nyuuzyou/wikis huggingface dataset:
General Logarithm/Examples/Base b of 1

== Theorem ==
Let $b \in \R_{>0}$ be a strictly positive real number such that $b \ne 1$.
Let $\log_b$ denote the logarithm to base $b$.
Then:
:$\log_b 1 = 0$
== Proof ==
By definition of logarithm:
== Sources ==
* : $0$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $17$
Category:Examples of General Logarithms
General Morphism Property for Groups

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group homomorphism.
Then:
:$\forall g_k \in H: \map \phi {g_1 \circ g_2 \circ \cdots \circ g_n} = \map \phi {g_1} * \map \phi {g_2} * \cdots * \map \phi {g_n}$
== Proof ==
A group is a semigroup.
The result then follows from the General Morphism Property for Semigroups.
== Sources ==
* : $\S 47 \ \text {(b)}$ Homomorphisms and their elementary properties
Category:Morphism Property
Category:Group Homomorphisms
General Morphism Property for Semigroups

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.
Let $\phi: S \to T$ be a homomorphism.
Then:
:$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
Hence it follows that:
:$\forall n \in \N_{>0}: \forall s \in S: \map \phi {s^n} = \paren {\map \phi s}^n$
== Proof ==
$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ can be proved by induction.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
$P(1)$ is true, as this just says:
:$\map \phi {s_1} = \map \phi {s_1}$.
=== Basis for the Induction ===
$P(2)$ is the case:
:$\map \phi {s_1 \circ s_2} = \map \phi {s_1} * \map \phi {s_2}$
This follows from the fact that $\phi$ is a homomorphism.
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:
:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k}$
Then we need to show:
:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k + 1} } = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k} * \map \phi {s_{k + 1} }$
=== 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:
:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
The result for $n \in \N_{>0}$ follows directly from the above, by replacing each occurrence of $s_k$ with $s$.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Example $133$
Category:Morphism Property
Category:Semigroup Homomorphisms
General Periodicity Property

== Theorem ==
Let $f: \R \to \R$ be a periodic real function.
Let $L$ be a periodic element of $f$.
Then:
:$\forall n \in \Z: \forall x \in X: \map f x = \map f {x + n L}$
That is, after every distance $L$, the function $f$ repeats itself.
=== Corollary ===
== Proof ==
There are two cases to consider: either $n$ is not negative, or it is negative.
Since the Natural Numbers are Non-Negative Integers, the case where $n \ge 0$ will be proved using induction.
=== Case 1 ===
==== Basis for the Induction ====
The case for which $n = 0$ is trivial, because:
:$x + 0 \cdot L = x$
==== Induction Hypothesis ====
For some $n \in \Z_{\ge 0}$, suppose that:
:$\map f x = \map f {x + n L}$
==== Induction Step ====
For the induction step, let $n \to n + 1$.
Then:
=== Case 2 ===
If $n < 0$, then:
Combining the results above, it is seen that for all $n \in \Z$:
:$\map f x = \map f {x + n L}$
Category:Periodic Functions
Category:Proofs by Induction
General Periodicity Property/Corollary

== Theorem ==
Let $f: \R \to \R$ be a real function.
Then $L$ is a periodic element of $f$ :
:$\forall x \in \R: \map f {x \bmod L} = \map f x$
where $x \bmod L$ is the modulo operation.
== Proof ==
=== Necessary Condition ===
Let $f: \R \to \R$ be a real function with a periodic element $L$.
Then:
=== Sufficient Condition ===
Let $f: \R \to \R$ be a real function such that for all $x \in \R$:
:$\map f {x \bmod L} = \map f x$
Let $n = \floor {\dfrac x L}$.
Then:
Category:Periodic Functions
General Positivity Rule in Ordered Integral Domain

