nyuuzyou/wikis huggingface dataset:
Injection is Open Mapping iff Image of Sub-Basis Set is Open

== Theorem ==
Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.
Let $\SS \subseteq \powerset {X_1}$ be a sub-basis of $\tau_1$.
Let $f: X_1 \to X_2$ be an injection. 
Then:
:$f$ is an open mapping 
:
:$\forall U \in \SS: f \sqbrk U \in \tau_2$
== Proof ==
=== Necessary Condition ===
Let $f$ be an open mapping.
By definition of open mapping:
:$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$
By definition of sub-basis:
:$\SS \subseteq \tau_1$
Hence:
:$\forall U \in \SS: f \sqbrk U \in \tau_2$
=== Sufficient Condition ===
Let $f$ satisfy:
:$\forall U \in \SS: f \sqbrk U \in \tau_2$
Let $\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \FF \text{ is finite} }$
Let $B \in \BB$.
By definition of $\BB$:
:$\ds \exists \FF \subseteq \SS, \FF \text{ is finite} : B = \bigcap \FF$
From Image of Intersection under Injection:
By assumption:
:$\forall S \in \FF : f \sqbrk S \in \tau_2$
By :
:$f \sqbrk B \in \tau_2$
Hence we have shown that:
Let $W \in \tau_1$.
By definition of sub-basis:
:$\exists \AA \subseteq \BB : W = \bigcup \AA$
We have:
From $(1)$ above:
:$\forall B \in \AA : f \sqbrk B \in \tau_2$
By :
:$f \sqbrk W \in \tau_2$
Hence we have shown that:
:$\forall W \in \tau_1 : f \sqbrk W \in \tau_2$
Hence by definition, $f$ is an open mapping 
Category:Open Mappings
Injection to Image is Bijection

== Theorem ==
Let $f: S \rightarrowtail T$ be an injection.
Let $X \subseteq T$ be the image of $f$.
Then the restriction $f {\restriction_{S \times X}}: S \to X$ of $f$ to the image of $f$ is a bijection of $S$ onto $X$.
== Proof ==
We have:
:Restriction of Injection is Injection
:Restriction of Mapping to Image is Surjection
Thus we have that:
:$f {\restriction_{S \times X}}: S \to X$ is an injection
and
:$f {\restriction_{S \times X}}: S \to X$ is a surjection
Hence the result by definition of bijection.
== Sources ==
* : $\S 8$: Example $8.2$
Category:Injections
Category:Bijections
Category:Restrictions
Injective Linear Transformation between Normed Vector Spaces sends Closed Unit Ball to Closed Unit Ball iff Isometric Isomorphism

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a injective linear transformation.
Let $B_X^-$ be the closed unit ball of $X$.
Let $B_Y^-$ be the closed unit ball of $Y$. 
Then $T$ is an isometric isomorphism :
:$T \sqbrk {B_X^-} = B_Y^-$
== Proof ==
=== Necessary Condition ===
Suppose that $T$ is an isometric isomorphism.
Then: 
:$\norm {T x}_Y = \norm x_X$ for each $x \in X$. 
Hence if $\norm x_X \le 1$, we have $\norm {T x}_Y \le 1$.
So $T \sqbrk {B_X^-} \subseteq B_Y^-$.
Conversely, let $y \in B_Y^-$.
Since $T$ is an isometric isomorphism:
:$T$ is bijective and there exists $x \in X$ such that $y = T x$. 
Since $T$ is a linear isometry, we have: 
:$\norm x_X = \norm {T x}_Y \le 1$
So we have $x \in B_X^-$. 
So we have $T \sqbrk {B_X^-} = B_Y^-$.
=== Sufficient Condition ===
Suppose that:
:$T \sqbrk {B_X^-} = B_Y^-$
We first show that $T$ is surjective.
We have: 
We now show that $T$ is a bounded linear transformation.
We have: 
So $T$ is a bounded linear transformation with $\norm T_{\map B {X, Y} } = 1$.
We move to proving that $T^{-1}$ is a bounded linear transformation.
Since $T$ is bijective, we have: 
:$T^{-1} \sqbrk {B_Y^-} = B_X^-$
from Preimage of Image of Subset under Injection equals Subset.
From Inverse of Linear Transformation is Linear Transformation, $T^{-1}$ is a linear transformation.
Repeating the previous calculation with $T$ swapped for $T^{-1}$, we obtain that: 
:$\ds \sup_{y \in B_Y^-} \norm {T^{-1} y}_X = \sup_{x \in B_X^-} \norm x_X = 1$ 
So $T^{-1}$ is a bounded linear transformation with $\norm {T^{-1} }_{\map B {Y, X} } = 1$.
Now, we have: 
:$\norm {T x}_Y \le \norm x_X$ for all $x \in X$
and:
:$\norm {T^{-1} y}_X \le \norm y_Y$ for all $y \in Y$.
In particular, taking $x \in X$ and $y = T x$ in the latter inequality, we obtain: 
:$\norm x_X \le \norm {T x}_Y$ for all $x \in X$.
Hence we obtain: 
:$\norm x_X \le \norm {T x}_Y \le \norm x_X$
Hence: 
:$\norm {T x}_Y = \norm x_X$
So $T$ is an surjective linear isometry.
From Linear Isometry is Injective: Corollary, we have that $T$ is an isometric isomorphism.
Category:Isometric Isomorphisms (Normed Vector Spaces)
Injective Module over Principal Ideal Domain

