and so from Intersection is Subset:
:$\partial B \subseteq B^-$
But from Closed Set Equals its Closure:
:$B = B^-$
and so:
:$\partial B \subseteq B$
That is:
:$\map \partial {\partial H} \subseteq \partial H$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Boundaries
Boundary of Boundary is not necessarily Equal to Boundary

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Let $\partial H$ denote the boundary of $H$.
While it is true that:
:$\map \partial {\partial H} \subseteq \partial H$
it is not necessarily the case that:
:$\map \partial {\partial H} = \partial H$
== Proof ==
From Boundary of Boundary is Contained in Boundary, we have that:
:$\map \partial {\partial H} \subseteq \partial H$
It remains to be proved that the equality does not always hold.
Proof by Counterexample:
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ such that $H \ne \O$ and $H \ne S$.
From Boundary of Subset of Indiscrete Space:
:$\partial H = S$
From Boundary of Boundary of Subset of Indiscrete Space:
:$\map \partial {\partial H} = \O$
The result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Boundaries
Boundary of Boundary of Subset of Indiscrete Space

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
Then:
:$\map \partial {\partial H} = \O$
where $\partial H$ denotes the boundary of $H$.
== Proof ==
From Boundary of Subset of Indiscrete Space, either $\partial H = S$ or $\partial H = \O$, depending on whether $H = \O$ or $H = S$ or not.
From Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed and open in $T$.
So from Set Clopen iff Boundary is Empty:
:$\map \partial {\partial H} = \O$
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $6$
Category:Indiscrete Topology
Category:Set Boundaries
Boundary of Cartesian Product of Subsets

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
:$\map \partial {H \times K} = \paren {\map \partial H \times \map \cl K} \cup \paren {\map \cl H \times \map \partial K}$
where:
:$\map \cl H$, for example, denotes the closure of $H$.
:$\map \partial H$, for example, denotes the boundary of $H$.
== Proof ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 35 \ \text {(iii)}$
Category:Set Boundaries
Category:Product Topology
Boundary of Compact Closed Set is Compact

== Theorem ==
Let $X$ be a topological space.
Let $K\subset X$ be a compact subspace of $X$.
Let $K$ be closed in $X$.
Then its boundary $\partial K$ is compact.
== Proof ==
By Boundary of Set is Closed, $\partial K$ is closed in $X$.
By Set is Closed iff it Contains its Boundary, $\partial K \subset K$.
By Closed Set in Topological Subspace, $\partial K$ is closed in $K$.
By Closed Subspace of Compact Space is Compact, $\partial K$ is compact.
== Also see ==
* Compact Subspace of Hausdorff Space is Closed
* Boundary of Compact Set in Hausdorff Space is Compact
Category:Set Boundaries
Category:Compact Spaces
Category:Closed Sets
Boundary of Compact Convex Set with Nonempty Interior is Homeomorphic to Sphere