== Theorem ==
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $S \subset D$ be a subset of $D$ such that:
:$\forall s \in S: \map P x$
Then the following are true:
:$\ds \forall n \in \N_{>0}: \forall s_i \in S: \map P {\sum_{i \mathop = 1}^n s_i}$
:$\ds \forall n \in \N_{>0}: \forall s_i \in S: \map P {\prod_{i \mathop = 1}^n s_i}$
where:
:$\ds \sum_{i \mathop = 1}^n s_i = s_1 + s_2 + \cdots + s_n$
:$\ds \prod_{i \mathop = 1}^n s_i = s_1 \times s_2 \times \cdots \times s_n$
That is, the ring sum and ring product of any number of elements in $D$ which have the (strict) positivity property also have the (strict) positivity property.
=== Corollary ===
== Proof ==
Trivially true for $n = 1$, and true by definition for $n = 2$.
Suppose it is true for all $n$ up to $n = k$ for some $k \in \N$.
Then:
:$\ds \forall s_i \in S: \map P {\sum_{i \mathop = 1}^k s_i}$
:$\ds \forall s_i \in S: \map P {\prod_{i \mathop = 1}^k s_i}$
Take any $\ds \sum_{i \mathop = 1}^{k + 1} s_i = \sum_{i \mathop = 1}^k s_i + s_{k + 1}$.
Then:
:$\ds \map P {\sum_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$
so:
:$\ds \map P {\sum_{i \mathop = 1}^{k + 1} s_i}$
because the (strict) positivity property holds for all $s_i \in S$.
Similarly:
:$\ds \map P {\prod_{i \mathop = 1}^k s_i}$ and $\map P {s_{k + 1} }$
so:
:$\ds \map P {\prod_{i \mathop = 1}^{k + 1} s_i}$
The result follows by the Principle of Complete Induction.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $11$
Category:Ordered Integral Domains
General Positivity Rule in Ordered Integral Domain/Corollary

== Corollary to General Positivity Rule in Ordered Integral Domain ==
Let $\struct {D, +, \times}$ be an ordered integral domain, whose (strict) positivity property is denoted $P$.
Let $\map P x$ where $x \in D$.
Then:
:$\map P {n \cdot x}$ and $\map P {x^n}$
== Proof ==
From the definition of power of an element:
:$\ds n \cdot x = \sum_{i \mathop = 1}^n x$
:$\ds x^n = \prod_{i \mathop = 1}^n x$
The result then follows directly from General Positivity Rule in Ordered Integral Domain.
Category:Ordered Integral Domains
General Reciprocity Law

== Theorem ==
:$\ds \sum_{0 \mathop \le j \mathop < \alpha n} \map f {\floor {\dfrac {m j} n} } = \sum_{0 \mathop \le r \mathop < \alpha m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} - \map f r} + \ceiling {\alpha n} \map f {\ceiling {\alpha m} - 1}$
for $\alpha \in \R$.
== Proof ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $46$
Category:Summations
Category:Floor Function
General Solution equals Particular Solution plus Complementary Function

== Theorem ==
Consider the linear first order ODE with constant coefficients:
:$(1): \quad \dfrac {\d y} {\d x} + a y = \map Q x$
The general solution to $(1)$ consists of:
:the particular solution to $(1)$ for which the constant of integration is $0$
plus:
:the complementary function to $(1)$.
== Proof ==
From Solution to Linear First Order ODE with Constant Coefficients, $(1)$ has the general solution:
:$\ds y = e^{-a x} \int e^{a x} \map Q x \rd x + C e^{-a x}$
Setting $C = 0$ we get:
:$\ds y = e^{-a x} \int e^{a x} \map Q x \rd x$
which is a particular solution to $(1)$
By definition, the complementary function to $(1)$ is the general solution to the reduced equation:
:$\dfrac {\d y} {\d x} + a y = 0$
From First Order ODE: $\dfrac {\d y} {\d x} = k y$, that general solution of $(1)$ is:
:$y = C e^{-a x}$
Hence the result.
== Warning ==
I am beginning to wonder whether the author of the source work from which this result is taken really understands the material he is presenting.
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 1$. The first order equation: $\S 1.3$ The form of the general solution
Category:Solution to Linear First Order ODE with Constant Coefficients
General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution

