nyuuzyou/wikis huggingface dataset:
Category of Relations is Category

== Theorem ==
Let $\mathbf{Rel}$ be the category of relations.
Then $\mathbf{Rel}$ is a metacategory.
== Proof ==
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
We have already defined the composition of morphisms.
For any set $X$, we have the diagonal relation $\operatorname{id}_X$.
By Diagonal Relation is Left Identity and Diagonal Relation is Right Identity it follows that this is the identity morphism for $X$.
Finally by Composition of Relations is Associative, the associative property is satisfied.
Hence $\mathbf{Rel}$ is a metacategory.
== Sources ==
* : $\S 1.9$: Exercise $1 \ \text{(a)}$
Category:Category of Relations
Category of Semilattices is Category

== Theorem ==
Let $\mathbf{SLat}$ denote the category of semilattices.
Then:
:$\mathbf{SLat}$ is a metacategory
== Proof ==
Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
For any two semilattice homomorphisms their composition (in the usual set theoretic sense) is again a semilattice homomorphism by Composite Semilattice Homomorphisms is Semilattice Homomorphism.
For any semilattice $\struct{S, \circ}$, we have the identity mapping $\operatorname{id}_S$.
From Identity Mapping is Semilattice Homomorphism we have $\operatorname{id}_S$ is a semilattice homomorphism.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity, this is the identity morphism for $\struct{S, \circ}$.
Finally by Composition of Mappings is Associative, the associative property is satisfied.
Hence $\mathbf{SLat}$ is a metacategory.
== Sources ==
* : Chapter $\text I$: Preliminaries, $\S 3.2$ Definition
Category:Category of Semilattices
Category of Sets is Category

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Then $\mathbf{Set}$ is a metacategory.
== Proof ==
Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
For any two mappings their composition (in the usual set theoretic sense) is again a mapping by Composite Mapping is Mapping.
For any set $X$, we have the identity mapping $\operatorname{id}_X$.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity, this is the identity morphism for $X$.
Finally by Composition of Mappings is Associative, the associative property is satisfied.
Hence $\mathbf{Set}$ is a metacategory.
Category:Examples of Categories
Category:Category of Sets
Category of Subobject Classes is Category

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\overline {\mathbf {Sub} }_{\mathbf C} } C$ be the category of subobject classes of $C$.
:$g \circ f : X \to Z$
called the composition of $f$ and $g$.
Then $\map {\overline {\mathbf {Sub} }_{\mathbf C} } C$ is a metacategory.
== Proof ==
Let us verify the axioms $(\text C 1)$ to $(\text C 3)$ for a metacategory.
=== $(\text C 1)$ Composition ===
We have that for every objects $X, Y, Z$ and morphisms $X \stackrel {f} {\longrightarrow} Y \stackrel {g} {\longrightarrow} Z$ with $\operatorname{cod} f = \operatorname{dom} \, g$, there exists a morphism inherited from $\map {\mathbf{Sub}_{\mathbf C} } C$:
:$g \circ f : X \to Z$
such that:
:$\eqclass g {} \circ \eqclass f {} := \eqclass {g \circ f} {}$
where $\eqclass {g \circ f} {}$ is a morphism class under $\sim$ for $g \circ f$.
Suppose there exists morphisms $f_1: m_1 \to n_1$ and $f_2: m_2 \to n_2$ of $\map {\mathbf{Sub}_{\mathbf C} } C$:
:$f_1 \sim f_2$ $m_1 \sim_S m_2$ and $n_1 \sim_S n_2$
Suppose there exists morphisms $g_1: m_1 \to n_1$ and $g_2: m_2 \to n_2$ of $\map {\mathbf{Sub}_{\mathbf C} } C$:
:$g_1 \sim g_2$ $m_1 \sim_S m_2$ and $n_1 \sim_S n_2$
where $\sim_S$ denotes the equivalence of subobjects.
Then:
:$g_1 \circ f_1 \sim g_2 \circ f_2$ $m_1 \sim_S m_2$ and $n_1 \sim_S n_2$
So:
Therefore:
:$\eqclass g {} \circ \eqclass f {} := \eqclass {g \circ f} {}$
is well-defined.
Hence the result.
=== $(\text C 2)$ Identity ===
Let the morphism class of the identity morphism of $m$ in $\map {\mathbf{Sub}_{\mathbf C} } C$ be defined by:
:$\operatorname{id}_{\eqclass m {} } := \eqclass {\operatorname{id}_m} {}$
It remains to be shown that the above operation is well-defined.
We have that for every object $X$, there is a morphism ${\operatorname{id}_X}: X \to X$, called the identity morphism, such that:
:$f \circ {\operatorname{id}_X} = f$, and ${\operatorname{id}_X} \circ g = g$
for any object $Y$ and morphisms $f : X \to Y$ and $g : Y \to X$
=== $(\text C 3)$ Associativity ===
We have that for any three morphisms $f, g, h$:
:$f \circ \paren {g \circ h} = \paren {f \circ g} \circ h$
whenever these compositions are defined (according to $(\text C 1)$).
Hence the result.
Hence the result.
Category:Categories of Subobjects
Category of Subobject Classes is Order Category