== Theorem ==
Let $n \in \N_{> 0}$.
Let $C \subseteq \R^n$ be a compact convex subset of real Euclidean $n$-space.
Suppose that the interior $C^\circ$ is non-empty.
Then, the boundary $\partial C$ is homeomorphic to $\Bbb S^{n - 1}$, the unit $n - 1$-sphere.
== Proof ==
Let $\bsx_0 \in C^\circ$ be an element of $C^\circ$.
Define $\phi : \R^n \setminus \set {\bsx_0} \to \Bbb S^{n - 1}$ as:
:$\map \phi \bsx = \dfrac 1 {\norm {\bsx - \bsx_0}} \paren {\bsx - \bsx_0}$
As Normed Vector Space is Hausdorff Topological Vector Space, it follows that $\phi$ is continuous.
Define $\phi^* : \partial C \to \Bbb S^{n - 1}$ as:
:$\map {\phi^*} \bsx = \map \phi \bsx$
which is well-defined as the boundary is disjoint from the interior by definition.
By Restriction of Continuous Mapping is Continuous:
:$\phi^*$ is continuous.
It remains to show that $\phi^*$ is a bijection.
Let $\bsy \in \Bbb S^{n - 1}$ be arbitrary.
We need to prove that there is a unique $\bsx \in \partial C$ such that:
:$\map \phi \bsx = \bsy$
By Ray from Interior of Compact Convex Set Meets Boundary Exactly Once, there is a unique $t \in \R_{> 0}$ such that:
:$\bsx_0 + t \bsy \in \partial C$
Therefore:
Now, $\bsx \in \partial C$ satisfies:
:$\map \phi \bsx = \bsy$.
Then:
Therefore:
:$\bsx_0 + \norm {\bsx - \bsx_0} \bsy \in \partial C$
so:
:$\norm {\bsx - \bsx_0} = t$
Hence:
:$\bsx = \bsx_0 + t \bsy$
Therefore, for every $\bsy \in \Bbb S^{n - 1}$, there is a unique $\bsx \in \partial C$ such that:
:$\map {\phi^*} \bsx = \map \phi \bsx = \bsy$
and so $\phi^*$ is a bijection by definition.
We have shown that $\phi^*$ is a continuous bijection.
By Boundary of Compact Set in Hausdorff Space is Compact, $\partial C$ is compact.
Therefore, by Continuous Bijection from Compact to Hausdorff is Homeomorphism:
:$\phi^*$ is a homeomorphism from $\partial C$ to $\Bbb S^{n - 1}$
Category:Homeomorphisms (Topological Spaces)
Boundary of Compact Set in Hausdorff Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff topological space.
Let $K \subset S$ be a compact subspace of $T$.
Then its boundary $\partial K$ is compact.
== Proof ==
By Compact Subspace of Hausdorff Space is Closed, $K$ is closed in $T$.
By Boundary of Compact Closed Set is Compact, $\partial K$ is compact.
Category:Set Boundaries
Category:Compact Spaces
Category:Closed Sets
Category:Hausdorff Spaces
Boundary of Empty Set is Empty/Proof 1

== Theorem ==
== Proof ==
By Boundary is Intersection of Closure with Closure of Complement:
:$\partial_T \O = \O^- \cap \relcomp T \O^-$
where $\O^-$ denotes the closure of $\O$.
By Closure of Empty Set is Empty Set:
:$\O^- = \O$
Thus the result follows by Intersection with Empty Set.
== Sources ==
* 
Category:Boundary of Empty Set is Empty
Boundary of Intersection is Subset of Union of Boundaries

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\map \partial {A \cap B} \subseteq \partial A \cup \partial B$
where $\partial A$ denotes the boundary of $A$.
== Proof ==
By Intersection is Subset:
:$A \cap B \subseteq A \land A \cap B \subseteq B$
Then by Topological Closure of Subset is Subset of Topological Closure:
:$\paren {A \cap B}^- \subseteq A^- \land \paren {A \cap B}^- \subseteq B^-$
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$\paren {A \cap B}^- \cap \paren {\relcomp S A}^- \subseteq \partial A \land \paren {A \cap B}^- \cap \paren {\relcomp S B}^- \subseteq \partial B$
Thus
== Sources ==
* 
Category:Set Boundaries
Boundary of Polygon as Contour

== Theorem ==
Let $P$ be a polygon embedded in the complex plane $\C$.
Denote the boundary of $P$ as $\partial P$.
Then there exists a simple closed contour $C$ such that:
:$\Img C = \partial P$
where $\Img C$ denotes the image of $C$.
== Proof ==
Let $n \in \N$ be the number of sides of $P$.
Denote the vertices of $P$ as $A_1, \ldots, A_n$.
From Complex Plane is Metric Space, it follows that $\C$ is homeomorphic to $\R^2$.
Then, we can consider $\partial P$ as a subset of $\R^2$.
From Boundary of Polygon is Jordan Curve, it follows that there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that $\Img \gamma = \partial P$.
The same theorem shows that $\gamma$ is a concatenation of $n$ line segments, parameterized as:
:$\map {\gamma_k} t = \paren {1 - t} A_k + t A_{k + 1}$
where $k \in \set {1, 2, \ldots, n}$.
We identify $A_{n + 1}$ with $A_1$.
Then, $\gamma_k: \closedint {a_{k - 1} } {a_k} \to \R^2$, where $a_{k - 1}, a_k \in \closedint 0 1$.
As $\gamma$ is a concatenation of $\gamma_1, \ldots, \gamma_n$, it follows that $\set {a_0, a_1, \ldots, a_n}$ is a subdivision of $\closedint 0 1$.
We have:
:$\dfrac \d {\d t} \map {\gamma_k} t = A_{k + 1} - A_k \ne \mathbf 0$
as $A_k$ and $A_{k + 1}$ are two different vertices.
As $\C$ is homeomorphic to $\R^2$, we can consider $\gamma$ as a continuous complex function $\gamma: \closedint 0 1 \to \C$.
Then $\gamma_k$ is complex-differentiable for all values of $t \in \openint {a_{k - 1} } {a_k}$ with its derivative $\gamma_k'$ defined by:
:$\map {\gamma_k'} t = \map x {\dfrac \d {\d t} \map {\gamma_k} t} + i \map y {\dfrac \d {\d t} \map {\gamma_k} t}$
where:
:$\map x {\dfrac \d {\d t} \map {\gamma_k} t}$ is the real part of $\dfrac \d {\d t} \map {\gamma_k} t$
:$\map y {\dfrac \d {\d t} \map {\gamma_k} t}$ is the imaginary part of $\dfrac \d {\d t} \map {\gamma_k} t$
As shown above:
:$\map {\gamma_k'} t \ne 0$
By definition, it follows that $\gamma_k = \gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ is a smooth path.
From Path as Parameterization of Contour, it follows that there exists a simple closed contour $C$ with $\Img C = \partial P$.
Category:Complex Contour Integrals
Boundary of Polygon is Jordan Curve

