nyuuzyou/wikis huggingface dataset:
Element of Natural Number is Natural Number

== Theorem ==
Let $n$ be a natural number.
Let $m \in n$.
Then $m$ is also a natural number.
== Proof ==
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:for all $m \in n$: $m$ is a natural number.
=== Basis for the Induction ===
$\map P 0$ is the case:
:for all $m \in 0$: $m$ is a natural number.
This is true vacuously, as $\O$ has no elements by definition of empty class.
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k^+}$ is true.
So this is the induction hypothesis:
:for all $m \in k$: $m$ is a natural number.
from which it is to be shown that:
:for all $m \in k^+$: $m$ is a natural number.
=== Induction Step ===
This is the induction step:
Let $m \in k^+$.
Then either:
:$m = k$
or: 
:$m \in k$
In the first case, $m$ is the natural number $k$.
In the second case, it follows by the induction hypothesis that $m$ is a natural number.
In both cases $m$ is a natural number.
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: \forall m \in n$: $m$ is a natural number.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.4$
Category:Natural Numbers
Element of Ordered Set of Topology is Dense iff is Everywhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \preceq}$ be an ordered set
where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$
Let $A \in \tau$.
Then $A$ is a dense element in $P$ $A$ is everywhere dense.
== Proof ==
=== Sufficient Condition ===
Assume that
:$A$ is a dense element in $P$.
By Bottom in Ordered Set of Topology:
:$\bot_P = \O$
We will prove that
:for every open subset $U$ of $S$: $U \ne \O \implies U \cap A \ne \O$
Let $U$ be an open subset of $S$ such that
:$U \ne \O$
By definition of open set:
:$U \in \tau$
By definition of topological space:
:$U \cap A \in \tau$
By Meet in Inclusion Ordered Set:
:$U \wedge A = U \cap A$
Thus by definition of dense element:
:$U \cap A \ne \O$
By Characterization of Closure by Open Sets:
:$A^- = S$
where $A^-$ denotes the topological closure of $A$.
Hence $A$ is everywhere dense.
=== Necessary Condition ===
Assume that
:$A$ is everywhere dense.
Let $B \in \tau$ such that
:$B \ne \bot_P$
By definition of non-empty set:
:$\exists x: x \in B$
By definition:
:$B$ is an open set.
By definition of everywhere dense:
:$A^- = S$
By Characterization of Closure by Open Sets:
:$B \cap A \ne \O$
By definition of topological space:
:$B \cap A \in \tau$
Thus by Meet in Inclusion Ordered Set:
:$B \wedge A \ne \bot_P$
== Sources ==
* 
* 
Category:Denseness
Element of Ordinal is Ordinal

== Theorem ==
Let $n$ be an ordinal.
Let $m \in n$.
Then $m$ is also an ordinal.
That is, the class of all ordinals $\On$ is a transitive class.
== Proof 1 ==
== Proof 2 ==
Category:Element of Ordinal is Ordinal
Category:Ordinals
Category:Transitive Classes
Element of Ordinal is Ordinal/Proof 1

== Theorem ==
== Proof ==
Let $\On$ denote the class of all ordinals.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.
By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$
Vacuously, every element of $\O$ is an ordinal.
Let $\alpha \in \On$ be an ordinal.
Let us assume that every element of $\alpha$ is an ordinal.
Let $\alpha^+$ denote the successor to $\alpha$.
Because $\On$ is superinductive:
:$\alpha \in \On \implies \alpha^+ \in \On$
By definition of successor mapping:
:$\alpha^+ = \alpha \cup \set \alpha$
Then every element of $\alpha^+$ is either an element of $\alpha$ or $\alpha$ itself.
We have that:
:$\alpha$ is an ordinal
:every element of $\alpha$ is an ordinal.
That is, $\alpha^+$ is an ordinal.
Let $C$ be a chain of ordinals such that:
:each element $\alpha$ of $C$ has the property that each element of $\alpha$ is itself an ordinal.
Let $\beta \in \bigcup C$.
Then $\beta \in \alpha$ for some $\alpha \in C$.
Hence, by definition of $C$, $\beta$ is an ordinal which has the property that each element of $\beta$ is itself an ordinal.
It follows by the Principle of Superinduction that every element of an ordinal is an ordinal.
Let $\alpha \in \On$ be an arbitrary ordinal.
Let $\beta \subseteq \alpha$.
From Class of All Ordinals is Well-Ordered by Subset Relation, $\beta$ is an ordinal
Hence from the above:
:$\beta \in \alpha$
As $\alpha$ is arbitrary:
:$\forall \alpha \in \On: \beta \subseteq \alpha \implies \beta \in \alpha$
proving that $\On$ is transitive.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.6$
Category:Element of Ordinal is Ordinal
Element of Ordinal is Ordinal/Proof 2