== Theorem ==
Consider the nonhomogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$
Let $\map {y_g} x$ be the general solution of the homogeneous linear second order ODE:
:$(2): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
Let $\map {y_p} x$ be a particular solution of $(1)$.
Then $\map {y_g} x + \map {y_p} x$ is the general solution of $(1)$.
== Proof ==
Let $\map {y_g} {x, C_1, C_2}$ be a general solution of $(2)$.
Note that $C_1$ and $C_2$ are the two arbitrary constants that are to be expected of a second order ODE.
Let $\map {y_p} x$ be a certain fixed particular solution of $(1)$.
Let $\map y x$ be an arbitrary particular solution of $(1)$.
Then:
We have that $\map {y_g} {x, C_1, C_2}$ is a general solution of $(2)$.
Thus:
:$\map y x - \map {y_p} x = \map {y_g} {x, C_1, C_2}$
or:
:$\map y x = \map {y_g} {x, C_1, C_2} + \map {y_p} x$
== Sources ==
* : Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.2$ The general equation
* : $\S 3.14$: Second Order Linear Equations: Introduction: Theorem $\text{B}$
Category:Linear Second Order ODEs
General Solution of Riccati Equation from Particular Solution

== Theorem ==
Consider the Riccati equation:
:$(1): \quad y' = \map p x + \map q x y + \map r x y^2$
Let $\map {y_1} x$ be a particular solution to $(1)$.
Then the general solution to $(1)$ has the form:
:$\map y x = \map {y_1} x + \map z x$
where $\map z x$ is the general solution to the Bernoulli equation:
:$z' - \paren {q - 2 r y_1} z = r z^2$
== Proof ==
Let $\map y x = \map {y_1} x + \map z x$ be a particular solution to $(1)$.
Then:
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Problem $28$
Category:Riccati Equation
General Solution to Mathieu's Equation

== Theorem ==
The general solution to Mathieu's equation:
:$\dfrac {\d^2 y} {\d x^2} + \paren {a + b \cos 2 x} y = 0$
is:
:$A e^{r x} \map \phi x + B e^{-r x} \map \phi {-x}$
where $r$ is a constant and $\phi$ a periodic function of period $2 \pi$.
== Proof ==
== Sources ==
* 
* 
Category:Mathieu's Equation
General Sphere is Subspace of Closed Ball of Greater Dimension

== Theorem ==
Let $S^{n - 1}$ be an $\paren {n - 1}$-sphere of the real Euclidean space $\R^n$.
Then $S^{n - 1}$ is a subspace of the closed Euclidean ball $\map { {B_1}^-} 0$ of $\R^n$.
== Proof ==
== Examples ==
== Sources ==
* 
* 
Category:General Sphere is Subspace of Closed Ball of Greater Dimension
Category:Closed Balls
Category:Spheres (Topology)
General Sphere is Subspace of Closed Ball of Greater Dimension/Examples/Dimension 1

== Examples of Use of General Sphere is Subspace of Closed Ball of Greater Dimension ==
Consider the closed ball $\map { {\overline B}_1} 0$ in the real Euclidean space $\R^2$.
The $1$-sphere $S^1$ is the closed real interval $\closedint {-1} 1$, and is a subspace of $\map { {\overline B}_1} 0$.
== Sources ==
* 
* 
Category:General Sphere is Subspace of Closed Ball of Greater Dimension
General Stokes' Theorem

