nyuuzyou/wikis huggingface dataset:
Cayley-Hamilton Theorem/Finitely Generated Module

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module.
Let $\mathfrak a$ be an ideal of $A$.
Let $\phi$ be an endomorphism of $M$ such that $\phi \sqbrk M \subseteq \mathfrak a M$.
Then $\phi$ satisfies an equation of the form:
:$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
with the $a_i \in \mathfrak a$.
== Proof ==
Recall that the set of endomorphisms $\operatorname{End}_A(M)$ is an $A$-module with unity.
Let $R \subseteq \operatorname{End}_A(M)$ be the $A$-submudole generated by $\set {\phi^n : n \in \N}$.
That is:
:$R = \set {a_0 + a_1 \phi + \cdots + a_k \phi^k : k \in \N, a_1,\ldots ,a_k \in A}$
Note that $R$ is a commutative ring with unity.
Recall also the ring of $n$-matrices $\map {\MM_R} n$.
Let $\set {m_1, \ldots, m_n}$ be a generator for $M$.
Then for each $i \in \set {1, \ldots, n}$, as $\map \phi {m_i} \in \mathfrak a M$, we can write:
:$\ds \map \phi {m_i} = \sum_{j \mathop = 1}^n a_j m_j$
where $a_j \in A$ for $j \in \set {1, \ldots, n}$.
Thus for each $i \in \set {1, \ldots, n}$:
:$(1): \quad \ds \sum_{j \mathop = 1}^n \sqbrk{\delta_{ij} \phi - a_{ij} \mathbf 1_R } m_i = 0$
where:
:$\delta_{i j}$ is the Kronecker delta with values in $A$
Now let $\Delta \in \map {\MM_R} n$ defined as:
:$\Delta := \paren {\phi \delta_{i j} - a_{i j} \mathbf 1_R }$
Then $\paren 1$ can be written as:
:$\Delta \cdot \begin{bmatrix} m_1 \\ \vdots \\ m_n \end{bmatrix} = \mathbf 0_{R^n}$
Let $\adj \Delta \in \map {\MM_R } n$ be the adjugate matrix of $\Delta$.
By Matrix Product with Adjugate Matrix:
where $\map \det \Delta \in R$ denotes the determinant of $\Delta$.
Multiplying the above equality by $\adj \Delta$ and applying Cramer's Rule:
:$\ds \map \det \Delta m_i = 0$
for each $i \in \set {1, \ldots , n}$.
Therefore:
:$\map \det \Delta = \mathbf 0_R$
On the other hand, $\map \det \Delta$ is a monic polynomial in $\phi$, since:
where:
:$\lambda$ goes over all permutations of $\set {1, 2, \ldots, n}$
:$\map \sgn \lambda$ is the sign of the permutation $\lambda$
:$a_0, a_1, \ldots, a_{n-1} \in \mathfrak a$
As $R \subseteq \operatorname{End}_A(M)$, this means:
:$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
Category:Cayley-Hamilton Theorem
Category:Commutative Algebra
Cayley-Hamilton Theorem/Matrix

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathbf N = \sqbrk {a_{i j} }$ be an $n \times n$ matrix with entries in $A$.
Let $\mathbf I_n$ denote the $n \times n$ unit matrix.
Let $\map {p_{\mathbf N} } x$ be the determinant $\map \det {x \cdot \mathbf I_n - \mathbf N}$.
Then:
:$\map {p_{\mathbf N} } {\mathbf N} = \mathbf 0$
as an $n \times n$ zero matrix.
That is:
:$\mathbf N^n + b_{n - 1} \mathbf N^{n - 1} + \cdots + b_1 \mathbf N + b_0 = \mathbf 0$
where the $b_i$ are the coefficients of $\map {p_{\mathbf N} } x$.
== Proof ==
Taking $\phi = \mathbf N$ in the proof of Cayley-Hamilton Theorem for Finitely Generated Modules we see that $\mathbf N$ satisfies:
:$\map {p_{\mathbf N} } x = \map \det {x \cdot \mathbf I_n - \mathbf N} = 0$
Take $\mathfrak a$ to be the ideal generated by the entries of $\mathbf N$.
== Sources ==
* 
* 
Category:Cayley-Hamilton Theorem
Category:Linear Algebra
Cayley Table/Examples/Arbitrary Order 3 Structure

