avewright/proofwiki-math huggingface dataset:
Theorem: Independent Sets of Dual Matroid

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ be the set of bases of the matroid $M$.
Then the independent subsets of the dual $M^*$ of $M$ is:
:$\mathscr I^* = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq \paren{S \setminus B}}$

Proof:
By definition of the dual matroid the set of bases $\mathscr B^*$ of the dual $M^*$ of $M$:
:$\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$
From Characterization of Matroid Independent Sets in Terms of Bases:
:$\mathscr I^* = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq \paren{S \setminus B}}$
{{qed}}
Category:Dual Matroids
Category:Matroid Independent Subsets
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Theorem: Characterization of Matroid Independent Sets in Terms of Bases

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ be the set of bases of the matroid $M$.
Then:
:$\mathscr I = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$

Proof:
Let $\mathscr I^\prime = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
From Independent Subset is Contained in Base:
:$\mathscr I \subseteq \mathscr I^\prime$
By definition of matroid base:
:$\mathscr B \subseteq \mathscr I$
By matroid axiom $\text I 2$:
:$\mathscr I^\prime \subseteq \mathscr I$
By set equality:
:$\mathscr I = \mathscr I^\prime$
{{qed}}
Category:Matroid Bases
Category:Matroid Independent Subsets
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Theorem: Dual of Dual Matroid Equals Matroid

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $M^*$ denote the dual of $M$.
Then:
:$\paren{M^*}^* = M$
That is, the dual $\paren{M^*}^*$ of $M^*$ is $M$.

Proof:
Let $\mathscr B$ denote the set of bases of $M$.
Let $\paren{\mathscr B^*}^*$ denote the set of bases of $\paren{M^*}^*$.
By definition of dual matroid:
:
We have:
{{begin-eqn}}
{{eqn | l = \paren{\mathscr B^*}^*
      | r = \set{S \setminus \paren{S \setminus B} : B \in \mathscr B}
      | c = {{Defof|Dual Matroid}}
}}
{{eqn | r = \set{B : B \in \mathscr B}
      | c = Complement of Complement
}}
{{eqn | r = \mathscr B
}}
{{end-eqn}}
Let $\paren{\mathscr I^*}^*$ denote the independent subsets of $\paren{M^*}^*$.
We have:
{{begin-eqn}}
{{eqn | l = \paren{\mathscr I^*}^*
      | r = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}
      | c = Characterization of Matroid Independent Sets in Terms of Bases
}}
{{eqn | r = \mathscr I
      | c = Characterization of Matroid Independent Sets in Terms of Bases
}}
{{end-eqn}}
By definition of matroid:
:$\paren{M^*}^* = M$
{{qed}}
Theorem: Trapezium Rule for Definite Integrals/Error Term

The error can be quantified as:
:$\dfrac {\paren {b - a}^3 \map {f' '} \xi} {12 n^2}$
where $\xi \in \closedint a b$.

Proof:
{{Proofread}}
{{tidy}}
first, through integration by parts: 
:<nowiki>$\int_{x_{i+1}}^{x_i} f(x)\ dx \\</nowiki> =\int_{0}^{h} f(t+x_i)\ dt    \\  = \left [ (t-\frac{h}{2})f(t+x_i) \right ]^h_0 -\int_{0}^{h} (t-h/2) f'(t+x_i)\ dt \\  = \left [ (t-\frac{h}{2})f(t+x_i) \right ]^h_0   -\left [ \left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right)f'(t+x_i) \right ]^h_0   + \int_{0}^{h} \left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) f<nowiki>''</nowiki>(t+x_i)\ dt\\  =\frac{h}{2}(f(x_{i+1})+f(x_i))-0+\int_{0}^{h} \left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) f<nowiki>''</nowiki>(t+x_i)\ dt\\  $
Note the left component when added for all $i=0...n-1$ is the approximation according to the trapezoidal rule. Now, to bound total error we consider the series of the right component in the above equation:
:$E_T\\ =\left | \sum^{n-1}_{i=0}\int_{0}^{h}\left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) f<nowiki>''</nowiki>(t+x_i)\ dt \right |\\  =\left | \int_{0}^{h}\left(\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right) \sum^{n-1}_{i=0} f<nowiki>''</nowiki>(t+x_i)\ dt \right |\\  \leq \int_{0}^{h}\left |\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right | \sum^{n-1}_{i=0}\left |f<nowiki>''</nowiki>(t+x_i) \right |\ dt\\  \leq nK\int_{0}^{h}\left |\frac{(t-h/2)^2}{2}-\frac{h^2}{8}\right |\ dt\\  =nK\frac{h^3}{12}=\frac{(b-a)^3}{12n^2} K$
Theorem: Possibility Operator in terms of Necessity Operator