== Theorem ==
Let $\omega$ be a smooth $\paren {n - 1}$-form with compact support on a smooth $n$-dimensional oriented manifold $X$.
Let the boundary of $X$ be $\partial X$.
Then:
:$\ds \int_{\partial X} \omega = \int_X \rd \omega$
where $\d \omega$ is the exterior derivative of $\omega$.
== Proof ==
=== Special Case ===
Let there exist a chart:
:$x = \tuple {x_1, \ldots, x_n}: V \subseteq X \to \R^n$
such that:
:$\map \supp \omega \subseteq V$
where:
:$\map \supp \omega = \overline {\set {p \in M : \map \omega p \ne 0} }$
We may suppose that $V$ is relatively compact.
Thus, by composing $x$ with a translation, we may suppose that:
:$\ds \map x V \subseteq \mathbb H^n = \set {\tuple {x_1, \ldots, x_n} \in \R^n : x_1 < 0}$
We have, in the coordinates $x$:
:$\ds \omega = \sum_{i \mathop = 1}^n f_i \rd x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$
The forms:
:$\hat {\d x}_i := \d x_1 \wedge \cdots \wedge \hat {\d x}_i \wedge \cdots \wedge \d x_n$ 
vanish on the tangent space to $\mathbb H^n$ for $i > 1$
Hence we have:
:$(1): \quad \ds \int_{\partial \mathbb H^n} \omega = \int_{\partial \mathbb H^n} f_1 \hat {\d x}_1$
Moreover:
so that:
:$\ds \int_{\mathbb H^n} \rd \omega = \sum_{i \mathop = 1}^n \int_{\mathbb H^n} \frac {\partial f_i} {\partial x_i} \rd x_1 \wedge \cdots \wedge \d x_n$
Let $i > 1$.
Then:
Let $i = 1$.
Then:
So:
:$\ds \int_{\mathbb H^n} \rd \omega = \int_{\partial \mathbb H^n} f_1 \, \hat {\d x}_1$
Together with $(1)$, this establishes the result.
=== General Case ===
Let $\omega$ be a smooth $\paren {n - 1}$-form with compact support on a smooth $n$-dimensional oriented manifold $X$.
Choose a finite family of relatively compact charts $V_1, \ldots, V_k$ on $X$ such that:
:$\ds \map \supp \omega \subseteq \bigcup_{i \mathop = 1}^k V_i$
Choose a partition of unity:
:$\chi_1, \ldots, \chi_k$
with $\chi_1 + \cdots + \chi_k = 1$ subordinate to the cover $\set {V_1, \ldots, V_k}$.
Put:
:$\omega_i = \chi_i \omega$
Then we have:
Moreover, $\map \supp {\omega_i} \subset V_i$ by definition.
Let the boundary of $X$ be $\partial X$.
Therefore, by the special case above, Stokes' theorem holds for each $\omega_i$.
Hence we have:
:$\ds \int_X \rd \omega = \sum^k_{i \mathop = 1} \int_x \rd \omega_i = \sum^k_{i \mathop = 1} \int_{\partial X} \omega_i = \int_{\partial X} \omega$
== Also see ==
* Implications of Stokes' Theorem
== Sources ==
* 
* 
Category:Stokes' Theorem
Category:Integral Calculus
Category:Manifolds
General Variation of Integral Functional/Dependent on N Functions