== Theorem ==
Let $D$ be a principal ideal domain.
Let $M$ be a $D$-module.
Then $M$ is injective it is divisible.
== Proof ==
By Principal Ideal Domain is Dedekind Domain $D$ is a Dedekind domain.
By Injective Module over Dedekind Domain $M$ is injective it is divisible.
Category:Principal Ideal Domains
Category:Commutative Algebra
Category:Homological Algebra
Category:Module Theory
Injective Quotient Mapping Equals Homeomorphism

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a mapping.
Then $f$ is an injective quotient mapping, $f$ is a homeomorphism.
== Proof ==
=== Sufficient condition ===
Suppose $f$ is an injective quotient mapping.
By definition of quotient mapping, $f$ is surjective.
Mapping is Injection and Surjection iff Inverse is Mapping shows that $f$ has an inverse $f^{-1}$.
To show continuity of $f^{-1}$, let $U \subseteq S_1$ be open in $T_1$.
As $f$ is bijective, it follows that $f^{-1} \sqbrk {f \sqbrk U} = U$.
By definition of quotient mapping, it follows that $f \sqbrk U$ is open in $T_2$.
As $f$ is bijective, it follows that the preimage of $U$ under $f^{-1}$ is equal to $f \sqbrk U$.
It follows that $f^{-1}$ is continuous.
By definition of homeomorphism, it follows that $f$ is a homeomorphism.
=== Necessary condition ===
Suppose $f$ is a homeomorphism.
By definition of homeomorphism, it follows that $f$ is injective.
Let $V \subseteq S_2$ such that $f^{-1} \sqbrk V$ is open in $T_1$.
As $f^{-1}$ is continuous, it follows that the preimage of $f^{-1} \sqbrk V$ under $f^{-1}$ is open in $T_2$.
As $f$ is bijective, it follows that the preimage of $f^{-1} \sqbrk V$ under $f^{-1}$ is equal to $V$.
By definition of quotient mapping, $f$ is a quotient mapping.
Category:Quotient Topology
Category:Homeomorphisms (Topological Spaces)
Injective Self-Adjoint Densely-Defined Linear Operator has Everywhere Dense Image

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a self-adjoint densely defined linear operator that is injective.
Then $\Img T$ is everywhere dense in $\HH$.
== Proof ==
From Kernel of Adjoint of Densely-Defined Linear Operator is Orthocomplement of Image, we have:
:$\map \ker {T^\ast} = \paren {\Img T}^\bot$ 
where $\tuple {\map D {T^\ast}, T^\ast}$ is the adjoint of $\tuple {\map D T, T}$.
Since $T$ is self-adjoint, we have $T^\ast = T$. 
Hence $\map \ker T = \paren {\Img T}^\bot$.
Since $T$ is injective, we have $\map \ker T = {\mathbf 0}_\HH$.
Hence:
:$\paren {\Img T}^\bot = \set { {\mathbf 0}_\HH}$
From Linear Subspace Dense iff Zero Orthocomplement we have that $\Img T$ is everywhere dense in $\HH$.
== Sources ==
* : $13.11$: Theorem
Category:Self-Adjoint Densely-Defined Linear Operators
Injective iff Projective in Dual Category