The possibility operator $\pos$ can be expressed in terms of the necessity operator $\nec$ thus:
:$\pos P \iff \lnot \nec \lnot P$

Proof:
{{begin-eqn}}
{{eqn | q = \forall P
      | o = 
      | r = \pos P
      | c = 
}}
{{eqn | o = \leadstoandfrom
      | r = \exists w: \map P w
      | c = {{Defof|Possibility Operator}}
}}
{{eqn | o = \leadstoandfrom
      | r = \lnot \forall w: \lnot \map P w
      | c = Assertion of Existence
}}
{{eqn | o = \leadstoandfrom
      | r = \lnot \nec \lnot P
      | c = {{Defof|Necessity Operator}}
}}
{{end-eqn}}
{{qed}}
Theorem: Natural Numbers under Multiplication form Monoid

The set of natural numbers under multiplication $\struct {\N, \times}$ is a monoid.

Proof:
Taking the monoid axioms in turn:
Theorem: Subsets Satisfy Formulation 1 Matroid Base Axiom Iff Complements Satisfy Formulation 5

Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Let $\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
Then:
:$\mathscr B$ satisfies formulation $1$ base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 1}}
{{iff}}
:$\mathscr B^*$ satisfies formulation $5$ base axiom:
{{begin-axiom}}
{{axiom | n = \text B 5
        | q = \forall B^*_1, B^*_2 \in \mathscr B^*
        | mr= x \in B^*_1 \setminus B^*_2 \implies \exists y \in B^*_2 \setminus B^*_1 : \paren {B^*_2 \setminus \set y}  \cup \set x \in \mathscr B
}}
{{end-axiom}}

Proof:
=== Necessary Condition ===
Let $\mathscr B$ satisfies formulation $1$ base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 1}}
Let $B^*_1, B^*_2 \in \mathscr B^*$.
By definition of $\mathscr B^*$:
:$\exists B_1, B_2 \in \mathscr B$: 
::$B^*_1 = S \setminus B_1$
::$B^*_2 = S \setminus B_2$
We have:
{{begin-eqn}}
{{eqn | l = B^*_1 \setminus B^*_2
      | r = \paren{S \setminus B_1} \setminus \paren{S \setminus B_2}
}}
{{eqn | r = B_2 \setminus B_1
      | c = Set Difference of Complements
}}
{{end-eqn}}
And similarly:
:$B^*_2 \setminus B^*_1 = B_1 \setminus B_2$
Let $x \in B^*_1 \setminus B^*_2$.
Hence:
:$x \in B_2 \setminus B_1$
From formulation $1$ base axiom:
:$\exists y \in B_1 \setminus B_2 : \paren{B_2 \setminus \set x} \cup \set y \in \mathscr B$
Hence:
:$y \in B^*_2 \setminus B^*_1$
We have:
{{begin-eqn}}
{{eqn | l = S \setminus \paren{\paren{B_2 \setminus \set x} \cup \set y}
      | r = S \setminus \paren{\paren{B_2 \setminus \set x} \cup \paren{S \cap \set y} }
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \paren{S \setminus \paren{B_2 \setminus \set x} } \setminus \set y
      | c = Set Difference with Set Difference is Union of Set Difference with Intersection
}}
{{eqn | r = \paren{\paren{S \setminus B_2}\cup \paren{ S \cap \set x} } \setminus \set y
      | c = Set Difference with Set Difference is Union of Set Difference with Intersection
}}
{{eqn | r = \paren{\paren{S \setminus B_2}\cup \set x} \setminus \set y
      | c = Intersection with Subset is Subset
}}
{{eqn | r = \paren{B^*_2 \cup \set x} \setminus \set y
      | c = Definition of $B^*_2$
}}
{{eqn | r = \paren{B^*_2 \setminus \set y} \cup \set x 
      | c = {{Corollary|Set Difference Then Union Equals Union Then Set Difference}}
}}
{{end-eqn}}
Hence:
:$\paren{B^*_2 \setminus \set y} \cup \set x \in \mathscr B^*$
It follows that $\mathscr B^*$ satisfies formulation 5 base axiom:
{{begin-axiom}}
{{axiom | n = \text B 5
        | q = \forall B^*_1, B^*_2 \in \mathscr B^*
        | mr= x \in B^*_1 \setminus B^*_2 \implies \exists y \in B^*_2 \setminus B^*_1 : \paren {B^*_2 \setminus \set y}  \cup \set x \in \mathscr B
}}
{{end-axiom}}
{{qed|lemma}}
Theorem: Subsets Satisfy Formulation 5 Matroid Base Axiom Iff Complements Satisfy Formulation 1

Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
Let $\mathscr B^* = \set{S \setminus B : B \in \mathscr B}$.
Then:
:$\mathscr B$ satisfies formulation $5$ base axiom:
{{:Axiom:Base Axiom (Matroid)/Formulation 5}}
{{iff}}
:$\mathscr B^*$ satisfies formulation $1$ base axiom:
{{begin-axiom}}
{{axiom | n = \text B 1
        | q = \forall B^*_1, B^*_2 \in \mathscr B^*
        | mr= x \in B^*_1 \setminus B^*_2 \implies \exists y \in B^*_2 \setminus B^*_1 : \paren {B^*_1 \setminus \set x} \cup \set y \in \mathscr B^*
}}
{{end-axiom}}

Proof:
Let $\mathscr B^{**} = \set{S \setminus B^* : B^* \in \mathscr B^*}$.
Then:
:$\mathscr B^{**} = \set{S \setminus \paren{S \setminus B} : B \in \mathscr B}$
From Relative Complement of Relative Complement:
:$\mathscr B^{**} = \mathscr B$
From Subsets Satisfy Formulation 1 Matroid Base Axiom Iff Complements Satisfy Formulation 5:
:$\mathscr B^*$ satisfies formulation $1$ base axiom
{{iff}}
:$\mathscr B^{**}$ satisfies formulation $5$ base axiom
That is:
:$\mathscr B^*$ satisfies formulation $1$ base axiom
{{iff}}
:$\mathscr B$ satisfies formulation $5$ base axiom
{{qed}}
Theorem: Matroid is Uniquely Defined by Bases

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\mathscr B$ be the set of bases of the matroid $M$.
Then:
:$M$ is the only matroid on $S$ whose set of bases is $\mathscr B$.

Proof:
Let $M^\prime = \struct {S, \mathscr I^\prime}$ whose set of bases is $\mathscr B$.
From Characterization of Matroid Independent Sets in Terms of Bases:
:$\mathscr I^\prime = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
Similarly:
:$\mathscr I = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
Hence:
:$\mathscr I^\prime = \mathscr I$
and so:
:$M^\prime = M$
{{qed}}
Category:Matroid Theory
Category:Matroid Bases
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Theorem: Set Difference of Matroid Dependent Set with Independent Set is Non-empty

Let $I$ be an independent subset of $M$.
Let $D$ be a dependent subset of $M$.
Then:
:$D \setminus I \ne \O$

Proof:
From Independent Subset Contains No Dependent Subset:
:$D \nsubseteq I$
By definition of subset:
:$\exists x \in D : x \notin I$
By definition of set difference:
:$\exists x \in D \setminus I$
The result follows.
{{qed}}
Category:Matroid Dependent Subsets
Category:Matroid Independent Subsets
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
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Theorem: Set Difference of Matroid Dependent Set with Independent Set is Non-empty/Corollary 2

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $I$ be an independent subset of $M$.
Let $C$ be a circuit of $M$.
Then:
:$C \setminus I \ne \O$