== Example of Cayley Table ==
The Cayley table of an algebraic structure of order $3$ can be presented:
$\qquad \begin{array}{c|cccc}
\circ & a & b & c \\
\hline
a & b & c & b \\
b & b & a & c \\
c & a & c & c \\
\end{array}$
== Sources ==
* : $\S 4.1$. Binary operations on a set: Example $58$
Category:Examples of Cayley Tables
Cayley Table/Examples/C4

== Example of Cayley Table ==
The Cayley table of the cyclic group of order $4$ can be written:
$\qquad \begin{array}{c|cccc}
& e & a & b & c \\
\hline
e & e & a & b & c \\
a & a & b & c & e \\
b & b & c & e & a \\
c & c & e & a & b \\
\end{array}$
Category:Examples of Cayley Tables
Category:Cyclic Group of Order 4
Cayley Table/Examples/S3

== Example of Cayley Table ==
The Cayley table of the symmetric group on $3$ letters can be written:
$\qquad \begin{array}{c|cccccc}
\circ & e & p & q & r & s & t \\
\hline
e & e & p & q & r & s & t \\
p & p & q & e & s & t & r \\
q & q & e & p & t & r & s \\
r & r & t & s & e & q & p \\
s & s & r & t & p & e & q \\
t & t & s & r & q & p & e \\
\end{array}$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table
Category:Examples of Cayley Tables
Cayley Table/Examples/Self-Maps on Doubleton

== Example of Cayley Table ==
Let $S$ be the set of self-maps on the doubleton $D = \set {a, b}$.
Let these be enumerated:
$\qquad \epsilon := \begin{pmatrix} a & b \\ a & b \end{pmatrix}
\quad \alpha := \begin{pmatrix} a & b \\ b & a \end{pmatrix}
\quad \beta := \begin{pmatrix} a & b \\ a & a \end{pmatrix}
\quad \gamma := \begin{pmatrix} a & b \\ b & b \end{pmatrix}$
Let $\struct {S, \circ}$ be the semigroup of self-maps under composition of mappings.
The Cayley table of $\struct {S, \circ}$ can be written:
$\qquad \begin{array}{c|cccc}
\circ & \epsilon & \alpha & \beta & \gamma \\
\hline
\epsilon & \epsilon & \alpha & \beta & \gamma \\
\alpha & \alpha & \epsilon & \gamma & \beta \\
\beta & \beta & \beta & \beta & \beta \\
\gamma & \gamma & \gamma & \gamma & \gamma \\
\end{array}$
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups
Category:Examples of Cayley Tables
Category:Self-Maps
Cayley Table for Commutative Operation is Symmetrical about Main Diagonal

== Definition ==
Let $\struct {S, \circ}$ be an algebraic structure.
Then:
:the Cayley table for $\struct {S, \circ}$ is symmetrical about the main diagonal
:
:$\circ$ is a commutative operation.
== Proof ==
Let $\CC$ denote the Cayley table for $\struct {S, \circ}$.
Let $\sqbrk c_{a b}$ denote the entry of $\CC$ corresponding to the element $a \circ b$ of $S$.
That is, $\sqbrk c_{a b}$ is the entry in the row headed by $a$ and the column headed by $b$.
=== Necessary Condition ===
Let $\circ$ be a commutative operation.
Then by definition:
:$\forall a, b \in S: a \circ b = b \circ a$
Thus:
:$\sqbrk c_{a b} = \sqbrk c_{b a}$
and so:
:the entry in the row headed by $a$ and the column headed by $b$
is the same as:
:the entry in the row headed by $b$ and the column headed by $a$.
This applies to all $a, b \in \S$.
Hence for all $a \in S$, the row headed by $a$ is the same as the column headed by $a$.
It follows that $\CC$ is symmetrical about the main diagonal.
=== Sufficient Condition ===
Let $\CC$ is symmetrical about the main diagonal.
Then:
:$\forall a, b \in S: \sqbrk c_{a b} = \sqbrk c_{b a}$
That is, by the definition of the Cayley table:
:$\forall a, b \in S: a \circ b = b \circ a$
Thus, by definition, $\circ$ is a commutative operation.
== Sources ==
* : Exercise $1.4: \ 6$
* : Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions
* : $\S 1.4$
* : Chapter $\text{II}$: Groups: Exercise $\text{A}$
Category:Cayley Tables
Ceiling Function/Examples/Ceiling of -1.1

