nyuuzyou/wikis huggingface dataset:
Eigenstates of Hamiltonian have Well Defined Symmetry only if Potential is Even

== Theorem ==
Given a certain Hamiltonian describing a quantum system, its eigenstates will have a well-defined symmetry in space (even or odd) only if the potential said system is subjected to is an even function of space
== Proof ==
Given some Hamiltonian $H$, we can find its eigenfunctions in order to determine the eigenstates of the system, thus forming an orthogonal (or orthonormal, if normalized) base of the corresponding Hilbert space. If these eigenfunctions are to have a well-defined symmetry, then they must also be eigenfunctions of the parity operator.
The only way the eigenstates can be eigenfunctions of both the Hamiltonian and the parity operator simultaneously is if their commutator is zero. Therefore, the proof of this lemma is reduced to finding the minimal set of conditions necessary for $[P, H]=0$ to be true. Let us examine how this commutator acts on some wave function $\phi(x)$ (this proof is carried out in one dimension in space for simplicity, but it can be extrapolated to an arbitrary dimension trivially).
$[P, H]\phi = (PH-HP)\phi$
$= P\left(\frac{p²}{2m}\phi(x)+V(x)\phi(x)\right)-\left(\frac{p²}{2m}P\phi(x)+V(x)P\phi(x)\right)$Taking into account the parity operator acts as $P\phi(x) = \phi(-x)$ and that the kinetic term is even (because $p²=-\hbar²\frac{d²}{dx²}$, so trivially $p²$ will remain unchanged under the parity operator).
$= \frac{p²}{2m}\phi(-x)+V(-x)\phi(-x)-\frac{p²}{2m}\phi(-x)-V(x)\phi(-x)$
$= \left[V(-x)-V(x)\right]\phi(-x)=0$Therefore, the condition we must impose for the commutator of the Hamiltonian and the parity operator to be zero $\forall x \in Dom(\phi)$ is that $V(x)=V(-x)$, that is to say, for $V(x)$ to be an even function of space $\blacksquare$
Eigenvalue is Instance of Generalized Eigenvalue

== Theorem ==
Let $\mathbf A$ be a square matrix of order $n$. 
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Then:
:$\lambda$ is a generalized eigenvalue of $\mathbf A$
:the corresponding eigenvector $\mathbf x$ corresponding to $\lambda$ is the generalized eigenvector of $\mathbf A$ corresponding to $\lambda$.
== Proof ==
By the definition of eigenvalue of $\mathbf A$:
:$\map \det {\mathbf I_n \mathbf x - \mathbf A} = 0$
for some non-zero vector $\mathbf x$.
Recall the definition of generalized eigenvalue of $\mathbf A$:
:$\mathbf A \mathbf x = \lambda \mathbf B \mathbf x$
for:
:some non-zero vector $\mathbf x$
:some square matrix $\mathbf B$ of order $n$.
Then from Generalized Eigenvalues as Roots of Equation:
:$\map \det {\mathbf A - \lambda \mathbf B} = 0$
where:
:$\mathbf B$ is another square matrix of order $n$
:$\det$ denotes the determinant.
Hence setting:
:$\mathbf B = \mathbf I_n$
it is seen that $\lambda$ is a generalized eigenvalue of $\mathbf A$
Hence, by definition, the eigenvector $\mathbf x$ corresponding to $\lambda$ is the generalized eigenvector of $\mathbf A$ corresponding to $\lambda$.
== Sources ==
* 
Category:Generalized Eigenvalues
Category:Eigenvalues
Category:Generalized Eigenvectors
Category:Eigenvectors
Eigenvalue of Bounded Linear Operator is Approximate Eigenvalue