== Theorem ==
Let $\mathbf A$ be an abelian category.
Let $X$ be an object in $\mathbf A$.
Then:
:$X$ is injective in $\mathbf A$
:
:$X$ is projective in the dual category $\mathbf A^{\mathrm{op}}$ of $\mathbf A$.
== Proof ==
=== Sufficient Condition ===
Let $X$ be injective in $\mathbf A$.
Let $f : B \to A$ be an epimorphism in $\mathbf A^{\mathrm {op} }$.
By Monomorphism iff Epimorphism in Dual Category:
:$f : A \to B$ is a monomorphism in $\mathbf A$.
By definition of injective object, the mapping:
:$\alpha: \map {\mathrm {Hom}_{\mathbf A} } {B, X} \to \map {\mathrm {Hom}_{\mathbf A} } {A, X}, \quad g \mapsto g \circ f$ 
is surjective.
In the above, $\map {\mathrm {Hom}_{\mathbf A} } {B, X}$ is the set of morphisms in $\mathbf A$ from $B$ to $X$, and $g$ is an arbitrary element of the domain of $\alpha$.
Further, by definition of dual category:
:$\map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, B} = \map {\mathrm {Hom}_{\mathbf A} } {B, X}$
:$\map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, A} = \map {\mathrm {Hom}_{\mathbf A} } {A, X}$
So:
:$\alpha : \map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, B} \to \map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, A}, \quad g \mapsto g \circ f = f \circ_{\mathbf A^{\mathrm{op}}} g$ 
is surjective.
In the above, $\circ_{\mathbf A^{\mathrm{op}}}$ denotes composition in $\mathbf A^{\mathrm {op} }$.
By definition of projective object, $X$ is projective in $\mathbf A^{\mathrm {op} }$.
=== Necessary Condition ===
== Also see ==
* Dual Category of Abelian Category is Abelian
* Definition:Hom Functor
Category:Category Theory
Category:Homological Algebra
Injectivity Radius of Compact Riemannian Manifold is Positive

== Theorem ==
Let $\struct {M, g}$ be a compact Riemannian manifold.
Let $\map {\operatorname{inj} } M$ be the injectivity radius of $M$.
Then $\map {\operatorname{inj} } M > 0$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Uniformly Normal Neighborhoods
Category:Riemannian Manifolds
Injectivity of Laplace Transform

== Theorem ==
Let $f$, $g$ be functions from $\R_{\ge 0} \to K$ of a real variable $t$, where $K \in \set {\R, \C}$.
Further let $f$ and $g$ be continuous everywhere on their domains.
Let $f$ and $g$ both admit Laplace transforms.
Suppose that the Laplace transforms $\laptrans f$ and $\laptrans g$ satisfy:
:$\forall t \in \R_{\ge 0}: \laptrans {\map f t} = \laptrans {\map g t}$
Then $f = g$ everywhere on $\R_{\ge 0}$.
=== Corollary ===
== Proof ==
Category:Laplace Transforms‎
Injectivity of Laplace Transform/Corollary

== Corollary to Injectivity of Laplace Transform ==
Let $f$, $g$ be functions from $\R_{\ge 0} \to K$ of a real variable $t$, where $K \in \set {\R, \C}$.
Let $f$ and $g$ both admit Laplace transforms.
Suppose that the Laplace transforms $\laptrans f$ and $\laptrans g$ satisfy:
:$\forall t \ge 0: \laptrans {\map f t} = \laptrans {\map g t}$
Let $f$ and $g$ be continuous everywhere on their domains, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\R_{\ge 0}$.
Then $f = g$ everywhere on $\R_{\ge 0}$, except possibly where $f$ or $g$ have discontinuities of the first kind.
== Proof ==
Category:Laplace Transforms‎
Inner Automorphism Group is Isomorphic to Quotient Group with Center