== Theorem ==
:$\ceiling {-1 \cdotp 1} = -1$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
We have that:
:$-2 < -1 \cdotp 1 \le -1$
Hence $-1$ is the ceiling of $-1 \cdotp 1$ by definition.
== Also see ==
* Floor of $1 \cdotp 1$: $\floor {1 \cdotp 1} = 1$
* Floor of $-1 \cdotp 1$: $\floor {-1 \cdotp 1} = -2$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $1$
Category:Examples of Ceiling Function
Ceiling Function/Examples/Ceiling of 4.35

== Theorem ==
:$\ceiling {4 \cdotp 35} = 5$
where $\ceiling x$ denotes the floor of $x$.
== Proof ==
We have that:
:$4 < 4 \cdotp 35 \le 5$
Hence $5$ is the ceiling of $4 \cdotp 35$ by definition.
== Also see ==
* Floor of $4 \cdotp 35$: $\floor {4 \cdotp 35} = 4$
== Sources ==
* 
* 
Category:Examples of Ceiling Function
Ceiling Function/Examples/Ceiling of Minus One Half

== Theorem ==
:$\ceiling {-\dfrac 1 2} = 0$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
We have:
:$0 \ge -\dfrac 1 2$
and:
:$-1 = -\dfrac 2 2 < -\dfrac 1 2$
Hence $0$ is the ceiling of $-\dfrac 1 2$ by definition.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Examples of Ceiling Function
Ceiling Function/Examples/Ceiling of Root 2

== Theorem ==
:$\ceiling {\sqrt 2} = 2$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
The decimal expansion of $\sqrt 2$ is:
:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
Thus:
:$1 < \sqrt 2 \le 2$
Hence $2$ is the ceiling of $\sqrt 2$ by definition.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Examples of Ceiling Function
Ceiling Function is Idempotent

== Theorem ==
Let $x \in \R$ be a real number.
Let $\ceiling x$ denote the ceiling of $x$.
Then:
:$\ceiling {\ceiling x} = \ceiling x$
That is, the ceiling function is idempotent.
== Proof ==
Let $y = \ceiling x$.
By Ceiling Function is Integer, $y$ is an integer.
Then from Real Number is Integer iff equals Ceiling:
:$\ceiling y = y$
So:
:$\ceiling {\ceiling x} = \ceiling x$
== Also see ==
* Floor Function is Idempotent
* Floor of Ceiling is Ceiling
* Ceiling of Floor is Floor
Category:Ceiling Function
Ceiling defines Equivalence Relation

== Theorem ==
Let $\RR$ be the relation defined on $\R$ such that:
:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \ceiling x = \ceiling y$
where $\ceiling x$ is the ceiling of $x$.
Then $\RR$ is an equivalence, and $\forall n \in \Z$, the $\RR$-class of $n$ is the half-open interval $\hointl {n - 1} n$.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
:$\forall x \in \R: \ceiling x = \ceiling x$
Thus the ceiling function is reflexive.
=== Symmetry ===
:$\forall x, y \in \R: \ceiling x = \ceiling y \implies \ceiling y = \ceiling x$
Thus the ceiling function is symmetric.
=== Transitivity ===
Let:
:$\ceiling x = \ceiling y$
:$\ceiling y = \ceiling z$
Let:
:$n = \ceiling x = \ceiling y = \ceiling z$
which follows from transitivity of $=$.
Thus:
:$x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \hointr 0 1$
from Real Number is Ceiling minus Difference‎.
So:
:$x = n - t_x$
and:
:$z = n - t_z$
and so:
:$\ceiling x = \ceiling z$
Thus the ceiling function is transitive.
Thus we have shown that $\RR$ is an equivalence.
Now we show that the $\RR$-class of $n$ is the interval $\hointl {n - 1} n$.
Defining $\RR$ as above, with $n \in \Z$:
== Also see ==
* Floor defines Equivalence Relation
Category:Ceiling Function
Category:Examples of Equivalence Relations
Ceiling is between Number and One More

== Theorem ==
:$x \le \ceiling x < x + 1$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
From Number is between Ceiling and One Less:
:$\ceiling x - 1 < x \le \ceiling x$
Thus by adding $1$:
:$x + 1 > \paren {\ceiling x - 1} + 1 = \ceiling x$
So:
:$x \le \ceiling x$
and:
:$\ceiling x < x + 1$
as required.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Ceiling Function
Ceiling of Floor is Floor