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\lambda \in \C$ be an eigenvalue of $T$.
Then $\lambda$ is an approximate eigenvalue of $T$.
== Proof ==
Since $\lambda$ is an eigenvalue of $T$, there exists $y \in \HH$ with $y \ne {\mathbf 0}_\HH$ such that:
:$T y = \lambda y$
Setting:
:$\ds x_n = \frac y {\norm y}$
we have $\norm {x_n} = 1$ by and:
:$T x_n = \lambda x_n$ 
We then have:
:$\norm {T x_n - \lambda x_n} = 0$ for each $n \in \N$.
So:
:$\norm {T x_n - \lambda x_n} \to 0$
Hence $\lambda$ is an approximate eigenvalue of $T$.
Category:Approximate Eigenvalues (Bounded Linear Operators)
Eigenvalue of Densely-Defined Linear Operator is Approximate Eigenvalue

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a densely-defined linear operator.
Let $\lambda$ be an eigenvalue of $T$.
Then $\lambda$ is an approximate eigenvalue of $T$.
== Proof ==
Since $\lambda$ is an eigenvalue of $T$, there exists $x \in \map D T \setminus \set 0$ such that: 
:$\paren {T - \lambda I} x = 0$
Then setting:
:$\ds x_n = \frac x {\norm x}$
we have: 
:$\paren {T - \lambda I} x_n = 0$
for each $n \in \N$, while $\norm {x_n} = 1$.
Then, we have: 
:$\paren {T - \lambda I} x_n \to 0$
So $\lambda$ is an approximate eigenvalue of $T$.
Category:Approximate Eigenvalues (Densely-Defined Linear Operators)
Eigenvalue of Matrix Powers

== Theorem ==
Let $A$ be a square matrix.
Let $\lambda$ be an eigenvalue of $A$ and $\mathbf v$ be the corresponding eigenvector.
Then:
:$A^n \mathbf v = \lambda^n \mathbf v$
holds for each positive integer $n$.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$A^n \mathbf v = \lambda^n \mathbf v$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$A \mathbf v = \lambda \mathbf v$
which follows by definition of eigenvector.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$A^k \mathbf v = \lambda^k \mathbf v$
Then we need to show:
:$A^{k + 1} \mathbf v = \lambda^{k + 1} \mathbf v$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{> 0}: A^n \mathbf v = \lambda^n \mathbf v$
Category:Matrix Algebra
Eigenvalues of Compact Riemannian Manifold without Boundary are Nonnegative

== Theorem ==
Let $\struct {M, g}$ be a compact Riemannian manifold without boundary.
Let $\lambda \in \R$ be an eigenvalue of $M$.
Then $0$ is an eigenvalue of $M$, and the rest of eigenvalues are strictly positive.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry
Eigenvalues of Companion Matrix are Zeroes of Polynomial

== Definition ==
Let $P$ be the polynomial of degree $n$ presented in the form:
:$\map P x = x^n - a_{n - 1} x^{n - 1} - \cdots - a_1 x - a_0$
Let $C$ be the companion matrix of $P$.
The eigenvalues of $C$ are the zeroes of $P$.
== Proof ==
== Sources ==
* 
Category:Companion Matrices
Category:Eigenvalues
Eigenvalues of Correlation Matrix are Non-Negative

== Definition ==
Let $\sequence a_n$ and $\sequence b_n$ be sequences of $n$ observations.
Let $\mathbf C$ be the correlation matrix $\sequence a_n$ and $\sequence b_n$.
Then the eigenvalues of $\mathbf C$ are non-negative.
== Proof ==
== Sources ==
* 
Category:Correlation Matrices
Category:Eigenvalues
Eigenvalues of G-Representation are Roots of Unity