== Theorem ==
Let $J$ be a (real) functional of the form:
$\ds J \sqbrk {\ldots y_i \ldots} = \int_{x_0}^{x_1} \map F {x, \ldots y_i \ldots, \ldots y_i' \ldots} \rd x, i = \openint 1 n$
Then:
$\ds \delta J = \int_{x_0}^{x_1} \sum_{i \mathop = 1}^n \paren {F_{y_i} - \frac \d {\d x} F_{y_i'} } \map {h_i} x + \intlimits {\sum_{i \mathop = 1}^n F_{y_i'} \delta y_i} {x \mathop = x_0} {x \mathop = x_1} + \intlimits {\paren {F - \sum_{i \mathop = 1}^n y_i'F_{y_i'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$
== Proof ==
Let $y = \map y x, y = \map {y^*} x$ be smooth real functions.
Let $\map h x = \map {y^*} x - \map y x$
Let the endpoints of the curve $y_i = \map {y_i} x, i = \openint 1 n$ be:
:$P_0 = \tuple {x_0, \dotsc, y_i^0 \dotsc}$
:$P_1 = \tuple {x_1, \dotsc, y_i^1 \dotsc}$
Let the endpoints of the curve $y_i = y_i^* = \map {y_i} x + \map {h_i} x, i = \openint 1 n$ be:
:$P_0^* = \tuple {x_0 + \delta x_0, \dotsc, y_i^0 + \delta y_i^0 \dotsc}$
:$P_1^* = \tuple {x_1 + \delta x_1, \dotsc, y_i^1 + \delta y_i^1 \dotsc}$
Note that the endpoints of both functions may not necessarily match, thus making the functions defined on different intervals.
We choose to extend both curves in such a way that:
:if there is a difference between original endpoints of intervals at the same end, then the curve that is not defined in the given interval is extended linearly by drawing a straight line along the tangent of the curve at the point, closest to that interval.
Now both functions $\map {y_i} x$ and $\map {y^*_i} x$ are defined in $\closedint {x_0} {x_1 + \delta x_1}$.
The corresponding variation $\delta J$ of $J \sqbrk {\ldots y_i \ldots}$ is defined as the expression which is:
:linear in $\delta x_0, \delta x_1$ and $h_i, h_i', y_i^0, y_i^1$ for $i = \openint 1 n$
and which:
:differs from the increment by a quantity of order higher than 1 relative to $\ds \sum_{i \mathop = 1}^n \map \rho {y_i, y_i^*}$.
Here $\map \rho {y_i,y_i^*}$ is defined as:
:$\map \rho {y_i, y_i^*} = \max \size {y_i - y_i^*} + \max \size {y_i' - {y_i^*}'} + \map {d_2} {P_0, P_0^*} + \map {d_2} {P_1, P_1^*}$
where $d_2$ is the Euclidean metric.
By using Taylor's theorem, $\Delta J$ can be rewritten as:
where $h_i$ terms were integrated by parts.
In the same manner, $\map h x$ can be expanded as:
:$\ds \map {h_i} {x_0} = \delta y_1^0 - \map {y_i'} {x_0} \delta x_0 + \map \OO {\map \rho {y, y + h} }$
where $\map \OO {\map {\rho} {y, y + h} }$ is big-O notation.
:$\map {h_i} {x_1} = \delta y_1^1 - \map {y_i'} {x_1} \delta x_1 + \map \OO {\map \rho {y, y + h} }$
Thus:
where:
:$\bigvalueat {\delta x} {x \mathop = x_j} = \delta x_j$
:$\bigvalueat {\delta y_i} {x \mathop = x_j} = \delta y_i^j, j \in \set {0, 1}$
== Sources ==
* : $\S 3.13$: Derivation of the Basic Formula
Category:Calculus of Variations
General Variation of Integral Functional/Dependent on N Functions/Canonical Variables

== Theorem ==
Let $\delta J$ be a general variation of integral functional dependent on n functions.
Suppose a following coordinate transformation is done:
:$\set {x, \ldots y_i, \ldots, \ldots, y_i', \ldots, F} \to \set {x, \ldots, y_i, \ldots, \ldots p', \ldots, H}, i = \tuple {1, \ldots, n}$
Then, in canonical variables:
:$\ds \delta J = \int_{x_0}^{x_1} \sum_{i \mathop = 1}^n \paren {F_{y_i} - \dfrac {\d {p_i} } {\d x} } \map {h_i} x \rd x + \intlimits {\sum_{i \mathop = 1}^n p_i \delta y_i - H \delta x} {x \mathop = x_0} {x \mathop = x_1}$
where:
== Proof ==
== Sources ==
* : $\S 3.13$: Derivation of the Basic Formula
Category:Calculus of Variations
General Vector Solution of Fundamental Matrix