== Theorem ==
== Proof ==
By the definition of ordinal, $n$ is transitive.
Thus $m \subseteq n$.
By Subset of Strictly Well-Ordered Set is Strictly Well-Ordered, it follows that $m$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_m}$.
It is now to be shown that $m$ is transitive.
If $m = \O$ then the result follows by Empty Class is Transitive.
If $m \ne \O$, then let $x \in m$.
If $x = \O$, then $x \subseteq m$ by Empty Set is Subset of All Sets.
If $x \ne \O$, then let $y \in x$.
It suffices to show that $y \in m$.
Since $m \subseteq n$, it follows that $x \in n$.
Also, $y \in x \land x \in n \implies y \in n$ because $n$ is transitive.
And so $x \in n$, $y \in n$, and $m \in n$.
A strict well-ordering is transitive by definition.
Therefore:
:$y \in x \land x \in m \implies y \in m$
Hence the result.
Category:Element of Ordinal is Ordinal
Element of Pascal's Triangle is Sum of Diagonal or Column starting above it going Upwards

== Theorem ==
Consider Pascal's triangle:
:
Let $\tuple {n, m}$ be the element in the $n$th row and $m$th column.
Then:
:$\tuple {n, m} = \ds \sum_{k \mathop \ge 0} \tuple {n - k - 1, m - 1}$
and:
:$\tuple {n, m} = \ds \sum_{k \mathop \ge 0} \tuple {n - k - 1, m - k - 1}$
== Proof ==
We have $\tuple {n, m} = \dbinom n m$, and we have:
== Sources ==
* : $35$
* : $35$
Category:Pascal's Triangle
Element of Pascal's Triangle is Sum of Diagonal or Column starting above it going Upwards/Mistake

== Source Work ==
:
:The Dictionary
::$35$
== Mistake ==
:In Pascal's Triangle, each number is the sum of either of the diagonals starting immediately above it, and taking the long way to the edge: for example, $35 = 15 + 10 + 6 + 3 + 1$.
It is not necessary for the route to the edge to be the long way. The short way also works:
:$35 = 20 + 10 + 4 + 1$
== Sources ==
* : $35$
* : $35$
Category:Mistakes/Curious and Interesting Numbers
Element of Principal Ideal Domain is Finite Product of Irreducible Elements

== Theorem ==
Let $R$ be a principal ideal domain.
Let $p \in R$ such that $p \ne 0$ and $p$ is not a unit.
Then there exist irreducible elements $p_1, \ldots, p_n$ such that $p = p_1 \cdots p_n$.
== Proof ==
If $p$ is irreducible, it is proven.
Suppose $p$ is not irreducible.
Then $p = r_1 r_2$ where neither $r_1$ nor $r_2$ are units.
If $r_1$ and $r_2$ are irreducible, then the proof is complete.
If $r_1$ is not irreducible, then $r_1 = r_{11} r_{12}$ where neither $r_{11}$ nor $r_{12}$ are units.
If we continue, we get a proper inclusion of ideals $\ideal r \subset \ideal {r_1} \subset \ideal {r_{11} } \subset \cdots \subset R$.
If this process finishes in a finite number of steps, the proof is complete.
 the process does not complete in a finite number of steps.
Thus the chain $\ideal r \subset \ideal {r_1} \subset \ideal {r_{11} } \subset \cdots \subset R$ is infinite.
From Principal Ideal Domain fulfills Ascending Chain Condition, this cannot happen.
Thus, the process ends in a finite number of steps.
Category:Factorization
Category:Ideal Theory
Category:Principal Ideal Domains
Element of Simple Algebraic Field Extension of Degree n is Polynomial in Algebraic Number of Degree Less than n