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor of $x$, and $\ceiling x$ denote the ceiling of $x$.
Then:
:$\ceiling {\floor x} = \floor x$
That is, the ceiling of the floor is the floor.
== Proof ==
Let $y = \floor x$.
By Floor Function is Integer, we have that $y \in \Z$.
Then from Real Number is Integer iff equals Ceiling, we have:
:$\ceiling y = y$
So:
:$\ceiling {\floor x} = \floor x$
== Also see ==
* Floor of Ceiling is Ceiling
* Floor Function is Idempotent
* Ceiling Function is Idempotent
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $2$
Category:Floor Function
Category:Ceiling Function
Ceiling of Half of n+m plus Ceiling of Half of n-m+1

== Theorem ==
Let $n, m \in \Z$ be integers.
:$\ceiling {\dfrac {n + m} 2} + \ceiling {\dfrac {n - m + 1} 2} = n + 1$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
Either $n + m$ or $n - m + 1$ is even.
Thus either $\dfrac {n + m} 2$ or $\dfrac {n - m + 1} 2$ is an integer.
So:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $33 \ \text{(b)}$
Category:Ceiling Function
Ceiling of Negative equals Negative of Floor

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ be the floor of $x$, and $\ceiling x$ be the ceiling of $x$.
Then:
:$\ceiling {-x} = -\floor x$
== Proof ==
From Integer equals Floor iff between Number and One Less we have:
:$x - 1 < \floor x \le x$
and so, by multiplying both sides by -1:
:$-x + 1 > -\floor x \ge -x$
From Integer equals Ceiling iff between Number and One More we have:
:$\ceiling x = n \iff x \le n < x + 1$
Hence:
:$-x \le -\floor x < -x + 1 \implies \ceiling {-x} = -\floor x$
== Also see ==
* Floor of Negative equals Negative of Ceiling
Category:Floor Function
Category:Ceiling Function
Ceiling of Non-Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $x \notin \Z$.
Then:
:$\left \lceil{x}\right \rceil > x$
where $\left \lceil{x}\right \rceil$ denotes the ceiling of $x$.
== Proof ==
From Ceiling is between Number and One More:
:$\left \lceil{x}\right \rceil \ge x$
From Real Number is Integer iff equals Ceiling:
:$x = \left \lceil {x} \right \rceil \iff x \in \Z$
But we have $x \notin \Z$.
So:
:$\left \lceil {x} \right \rceil \ne x$
and so:
:$\left \lceil {x} \right \rceil > x$
== Also see ==
* Floor of Non-Integer
Category:Ceiling Function
Ceiling of Root of Ceiling equals Ceiling of Root

== Theorem ==
Let $x \in \R_{\ge 0}$ be a positive real number.
Let $\ceiling x$ denote the ceiling of $x$.
Then:
:$\ds \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Floor of Root of Floor equals Floor of Root
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $6 \ \text{(b)}$
Category:Ceiling Function
Category:Ceiling of Root of Ceiling equals Ceiling of Root
Ceiling of Root of Ceiling equals Ceiling of Root/Proof 2

== Theorem ==
== Proof ==
The square root is defined on the interval $\hointr 0 \to$.
We have that Square Root is Strictly Increasing.
From Continuity of Root Function, the square root is continuous.
Hence the conditions are fulfilled for for McEliece's Theorem (Integer Functions) to be applied:
:$\forall x \in A: \paren {\map f x \in \Z \implies x \in \Z} \iff \ceiling {\map f x} = \ceiling {\map f {\ceiling x} }$
It remains to be proved that $\sqrt x \in \Z \implies x \in \Z$.
Let $y = \sqrt x$ such that $y \in \Z$.
By definition of square root:
:$x = y^2$
From Integer Multiplication is Closed:
:$x \in \Z$
as required.
Category:Ceiling of Root of Ceiling equals Ceiling of Root
Ceiling of x+m over n

== Theorem ==
Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $\ceiling x$ denotes the ceiling of $x$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$
Category:Ceiling Function
Category:Ceiling of x+m over n
Ceiling of x+m over n/Corollary