== Theorem ==
Let $G$ be a group.
Let $\Inn G$ be the inner automorphism group of $G$.
Let $\map Z G$ be the center of $G$.
Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.
Then $G / \map Z G \cong \Inn G$.
== Proof ==
Let $G$ be a group.
Let the mapping $\kappa: G \to \Inn G$ be defined as:
:$\map \kappa a = \kappa_a$
where $\kappa_a$ is the inner automorphism of $G$ given by $a$.
From Kernel of Inner Automorphism Group is Center:
:$\map \ker \kappa = \map Z G$
and also that:
:$\Img \kappa = \Inn G$
From the First Isomorphism Theorem:
:$\Img \kappa \cong G / \map \ker \kappa$
Thus, as $\map \ker \kappa = \map Z G$ and $\Img \kappa = \Inn G$:
:$G / \map Z G \cong \Inn G$
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $10$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \beta$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $25$
Category:Quotient Groups
Category:Inner Automorphisms
Category:Group Isomorphisms
Inner Automorphism Maps Subgroup to Itself iff Normal

== Theorem ==
Let $G$ be a group.
For $x \in G$, let $\kappa_x$ denote the inner automorphism of $x$ in $G$.
Let $H$ be a subgroup of $G$.
Then:
:$\forall x \in G: \kappa_x \sqbrk H = H$
:
:$H$ is a normal subgroup of $G$.
== Proof ==
=== Sufficient Condition ===
=== Necessary Condition ===
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.10$
Category:Normal Subgroups
Category:Inner Automorphisms
Category:Inner Automorphism Maps Subgroup to Itself iff Normal
Inner Automorphism Maps Subgroup to Itself iff Normal/Necessary Condition

== Theorem ==
Let $G$ be a group.
For $x \in G$, let $\kappa_x$ denote the inner automorphism of $x$ in $G$.
Suppose that:
:$\forall x \in G: \kappa_x \sqbrk H = H$
Then $H$ is a normal subgroup of $G$.
== Proof ==
Suppose that:
:$\forall x \in G: \kappa_x \sqbrk H = H$
Let $x \in G$ be arbitrary.
By definition of inner automorphism of $x$ in $G$:
:$\forall h \in H: x h x^{-1} \in H$
So, by definition, $H$ is a normal subgroup of $G$
Category:Inner Automorphism Maps Subgroup to Itself iff Normal
Inner Automorphism Maps Subgroup to Itself iff Normal/Sufficient Condition

== Theorem ==
Let $G$ be a group.
For $x \in G$, let $\kappa_x$ denote the inner automorphism of $x$ in $G$.
Let $H$ be a normal subgroup of $G$.
Then:
:$\forall x \in G: \kappa_x \sqbrk H = H$
== Proof ==
Let $H$ be a normal subgroup of $G$.
Let $x \in G$ be arbitrary.
By definition, $\kappa_x: G \to G$ is a mapping defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Let $n \in N$.
Then:
== Sources ==
* : $\S 1.10$: Theorem $27$
Category:Inner Automorphism Maps Subgroup to Itself iff Normal
Inner Automorphism is Automorphism

== Theorem ==
Let $G$ be a group.
Let $x \in G$.
Let $\kappa_x$ be the inner automorphism of $x$ in $G$.
Then $\kappa_x$ is an automorphism of $G$.
== Proof ==
By definition, $\kappa_x: G \to G$ is a mapping defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
We need to show that $\kappa_x$ is an automorphism.
First we show $\kappa_x$ is a homomorphism.
Thus the morphism property is demonstrated.
Next we show that $\kappa_x$ is injective.
So $\kappa_x$ is injective.
Finally we show that $\kappa_x$ is surjective.
Note that $\forall h \in G: x^{-1} h x \in G$ from fact that $G$ is a group and therefore closed. So:
Thus every element of $G$ is the image of some element of $G$ under $\kappa_x$ (that is, of $x^{-1} h x$), and surjectivity is proved.
Since $\kappa_x$ is injective and surjective, it is bijective.
Since $\kappa_x$ is a homomorphism that is bijective, it is a (group) isomorphism.
Hence, $\kappa_x: G \to G$ is a (group) isomorphism from $G$ to itself, which is by definition, an automorphism of $G$.
== Also see ==
* Conjugate of Subgroup is Subgroup: Performing an inner automorphism of a subgroup
* Definition:Inner Automorphism Group
== Sources ==
* : $\S 7.1$. Homomorphisms: Example $131$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.8 \ \text{(a)}$
* : $\S 1.10$: Theorem $27$
* : Chapter $\text{II}$: Groups: Problem $\text{AA}$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $25$
* : $\S 3.2$: Groups; the axioms: Exercise $(10)$
* : Chapter $8$: The Homomorphism Theorem: Proposition $8.17$
Category:Inner Automorphisms
Inner Automorphisms form Normal Subgroup of Automorphism Group

