nyuuzyou/wikis huggingface dataset:
Fundamental Theorem of Calculus/First Part/Proof 1

== Theorem ==
== Proof ==
To show that $F$ is a primitive of $f$ on $\closedint a b$, we need to establish the following:
:$F$ is continuous on $\closedint a b$
:$F$ is differentiable on the open interval $\openint a b$
:$\forall x \in \closedint a b: \map {F'} x = \map f x$.
=== Proof that $F$ is Continuous ===
We have that $f$ is continuous on $\closedint a b$.
It follows from Continuous Image of Closed Interval is Closed Interval that $f$ is bounded on $\closedint a b$.
Suppose that:
:$\forall t \in \closedint a b: \size {\map f t} < \kappa$
Let $x, \xi \in \closedint a b$.
From Sum of Integrals on Adjacent Intervals for Continuous Functions‎, we have that:
:$\ds \int_a^x \map f t \rd t + \int_x^\xi \map f t \rd t = \int_a^\xi \map f t \rd t$
That is:
:$\ds \map F x + \int_x^\xi \map f t \rd t = \map F \xi$
So:
:$\ds \map F x - \map F \xi = -\int_x^\xi \map f t \rd t = \int_\xi^x \map f t \rd t$
From :
:$\size {\map F x - \map F \xi} < \kappa \size {x - \xi}$
Thus it follows that $F$ is continuous on $\closedint a b$.
=== Proof that $F$ is Differentiable and $f$ is its Derivative ===
It is now to be shown that that $F$ is differentiable on $\openint a b$ and that:
:$\forall x \in \closedint a b: \map {F'} x = \map f x$
Let $x, \xi \in \closedint a b$ such that $x \ne \xi$.
Then:
Now, let $\epsilon > 0$.
If $\xi \in \openint a b$, then $f$ is continuous at $\xi$.
So for some $\delta > 0$:
:$\size {\map f t - \map f \xi} < \epsilon$
provided $\size {t - \xi} < \delta$.
So provided $\size {x - \xi} < \delta$ it follows that:
:$\size {\map f t - \map f \xi} < \epsilon$
for any $t$ in an interval whose endpoints are $x$ and $\xi$.
So from :
provided $0 < \size {x - \xi} < \delta$.
But that is what this means:
:$\dfrac {\map F x - \map F \xi} {x - \xi} \to \map f \xi$ as $x \to \xi$
So $F$ is differentiable on $\openint a b$, and:
:$\forall x \in \closedint a b: \map {F'} x = \map f x$
== Sources ==
* : $\S 13.12$
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/First Part/Proof 2

== Theorem ==
== Proof ==
Suppose $\Delta x > 0$ and both $x$ and $x + \Delta x$ are in the closed interval $\closedint a b$.
By hypothesis, $f$ is continuous on the closed interval $\closedint a b$
Thus it is also continuous on the closed interval $\closedint x {x + \Delta x}$.
Thus the conditions of the Mean Value Theorem for Integrals are fulfilled.
So, by the Mean Value Theorem for Integrals, there exists some $k \in \closedint x {x + \Delta x}$ such that:
:$\ds \int_x^{x + \Delta x} \map f x \rd x = \map f k \paren {x + \Delta x - x} = \map f k \Delta x$
Then:
By the definition of $k$:
:$x \le k \le x + \Delta x$
which means that $k \to x$ as $\Delta x \to 0$.
So:
For $\Delta x < 0$, consider $k \in \closedint {x + \Delta x} x$, and the argument is similar.
Hence the result, by the definition of primitive.
== Sources ==
* : $\S 4.4$
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/Historical Note

== Historical Note on the Fundamental Theorem of Calculus ==
In $1668$, published Geometriae Pars Universalis, in which the Fundamental Theorem of Calculus first makes its appearance, although only for a limited class of functions.
It is believed that the earliest complete statement and proof was made by .
This can be seen in a letter to from $1676$ or $1677$, collected as item $190$ of .
 is also cited by some as being the first to establish it.
 himself, in his own turn, claimed to have made the same startling realisation on reading 's $1658$ work .
In 's $1684$ article Nova Methodus pro Maximis et Minimis, published in , he takes this result as given, stating that $\int$ and $\d$ are each other's converse, with no attempt at proof.
== Sources ==
* : Chapter $\text{VI}$: On the Seashore
* : Chapter $\text{VII}$: Master of All Trades
* : $\S 3$: Appendix $\text B$: Newton: Footnote $1$
* : Chapter $\text {A}.18$: Newton (): Footnote $2$
* : Chapter $\text {A}.19$: Leibniz ()
Category:Historical Notes
Fundamental Theorem of Calculus/Motivation