== Theorem ==
Let $F$ be a field.
Let $\theta \in \C$ be algebraic over $F$ of degree $n$.
Let $\map F \theta$ be the simple field extension of $F$ by $\theta$.
Then any element of $\map F \theta$ can be written as $\map f \theta$, where $\map f x$ is a polynomial over $F$ of degree at most $n - 1$.
== Proof ==
From Simple Algebraic Field Extension consists of Polynomials in Algebraic Number, an arbitrary element of $\map F \theta$ can be written as $\map f \theta$.
But:
:$\map f x = \map m x \, \map q x + \map r x$
where:
:$\map m x$ is minimal polynomial in $\theta$
:$\map q x$ is a polynomial in $\map F \theta$
:$\map r x$ is a polynomial in $\map F \theta$ such that either:
::$\map \deg {\map r x} < \map \deg {\map m x}$
:or:
::$\map r x = 0$
Thus:
:$\map f \theta = \map m \theta \, \map q \theta + \map r \theta$
and as $\map m \theta = 0$ we have:
:$\map f \theta = \map r \theta$
So $\map f \theta$ can be expressed as $\map r \theta$ instead, which is of degree strictly less than that of $\map m \theta$.
Hence the result.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $73$
Category:Field Extensions
Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a self-adjoint densely-defined linear operator.
Let $\map \sigma T$ be the spectrum of $\struct {\map D T, T}$.
Let $\map {\sigma_r} T$ be the residual spectrum of $\struct {\map D T, T}$.
Let $\lambda \in \map \sigma T \setminus \map {\sigma_r} T$. 
Then $\lambda$ is an approximate eigenvalue of $\struct {\map D T, T}$.
=== Corollary ===
== Proof ==
Let $\lambda \in \map \sigma T$.
From Partition of Spectrum of Densely-Defined Linear Operator, $\lambda$ is contained in either the point spectrum or continuous spectrum of $T$.
If $\lambda$ is contained in the point spectrum, we have the result from Eigenvalue of Densely-Defined Linear Operator is Approximate Eigenvalue.
Otherwise, $\lambda$ is contained in the continuous spectrum, in which case $\paren {T - \lambda I}^{-1}$ is not bounded.
That is, there exists a sequence $\sequence {y_n}_{n \mathop \in \N}$ in $\map D {\paren {T - \lambda I}^{-1} }$ with $\norm {y_n} = 1$ and: 
:$\norm {\paren {T - \lambda I}^{-1} y_n} \to \infty$
Now set: 
:$\ds x_n = \paren {T - \lambda I}^{-1} y_n$
for each $n \in \N$ and then: 
:$\ds z_n = \frac {x_n} {\norm {x_n} }$
We then have:
Then: 
so we have: 
:$\paren {T - \lambda I} z_n \to 0$ 
So $\lambda$ is an approximate eigenvalue of $T$.
== Sources ==
* $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators
Category:Approximate Eigenvalues (Densely-Defined Linear Operators)
Category:Spectra (Densely-Defined Linear Operators)
Category:Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue
Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue/Corollary

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a self-adjoint densely-defined linear operator.
Let $\map \sigma T$ be the spectrum of $\struct {\map D T, T}$.
Let $\lambda \in \map \sigma T$. 
Then $\lambda$ is an approximate eigenvalue of $\struct {\map D T, T}$.
== Proof ==
Let $\map {\sigma_r} T$ be the residual spectrum of $\struct {\map D T, T}$.
From Self-Adjoint Densely-Defined Linear Operator has Empty Residual Spectrum, we have $\map {\sigma_r} T = \emptyset$. 
Hence we have $\lambda \in \map \sigma T \setminus \map {\sigma_r} T$.
From Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue, we therefore have that $\lambda$ is an approximate eigenvalue of $\struct {\map D T, T}$.
Category:Element of Spectrum of Densely-Defined Linear Operator not in Residual Spectrum is Approximate Eigenvalue
Element of Swelled Set with no Immediate Extension is Maximal

== Theorem ==
Let $S$ be a swelled set of sets.
Let $x \in S$ have no immediate extension.
Then $x$ is a maximal element of $S$ the subset relation.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles
Category:Swelled Classes
Element of Topological Space may or may not be Limit Point