== Theorem ==
Let $G$ be a group.
Then the set $\Inn G$ of all inner automorphisms of $G$ is a normal subgroup of the automorphism group $\Aut G$ of $G$:
:$\Inn G \lhd \Aut G$
== Proof ==
Let $G$ be a group whose identity is $e$.
From Inner Automorphisms form Subgroup of Automorphism Group, $\Inn G$ forms a subgroup of $\Aut G$.
It remains to be shown that $\Inn G$ is normal in $\Aut G$.
Let $\kappa_x: G \to G$ be the inner automorphism defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
Let $\phi \in \Aut G$.
If we can show that:
:$\forall \phi \in \Aut G: \forall \kappa_x \in \Inn G: \phi \circ \kappa_x \circ \phi^{-1} \in \Inn G$
then by the Normal Subgroup Test:
:$\Inn G \lhd \Aut G$
Fix $\kappa_x \in \Inn G$.
We claim $\phi \circ \kappa_x \circ \phi^{-1} = \kappa_{\map \phi x}$.
Since $\phi \in \Aut G$ then $\phi$ is, in particular, a homomorphism.
Therefore:
Therefore:
:$\phi \circ \kappa_g \circ \phi^{-1} = \kappa_{\map \phi g} \in \Inn G$
Since $\kappa_x \in \Inn G$ and $\phi \in \Aut G$ were arbitrary:
:$\Inn G \lhd \Aut G$
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $3$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.8 \ \text{(b)}$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \alpha$
Category:Normal Subgroups
Category:Automorphism Groups
Category:Inner Automorphisms
Inner Automorphisms form Subgroup of Automorphism Group

== Theorem ==
Let $G$ be a group.
Then the set $\Inn G$ of all inner automorphisms of $G$ forms a normal subgroup of the automorphism group $\Aut G$ of $G$:
:$\Inn G \le \Aut G$
== Proof ==
Let $G$ be a group whose identity is $e$.
Let $\kappa_x: G \to G$ be the inner automorphism defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
We see that:
:$\Inn G \ne \O$
as $\kappa_x$ is defined for all $x \in G$.
We show that:
:$\kappa_x, \kappa_y \in \Inn G: \kappa_x \circ \paren {\kappa_y}^{-1} \in \Inn G$
So:
As $x y^{-1} \in G$, it follows that:
:$\kappa_{x y^{-1} } \in \Inn G$
By the One-Step Subgroup Test:
:$\Inn G \le \Aut G$
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Example $134$
* : Chapter $\text{II}$: Groups: Problem $\text{AA}$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $25$
Category:Automorphism Groups
Category:Inner Automorphisms
Inner Automorphisms form Subgroup of Symmetric Group

== Theorem ==
Let $G$ be a group.
Let $\struct {\map \Gamma G, \circ}$ be the symmetric group on $G$.
Let $\Inn G$ denote the inner automorphism group of $G$.
Then:
:$\Inn G \le \struct {\map \Gamma G, \circ}$
where $\le$ denotes the relation of being a subgroup.
== Proof ==
An inner automorphism is a permutation on $G$ by definition.
From Inner Automorphisms form Subgroup of Automorphism Group:
:$\Inn G \le \Aut G$
where $\Aut G$ denotes the set of automorphisms of $G$.
From Automorphism Group is Subgroup of Symmetric Group:
:$\Aut G \le \struct {\map \Gamma G, \circ}$
Thus $\Inn G \le \struct {\map \Gamma G, \circ}$ as required.
== Sources ==
* : $\S 7.1$. Homomorphisms: Example $131$
Category:Inner Automorphisms
Category:Symmetric Groups
Inner Jordan Content is Monotone