== Motivation for Fundamental Theorem of Calculus ==
It can be seen that, to all intents and purposes, the first part and the second part of the Fundamental Theorem of Calculus are converses of each other.
What it tells us is that, in general, to work out the value of a definite integral, we do not have to flog through the difficult and tedious work of calculating it from first principles.
All we need to do is work out the formula for the antiderivative.
This of course can only be done if the function in question does in fact have an antiderivative.
In cases where it does not, or it can not be calculated, then it may well be necessary to go back to first principles after all.
However, what it does allow us to do is to define such functions as definite integrals, for example:
* the error function:
::$\map {\erf} x = \ds \dfrac 2 {\sqrt \pi} \int_0^x \map \exp {-t^2} \rd t$
* the sine integral function
::$\map \Si x = \ds \int_{t \mathop \to 0}^{t \mathop = x} \frac {\sin t} t \rd t$
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $2$ Fundamental Theorem of the Calculus
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/Second Part

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then:
:$(1): \quad f$ has a primitive on $\closedint a b$
:$(2): \quad$ If $F$ is any primitive of $f$ on $\closedint a b$, then:
:::$\ds \int_a^b \map f t \rd t = \map F b - \map F a = \bigintlimits {\map F t} a b$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
== Sources ==
* : $\text {II}$. Calculus: Definite Integrals
* : $\S 15$: Definition of a Definite Integral: $15.2$
* 
* : $\S 18$: Definite Integrals: Definition of a Definite Integral: $18.2$
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/Second Part/Also presented as

== Theorem ==
The Fundamental Theorem of Calculus can also be presented in the following form:
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ expressed as a normal first order ODE:
:$(1): \quad \dfrac {\d y} {\d x} = \map f x$
Let $c \in \R$ be an arbitrary real number.
Then there exists a unique solution $\map F x$ to $(1)$ on $\closedint a b$ such that $\map F a = c$, given by the definite integral:
:$\ds \map F x = c + \int_a^x \map f t \rd t$
== Sources ==
* : Chapter $1$ First-Order Differential Equations: $2$ Fundamental Theorem of the Calculus
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/Second Part/Proof 1

== Theorem ==
== Proof ==
Let $G$ be defined on $\closedint a b$ by:
:$\ds \map G x = \int_a^x \map f t \rd t$
We have:
:$\ds \map G a = \int_a^a \map f t \rd t = 0$ from Integral on Zero Interval
:$\ds \map G b = \int_a^b \map f t \rd t$ from the definition of $G$ above.
Therefore, we can compute:
By the first part of the theorem, $G$ is a primitive of $f$ on $\closedint a b$.
By Primitives which Differ by Constant‎, we have that any primitive $F$ of $f$ on $\closedint a b$ satisfies $\map F x = \map G x + c$, where $c$ is a constant.
Thus we conclude:
== Sources ==
* : $\S 13.14$
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/Second Part/Proof 2

== Theorem ==
== Proof ==
As $f$ is continuous, by the first part of the theorem, it has a primitive. Call it $F$.
$\closedint a b$ can be divided into any number of closed subintervals of the form $\closedint {x_{k - 1} } {x_k}$ where:
:$a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$
Fix such a finite subdivision of the interval $\closedint a b$; call it $P$.
Next, we observe the following telescoping sum identity:
Because $F' = f$, $F$ is differentiable.
By Differentiable Function is Continuous, $F$ is also continuous.
Therefore we can apply the Mean Value Theorem on $F$.
It follows that in every closed subinterval $I_i = \closedint {x_{i - 1} } {x_i}$ there is some $c_i$ such that:
:$\map {F'} {c_i} = \dfrac {\map F {x_i} - \map F {x_{i - 1} } } {x_i - x_{i - 1} }$ 
It follows that:
From the definitions of supremum and infimum, we have for all $i$ (recall $I_i = \closedint {x_{i - 1} } {x_i}$):
:$\displaystyle \inf_{x \mathop \in I_i} \map f x \le \map f {c_i} \le \sup_{x \mathop \in I_i} \map f x$
From the definitions of upper Darboux sum and lower Darboux sum, we conclude for any finite subdivision $P$:
:$\displaystyle \map L P \le \sum_{i \mathop = 1}^k \map f {c_i} \paren {x_i - x_{i - 1} } \le \map U P$
Lastly, from the definition of a definite integral and from $(2)$, we conclude:
:$\displaystyle \map F b - \map F a = \int_a^b \map f t \rd t$
== Sources ==
* : $\S 4.4$
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus/Second Part/Proof 3