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\overline{\mathbf{Sub}}_{\mathbf C} \left({C}\right)$ be the category of subobject classes of $C$.
Then $\overline{\mathbf{Sub}}_{\mathbf C} \left({C}\right)$ is an order category.
== Proof ==
Category:Categories of Subobjects
Category:Order Categories
Category of Subobjects is Category

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf{Sub}_{\mathbf C}} C$ be the category of subobjects of $C$.
Then $\map {\mathbf{Sub}_{\mathbf C}} C$ is a metacategory.
== Proof ==
Let us verify the axioms $(C1)$ up to $(C3)$ for a metacategory.
Let $f: m_1 \to m_2$ and $g: m_2 \to m_3$ be morphisms of $\map {\mathbf{Sub}_{\mathbf C}} C$.
That $g \circ f: m_1 \to m_3$ is again a morphism follows from the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}\xymatrix@+2em{
\operatorname{dom} m_1
\ar[r]_*+{f}
\ar[rd]_*+{m_1}
\ar@/^1pc/[rr]^*+{g \circ f}
&
\operatorname{dom} m_2
\ar[r]_*+{g}
\ar[d]^*+{m_2}
&
\operatorname{dom} m_3
\ar[ld]^*+{m_3}
\\
&
C
}\end{xy}$
For every subobject $m$ of $C$, the diagram:
$\quad\quad \begin{xy}\xymatrix{
\operatorname{dom} m
\ar[r]^*+{\operatorname{id}_m}
\ar[rd]_*+{m}
&
\operatorname{dom} m
\ar[d]^*+{m}
\\
&
C
}\end{xy}$
in $\mathbf C$ commutes, for $\operatorname{id}_m = \operatorname{id}_{\operatorname{dom} m}$, and the latter is an identity morphism in $\mathbf C$.
This last property also immediately proves the required behaviour of $\operatorname{id}_m$ with respect to other morphisms of $\map {\mathbf{Sub}_{\mathbf C}} C$.
Since the composition is inherited from $\mathbf C$, it is necessarily associative.
Hence $\map {\mathbf{Sub}_{\mathbf C}} C$ is a metacategory.
== Also see ==
* Category of Subobjects is Preorder Category
Category:Categories of Subobjects
Category of Subobjects is Preorder Category

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf{Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.
Then $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.
== Proof ==
By Category of Subobjects is Category, we know $\map {\mathbf{Sub}_{\mathbf C} } C$ is a metacategory.
By definition of preorder category, it suffices to show that if $f, g: m \to m'$ are morphisms with the same domain and codomain, then $f = g$.
The situation is sketched by the following commutative diagram in $\mathbf C$:
$\quad\quad \begin{xy}\xymatrix@+1em{
\Dom m
\ar[r]<2pt>^*+{f}
\ar[r]<-2pt>_*+{g}
\ar[rd]_*+{m}
&
\Dom {m'}
\ar[d]^*+{m'}
\\
&
C
}\end{xy}$
Thus, we see that $m' \circ f = m = m' \circ g$.
Now $m'$ is a subobject, and a fortiori a monomorphism.
Hence $f = g$, and $\map {\mathbf{Sub}_{\mathbf C} } C$ is a preorder category.
== Also see ==
* Inclusion Relation on Subobjects
== Sources ==
* : $\S 5.1$
Category:Categories of Subobjects
Category:Preorder Categories
Catenary is Symmetric about Y-Axis