== Theorem ==
Let $A, B \subseteq \R^n$ be bounded subspaces of Euclidean $n$-space.
Suppose that $A \subseteq B$.
Then:
:$\map {m_*} A \le \map {m_*} B$
where $m_*$ denotes the inner Jordan content.
== Proof ==
Let $R \subseteq \R^n$ be a closed $n$-rectangle that contains $B$.
Then, by Subset Relation is Transitive, $R$ contains $A$ as well.
By Set Difference with Subset is Superset of Set Difference:
:$R \setminus B \subseteq R \setminus A$
So, by Outer Jordan Content is Monotone:
:$\map {m^*} {R \setminus B} \le \map {m^*} {R \setminus A}$
Therefore:
:$\map V R - \map {m^*} {R \setminus B} \ge \map V R - \map {m^*} {R \setminus A}$
Hence the result, by definition of inner Jordan content.
Category:Measure Theory
Inner Jordan Content is Well-Defined

== Theorem ==
Let $M \subseteq \R^n$ be a bounded subspace of Euclidean $n$-space.
Let:
:$\ds R = \prod_{i \mathop = 1}^n \closedint {a_i} {b_i}$
:$\ds R' = \prod_{i \mathop = 1}^n \closedint {a'_i} {b'_i}$
be closed $n$-rectangles that contain $M$.
Let $V_R, V_{R'} \in \R_{\ge 0}$ be defined as:
:$\ds V_R = \prod_{i \mathop = 1}^n \paren {b_i - a_i}$
:$\ds V_{R'} = \prod_{i \mathop = 1}^n \paren {b'_i - a'_i}$
Then:
:$V_R - \map {m^*} {R \setminus M} = V_{R'} - \map {m^*} {R' \setminus M}$
Therefore, the inner Jordan content of $M$ is independent of the rectangle $R$ used to define it.
== Proof ==
For each $k \in \set {0, 1, \dotsc, n}$, define:
:$\ds R_k = \prod_{i \mathop = 1}^k \closedint {a_i} {b_i} \times \prod_{i \mathop = k + 1}^n \closedint {a'_i} {b'_i}$
:$\ds V_k = \prod_{i \mathop = 1}^k \paren {b_i - a_i} \times \prod_{i \mathop = k + 1}^n \paren {b'_i - a'_i}$
In particular, we have:
:$R_0 = R$
:$R_n = R'$
:$V_0 = V_R$
:$V_n = V_{R'}$
We will prove that, for every $k \in \set {1, 2, \dotsc, n}$:
:$\paren 1 \quad V_{k - 1} - \map {m^*} {R_{k - 1} \setminus M} = V_k - \map {m^*} {R_k \setminus M}$
From this, we would have:
which is the stated result.
Let $k \in \set {1, 2, \dotsc, n}$ be arbitrary.
We want to show that:
:$V_{k - 1} - \map {m^*} {R_{k - 1} \setminus M} = V_k - \map {m^*} {R_k \setminus M}$
Or equivalently:
:$\map {m^*} {R_k \setminus M} - \map {m^*} {R_{k - 1} \setminus M} = V_k - V_{k - 1}$
We can compute:
Category:Jordan Content
Inner Limit in Hausdorff Space by Open Neighborhoods