Proof:
By definition of matroid circuit:
:$C$ is a dependent subset of $M$
From Set Difference of Matroid Dependent Set with Independent Set is Non-empty:
:$C \setminus I \ne \O$
{{qed}}
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
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Theorem: Set Difference of Matroid Dependent Set with Independent Set is Non-empty/Corollary 1

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $D$ be a dependent subset of $M$.
Let $B$ be a base of $M$.
Then:
:$D \setminus B \ne \O$

Proof:
By definition of matroid base:
:$B$ is an independent subset of $M$
From Set Difference of Matroid Dependent Set with Independent Set is Non-empty:
:$D \setminus B \ne \O$
{{qed}}
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
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Theorem: Set Difference of Matroid Dependent Set with Independent Set is Non-empty/Corollary 3

Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B$ be an base of $M$.
Let $C$ be a circuit of $M$.
Then:
:$C \setminus B \ne \O$

Proof:
By definition of matroid base:
:$B$ is an independent subset of $M$
By definition of matroid circuit:
:$C$ is a dependent subset of $M$
From Set Difference of Matroid Dependent Set with Independent Set is Non-empty:
:$C \setminus B \ne \O$
{{qed}}
Category:Set Difference of Matroid Dependent Set with Independent Set is Non-empty
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Theorem: Preimage of Subset under Identity Mapping

Let $S$ be a set.
Let $\iota_S: S \to S$ be the identity mapping on $S$.
Let $T \subseteq S$.
Then:
:$\iota_S^{-1} \sqbrk T = T$
where $\iota_S^{-1} \sqbrk T$ is the preimage of $T$ under $\iota_S$.

Proof:
By definition of identity mapping:
:$\iota_S: S \to S: \forall x \in S: \map {\iota_S} x = x$
Let $i_S: S \to S$ be the inclusion mapping of $S$ into $S$.
By definition of inclusion mapping:
:$i_S: S \to S: \forall x \in S: \map {i_S} x = x$
From Equality of Mappings:
:$\iota_S = i_S$
From Preimage of Subset under Inclusion Mapping:
:$\forall T\subseteq S : \iota_S^{-1} \sqbrk T = T \cap S$
From Intersection with Subset is Subset:
:$\forall T\subseteq S : T \cap S = T$
Hence:
:$\forall T\subseteq S : \iota_S^{-1} \sqbrk T = T$
{{qed}}
Category:Identity Mappings
Category:Subsets
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Theorem: Extension Theorem for Positive Linear Functional defined on Cofinal Linear Subspace

Let $\struct {X, \succeq}$ be a preordered vector space over $\R$.
Let $Y$ be a linear subspace of $X$ that is cofinal in $\struct {X, \succeq}$.
Let $f_0 : Y \to \R$ be a positive linear functional.
Then there exists a positive linear functional $f : X \to \R$ such that:
:$\map f y = \map {f_0} y$ for each $y \in Y$.

Proof:
Let:
:$P = \set {x \in X : x \succeq {\mathbf 0}_X}$
Let:
:$Z = Y + P - P = \set {y + z - w : y \in Y, \, z, w \in P}$
be the Minkowski sum of $Y$, $P$ and $-P$.
Theorem: Extension Theorem for Positive Linear Functional defined on Cofinal Linear Subspace/Lemma 2

:$q$ is a sublinear functional.

Proof:
Let $x, z \in Z$. 
Note that if $y, w \in Y$ have $y \succeq x$ and $w \succeq z$, we have $y + w \succeq x + z$.
We are then able to deduce that:
:$\set {\map {f_0} {y + w} : y, w \in Y \text { and } y \succeq x, \, w \succeq z} \subseteq \set {\map {f_0} y : y \in Y \text { and } y \succeq x + z}$
From Infimum of Subset, we have:
:$\map q {x + z} \le \inf \set {\map {f_0} {y + w} : y, w \in Y \text { and } y \succeq x, \, w \succeq z}$
We further have from linearity:
:$\inf \set {\map {f_0} {y + w} : y, w \in Y \text { and } y \succeq x, \, w \succeq z} = \inf \set {\map {f_0} y : y \in Y \text { and } y \succeq x} + \inf \set {\map {f_0} w : w \in Y \text { and } w \succeq z}$
That is:
:$\map q {x + z} \le \map q x + \map q z$
So we have subadditivity. 
We just need to establish positive homogeneity. 
Let $x \in X$ and $\lambda > 0$. 
The key to this is the fact that for $\lambda > 0$ we have:
:$\set {y \in Y : y \succeq \lambda x} = \lambda \set {y \in Y : y \succeq x}$
From the definition of a preordered vector space, we have:
:$y \succeq \lambda x$ {{iff}} $\lambda^{-1} y \succeq x$ 
giving the equality.
Taking infimums and using Multiple of Infimum, we have:
:$\map q {\lambda x} = \lambda \map q x$ for each $x \in X$ and $\lambda > 0$.
We conclude that $q$ is a sublinear functional.
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Theorem: Proper Modular Ideal of Algebra is Contained in Maximal Ideal