Let $\closedint a b$ be the closed (real) interval.
We claim that closed (real) interval is a smooth 1-dimensional oriented manifold.
By Classification of Compact One-Manifolds, every compact connected 1-dimensional manifold is diffeomorphic to either a circle or a closed interval.
Therefore, the closed interval is a 1-dimensional manifold.
By Subset of Real Numbers is Interval iff Connected, since $\closedint a b$ is an interval of $\R$, it follows that $\closedint a b$ is connected.
Let $\F$ be a smooth 0-form with compact support on the $\closedint a b$.
Let the boundary of $\closedint a b$ be:
:$\partial \closedint a b$
Since the manifold is oriented, and has compact support, the integrals:
:$\ds \int_{\partial \closedint a b} \F$
and:
:$\int_{\closedint a b} \rd \F$
are well-defined.
Then, by General Stokes' Theorem:
:$\ds \int_{\partial \closedint a b} \F = \int_{\closedint a b} \rd \F$
where $\d \F$ is the exterior derivative of 0-form:
:$\F = \map f x dx$
It follows that:
as required.
Fundamental Theorem of Calculus for Complex Riemann Integrals

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $F, f: \closedint a b \to \C$ be complex functions.
Suppose that $F$ is a primitive of $f$.
Then the complex Riemann integral of $f$ satisfies:
:$\ds \int_a^b \map f t \rd t = \map F b - \map F a$
== Proof ==
Let $u, v: \closedint a b \times \set 0 \to \R$ be defined as in the Cauchy-Riemann Equations:
:$\map u {t, y} = \map \Re {\map F z}$
:$\map v {t, y} = \map \Im {\map F z}$
where:
:$\map \Re {\map F z}$ denotes the real part of $\map F z$
:$\map \Im {\map F z}$ denotes the imaginary part of $\map F z$.
Then:
== Sources ==
* : $\S 2.3$
Category:Complex Analysis
Category:Fundamental Theorem of Calculus
Fundamental Theorem of Calculus for Contour Integrals

== Theorem ==
Let $F, f: D \to \C$ be complex functions, where $D$ is a connected domain.
Let $C$ be a contour that is a concatenation of the directed smooth curves $C_1, \ldots, C_n$.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to D$ for all $k \in \set {1, \ldots, n}$.
Suppose that $F$ is an antiderivative of $f$.
If $C$ has start point $z$ and end point $w$, then:
:$\ds \int_C \map f z \rd z = \map F w - \map F z$
If $C$ is a closed contour, then:
:$\ds \oint_C \map f z \rd z = 0$
=== Corollary ===
== Proof ==
If $C$ is a closed contour, then $z = w$.
It follows that:
:$\map F w - \map F z = 0$
== Also see ==
* Primitive of Function on Connected Domain, for the converse of this result.
== Sources ==
* : $\S 2.3$
Category:Fundamental Theorem of Calculus for Contour Integrals
Category:Fundamental Theorem of Calculus
Category:Complex Contour Integrals
Category:Fundamental Theorems
Fundamental Theorem of Calculus for Contour Integrals/Corollary

== Theorem ==
Let $D \subseteq \C$ be an open set.
Let $f: D \to \C$ be a continuous function.
Suppose that $F: D \to \C$ is an antiderivative of $f$.
Let $\gamma: \closedint a b \to D$ be a contour that consists of one directed smooth curve.
Then the contour integral:
:$\ds \int_\gamma \map f z \rd z = \map F {\map \gamma b} - \map F {\map \gamma a}$
== Proof ==
By the chain rule:
:$\dfrac \d {\d t} \map F {\map \gamma t} = \map {F'} {\map \gamma t} \map {\gamma'} t = \map f {\map \gamma t} \map {\gamma'} t$
Thus:
Category:Fundamental Theorem of Calculus for Contour Integrals
Fundamental Theorem of Finite Abelian Groups