== Corollary to Ceiling of $\frac {x + m} n$ ==
Let $n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ceiling {\dfrac x n} = \ceiling {\dfrac {\ceiling x} n}$
where $\ceiling x$ denotes the ceiling of $x$.
== Proof ==
This is a special case of Ceiling of $\dfrac {x + m} n$:
:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $m = 0$.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$
Category:Ceiling Function
Category:Ceiling of x+m over n
Ceiling of x+m over n/Proof 2

== Theorem ==
== Proof ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac {x + m} n$
It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$.
Let $\dfrac {x + m} n \in \Z$.
Then:
Thus:
:$\forall x \in \R: \map f x \in \Z \implies x \in \Z$
So the conditions are fulfilled for McEliece's Theorem (Integer Functions) to be applied:
:$\ceiling {\map f x} = \ceiling {\map f {\ceiling x} } \iff \paren {\map f x \in \Z \implies x \in \Z}$
Hence the result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $35$ (Answers to Exercises)
Category:Ceiling of x+m over n
Celestial Body moves along Parallel of Declination

== Theorem ==
Consider the celestial sphere $C$ whose observer is $O$.
Let $B$ be a celestial body on $C$.
Let $\delta$ be the declination of $B$.
Let $P_\delta$ be the parallel of declination whose declination is $\delta$.
Then $B$ appears to move along the path of $P_\delta$, in the direction from the north to east to south to west.
It takes $1$ sidereal day for $B$ to travel around $P_\delta$.
== Proof ==
The apparent movement of $C$ is caused by the rotation of Earth about its axis.
This rotation takes $1$ sidereal day, by definition, for the principal vertical circle to be in the same position again.
As the Earth rotates, the observer $O$ moves in an easterly direction.
Hence the celestial sphere $C$ appears to move in the opposite direction, so from north to east and so on.
$B$ remains at the same distance from the north celestial pole as it did.
Hence its declination $\delta$ stays the same.
Hence the result.
== Sources ==
* : Chapter $\text {II}$. The Celestial Sphere: $19$. Declination and hour angle.
Category:Spherical Astronomy
Celestial Equator is Parallel to Geographical Equator

== Theorem ==
Consider the celestial sphere with observer $O$.
The plane of the celestial equator is parallel to the plane of the geographical equator.
== Proof ==
Let $P$ denote the north celestial pole.
The celestial equator is by definition perpendicular to $OP$.
Also by definition, $OP$ is parallel to Earth's axis.
Also by definition, Earth's axis is perpendicular to the geographical equator.
Hence $OP$ is also perpendicular to the geographical equator.
The result follows from Planes Perpendicular to same Straight Line are Parallel.
== Sources ==
* : Chapter $\text {II}$. The Celestial Sphere: $19$. Declination and hour angle.
Category:Spherical Astronomy
Center is Characteristic Subgroup

== Theorem ==
Let $G$ be a group.
Then its center $\map Z G$ is characteristic in $G$.
== Proof ==
By Identity Mapping is Group Automorphism, there exists at least one automorphism of $G$.
Let $\phi$ be an automorphism of $G$.
Let $x \in \map Z G, y \in G$.
Then:
Hence $\map \phi x \in \map Z G$.
So we have $\phi \sqbrk {\map Z G} \subseteq \map Z G$.
Since $\phi^{-1}$ is also an automorphism:
:$\phi^{-1} \sqbrk {\map Z G} \subseteq \map Z G$
Since $\phi$ is a bijection:
:$\map Z G = \phi \sqbrk {\phi^{-1} \sqbrk {\map Z G}} \subseteq \phi \sqbrk {\map Z G}$
Therefore we conclude that $\phi \sqbrk {\map Z G} = \map Z G$.
Hence $\map Z G$ is characteristic in $G$.
Category:Characteristic Subgroups
Category:Centers of Groups
Center is Element of Closed Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon \in \R_{>0}$ be a positive real number.
Let $\map { {B_\epsilon}^-} a$ be the closed $\epsilon$-ball of $a$ in $M$.
Then:
:$a \in \map {{B_\epsilon}^-} a$
=== Normed Division Ring ===
=== P-adic Numbers ===
== Proof ==
By metric axiom $(\text M 1)$:
:$\map d {a, a} = 0$
By assumption:
:$\epsilon > 0$
Hence:
:$\map d {a, a} < \epsilon$
By definition of the closed $\epsilon$-ball of $a$ $\map {{B_\epsilon}^-} a$ in $M$:
:$a \in \map { {B_\epsilon}^-} a$
Category:Metric Spaces
Center is Element of Closed Ball/Normed Division Ring