== Theorem ==
Let $P$ be a polygon embedded in $\R^2$.
Then there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that the image of $\gamma$ is equal to the boundary $\partial P$ of $P$.
== Proof ==
The polygon $P$ has $n$ sides, where $n \in \N$.
Denote the vertices of $P$ as $A_1, \ldots, A_n$ and its sides as $S_1, \ldots, S_n$, such that each vertex $A_i$ has adjacent sides $S_{i - 1}$ and $S_i$.
We use the conventions that $S_0 = S_n$, and $A_{n + 1} = A_1$.
As each side $S_i$ is a line segment joining $A_i$ and $A_{i + 1}$, when we define the path $\gamma_i: \closedint 0 1 \to \R^2$ by:
:$\map {\gamma_i} t = \paren {1 - t} A_i + t A_{i + 1}$
the image of $\gamma_i$ is equal to the side $S_i$.
Define $\gamma: \closedint 0 1 \to \R^2$ as the concatenation $\paren {\cdots \paren {\paren {\gamma_1 * \gamma_2} * \gamma_3} * \ldots * \gamma_{n - 1} } * \gamma_n$.
Then each point in the image of $\gamma$ corresponds to a point in a side of $P$.
As $\map \gamma 0 = A_1 = \map \gamma 1$ by our definition of $A_{n + 1}$, it follows that $\gamma$ is a closed path.
It follows from the definition of polygon that the sides of $P$ do not intersect, except at the vertices.
For $i \ne 1$, each vertex $A_i$ is the initial point of $\gamma_i$ and the final point of $\gamma_{i - 1}$, and is equal to exactly one point $\map \gamma {2^{-n - 1 + i} }$ in the image of $\gamma$.
Then we have that $\gamma$ restricted to $\hointr 0 1$ is injective.
Hence $\gamma$ is a Jordan curve.
Category:Jordan Curves
Boundary of Set is Closed

== Theorem ==
Let $T$ be a topological space, and let $H \subseteq T$.
Let $\partial H$ denote the boundary of $H$.
Then $\partial H$ is closed in $T$.
== Proof ==
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {T \setminus H}^-$
where $H^-$ is the closure of $H$
From Topological Closure is Closed, both $H^-$ and $\paren {T \setminus H}^-$ are closed in $T$.
From Topology Defined by Closed Sets, the intersection of arbitrarily many (in particular two) closed sets of $T$ is a closed set of $T$.
As $\partial H$ is the intersection of $H^-$ and $\paren {T \setminus H}^-$ the result follows.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 29$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Closed Sets
Category:Set Boundaries
Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra

== 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.
Then:
:$\partial \map {\sigma_B} x \subseteq \partial \map {\sigma_A} x$
where $\partial$ denotes topological boundary.
== Proof 1 ==
 
== Proof 2 ==
Category:Spectra (Spectral Theory)
Category:Unital Banach Algebras
Category:Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra
Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra/Proof 1