== Theorem ==
Let $S$ be a set.
Let $H \subseteq S$ be a subset of $S$.
Let $T = \struct {H, \tau}$ be a topological space on the underlying set $H$.
Let $a \in H$.
Then $a$ may or may not be a limit point of $T$.
Whether it is or not depends upon the nature of both $a$ and $T$.
== Proof ==
Let $\R$ be the set of real numbers.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
:$H = \set 0 \cup \openint 1 2$
From Limit Point Examples: Union of Singleton with Open Real Interval, $0$ is not a limit point of $H$, although $0 \in H$.
From Limit Point Examples: End Points of Real Interval, $a$ is a limit point of $\closedint a b$ where $a \in \closedint a b$.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.10 \ \text {(b)}$
Category:Limit Points
Element of Transitive Class

== Theorem ==
Let $B$ be a transitive class.
Then:
:$A \in B \implies A \subsetneq B$
where $\subsetneq$ denotes a proper subset).
== Proof ==
By the definition of a transitive class:
:$A \in B \implies A \subseteq B$
But $A \ne B$ because $\paren {A = B \land A \in B} \implies A \in A$, which by No Membership Loops is a contradiction.
Therefore $A \subsetneq B$.
== Sources ==
* : $\S 7.2$
Category:Class Theory
Category:Transitive Classes
Element of Unital Banach Algebra Close to Identity is Invertible

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {A, \norm \cdot}$ be a unital Banach algebra over $\Bbb F$ with identity element $\mathbf 1_A$. 
Let $a \in A$ be such that: 
:$\norm {\mathbf 1_A - a} < 1$ 
Then $a$ is invertible with inverse element $a^{-1}$ satisfying: 
:$\ds \norm {a^{-1} } \le \frac 1 {1 - \norm {\mathbf 1_A - a} }$
== Proof ==
Let:
:$x = \mathbf 1_A - a$
From Bound on Norm of Power of Element in Normed Algebra, we have: 
:$\norm {x^n} \le \norm x^n$
for each $n \in \Z_{\ge 0}$.
Then we have: 
:$\ds \sum_{n \mathop = 0}^\infty \norm {x^n} \le \sum_{n \mathop = 0}^\infty \norm x^n$ 
Since $\norm x < 1$, we have: 
:$\ds \sum_{n \mathop = 0}^\infty \norm x^n$ converges
from Sum of Infinite Geometric Sequence. 
So:
:$\ds \sum_{n \mathop = 0}^\infty \norm {x^n}$ converges.
Since $A$ is a Banach space, we have: 
:$\ds \sum_{n \mathop = 0}^\infty x^n$ converges.
by Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach.
Let: 
:$\ds b = \sum_{n \mathop = 0}^\infty x^n$
We show that: 
:$\ds a b = \mathbf 1_A$
and:
:$\ds b a = \mathbf 1_A$
Note that: 
:$a = \mathbf 1_A - x$ 
Note that for each $N \in \N$, we have: 
and similarly: 
We have: 
so that: 
:$\mathbf 1_A - x^{N + 1} \to \mathbf 1_A$
from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence.
So taking $N \to \infty$ in: 
:$\ds a \sum_{n \mathop = 0}^N x^n = \mathbf 1_A - x^{N + 1}$
we obtain: 
:$a b = \mathbf 1_A$ 
from Product Rule for Sequence in Normed Algebra. 
Similarly taking $N \to \infty$ in: 
:$\ds \paren {\sum_{n \mathop = 0}^N x^n} a = \mathbf 1_A - x^{N + 1}$
we obtain: 
:$b a = \mathbf 1_A$
So $a$ is invertible with inverse element $b$. 
It remains to show that: 
:$\ds \norm b \le \frac 1 {1 - \norm x}$
For each $N \in \N$, we have that:
Taking $N \to \infty$, we have: 
:$\ds \norm b \le \frac 1 {1 - \norm x} = \frac 1 {1 - \norm {\mathbf 1_A - a} }$ 
from Modulus of Limit: Normed Vector Space.
== Sources ==
* : $4.4$: The group of units
Category:Unital Banach Algebras
Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Let $B$ be a closed subalgebra of $A$.
Let $\map G B$ be the group of units of $A$.
Let $x \in \partial \map G B$, where $\partial \map G B$ is the topological boundary of $\map G B$.
Then $x$ is not invertible in $A$. 
== Proof ==
=== Lemma ===
 that there exists $y \in A$ such that $x y = 1$.
From the lemma, there exists a sequence $\sequence {z_n}_{n \in \N}$ in $A$ such that $\norm {z_n} = 1$ for each $n \in \N$, and: 
:$z_n x \to 0$ as $n \to \infty$
From Product Rule for Sequence in Normed Algebra, we have: 
:$z_n x y \to 0$ as $n \to \infty$.
Since $x y = 1$, this gives $z_n \to 0$ as $n \to \infty$. 
From Modulus of Limit: Normed Vector Space, we then have $\norm {z_n} \to 0$ as $n \to \infty$. 
But $\norm {z_n} = 1$ for each $n \in \N$, so this is impossible.
So we have a contradiction and there exists no such $y \in A$.
Category:Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra
Category:Unital Banach Algebras
Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra/Lemma