Let $R$ be a ring.
Let $\struct {A, \ast}$ be an $R$-algebra.
Let $\struct {J, \ast}$ be a proper modular ideal of $\struct {A, \ast}$. 
Then there exists a maximal ideal $M$ of $\struct {A, \ast}$ such that $J \subseteq M$.

Proof:
Let $u \in A$ be such that:
:$a - u a \in J$ and $a - a u \in J$
for each $a \in A$.
Let $S$ be the set of ideals $I$ of $\struct {A, \ast}$ such that:
:$u \not \in I$ and $J \subseteq I$.
Consider the partially ordered set $\struct {S, \subseteq}$ where $\subseteq$ is the subset relation.
We use Zorn's Lemma on $S$.
Let $\CC$ be a chain in $\struct {S, \subseteq}$.
We show that $\bigcup \CC$ is an upper bound for $\CC$. 
From Union of Chain of Ideals is Ideal: Algebra over Ring, we have:
:$\bigcup \CC$ is an ideal. 
From Set Union Preserves Subsets, we have $J \subseteq \bigcup \CC$. 
Since $u \not \in I$ for each $I \in \CC$, we also have $u \not \in \bigcup \CC$.
So every chain in $\struct {S, \subseteq}$ has an upper bound. 
Hence from Zorn's Lemma, $\struct {S, \subseteq}$ has a maximal element $I$. 
We show that $I$ is a maximal ideal. 
We know that there is no proper ideal $K$ with $u \not \in K$ such that $I \subsetneq K$ since $I$ is a maximal element of $\struct {S, \subseteq}$.
{{AimForCont}} that there exists an ideal $K$ with $u \in K$ and $I \subseteq K$.
We then have $a u \in K$ for each $a \in A$ since $K$ is an ideal.
Since $J$ is modular and $J \subseteq K$, we also have $a - a u \in K$ for each $a \in A$.
Hence for each $a \in A$ we have $a = \paren {a - a u} + a u \in K$.
Hence $K = A$.
So there is no proper ideal $K$ such that $I \subsetneq K$.
Hence $I$ is a maximal ideal.
{{qed}}
Theorem: Union of Chain of Submodules is Submodule

Let $R$ be a ring.
Let $M$ be an $R$-module.
Let $\NN$ be a chain of submodules with respect to inclusion.
Let:
:$N = \bigcup \NN$
Then $N$ is a submodule of $M$.

Proof:
We use the Submodule Test.
Theorem: Quotient Algebra is Algebra

Let $R$ be a ring. 
Let $A$ be a $R$-algebra.
Let $I$ be an ideal of $A$.
Let $A/I$ be the quotient module of $A$ modulo $I$.
For $a, b \in A$, define $\ast : \paren {A/I}^2 \to A/I$ by:
:$\paren {a + I} \ast \paren {b + I} = a b + I$ 
Then $\ast$ is well-defined and $\struct {A/I, \ast}$ is an $R$-algebra.