== Theorem ==
Let $\map \Phi t$ be a fundamental matrix of the system $x' = \map A t x$.
Then:
:$\map \Phi t c$ is a general solution of $x' = \map A t x$.
== Proof ==
By definition, $\map \Phi t$ is non-singular, and therefore has an inverse $\map {\Phi^{-1} } t$.
If $z$ is an arbitrary solution, then $\map \Phi t \, \map {\Phi^{-1} } {t_0} \, \map z {t_0}$ also solves the system and has the same initial condition.
Hence by Existence and Uniqueness Theorem for 1st Order IVPs $\map \Phi t \, \map {\Phi^{-1} } {t_0} \, \map z {t_0}$ equals $z$.
Letting $c = \map {\Phi^{-1} } {t_0} \, \map z {t_0}$ finishes the proof.
Category:Differential Equations
Generalization to Dudeney's Distribution Problem

== Theorem ==
Suppose $n > 1$ people, each with $a_1, a_2, \dots, a_n$ dollars, redistribute their wealth as follows:
Going in order of their index, each person doubles everyone else's money by giving out their own money.
If they all have the same amount of money after the process, the least (strictly) positive integral solution is:
:$a_j = 2^{j - 1} n + 1, \quad j = 1, \dots, n$
and each person will end up with $2^n$ dollars each.
== Proof ==
Let us define the total amount of money $S = \ds \sum_{j \mathop = 1}^n a_j$.
Consider the wealth of the $j$th person.
They would see their wealth doubled $j - 1$ times:
:$a_j \to 2^{j - 1} a_j$
Then they would have to give out money to double everyone else's money:
:$2^{j - 1} a_j \to 2^{j - 1} a_j - \paren {S - 2^{j - 1} a_j} = 2^j a_j - S$
since the total amount of money never changes.
Finally, their money will be doubled for $n - j$ more times:
:$2^j a_j - S \to 2^{n - j} \paren {2^j a_j - S} = 2^n a_j - 2^{n - j} S$
Since everyone ends up with the same amount of money, we have:
Since $a_1 \in \mathbb N$ and $n 2^{n - 1} + 1$ is coprime to both $n$ and $2$ for $n > 1$, by Euclid's Lemma, $S$ must be a multiple of $n 2^n$.
Hence the least positive integral solution is attained at $S = n2^n$.
This shows that $a_j = 2^{j - 1} n + 1$ is the least positive integral solution, and each person will end up with $\dfrac S n = 2^n$ dollars.
== Also see ==
* Henry Ernest Dudeney/Puzzles and Curious Problems/11 - Distribution: Case of $n = 9$
* Henry Ernest Dudeney/Modern Puzzles/12 - A Weird Game: Case of $n = 7$
Generalized Continuum Hypothesis

== Hypothesis ==
The Generalized Continuum Hypothesis is the proposition:
Let $x$ and $y$ be infinite sets.
Suppose:
:$\phi_1: x \to y$ is injective
and:
:$\phi_2: y \to \powerset x$ is injective
Then:
:$y \sim x$ or $y \sim \powerset x$
In other words, there are no infinite cardinals between $x$ and $\powerset x$.
== Notation ==
== Historical Note ==
== Sources ==
* : Chapter $1$: General Background: $\S 5$ The continuum problem
Category:Set Theory
Category:Named Theorems
Category:Open Questions
Category:Continuum Hypothesis
Generalized Eigenvalues as Roots of Equation