== Theorem ==
Let $G$ be a finite group.
Let $\left({K, +, \cdot}\right)$ be a field.
Let $V$ be a $G$-module over $K$ (i.e. $V$ is a $K \left[{G}\right]$-module).
Then $\forall g \in G$, the eigenvalues of the action by the vector $g \in K \left[{G}\right]$ on $V$ are roots of unity.
== Proof ==
Fix an arbitrary $g \in G$ and consider the corresponding vector $g \in K \left[{G}\right]$.
Let $\lambda$ be an eigenvalue of $g$, that is $\lambda$ is an eigenvalue of the map in $\operatorname{Aut} \left({V}\right): \vec v \mapsto g \vec v$.
Then by definition of an eigenvalue we have:
:$\exists \vec v_\lambda \in V : g \vec v_\lambda = \lambda \vec v_\lambda$
Let $n$ be the order of $g$ in $G$.
Then:
Thus:
:$\vec v_\lambda = \lambda^n \vec v_\lambda$
which means:
:$\lambda^n = 1$
and thus by definition, $\lambda$ is an $n$th root of unity.
Eigenvalues of Hermitian Operator are Real

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A \in \map B \HH$ be a Hermitian operator.
Then all eigenvalues of $A$ are real.
== Proof ==
Let $\lambda$ be an eigenvalue of $A$.
Let $v \in H$ be an eigenvector for $\lambda$.
That is:
:$A v = \lambda v$
Now compute:
Now $v$, being an eigenvector, is non-zero.
By property $(5)$ of an inner product, this implies $\innerprod v v \ne 0$.
Thence, dividing out $\innerprod v v$, obtain $\lambda = \bar \lambda$.
From Complex Number equals Conjugate iff Wholly Real, $\lambda \in \R$.
== Sources ==
* $\text {II}.5.8$
Category:Hermitian Operators
Eigenvalues of Hermitian Operator have Orthogonal Eigenspaces

== Theorem ==
The eigenvectors of a Hermitian operator have eigenspaces which are orthogonal.
== Proof ==
Directly follows from Hermitian Operator is Normal and Eigenvalues of Normal Operator have Orthogonal Eigenspaces.
Category:Linear Algebra
Category:Linear Operators
Category:Linear Transformations on Hilbert Spaces
Eigenvalues of Normal Operator have Orthogonal Eigenspaces

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $\mathbf T: \HH \to \HH$ be a normal operator.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $\mathbf T$.
Then:
:$\map \ker {\mathbf T - \lambda_1} \perp \map \ker {\mathbf T - \lambda_2}$
where:
:$\ker$ denotes kernel
:$\perp$ denotes orthogonality.
== Proof ==
Requisite knowledge: $\mathbf T^*$ is the adjoint of $\mathbf T$ and is defined by the fact that for any $\mathbf u, \mathbf w \in \HH$, we have
:$\innerprod {\mathbf {T u} } {\mathbf w} = \innerprod {\mathbf u} {\mathbf T^* \mathbf w}$
It is important to note the existence and uniqueness of adjoint operators.
Claim: We know that for $\mathbf v \in \HH$:
:$\mathbf {T v} = \lambda \mathbf v \iff \mathbf T^* \mathbf v = \overline \lambda \mathbf v$
This is true because for all normal operators, by definition:
:$\mathbf T^* \mathbf T = \mathbf T {\mathbf T^*}$
and so:
Since $\mathbf T$ is normal, $\mathbf T - \lambda \mathbf I$ is also normal.
Thus:
Let $\mathbf v_1$ and $\mathbf v_2$ be non-zero eigenvectors of $\mathbf T$ with corresponding eigenvalues $\lambda_1$ and $\lambda_2$, respectively.
Then:
Since $\lambda_1 \ne \lambda_2$, this is only possible if $\innerprod {\mathbf v_1} {\mathbf v_2} = 0$, which means the eigenvectors of our normal operator are orthogonal.
== Sources ==
* : $\text{II}.5.7$
Category:Linear Transformations on Hilbert Spaces
Eigenvalues of Schur Decomposition

== Theorem ==
Let $\mathbf A$ be a square matrix.
Let $\mathbf A = \mathbf Q \mathbf T \mathbf Q^*$ be a Schur decomposition for $\mathbf A$.
Then the eigenvalues of $\mathbf A$ appear on the main diagonal of $\mathbf T$.
== Proof ==
== Sources ==
* 
Category:Schur Decompositions
Category:Square Matrices
Eigenvalues of Symmetric Matrix are Real