== Theorem ==
Consider a catenary $\CC$.
Let a cartesian plane be arranged so that the $y$-axis passes through the lowest point of the catenary.
$\CC$ exhibits reflectional symmetry in that $y$-axis.
</onlyinclude>
== Proof ==
From Cartesian Equation of Catenary: Formulation $2$, we have the equation of $\CC$:
:$y = \dfrac a 2 \paren {e^{x / a} + e^{-x / a} } = a \cosh \dfrac x a$
The result follows directly from Hyperbolic Cosine Function is Even.
== Sources ==
* 
* 
Category:Catenary
Catesian Product of Open Real Intervals is Open in Real Number Plane

== Theorem ==
Let $\struct {\R^2, d}$ denote the real number plane with the Euclidean metric.
Let $\openint a b$ and $\openint c d$ be open real intervals.
Then their Cartesian product:
:$S = \openint a b \times \openint c d$
is an open set of $\struct {\R^2, d}$.
== Proof ==
Let $P = \tuple {x, y} \in S$.
Then by definition:
:$a < x < b$
and:
:$c < y < d$
Let $\epsilon = \min \set {x - a, b - x, y - c, d - y}$
By definition, $\epsilon > 0$.
Consider the open ball $\map {B_\epsilon} P$.
Let $Q = \tuple {x_0, y_0} \in \map {B_\epsilon} P$ be an arbitrary point in $\map {B_\epsilon} P$.
By definition of $\map {B_\epsilon} P$:
:$\size {x - x_0} < \epsilon$
and:
:$\size {y - y_0} < \epsilon$
Hence we have:
:$\size {x - x_0} < x - a$
:$\size {x - x_0} < b - x$
:$\size {y - y_0} < y - c$
:$\size {y - y_0} < d - y$
and so:
:$Q \in S$
Thus we have that:
:$\map {B_\epsilon} P \subseteq S$
Hence the result by definition of open set.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 11$
Category:Open Sets
Cauchy's Convergence Criterion/Complex Numbers

== Theorem ==
Let $\sequence {z_n}$ be a complex sequence.
Then $\sequence {z_n}$ is a Cauchy sequence it is convergent.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Cauchy's Convergence Criterion on Real Numbers
Category:Convergent Complex Sequences
Category:Cauchy Sequences
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Complex Numbers/Lemma 1

== Lemma for Cauchy's Convergence Criterion on Complex Numbers ==
Let $\sequence {z_n}$ be a complex sequence.
Let $\NN$ be the domain of $\sequence {z_n}$.
Let $x_n = \map \Re {z_n}$ for every $n \in \NN$.
Let $y_n = \map \Im {z_n}$ for every $n \in \NN$.
Then $\sequence {z_n}$ is a (complex) Cauchy sequence $\sequence {x_n}$ and $\sequence {y_n}$ are (real) Cauchy sequences.
== Proof ==
=== Necessary Condition ===
Let $\sequence {z_n}$ be a Cauchy sequence.
This means that, for a given $\epsilon > 0$:
:$\exists N: \forall m, n \in \NN: m, n \ge N: \cmod {z_n - z_m} < \epsilon$
We have, for every $m, n \ge N$:
Thus $\sequence {x_n}$ is a Cauchy sequence by definition.
A similar argument shows that $\sequence {y_n}$ is a Cauchy sequence.
=== Sufficient Condition ===
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences.
This means for $\sequence {x_n}$ that, for a given $\epsilon > 0$:
:$\exists N_1: \forall m, n \in \NN: m, n \ge N_1: \cmod {x_n - x_m} < \dfrac \epsilon 2$
Also, for $\sequence {y_n}$:
:$\exists N_2: \forall m, n \in \NN: m, n \ge N_2: \cmod {y_n - y_m} < \dfrac \epsilon 2$
Let $N = \max \paren {N_1, N_2}$.
Let $i = \sqrt {-1}$ denote the imaginary unit.
We have, for every $m, n \ge N$:
Thus $\sequence {z_n}$ is a Cauchy sequence by definition.
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Complex Numbers/Proof 1