== Theorem ==
Let $\mathbf A$ be a square matrix of order $n$.
Let $\lambda$ be a generalized eigenvalue of $\mathbf A$.
Then:
:$\map \det {\mathbf A - \lambda \mathbf B} = 0$
where:
:$\mathbf B$ is another square matrix of order $n$
:$\det$ denotes the determinant.
== Proof ==
By definition of generalized eigenvalue:
We have that $\mathbf x$ is non-zero .
Hence from Matrix is Singular iff Product with Non-Zero Vector is Zero, the matrix $\mathbf A - \lambda \mathbf B$ is singular.
Hence $\map \det {\mathbf A - \lambda \mathbf B} = 0$ by definition of singular matrix.
== Sources ==
* 
Category:Generalized Eigenvalues
Generalized Hilbert Sequence Space is Metric Space

== Theorem ==
Let $\alpha$ be an infinite cardinal number.
Let $I$ be an indexed set of cardinality $\alpha$.
Let $H^\alpha$ be the generalized Hilbert sequence space of weight $\alpha$.
Then:
:$H^\alpha$ is a metric space.
== Proof ==
Recall $H^\alpha$ is the structure $\struct{A, d_2}$ where:
:$A$ is the set of all real-valued functions $x : I \to \R$ such that:
::$(1)\quad \set{i \in I: x_i \ne 0}$ is countable
::$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.
:$d_2: A \times A \to \R$ is the real-valued function defined as:
::$\ds \forall x = \family {x_i}, y = \family {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$
=== $d_2$ is Well-defined ===
Let $H = \struct{\ell^2, d_{\ell^2}}$ denote the Hilbert sequence space, where:
:$\ell^2$ denotes the real $2$-sequence space, that is, the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{n \mathop = 0}^\infty x_n^2$ is convergent
:$d_{\ell^2}$ denotes the real-valued function $d_{\ell^2}: \ell^2 \times \ell^2: \to \R$ defined as:
::$\ds \forall x = \sequence {x_n}, y = \sequence {y_n} \in \ell^2: \map {d_{\ell^2}} {x, y} := \paren {\sum_{n \mathop = 0}^\infty \paren {x_n - y_n}^2}^{\frac 1 2}$
=== Lemma ===
=== $d_2$ satisfies Metric Space Axiom $(M1)$ ===
=== $d_2$ satisfies Metric Space Axiom $(M2)$ ===
=== $d_2$ satisfies Metric Space Axiom $(M3)$ ===
=== $d_2$ satisfies Metric Space Axiom $(M4)$ ===
Category:Generalized Hilbert Sequence Spaces
Category:Generalized Hilbert Sequence Space is Metric Space
Generalized Hilbert Sequence Space is Metric Space/Metric Space Axiom M1

== Theorem ==
Let $\alpha$ be an infinite cardinal number.
Let $I$ be an indexed set of cardinality $\alpha$.
Let $A$ be the set of all real-valued functions $x : I \to \R$ such that:
:$(1)\quad \set{i \in I: x_i \ne 0}$ is countable
:$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.
Let $d_2: A \times A \to \R$ be the real-valued function defined as:
:$\ds \forall x = \family {x_i}, y = \family {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$
Then $d_2$ satisfies .
== Proof ==
Let $H = \struct{\ell^2, d_{\ell^2}}$ denote the Hilbert sequence space, where:
:$\ell^2$ denotes the real $2$-sequence space, that is, the set of all real sequences $\sequence {x_n}$ such that the series $\ds \sum_{n \mathop = 0}^\infty x_n^2$ is convergent
:$d_{\ell^2}$ denotes the real $2$-sequence metric, that is, the real-valued function $d_{\ell^2}: \ell^2 \times \ell^2: \to \R$ defined as:
::$\ds \forall x = \sequence {x_n}, y = \sequence {y_n} \in \ell^2: \map {d_{\ell^2}} {x, y} := \paren {\sum_{n \mathop = 0}^\infty \paren {x_n - y_n}^2}^{\frac 1 2}$
=== Lemma ===
Let $x \in A$.
From Lemma:
:$\exists y \in \ell^2 :$
::$\map {d_2} {x, x} = \map {d_{\ell^2}} {y, y}$
We have:
The result follows.
Category:Generalized Hilbert Sequence Space is Metric Space