== Corollary to Hermitian Matrix has Real Eigenvalues ==
Every real symmetric matrix has eigenvalues which are all real numbers.
== Proof ==
We have that a Real Symmetric Matrix is Hermitian.
The result follows from Hermitian Matrix has Real Eigenvalues.
== Examples ==
== Sources ==
* 
Category:Eigenvalues of Symmetric Matrix are Real
Category:Hermitian Matrix has Real Eigenvalues
Category:Symmetric Matrices
Category:Real Matrices
Category:Eigenvalues
Eigenvalues of Symmetric Matrix are Real/Examples/Arbitrary Example 1

== Examples of Eigenvalues of Symmetric Matrix are Real ==
Let $\mathbf A$ be the real symmetric matrix defined as:
:$\mathbf A = \begin {pmatrix} 1 & -2 & 5 \\ -1 & 6 & -1 \\ 5 & -2 & 1 \end {pmatrix}$
The eigenvalues of $\mathbf A$ are:
Their corresponding eigenvectors are:
== Proof ==
== Sources ==
* 
Category:Eigenvalues of Symmetric Matrix are Real
Eigenvectors of Symmetric Matrix are Orthogonal

== Theorem ==
Let $K$ be a ring. 
Let $A$ be a symmetric matrix over $K$.
Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $A$.
Let $\mathbf v_1, \mathbf v_2$ be eigenvectors of $A$ corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$ respectively.
Let $\innerprod \cdot \cdot$ be the dot product on $K$. 
Then $\mathbf v_1$ and $\mathbf v_2$ are orthogonal with respect to $\innerprod \cdot \cdot$.
== Proof ==
We have:
:$A \mathbf v_1 = \lambda_1 \mathbf v_1$
and:
:$A \mathbf v_2 = \lambda_2 \mathbf v_2$
We also have: 
and:
We therefore have: 
:$0 = \paren {\lambda_2 - \lambda_1} \innerprod {\mathbf v_1} {\mathbf v_2}$
Since $\lambda_1 \ne \lambda_2$, we have:
:$\innerprod {\mathbf v_1} {\mathbf v_2} = 0$
hence $\mathbf v_1$ and $\mathbf v_2$ are orthogonal with respect to $\innerprod \cdot \cdot$.
Category:Eigenvectors
Category:Symmetric Matrices
Eight Convex Deltahedra

== Theorem ==
There exist exactly $8$ distinct convex deltahedra:
:$4$ faces: regular tetrahedron
:$6$ faces: triangular bipyramid
:$8$ faces: regular octahedron
:$10$ faces: pentagonal bipyramid
:$12$ faces: snub disphenoid (split a regular tetrahedron into two wedges and join them with a band of $8$ equilateral triangles)
:$14$ faces: triaugmented triangular prism (attach $3$ square pyramids to a triangular prism)
:$16$ faces: gyroelongated square bipyramid (attach $2$ square pyramids to a square antiprism)
:$20$ faces: regular icosahedron.
== Proof ==
== Sources ==
* : $8$
* : $8$
Category:Deltahedra
Einstein's Law of Motion

== Physical Law ==
The force and acceleration on a body of constant rest mass are related by the equation:
:$\mathbf F = \dfrac {m_0 \mathbf a} {\paren {1 - \dfrac {v^2} {c^2} }^{\tfrac 3 2} }$
where:
:$\mathbf F$ is the force on the body
:$\mathbf a$ is the acceleration induced on the body
:$v$ is the magnitude of the velocity of the body
:$c$ is the speed of light
:$m_0$ is the rest mass of the body.
== Proof ==
Into Newton's Second Law of Motion:
:$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$
we substitute Einstein's Mass-Velocity Equation:
:$m = \dfrac {m_0} {\sqrt {1 - \dfrac {v^2} {c^2} } }$
to obtain:
:$\mathbf F = \map {\dfrac \d {\d t} } {\dfrac {m_0 \mathbf v} {\sqrt {1 - \dfrac {v^2} {c^2} } } }$
Then we perform the differentiation time:
Thus we arrive at the form:
:$\mathbf F = \dfrac {m_0 \mathbf a} {\paren {1 - \dfrac{v^2} {c^2} }^{\tfrac 3 2} }$
== Comment ==
Thus we see that at low velocities (that is, much less than that of light), the well-known equation $\mathbf F = m \mathbf a$ holds to a high degree of accuracy.
== Sources ==
* : Chapter $\text {B}.7$: A Simple Approach to $E = M c^2$
Category:Physics
Einstein's Mass-Energy Equation