== Lemma ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$. 
Let $x \in \partial \map G A$, where $\partial \map G A$ is the topological boundary of $\map G A$.
Then there exists a sequence $\sequence {z_n}_{n \in \N}$ in $A$ such that $\norm {z_n} = 1$ for each $n \in \N$, and: 
:$z_n x \to 0$ as $n \to \infty$ 
and:
:$x z_n \to 0$ as $n \to \infty$.
== Proof ==
From Group of Units in Unital Banach Algebra is Open, we have: 
:$\partial \map G A = \map G A^- \setminus \map G A$ 
So if $x \in \partial \map G A$, then $x \in A \setminus \map G A$ and there exists a sequence $\sequence {x_n}_{n \in \N}$ with $x_n \in \map G A$ for each $n \in \N$, such that: 
:$x_n \to x$ as $n \to \infty$.
Let: 
:$\ds z_n = \frac {x_n^{-1} } {\norm {x_n^{-1} } }$
We have:
:$\ds \norm {z_n} = \norm {\frac {x_n^{-1} } {\norm {x_n^{-1} } } } = \frac {\norm {x_n^{-1} } } {\norm {x_n^{-1} } } = 1$
from , for each $n \in \N$. 
Now, we have: 
We know by hypothesis that $\norm {x - x_n} \to 0$, while from Norm of Inverse of Sequence of Invertible Elements Converging to Non-Invertible Element in Unital Banach Algebra, we have: 
:$\ds \frac 1 {\norm {x_n^{-1} } } \to 0$ 
So, we have: 
:$z_n x \to 0$ as $n \to \infty$. 
Similarly, we have:
So $\sequence {z_n}_{n \in \N}$ satisfies the conditions given in the theorem. 
Category:Element of Unital Banach Algebra on Boundary of Group of Units of Subalgebra is Not Invertible in Algebra
Element of Unital C*-Algebra Bounded below by Non-Negative Multiple of Identity Element is Invertible and Positive

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering of $A$.
Let $a \in A$ be such that:
:$a \ge_A \alpha {\mathbf 1}_A$
for some $\alpha > 0$.
Then $a$ is invertible and positive element.
== Proof ==
From Element of Unital C*-Algebra Bounded by Multiple of Identity is Hermitian, $a$ is Hermitian.
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we then have:
:$\map {\sigma_A} a \subseteq \hointr \alpha \infty \subseteq \hointr 0 \infty$
So $a$ is positive.
We have $\alpha > 0$ and hence $0 \not \in \map {\sigma_A} a$.
Hence $a$ is invertible.
Category:Canonical Preorderings of C*-Algebras
Element of Unital C*-Algebra Bounded by Multiple of Identity is Hermitian