== Theorem ==
Let $\sequence {C_n}_{n \mathop \in \N}$ be a sequence of sets in a Hausdorff topological space $\struct {\XX, \tau}$.
Let $x \in \XX$.
Let $\map \mho x := \set {V \in \tau:\ x \in V}$ denote the set of open neighborhoods of $x$.
Let $\NN_\infty$ denote the set of cofinite subsets of $\N$:
:$\NN_\infty := \set {N \subset \N: \N \setminus N \text{ is finite} }$
Then the inner limit of $\sequence {C_n}_{n \mathop \in \N}$ is:
:$\ds \liminf_n C_n = \set {x \in \XX: \forall V \in \map \mho x: \exists N \in \NN_\infty: \forall n \in N: C_n \cap V \ne \O}$
or equivalently:
:$\ds \liminf_n C_n = \set {x \in \XX: \forall V \in \map \mho x: \exists N_0 \in \N: \forall n \ge N_0: C_n \cap V \ne \O}$
== Proof == 
If $x \in \liminf_n C_n$ then there exist a sequence $\sequence {x_k}_{n \mathop \in \N}$ such that $x_k \to x$ while:
:$x_k \in C_{n_k}$ 
and 
:$\sequence {n_k}_{k \mathop \in \N} \subseteq \N$ is a strictly increasing sequence of indices.
For any $V \in \map \mho x$ there exists $N_0 \in\N$ such that for all $i \ge N_0$: 
:$x_i \in V$
and:
:$x_i \in C_{n_i}$
Thus:
:$C_{n_i} \cap V \ne \O$
Therefore $x \in \set {x \in \XX: \forall V \in \map \mho x: \exists N_0 \in \N: \forall n \ge N_0: C_n \cap V \ne \O}$.
Let $x \in \set {x \in \XX: \forall V \in \map \mho x: \exists N \in \NN_\infty: \forall n \in N: C_n \cap V \ne \O}$.
Thus:
:$\forall V \in \map \mho x: \exists N \in \NN_\infty: \forall n \in N: C_n \cap V \ne \O$
Then there exists a strictly increasing sequence:
:$\sequence {n_k}_{k \mathop \in \N} \subseteq \N$
such that for every $V \in \map \mho x$:
:$\exists x_k \in C_{n_k} \cap V$.
Hence $x_k \to x$ in the topology $\tau$.
== Also see ==
* Inner Limit of Sequence of Sets in Normed Space described via the point-to-set distance function induced by the norm of the space
* Inner Limit in Normed Spaces by Open Balls
* Inner Limit in Hausdorff Space by Set Closures
* Inner Limit is Closed Set
Category:Set Theory
Category:Hausdorff Spaces
Category:Measure Theory
Category:Limits of Sequence of Sets
Inner Limit in Hausdorff Space by Set Closures

== Theorem ==
Let $\struct {\XX, \tau}$ be a Hausdorff space.
Let $\sequence {C_n}_{n \mathop \in \N}$ be a sequence of sets in $\XX$.
Then:
:$\ds \liminf_n C_n = \bigcap_{N \mathop \in \NN_\infty^\#} \map \cl {\bigcup_{n \mathop \in N} C_n}$
where:
:$\cl$ denotes set closure
:$\NN_\infty^\#$ denotes the set of cofinal subsets of $\N$.
== Proof ==
$(1)$: Let:
:$\ds x \in \liminf_n \ C_n$
Let:
:$\Sigma \in \NN_\infty^\#$
Let $W$ be an open neighborhood of $x$. 
Then there exists $N_0 \in \N$ such that for all $n \ge N_0$ such that $n \in \Sigma$:
:$W \cap C_n \ne \O$
Thus:
:$\ds x \in \map \cl {\bigcup_{n \mathop \in \Sigma} C_n}$
$(2)$: Let:
:$\ds x \notin \liminf_n C_n$
Then there exists an open neighborhood of $x$.
Let $\map \mho x := \set {V \in \tau: x \in V}$ denote the set of open neighborhoods of $x$.
Let $W \in \map \mho x$ such that:
:$\Sigma_0 := \set {n \in \N: W \cap C_n = \O}$
is cofinal.
Then:
:$\ds x \notin \map \cl {\bigcup_{n \mathop \in \Sigma_0} C_n}$
This completes the proof.
== Also see ==
* Inner Limit is Closed Set: a corollary of this theorem
* Inner Limit in Hausdorff Space by Open Neighborhoods
* Inner Limit of Sequence of Sets in Normed Space
Category:Set Theory
Category:Hausdorff Spaces
Category:Measure Theory
Category:Limits of Sequence of Sets