== Theorem ==
Every finite abelian group is an internal group direct product of cyclic groups whose orders are prime powers.
The number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.
== Proof ==
Let $G$ be a finite abelian group.
By means of Abelian Group is Product of Prime-power Order Groups, we factor it uniquely into groups of prime-power order.
Then Abelian Group of Prime-power Order is Product of Cyclic Groups applies to each of these factors.
Hence we conclude $G$ factors into prime-power order cyclic groups.
The factorisation of $G$ into prime-power order factors is already unique.
Therefore, a demonstration of the uniqueness of the secondary factorisation suffices:
Suppose $\order G = p^k$ with $p$ a prime.
Let $G$ have the following two factorisations:
:$G = H_1 \times H_2 \times \cdots \times H_m = K_1 \times K_2 \times \dotsb \times K_n$
where the $H_i$'s and $K_i$'s are non-trivial cyclic subgroups with
:$\order {H_1} \ge \order {H_2} \ge \dotsb \ge \order {H_m}|$
:$\order {K_1} \ge \order {K_2} \ge \dotsb \ge \order {K_n}$
We proceed through complete induction on $k$.
=== Basis for the induction ===
For $k = 1$, the statement follows from Prime Group is Cyclic.
=== Induction Hypothesis ===
Now we assume the theorem is true for all abelian groups of order $p^l$, where $l < k$.
=== Induction Step ===
By Power of Elements is Subgroup, $G^p = \set {x^p: x \in G}$ is a proper subgroup of $G$.
It follows that:
:$G^p = H_1^p \times \cdots \times H_{m'}^p = K_1^p \times \cdots \times K_{n'}^p$
where:
:$m'$ is the largest integer $i$ such that $\order {H_i} > p$
:$n'$ is the largest integer $j$ such that $\order {K_j} > p$
This $m'$ and $n'$ ensure that the direct products above do not have trivial factors.
Also, by Cauchy's Group Theorem, we have:
:$\order {G^p} < \order G$
This means that we can apply the induction hypothesis.
It follows that:
:$m' = n'$
and:
:$\order {H_i^p} = \order {K_i^p}$
for $i = 1, 2, \ldots, m'$.
We know that $\order {H_i} = p \order {H_i^p}$, and the same for $K_i$ and $K_i^p$.
It follows that $\order {H_i} = \order {K_i}$ for all $i = 1, 2, \ldots, m'$. 
Now by construction of $m'$:
:$i > m' \implies \order {H_i} = p = \order {K_i}$
It follows that we have:
:$\order {H_1} \order {H_2} \dotsm \order {H_{m'} } p^{m - m'} = \order G = \order {K_1} \order {K_2} \dotsm \order {K_{n'} } p^{n - n'}$
Therefore:
:$m - m' = n - n'$
and as $m' = n'$, it must follow that $m = n$.
The result follows by induction.
== Examples ==
== Also known as ==
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62 \zeta$
* 
Category:Fundamental Theorem of Finite Abelian Groups
Category:Fundamental Theorems
Category:Abelian Groups
Category:Proofs by Induction
Fundamental Theorem of Finite Abelian Groups/Examples/Order 240

== Examples of Use of Fundamental Theorem of Finite Abelian Groups ==
There are $5$ distinct abelian groups of order $240$:
:$C_{240}$
:$C_2 \times C_{120}$
:$C_4 \times C_{60}$
:$C_2 \times C_2 \times C_{60}$
:$C_2 \times C_2 \times C_2 \times C_{30}$
== Proof ==
== Sources ==
* 
Category:Fundamental Theorem of Finite Abelian Groups
Fundamental Theorem of Line Integrals

== Theorem ==
Let $\CC$ be a smooth curve given by the vector function $\map {\mathbf r} t$ for $a \le t \le b$.
Let $f$ be a differentiable function of two or three variables whose gradient vector $\nabla f$ is continuous on $\CC$.
Then:
:$\ds \int_\CC \nabla f \cdot d \mathbf r = \map f {\map {\mathbf r} b} - \map f {\map {\mathbf r} a}$
== Proof ==
Category:Vector Calculus
Category:Fundamental Theorems
Fundamental Theorem of Riemannian Geometry