== Theorem ==
== Proof ==
=== Lemma ===
Let $\sequence {x_n}$ be a real sequence where:
:$x_n = \map \Re {z_n}$ for every $n$
:$\map \Re {z_n}$ is the real part of $z_n$
Let $\sequence {y_n}$ be a real sequence where:
:$y_n = \map \Im {z_n}$ for every $n$
:$\map \Im {z_n}$ is the imaginary part of $z_n$
=== Necessary Condition ===
Let $\sequence {z_n}$ be a Cauchy sequence.
We aim to prove that $\sequence {z_n}$ is convergent.
We find:
:$\sequence {z_n}$ is a Cauchy sequence
:$\leadsto \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Lemma
:$\leadsto \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Cauchy's Convergence Criterion on Real Numbers
:$\leadsto \sequence {z_n}$ is convergent by definition of convergent complex sequence.
=== Sufficient Condition ===
Let $\sequence {z_n}$ be convergent.
We aim to prove that $\sequence {z_n}$ is a Cauchy sequence.
We find:
:$\sequence {z_n}$ is convergent
:$\leadsto \sequence {x_n}$ and $\sequence {y_n} $ are convergent by definition of convergent complex sequence
:$\leadsto \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Cauchy's Convergence Criterion on Real Numbers
:$\leadsto \sequence {z_n}$ is a Cauchy sequence by Lemma
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Complex Numbers/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
Let $\sequence {x_n}$ be a real sequence where:
:$x_n = \map \Re {z_n}$ for every $n$
:$\map \Re {z_n}$ is the real part of $z_n$
Let $\sequence {y_n}$ be a real sequence where
:$y_n = \map \Im {z_n}$ for every $n$
:$\map \Im {z_n}$ is the imaginary part of $z_n$
We find:
:$\sequence {z_n}$ is a Cauchy sequence
:$\leadstoandfrom \sequence {x_n}$ and $\sequence {y_n}$ are Cauchy sequences by Lemma
:$\leadstoandfrom \sequence {x_n}$ and $\sequence {y_n}$ are convergent by Cauchy's Convergence Criterion on Real Numbers
:$\leadstoandfrom \sequence {z_n}$ is convergent by definition of convergent complex sequence
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/General

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$ or $\C$.
Then $\sequence {x_n}$ is a Cauchy sequence :
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall r \in \N: r \ge N: \forall k > 0: \size {\ds \sum_{i \mathop = 1}^k x_{r + i} } < \epsilon$
== Proof ==
== Sources ==
* 
* 
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Then $\sequence {x_n}$ is a Cauchy sequence $\sequence {x_n}$ is convergent.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
The conditions are shown to be equivalent.
Hence the result.
== Also known as ==
== Also see ==
* Real Number Line is Banach Space
* Cauchy's Convergence Criterion on Complex Numbers
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.9$
* 
Category:Real Analysis
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Necessary Condition