Proof:
=== $\ast$ is Well-Defined ===
It suffices to show that if:
:$a - a' \in I$
and:
:$b - b' \in I$
then:
:$ab - a' b' \in I$ 
Suppose that:
:$a - a' = i_1 \in I$
and:
:$b - b' = i_2 \in I$
Then we have:
:$a' b' = \paren {a + i_1} \paren {b + i_2} = a b + i_1 b + a i_2 + i_1 i_2$
From the definition of an ideal, we have $-\paren {i_1 b + a i_2 + i_1 i_2} \in I$.
Hence $a b - a' b' = -\paren {i_1 b + a i_2 + i_1 i_2} \in I$, which is what we wanted to show. 
{{qed|lemma}}
Theorem: Quotient Algebra is Unital iff Quotienting Ideal is Modular

Let $R$ be a ring. 
Let $A$ be a $R$-algebra.
Let $I$ be an proper ideal of $A$.
Let $A/I$ be the quotient algebra of $A$ modulo $I$.
Then $A/I$ is unital {{iff}} $I$ is modular.

Proof:
=== Necessary Condition ===
Suppose that $A/I$ is unital.
That is, there exists $u \in I$ such that:
:$\paren {a + I} \paren {u + I} = \paren {u + I} \paren {a + I} = a + I$
for each $a \in A$.
That is:
:$a u + I = u a + I = a + I$
Hence, we have:
:$a - a u \in I$
and:
:$a - u a \in I$
for each $a \in A$.
So $I$ is a modular ideal.
Theorem: Ideal of Unital Algebra is Modular

Let $R$ be a ring. 
Let $A$ be a unital $R$-algebra.
Let $I$ be an ideal of $A$.
Then $I$ is a modular ideal of $A$.

Proof:
Let ${\mathbf 1}_A$ be the identity element of $A$.
Let ${\mathbf 0}_A$ be the zero vector of $A$.
Then we have:
:$a {\mathbf 1}_A = {\mathbf 1}_A a$
for each $a \in A$. 
Hence we have:
:$a - {\mathbf 1}_A a = {\mathbf 0}_A$
and:
:$a - a {\mathbf 1}_A = {\mathbf 0}_A$
Because $I$ is a submodule of $A$:
:${\mathbf 0}_A \in A$
Hence:
:$a - u a \in I$
and so:
:$a - a u \in I$
for $u = {\mathbf 1}_A$.
Hence $I$ is a modular ideal of $A$ by definition.
{{qed}}
Theorem: Quotient Normed Algebra is Normed Algebra

Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra.
Let $I$ be a closed ideal of $A$.
Let $A/I$ be the quotient algebra of $A$ modulo $I$.
Let $\norm {\, \cdot \,}_{A/I}$ be the quotient norm.
Then $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ is a normed algebra.

Proof:
Since $I$ is a closed linear subspace of $A$, Quotient Norm is Norm shows that $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ is a normed vector space.
It remains to show that for $a, b \in A$ we have:
:$\norm {a b + I}_{A/I} \le \norm {a + I}_{A/I} \norm {b + I}_{A/I}$
Let $\epsilon > 0$.
By the definition of the quotient norm, there exists $a', b' \in I$ such that:
:$\norm {a + a'}_{A/I} < \epsilon + \norm {a + I}_{A/I}$
and:
:$\norm {b + b'}_{A/I} < \epsilon + \norm {b + I}_{A/I}$
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {a b + a' b + a b' + a' b'}
		| r = \norm {\paren {a + a'} \paren {b + b'} }
}}
{{eqn	| o = \le
		| r = \norm {a + a'} \paren {b + b'}
		| c = {{Defof|Norm on Algebra}}
}}
{{eqn	| o = <
		| r = \paren {\epsilon + \norm {a + I}_{A/I} } \paren {\epsilon + \norm {b + I}_{A/I} }
}}
{{end-eqn}}
Since $I$ is an ideal, we have:
:$a' b + a b' + a' b' \in I$
Hence we have:
:$\ds \norm {a b + I}_{A/I} = \inf_{i \mathop \in I} \norm {a b + i} < \paren {\epsilon + \norm {a + I}_{A/I} } \paren {\epsilon + \norm {b + I}_{A/I} }$
Taking $\epsilon \to 0^+$, we obtain:
:$\norm {a b + I}_{A/I} \le \norm {a + I}_{A/I} \norm {b + I}_{A/I}$
So $\norm {\, \cdot \,}_{A/I}$ is an algebra norm on $A/I$. 
So $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ is a normed algebra.
{{qed}}
Theorem: Norm of Identity Element in Normed Algebra Lower Bounded by One

Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring.
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $R$ with identity element ${\mathbf 1}_A \ne {\mathbf 0}_A$. 
Then $1 \le \norm { {\mathbf 1}_A}$.