== Theorem ==
Let $\struct {M, g}$ be a Riemannian or pseudo-Riemannian manifold with or without boundary.
Let $\nabla$ be a Levi-Civita connection.
Then $\nabla$ is unique.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Symmetric Connections
Category:Connections
Category:Fundamental Theorems
Fundamental Theorem of Symmetric Polynomials

== Theorem ==
Let $K$ be a field.
Let $f$ a symmetric polynomial over $K$.
Then $f$ can be written uniquely as a polynomial in the elementary symmetric polynomials.
== Proof ==
Let $p \in R := K \sqbrk {x_1, \dotsc, x_n}$ be a symmetric polynomial.
Let $\sigma_1, \dotsc, \sigma_n$ denote the $n$ elementary symmetric polynomials.
Let $<$ be the degree reverse lexicographic monomial order.
Let us denote by $\map {LM} f$ the leading monomial of any $f \in R$, that is to say the unique $<$-largest monomial appearing in $f$.
Then it is straightforward to verify that $\map {LM} {\sigma_k} = x_1 \cdots x_k$ for $1 \le k \le n$:
For $k < n$, any other monomial appearing in $\sigma_k$ must contain a factor $x_j^{\alpha_1}$ with $j > k$ and $\alpha_1 > 0$.
This makes it smaller than $x_1 \cdots x_k$ with respect to $<$.
The case $k = n$ is trivial because $\sigma_n$ consists of only one monomial.
It follows the proof that $p$ may be written as a polynomial in $\sigma_1, \dotsc, \sigma_k$.
Firstly, since $p$ is symmetric, it must have the property:
:$\map {LM} p = x_1^{a_1} \dotsm x_n^{a_n}$
with $a_1 \ge a_2 \ge \ldots \ge a_n$.
To see this, assume that $a_i > a_j$ for some $1 \le i < j \le n$.
The monomial:
:$x_1^{a_1} \dotsm x_{i - 1}^{a_{i - 1} } \cdot x_i^{a_j} x_j^{a_i} \cdot x_{j + 1}^{a_{j + 1} } \cdots x_n^{a_n}$
must also appear in $p$, and if $a_i > a_j$ then it would be larger than $\map {LM} p$, a contradiction.
Now consider:
:$\sigma := c \sigma_1^{a_1 - a_2} \cdot \sigma_2^{a_2 - a_3} \cdot \dotsm \cdot \sigma_{n - 1}^{a_{n - 1} - a_n} \cdot \sigma_n^{a_n}$
where $c$ is the coefficient of $\map {LM} p$ in $p$.
Define $\bar p := p - \sigma$.
Because:
the leading monomials in the difference $\bar p$ cancel.
Therefore:
:$\map {LM} {\bar p} < \map {LM} p$
Iterate this method, and after finitely many steps one has decomposed $p$ into a polynomial $F \in K \sqbrk {y_1, \dotsc, y_n}$ with $\map F {\sigma_1, \dotsc, \sigma_n} = p$.
The termination of this algorithm follows immediately from Dickson's Lemma.
Category:Elementary Symmetric Polynomials
Category:Symmetric Polynomials
Category:Fundamental Theorems
Fundamental Theorem on Equivalence Relations

== Theorem ==
Let $\RR \subseteq S \times S$ be an equivalence on a set $S$.
Then the quotient set $S / \RR$ of $S$ by $\RR$ forms a partition of $S$.
== Proof ==
To prove that $S / \RR$ is a partition of $S$, we have to prove:
:$(1): \quad \ds \bigcup {S / \RR} = S$
:$(2): \quad \eqclass x \RR \ne \eqclass y \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$
:$(3): \quad \forall \eqclass x \RR \in S / \RR: \eqclass x \RR \ne \O$
Taking each proposition in turn:
=== Union of Equivalence Classes is Whole Set ===
The set of $\RR$-classes constitutes the whole of $S$:
=== Equivalence Classes are Disjoint ===
=== Equivalence Class is not Empty ===
Thus all conditions for $S / \RR$ to be a partition are fulfilled.
== Examples ==
== Also see ==
* Relation Induced by Partition is Equivalence for the converse.
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : Chapter $0$: Relations: Theorem $6$
* : Chapter $\text {II}$: Topological Spaces: $\S 10$. Equivalence Relations
* : $\S 7$: Relations
* : $\S 1.2$: Theorem $1.10$
* : $\S 2.4$. Equivalence classes
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.1$
* : $\S 0.3$: Theorem $5$ Part $\text{I}$
* : $\S 1.4$: Decomposition of a set into classes. Equivalence relations: Theorem $4$
* : Appendix: Elementary set and number theory
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.4$
* : $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces
* : Appendix $\text{A}.3$: Equivalence Relations: Theorem $\text{A}.2$
* : $\S 17.5$: Equivalence classes
* : Chapter $1$: Preliminaries: $1.5$ Relations: Equivalence Relations
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
* : $\S 1$: Naive Set Theory: $\S 1.5$: Relations: Exercise $1.5.2$
* : Chapter $2$: Maps and relations on sets: Proposition $2.29$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $10 \ \text {(a)}$
* : $1$: Set Theory and Logic: $\S 3$: Relations
* : Appendix $\text{A}$: Set Theory: Equivalence Relations
* : $\S 2.3$: Relations: Theorem $2.3.1$
* : $\S 3.4$
* 
* 
Category:Equivalence Relations
Category:Quotient Sets
Category:Fundamental Theorems
Category:Fundamental Theorem on Equivalence Relations
Fundamental Theorem on Equivalence Relations/Examples/Arbitrary Equivalence on Set of 6 Elements 1