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be convergent.
Then $\sequence {x_n}$ is a Cauchy sequence.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be convergent.
Let $\struct {\R, d}$ be the metric space formed from $\R$ and the usual (Euclidean) metric:
:$\map d {x_1, x_2} = \size {x_1 - x_2}$
where $\size x$ is the absolute value of $x$.
This is proven to be a metric space in Real Number Line is Metric Space.
From Convergent Sequence in Metric Space is Cauchy Sequence, we have that every convergent sequence in a metric space is a Cauchy sequence.
Hence $\sequence {x_n}$ is a Cauchy sequence.
== Also known as ==
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.9$
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Necessary Condition/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be a sequence in $\R$ that converges to the limit $l \in \R$.
Let $\epsilon > 0$. 
Then also $\dfrac \epsilon 2 > 0$.
Because $\sequence {x_n}$ converges to $l$, we have:
:$\exists N: \forall n > N: \size {x_n - l} < \dfrac \epsilon 2$
So if $m > N$ and $n > N$, then:
Thus $\sequence {x_n}$ is a Cauchy sequence.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.9$
* : Appendix: $\S 18.4$: Subsequences
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be a Cauchy sequence.
Then $\sequence {x_n}$ is convergent.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also known as ==
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
We have the result Real Number Line is Metric Space.
Hence by Convergent Subsequence of Cauchy Sequence, it is sufficient to show that $\sequence {a_n}$ has a convergent subsequence.
Since $\sequence {a_n}$ is Cauchy, by Real Cauchy Sequence is Bounded, it is also bounded.
By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence.
Hence the result.
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded.
By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence $\sequence {a_{n_r} }$.
Let $a_{n_r} \to l$ as $r \to \infty$.
It is to be shown that $a_n \to l$ as $n \to \infty$.
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then $\dfrac \epsilon 2 > 0$.
Hence:
:$(1): \quad \exists R \in \R: \forall r > R: \size {a_{n_r} - l} < \dfrac \epsilon 2$
We have that $\sequence {a_n}$ is a Cauchy sequence.
Hence:
:$(2): \quad \exists N \in \R: \forall m > N, n > N: \size {x_m - x_n} \le \dfrac \epsilon 2$
Let $n > N$.
Let $r \in \N$ be sufficiently large that:
:$n_r > N$
and:
:$r > R$
Then $(1)$ is satisfied, and $(2)$ is satisfied with $m = n_r$.
So:
So, given $\epsilon > 0$, we have found $n \in \R$ such that:
:$\forall n > N: \size {a_n - l} < \epsilon$
Thus:
:$x_n \to l$ as $n \to \infty$.
== Sources ==
* : $\S 5$: Subsequences: Cauchy sequences: $\S 5.19$
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 4

== Theorem ==
== Proof ==
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded.
Thus $\sequence {a_n}$ is both bounded above and bounded below.
Let us create a monotone subsequence $\sequence {b_n}$ of $\sequence {a_n}$ using the following construction:
For each $m \in \N$, let $S_m$ denote the set of elements of $\sequence {a_n}$ from $m$ onwards:
:$S_m = \set {a_n: n \ge m}$
From Real Cauchy Sequence is Bounded, we have that $S_m$ is also bounded.
Hence, by the Continuum Property, $\sup S_m$ exists.
Let $b_m := \sup S_m$.
Because $S_{m + 1} \subseteq S_m$, it follows from Supremum of Set of Real Numbers is at least Supremum of Subset that:
:$\sup S_{m + 1} \le \sup S_m$
Thus $\sequence {b_m}$ is decreasing.
By definition of $b_m$, we also have that:
:$b_m \ge a_m$
So, as $\sequence {a_n}$ is bounded below, so is $\sequence {b_m}$.
So, by the Monotone Convergence Theorem (Real Analysis), $\sequence {b_m}$ is convergent.
Let $\sequence {b_m}$ converge to $l$.
Let $\epsilon \in \R_{>0}$ be arbitrary.
We have that $\sequence {a_n}$ is a Cauchy sequence in $\R$.
We also have from Subsequence of Real Cauchy Sequence is Cauchy that $\sequence {b_n}$ is also a Cauchy sequence in $\R$.
Then there exists:
:$N_1 \in \N$ such that $\size {a_m - a_n} < \epsilon$ for $m, n \ge N_1$
:$N_2 \in \N$ such that $\size {l - b_n} < \epsilon$ for $m \ge N_2$
Let $N = \max \set {N_1, N_2}$.
By definition of $b_N$, we have that $b_N - \epsilon$ is not an upper bound of $S_N = \set {a_n: n \ge N}$.
Thus there exists $M \ge N$ such that $a_M > b_N - \epsilon$.
We note also that $a_M \le b_N$ because $b_N$ is an upper bound for $S_N$.
Now let $n \ge N$.
We have that:
Hence by Convergence by Multiple of Error Term we have that $\sequence {a_n}$ converges to $l$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.9$
Category:Cauchy's Convergence Criterion
Cauchy's Convergence Criterion for Series

== Theorem ==
A series $\ds \sum_{i \mathop = 0}^\infty a_i$ is convergent for every $\epsilon > 0$ there is a number $N \in \N$ such that:
:$\size {a_{n + 1} + a_{n + 2} + \cdots + a_m} < \epsilon$
holds for all $n \ge N$ and $m > n$.
== Proof ==
Let:
:$\ds s_n = \sum_{i \mathop = 0}^n a_i$
 