== Theorem ==
The energy imparted to a body to cause that body to move causes the body to increase in mass by a value $M$ as given by the equation:
:$E = M c^2$
where $c$ is the speed of light.
== Proof ==
From Einstein's Law of Motion, we have:
:$\mathbf F = \dfrac {m_0 \mathbf a} {\paren {1 - \dfrac {v^2} {c^2} }^{\tfrac 3 2} }$
where:
:$\mathbf F$ is the force on the body
:$\mathbf a$ is the acceleration induced on the body
:$v$ is the magnitude of the velocity of the body
:$c$ is the speed of light
:$m_0$ is the rest mass of the body.
, assume that the body is starting from rest at the origin of a cartesian plane.
Assume the force $\mathbf F$ on the body is in the positive direction along the $x$-axis.
To simplify the work, we consider the acceleration as a scalar quantity and write it $a$.
Thus, from the Chain Rule for Derivatives:
:$a = \dfrac {\d v} {\d t} = \dfrac {\d v} {\d x} \dfrac {\d x} {\d t} = v \dfrac {\d v} {\d x}$
Then from the definition of energy:
:$\ds E = \int_0^x F \rd x$
which leads us to:
== Also known as ==
== Also see ==
* Einstein's Mass-Velocity Equation
== Historical Note ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $32 \ \text{(b)}$
* : Chapter $\text {B}.7$: A Simple Approach to $E = M c^2$
* 
* 
* 
* 
* 
* 
Category:Einstein's Mass-Energy Equation
Category:Mass
Category:Energy
Category:Relativistic Mechanics
Category:Physics
Einstein's Mass-Energy Equation/Also known as

== Einstein's Mass-Energy Equation: Also known as ==
Einstein's Mass-Energy Equation is usually known as Einstein's Equation, but there are a number of such equations that deduced.
However, this is the most famous one, and has caught the imagination of the general public.
Hence, if you refer to Einstein's Equation at a party, for example, everyone will know which one you mean, and it's this one.
Some sources refer to it boringly as the Mass-Energy Equation.
== Sources ==
* : Chapter $\text {B}.7$: A Simple Approach to $E = M c^2$
* 
* 
* 
* 
* 
* 
Category:Einstein's Mass-Energy Equation
Eisenstein's Conjecture

== False Conjecture ==
Every natural number of the form:
:$2^2 + 1, 2^{2^2} + 1, 2^{2^{2^2} } + 1, \ldots$
is prime.
== Refutation ==
$2^{2^{16} } + 1$ is composite:
:$2^{2^{2^{2^2} } } + 1 = 2^{2^{16} } + 1 = 825 \, 753 \, 601 \times 188 \, 981 \, 757 \, 975 \, 021 \, 318 \, 420 \, 037 \, 633 \times c$
where $c$ is a composite number with $19 \, 694$ digits.
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Famous False Conjectures
Eisenstein's Conjecture/Mistake

== Source Work ==
:
:Eisenstein, Ferdinand Gotthold Max (1823-52)
== Mistake ==
:German mathematician who proposed the still unproved conjecture that numbers of the form $2^2 + 1, 2^{2^2} + 1, 2^{2^{2^2} } + 1, \ldots$ are prime.
== Correction ==
This conjecture had been disproved in $1953$ by the discovery that $2^{2^{16} } + 1$ is composite, $45$ years before the publication of this dictionary.
This entry was updated by the $4$th edition, in which the above information was presented.
== Sources ==
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Eisenstein's Lemma