Proof:
By the definition of a algebra norm, we have:
:$\norm { {\mathbf 1}_A} = \norm { {\mathbf 1}_A^2} \le \norm { {\mathbf 1}_A}^2$
From {{NormAxiomVector|1}}, we have $\norm { {\mathbf 1}_A} \ne 0$, we have:
:$1 \le \norm { {\mathbf 1}_A}$
{{qed}}
Category:Normed Algebras
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Theorem: Quotient Normed Algebra of Unital Normed Algebra is Unital Normed Algebra

Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring.
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital normed algebra with identity element ${\mathbf 1}_A$. 
Let $I$ be a closed proper ideal of $A$. 
Let $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ be the quotient normed algebra of $A$ modulo $I$.
Then:
:$\norm { {\mathbf 1}_A + I}_{A/I} = 1$

Proof:
From Norm of Identity Element in Normed Algebra Lower Bounded by One, we have $1 \le \norm { {\mathbf 1}_A + I}_{A/I}$.
Since $I$ is a linear subspace of $A$, we have ${\mathbf 0}_A \in I$. 
Hence we have:
:$\ds \inf_{i \mathop \in I} \norm { {\mathbf 1}_A + i} \le \norm { {\mathbf 1}_A + {\mathbf 0}_A} = \norm { {\mathbf 1}_A}$
From the definition of a unital normed algebra, we have:
:$\norm { {\mathbf 1}_A} = 1$
Hence we obtain:
:$\norm { {\mathbf 1}_A + I}_{A/I} \le 1$
So:
:$\norm { {\mathbf 1}_A + I}_{A/I} = 1$ 
{{qed}}
Category:Quotient Normed Algebras
Category:Unital Normed Algebras
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Theorem: Maximal Ideal of Unital Commutative Banach Algebra is Kernel of Character

Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $M$ be a maximal ideal of $A$. 
Then there exists a character $\phi$ such that:
:$\ker \phi = M$

Proof:
From Maximal Ideal in Unital Banach Algebra is Closed, $M$ is closed.
Let $A/M$ be the quotient algebra of $A$ modulo $M$.
From Quotient Normed Algebra of Unital Normed Algebra is Unital Normed Algebra, $A/M$ is a unital normed algebra. 
From Quotient Algebra of Commutative Algebra is Commutative, $A/M$ is a commutative algebra.
Hence from the Gelfand-Mazur Theorem, there exists an isometric algebra isomorphism $q_2 : A/M \to \C$.
From Quotient Mapping on Quotient Algebra is Algebra Homomorphism, $q_1 : A \to A/M$ defined by:
:$\map {q_1} x = x + M$
for each $x \in A$ is a algebra homomorphism.
From Composition of Algebra Homomorphisms is Algebra Homomorphism, $q_2 \circ q_1 : A \to \C$ is an algebra homomorphism.
Since $q_2$ is an algebra isomorphism, for any $x + M \in A/M \setminus \set { {\mathbf 0}_{A/M} }$, we have $\map {q_2} {x + M} = \map {q_2} {\map {q_1} x} \ne 0$.
Hence $q_2 \circ q_1 : A \to \C$ is a non-zero algebra homomorphism.
Hence $\phi = q_2 \circ q_1$ is a character on $A$.
We just need to assure ourselves that $\ker \phi = M$. 
Since $q_2$ is an algebra isomorphism we have $\ker {q_2} = \set { {\mathbf 0}_A + M}$.
Hence if $\map {q_2} {\map {q_1} x} = 0$ for $x \in A$ we have $\map {q_1} x = {\mathbf 0}_{A/M}$. 
From Kernel of Quotient Mapping, we then have $x \in M$.
Conversely, if $x \in M$ then $\map {q_1} x = 0$ so $\map \phi x = 0$. 
So $\ker \phi = M$ as desired. 
{{qed}}
Category:Unital Banach Algebras
Category:Characters (Banach Algebras)
Category:Commutative Banach Algebras
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