== Theorem ==
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map { {B_\epsilon}^-} a$ be the closed $\epsilon$-ball of $a$ in $\struct{R, \norm {\,\cdot\,} }$.
Then:
:$a \in \map { {B_\epsilon}^-} a$
== Proof ==
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
From Closed Ball in Normed Division Ring is Closed Ball in Induced Metric, $\map { {B_\epsilon}^-} a$ is the closed $\epsilon$-ball of $a$ in the metric space $\struct{R,d}$.
From Center is Element of Closed Ball:
:$a \in \map { {B_\epsilon}^-} a$
Category:Normed Division Rings
Center is Element of Closed Ball/P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map { {B_\epsilon}^-} a$ be the closed $\epsilon$-ball of $a$ in $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
Then:
:$a \in \map { {B_\epsilon}^-} a$
== Proof ==
By definition, $\map { {B_\epsilon}^-} a$ is the closed $\epsilon$-ball of $a$ in the normed division ring $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
From Center is Element of Closed Ball in Normed Division Ring:
$a \in \map { {B_\epsilon}^-} a$
Category:Topology of P-adic Numbers
Center is Element of Open Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon \in \R_{>0}$ be a positive real number.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball of $a$ in $M$.
Then:
:$a \in \map {B_\epsilon} a$
=== Normed Division Ring ===
=== P-adic Numbers ===
== Proof ==
By :
:$\map d {a, a} = 0$
By assumption:
:$\epsilon > 0$
Hence:
:$\map d {a, a} < \epsilon$
By definition of the open $\epsilon$-ball of $a$ $\map {B_\epsilon} a$ in $M$:
:$a \in \map {B_\epsilon} a$
Category:Open Balls
Category:Center is Element of Open Ball
Center is Element of Open Ball/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball of $a$ in $\struct{R, \norm {\,\cdot\,} }$.
Then:
:$a \in \map {B_\epsilon} a$
== Proof ==
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
From Open Ball in Normed Division Ring is Open Ball in Induced Metric, $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in the metric space $\struct{R,d}$.
From Center is Element of Open Ball:
:$a \in \map {B_\epsilon} a$
Category:Normed Division Rings
Category:Center is Element of Open Ball
Center is Element of Open Ball/P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball of $a$ in $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
Then:
:$a \in \map {B_\epsilon} a$
== Proof ==
By definition, $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in the normed division ring $\struct {\Q_p, \norm {\,\cdot\,}_p}$.
From Center is Element of Open Ball in Normed Division Ring:
$a \in \map {B_\epsilon} a$
Category:Topology of P-adic Numbers
Category:Center is Element of Open Ball
Center is Intersection of Centralizers

== Theorem ==
The center of a group is the intersection of all the centralizers of the elements of that group:
:$\ds \map Z G = \bigcap_{g \mathop \in G} \map {C_G} g$
== Proof ==
Denote $Z = \map Z G$ and $C = \ds \bigcap_{g \mathop \in G} \map {C_G} g$ for simplicity.
By definition of set equality, it suffices to prove $Z \subseteq C$ and $C \subseteq Z$.
=== $Z$ is contained in $C$ ===
Suppose that $x \in Z$.
Then from the definition of center:
:$\forall g \in G: x g = g x$
By definition of centralizer, this corresponds to:
:$\forall g \in G: x \in \map {C_G} g$
Therefore we have, by definition of set intersection:
:$\ds x \in \bigcap_{g \mathop \in G} \map {C_G} g = C$
Hence $x \in C$.
It follows by definition of subset that $Z \subseteq C$.
=== $C$ is contained in $Z$ ===
Suppose now that $x \in C$.
Then, by definition of intersection:
:$\forall g \in G: x \in \map {C_G} g$
That is, using the definition of centralizer:
:$\forall g \in G: x g = g x$
By definition of the center, this means:
:$x \in \map Z G = Z$
Hence $x \in Z$.
It follows that $C \subseteq Z$.
Therefore, we have established that:
:$x \in Z \iff x \in C$
By definition of set equality:
:$Z = C$
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.13$
* : $\S 37$. Some important general examples of subgroups
Category:Centers of Groups
Category:Centralizers