== Theorem ==
Let $p$ be an odd prime.
Let $a \in \Z$ be an odd integer such that $p \nmid a$.
Let $\paren {\dfrac a p}$ be defined as the Legendre symbol.
Then:
:$\paren {\dfrac a p} = \paren {-1}^{\map \alpha {a, p} }$
where:
:$\ds \map \alpha {a, p} = \sum_{k \mathop = 1}^{\frac {p - 1} 2} \floor {\frac {k a} p}$
:$\floor {\dfrac {k a} p}$ is the floor function of $\dfrac {k a} p$.
== Proof ==
We will borrow some ideas and techniques from the proof of Gauss's Lemma.
Let $S = \set {a, 2 a, 3 a, \ldots, \dfrac {p - 1} 2 a}$.
Let us create $S'$ from $S$ by replacing each element of $S$ by its least positive residue modulo $p$.
Arranging $S'$ into increasing order, we get:
:$S' = \set {b_1, b_2, \ldots, b_m, c_1, c_2, \ldots, c_n}$
where $b_m < \dfrac p 2 < c_1$ and $m + n = \dfrac {p - 1} 2$.
According to the Division Theorem, we can divide any $k a \in S$ by $p$ and get:
:$k a = p \times \floor {\dfrac {k a} p} + r$
where $r \in S'$.
Let $k$ run from $1$ to $\dfrac {p - 1} 2$ and add up the resulting equations.
We get:
:$(1): \quad \ds \sum_{k \mathop = 1}^{\frac {p - 1} 2} k a = \sum_{k \mathop = 1}^{\frac {p - 1} 2} p \times \floor {\frac {k a} p} + \sum_{i \mathop = 1}^{m} b_i + \sum_{j \mathop = 1}^n c_j$
As in the proof of Gauss's Lemma, all the remainders are in fact distinct elements of $S'$.
From the proof of Gauss's Lemma again, we have that:
:$S = \set {b_1, b_2, \ldots, b_m, p - c_1, p - c_2, \ldots, p - c_n} = \set {1, 2, 3, \ldots, \dfrac {p - 1} 2}$
Adding up all the elements of $S$ we have:
:$\ds \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = \sum_{i \mathop = 1}^m b_i + \sum_{j \mathop = 1}^n \paren {p - c_j} = \sum_{i \mathop = 1}^m b_i + p n - \sum_{j \mathop = 1}^n c_j$
Subtracting this equation from $(1)$ above gives us:
:$\ds \sum_{k \mathop = 1}^{\frac {p - 1} 2} k a - \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = \sum_{k \mathop = 1}^{\frac {p - 1} 2} p \times \floor {\frac {k a} p} + 2 \sum_{j \mathop = 1}^n c_j - p n$
That is:
:$\ds \paren {a - 1} \sum_{k \mathop = 1}^{\frac {p - 1} 2} k = p \times \map \alpha {a, p} + 2 \sum_{j \mathop = 1}^n c_j - p n$
Now we are going to write this equation modulo $2$.
Note that:
:$\ds \paren {a - 1} \sum_{k \mathop = 1}^{\frac {p - 1} 2} k$ is even, as $a$ is odd
:$\ds 2 \sum_{j \mathop = 1}^n c_j$ is also of course even.
So:
:$p \map \alpha {a, p} - p n \equiv 0 \pmod 2$
Since $p$ is odd, this amounts to saying:
:$\map \alpha {a, p} \equiv n \pmod 2$
From the conclusion of Gauss's Lemma:
:$\paren {\dfrac a p} = \paren {-1}^n$
Hence the result.
Category:Number Theory
Eisenstein Integers form Integral Domain