Then $\sequence {s_n}$ is a sequence in $\R$. 
From Cauchy's Convergence Criterion on Real Numbers, $\sequence {s_n}$ is convergent it is a Cauchy sequence.
For $m > n$ we have:
:$\size {s_m - s_n} = \size {a_{n + 1} + a_{n + 2} + \cdots + a_m}$
Category:Cauchy's Convergence Criterion
Category:Real Analysis
Cauchy's Group Theorem

== Theorem ==
Let $G$ be a finite group whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $G$ has a subgroup of order $p$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* : Chapter $14$ Solubility and Simplicity: $\S3$ Cauchy's Theorem: Theorem $14.15$
* 
* 
* 
Category:Cauchy's Group Theorem
Category:Abelian Groups
Category:Subgroups
Category:Prime Groups
Cauchy's Group Theorem/Proof 1

== Theorem ==
== Proof ==
=== Lemma ===
Let $G$ be a finite abelian group whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $G$ has a subgroup of order $p$.
=== Proof of Lemma ===
Let $\order G$ be a prime number.
Then from Prime Group is Cyclic $G$ has a generator $\gen g$ such that $\order g = p$.
Now suppose $e \ne g \in G$.
Let $\order g = n$.
Let $p \divides n$.
Then by Subgroup of Finite Cyclic Group is Determined by Order the cyclic group $\gen g$ has an element $g$ of order $p$.
Suppose $p \nmid n$.
From Subgroup of Abelian Group is Normal, $\gen g$ is normal in $G$.
Consider the quotient group $G' = \dfrac G {\gen g}$.
As $p \nmid n$, $\gen e \subsetneq \gen g \subsetneq G$.
Thus $\order {G'} < \order G$.
But we have that $p \divides \order G$.
It follows by induction that $G'$ has an element $h'$ of order $p$.
Let $h$ be a preimage of $h'$ under the quotient epimorphism $\phi: G \to G'$.
Then:
:$\paren {h'}^p = e'$
where $e'$ is the identity of $G'$.
Thus $h^p \in \gen g$ so $\paren {h^p}^n = \paren {h^n}^p = e$.
Thus either $h^n$ has order $p$ or $h^n = e$.
If $h^n = e$ then $\paren {h'}^n = e'$.
But since $p$ is the order of $h'$ it would follow that $p \divides n$, contrary to assumption.
We prove the theorem by induction on the order $\order G$.
The first few cases $\order G = 1,2,3$ are obvious. For the induction step, start with the class equation
:$\order G = 1 + \order {C_2} + \cdots + \order {C_r}$
Since $p \divides \order G$, we must have $p \nmid \order {C_j}$ for some $j \geq 2$,
where $C_j$ are conjugacy classes in $G \setminus \set e$.
Let $x \in C_j$,
it follows that $p \divides \order {\map {C_G} x}$,
where $\map {C_G} x$ is the centralizer of $x$ in $G$,
since $\order {C_j} = \frac{\order G}{\order {\map {C_G} x}}$ by Number of Elements in Conjugacy Class Equals the Index of Centralizer.
If $\map {C_G} x \neq G$, then by induction $\map {C_G} x$ contains an element of order $p$, and this element also belongs to $G$.
Otherwise, $\map {C_G} x = G$, which implies that $x \in \map Z G$ (the center of $G$), and by choice $x \neq e$, so $\map Z G \neq \set e$.
Either $p \divides \order {\map Z G}$ or $p \nmid \order {\map Z G}$.
In the first case the proof reduces to lemma (the abelian case).
In the second case, by induction hypothesis to the quotient group $G / \map Z G$, there exists $x \in G$ such that the image $\bar{x} \in G / \map Z G$ has order $p$,
That is, $x^p \in \map Z G$ but $x \notin \map Z G$.
Let $X$ be the cyclic group generated by $x$.
Now the product $X \map Z G$ is abelian and has order divisible by $p$, so by lemma (the abelian case) it has an element of order $p$, and again this element also belongs to $G$.
This completes the induction step, and with it the proof.
== Sources ==
* : $\S 1.10$: Theorem $31$
Category:Cauchy's Group Theorem
Category:Proofs by Induction