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $a \in A$.
Let $\le_A$ be the canonical preordering of $A$.
Suppose that either:
:$a \le_A \beta {\mathbf 1}_A$ for some $\alpha \in \R$
or:
:$\alpha {\mathbf 1}_A \le_A a$ for some $\alpha \in \R$ 
Then $a$ is Hermitian.
== Proof ==
Suppose first that:
:$a \le_A \beta {\mathbf 1}_A$ for some $\alpha \in \R$
Then $\beta {\mathbf 1}_A - a$ is positive.
In particular, $\beta {\mathbf 1}_A - a$ is Hermitian.
That is:
:$\paren {\beta {\mathbf 1}_A - a}^\ast = \overline \beta {\mathbf 1}_A - a^\ast = \beta {\mathbf 1}_A - a$
from the definition of an involution and Identity Element in Unital *-Algebra is Hermitian.
So $a = a^\ast$.
So $a$ is Hermitian in this case.
Now suppose that:
:$\alpha {\mathbf 1}_A \le_A a$ for some $\alpha \in \R$ 
Then $a - \alpha {\mathbf 1}_A$ is positive.
In particular, $a - \alpha {\mathbf 1}_A$ is Hermitian..
That is:
:$\paren {a - \alpha {\mathbf 1}_A}^\ast = a^\ast - \overline \alpha {\mathbf 1}_A$ 
from the definition of an involution and Identity Element in Unital *-Algebra is Hermitian.
So $a = a^\ast$.
So $a$ is Hermitian in this case also.
Category:Canonical Preorderings of C*-Algebras
Category:Hermitian Elements of *-Algebras
Element of Unital Commutative Algebra Invertible iff not Contained in Maximal Ideal

== Theorem ==
Let $R$ be a ring.
Let $A$ be an unital commutative $R$-algebra.
Let $a \in A$.
Then $a$ is invertible there does not exist a maximal ideal $M$ with $a \in M$. 
== Proof ==
=== Necessary Condition ===
We prove the contrapositive.
Then, by Proof by Contraposition we will done. 
We will prove that if $a$ is in a maximal ideal then $a$ is not invertible.
Let $M$ be a maximal ideal with $a \in M$. 
 there exists $b \in A$ with $a b = {\mathbf 1}_A$.
Since $M$ is an ideal, we have $a b \in M$, hence ${\mathbf 1}_A \in M$.
This contradicts that $M$ is proper.
=== Sufficient Condition ===
We will prove that if $a$ is not invertible then there exists a maximal ideal $M$ with $a \in M$.
Let:
:$I = \set {a m : m \in A}$
We show that $I$ is a proper ideal containing $a$.
That $a \in I$ is immediate since $a {\mathbf 1}_A \in A$.
Let $a x, a y \in I$ and $\lambda, \mu \in R$.
Then we have:
:$\mu a x + \lambda a y = a \paren {\mu x + \lambda y} \in I$ 
So $I$ is a submodule of $A$. 
Further, if $a x \in I$ and $b \in A$, we have:
:$\paren {a x} b = a \paren {x b} \in I$ 
So $I$ is a right ideal.
Since $A$ is commutative, $I$ is an ideal.
From Ideal of Unital Algebra is Modular, $I$ is modular. 
Further, since $a$ is invertible there exists no $m \in A$ such that $a m = {\mathbf 1}_A$.
Hence ${\mathbf 1}_A \not \in \set {a m : m \in A}$.
From Proper Modular Ideal of Algebra is Contained in Maximal Ideal, there then exists a maximal ideal $M$ with $I \subseteq M$. 
Then $a \in M$.
Category:Commutative Algebras
Category:Maximal Ideals of Algebras
Element of Unital Commutative Banach Algebra is Invertible iff not in Kernel of Character

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $a \in A$.
Then $x$ is invertible :
:for every character $\phi$ of $A$, we have $a \not \in \ker \phi$. 
== Proof ==
From Element of Unital Commutative Algebra Invertible iff not Contained in Maximal Ideal, we have that $a$ is invertible :
:for every maximal ideal $M$ of $A$, we have $a \not \in M$. 
From Characterization of Maximal Ideal of Unital Commutative Banach Algebra, we have that:
:$M$ is a maximal ideal of $A$ there exists a character $\phi$ of $A$ with $M = \ker \phi$. 
Hence $a$ is invertible :
:for every character $\phi$ of $A$, we have $a \not \in \ker \phi$.
Category:Commutative Banach Algebras
Element of Universe

== Theorem ==
Let $A$ be a class, which may be either a set or a proper class.
Then:
:$\forall A: \paren {A \in U \iff \exists x: x = A}$
where $U$ is the universal class.
That is, if $A$ is an element of $U$, then $A$ is a set.
== Proof ==
== Sources ==
* : $\S 6.9$
Category:Class Theory