== Theorem ==
The ring of Eisenstein integers $\struct {\Z \sqbrk \omega, +, \times}$ is an integral domain.
== Proof ==
By Eisenstein Integers form Subring of Complex Numbers we know that $\struct {\Z \sqbrk \omega, +, \times}$ is a subring of the complex numbers $\C$.
Let $1_\C$ be the unity of $\C$.
Let $1_\omega$ be the unity of $\Z \sqbrk \omega$.
By the Subdomain Test it suffices to show that $1_\C = 1_\omega$.
By Unity of Ring is Unique it suffices to show that $1_\C$ is a unity of $\Z \sqbrk \omega$.
First we note that:
:$\Z \sqbrk \omega = \set {a + b\omega: a, b \in \Z}$
In particular:
:$1_\C \in \Z \sqbrk \omega$
Moreover, by definition, $\Z \sqbrk \omega$ inherits its ring product from $\C$.
For any $\alpha \in \Z \sqbrk \omega$:
:$1_\C \alpha = \alpha 1_\C = \alpha$
in $\C$.
Therefore this identity holds in $\Z \sqbrk \omega$ as well.
Category:Number Theory
Category:Examples of Integral Domains
Category:Eisenstein Integers
Eisenstein Integers form Subring of Complex Numbers

== Theorem ==
The set of Eisenstein integers $\Z \sqbrk \omega$, under the operations of complex addition and complex multiplication, forms a subring of the set of complex numbers $\C$.
== Proof ==
We will use the Subring Test.
This is valid, as the set of complex numbers $\C$ forms a field, which is by definition itself a ring.
We note that $\Z \sqbrk \omega$ is not empty, as (for example) $0 + 0 \omega \in \Z \sqbrk \omega$.
Let $a + b \omega, c + d \omega \in \Z \sqbrk \omega$.
Then we have $-\paren {c + d \omega} = -c - d \omega$, and so:
We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain.
Therefore by definition $\Z$ is a ring.
So it follows that $a - c \in \Z$ and $b - d \in \Z$.
Hence $\paren {a - c} + \paren {b - d} \omega \in \Z \sqbrk \omega$.
Now consider $\paren {a + b \omega} \paren {c + d \omega}$.
By the definition of complex multiplication, we have:
:$\paren {a + b \omega} \paren {c + d \omega} = \paren {a c - b d} + \paren {a d + b c} \omega$
As $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that $a c - b d \in \Z$ and $ad + bc \in \Z$.
Hence:
:$\paren {a + b \omega} \paren {c + d \omega} \in \Z \sqbrk \omega$
So by the Subring Test, $\Z \sqbrk \omega$ is a subring of $\C$.
== Also see ==
* Definition:Ring of Eisenstein Integers
Category:Examples of Integral Domains
Category:Subrings
Category:Complex Numbers
Category:Eisenstein Integers
Either-Or Topology is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a compact space.
== Proof ==
Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$.
So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets.
So $\CC$ has a subcover containing at most three sets.
Hence $T$ is a compact space by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $2$
Category:Compact Spaces
Category:Either-Or Topology
Either-Or Topology is First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a first-countable space.
== Proof ==
Let $x \in S$ such that $x \ne 0$.
Then $\set x$ is open in $T$ and so on its own forms a local basis of $x$ which is (trivially) countable.
Let $x = 0$.
Let $U \in \tau$ be open in $T$ such that $x \in U$.
Then by definition of the either-or space, $U$ contains the open set $\openint {-1} 1$.
So $\openint {-1} 1$ forms a local basis of $0$ which is (trivially) countable.
Hence the result by definition of first-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $3$
Category:First-Countable Spaces
Category:Either-Or Topology
Either-Or Topology is Lindelöf/Proof 1

== Theorem ==
== Proof ==
We have:
:Either-Or Topology is Compact
:Compact Space is Lindelöf
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $2$
Category:Either-Or Topology is Lindelöf
Either-Or Topology is Lindelöf/Proof 2

== Theorem ==
== Proof ==
Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$.
So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets.
So $\CC$ has a subcover containing at most three sets.
Hence $T$ is a Lindelöf space by definition.
Category:Either-Or Topology is Lindelöf