== Theorem ==
== Proof ==
Let $\map G B$ be the group of units of $B$. 
Let $\lambda \in \partial \map {\sigma_B} x$. 
From Boundary is Intersection of Closure with Closure of Complement we have:
:$\lambda \in \map \cl {\C \setminus \map {\sigma_B} x}$
So from Point in Closure of Subset of Metric Space iff Limit of Sequence, we have: 
:there exists a sequence $\sequence {\lambda_n}_{n \in \N}$ in $\C \setminus \map {\sigma_B} x$ such that $\lambda_n \to \lambda$.
Hence, we have:
:$\lambda_n {\mathbf 1}_A - x \to \lambda {\mathbf 1}_A - x$ as $n \to \infty$
where $\lambda_n {\mathbf 1}_A - x \in \map G B$ for each $n \in \N$ yet $\lambda {\mathbf 1}_A - x \not \in \map G B$.
From Group of Units in Unital Banach Algebra is Open:
:$\map G B$ is open in $B$.
Hence $\partial \map G B = \map G B^- \setminus \map G B$.
Since $\lambda {\mathbf 1}_A - x$ is the limit of a sequence in $\map G B$, yet is not contained in $\map G B$, we have $\lambda {\mathbf 1}_A - x \in \partial \map G B$. 
From Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra, we have:
:$\lambda {\mathbf 1}_A - x \not \in \map G A$
Hence $\lambda \in \map {\sigma_A} x$.
We show additionally that $\lambda \not \in \map {\sigma_A} x^\circ$. 
Suppose otherwise that $\lambda \in \map {\sigma_A} x^\circ$, then by the definition of a convergent sequence we would have $\lambda_n \in \map {\sigma_A} x$ for sufficiently large $n$. 
However, $\lambda_n \not \in \map {\sigma_A} x$ for all $n \in \N$. 
Hence $\lambda \not \in \map {\sigma_A} x^\circ$.
Hence by the definition of topological boundary, we have:
:$\lambda \in \partial \map {\sigma_A} x$
Category:Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra
Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra/Proof 2

== Theorem ==
== 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 $A$. 
 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 $\lambda \in \map {\sigma_A} x \cap \map \cl {\map {\rho_A} x} = \partial \map {\sigma_A} x$.
Category:Boundary of Spectrum of Element in Subalgebra of Unital Banach Algebra
Boundary of Subset of Discrete Space is Null

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $A \subseteq S$.
Then:
:$\partial A = \O$
where:
:$\partial A$ is the boundary of $A$ in $T$.
== Proof ==
Let $A \subseteq S$.
Then from Set in Discrete Topology is Clopen it follows that $A$ is both open and closed in $T$.
The result follows from Set Clopen iff Boundary is Empty.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $4$
Category:Discrete Topology
Category:Set Boundaries
Boundary of Subset of Indiscrete Space

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\O \subsetneq H \subsetneq S$ (that is, let $H$ be a non-empty proper subset of $T$).
Then:
:$\partial H = S$
where $\partial H$ denotes the boundary of $H$.
If $H = \O$ or $H = S$ then $\partial H = \O$.
== Proof ==
From Closure of Subset of Indiscrete Space, $H^- = S$, where $H^-$ denotes set closure.
From Interior of Subset of Indiscrete Space, $H^\circ = \O$, where $H^\circ$ denotes set interior.
By definition:
:$\partial H = H^- \setminus H^\circ = S \setminus \O = S$
From Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed and open in $T$.
So from Set Clopen iff Boundary is Empty:
:$\partial H = \O$
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $6$
Category:Indiscrete Topology
Category:Set Boundaries
Boundary of Union is Subset of Union of Boundaries

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\map \partial {A \cup B} \subseteq \partial A \cup \partial B$
where $\partial A$ denotes the boundary of $A$.
== Proof ==
By Intersection is Subset:
:$\relcomp S A \cap \relcomp S B \subseteq \relcomp S A \land \relcomp S A \cap \relcomp S B \subseteq \relcomp S B$
Then by Topological Closure of Subset is Subset of Topological Closure:
:$\paren {\relcomp S A \cap \relcomp S B}^- \subseteq \paren {\relcomp S A}^- \land \paren {\relcomp S A \cap \relcomp S B}^- \subseteq \paren {\relcomp S B}^-$
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$\paren {\relcomp S A \cap \relcomp S B}^- \cap A^- \subseteq \partial A \land \paren {\relcomp S A \cap \relcomp S B}^- \cap B^- \subseteq \partial B$
Thus
== Sources ==
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Category:Set Boundaries