== Example of Use of Fundamental Theorem on Equivalence Relations ==
Let $S = \set {1, 2, 3, 4, 5, 6}$.
Let $\RR \subset S \times S$ be a relation on $S$ defined as:
:$\RR = \set {\tuple {1, 1}, \tuple {1, 2}, \tuple {1, 3}, \tuple {2, 1}, \tuple {2, 2}, \tuple {2, 3}, \tuple {3, 1}, \tuple {3, 2}, \tuple {3, 3}, \tuple {4, 4}, \tuple {4, 5}, \tuple {5, 4}, \tuple {5, 5}, \tuple {6, 6} }$
Then $\RR$ is an equivalence relation which partitions $S$ into:
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations
Category:Fundamental Theorem on Equivalence Relations
Fundamental Theorem on Equivalence Relations/Examples/Arbitrary Equivalence on Set of 6 Elements 2

== Example of Use of Fundamental Theorem on Equivalence Relations ==
Let $S = \set {1, 2, 3, 4, 5, 6}$.
Let $\RR \subset S \times S$ be an equivalence relation on $S$ with the properties:
Then the equivalence classes of $\RR$ are:
== Proof ==
We have that:
:$1 \mathrel \RR 3$ and $3 \mathrel \RR 4$
As $\RR$ is an equivalence relation, it follows that $\RR$ is transitive and so:
:$1 \mathrel \RR 4$
Thus:
:$\eqclass 1 \RR \subseteq \set {1, 3, 4}$
but as:
:$\size {\eqclass a \RR} = 3$
it follows that:
:$\eqclass 1 \RR= \set {1, 3, 4}$
There are $6$ elements of $S$.
Thus the other $3$ elements must all be in the same equivalence class of $\RR$ which does not contain $1$, for example.
Thus we have:
:$\eqclass 2 \RR \subseteq \set {2, 5, 6}$
and the information we were given that $2 \mathrel \RR 6$ was superfluous.
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations: Problem Set $\text{A}.3$: $14$
Category:Fundamental Theorem on Equivalence Relations
Furstenberg Topology is Topology

== Theorem ==
Let $\struct {\Z, \tau}$ be the topological space formed by the Furstenberg topology on the set of integers $\Z$.
Then $\tau$ is indeed a topology on $\Z$.
== Proof ==
Recall the definition of the Furstenberg topology:
In view of Union from Synthetic Basis is Topology it suffices to show that $\BB$ is a synthetic basis on $\Z$.
Recall the definition of synthetic basis:
=== $(\text B 1)$ ===
$\BB$ is trivially a cover of $\Z$, since $\Z \in \BB$.
=== $(\text B 2)$ ===
Let $a_1 \Z + b_1, a_2 \Z + b_2 \in \BB$.
If:
:$\paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2} = \O$
then it is done, since $\O = \bigcup \O$ and $\O \subseteq \BB$.
Now, suppose that:
:$\exists x \in \paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2}$
Let $\lcm \set {a_1, a_2}$ be the lowest common multiple of $a_1$ and $a_2$.
Then:
That is:
:$\paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2} = \lcm \set {a_1, a_2} \Z + x$
This concludes the proof, because:
:$\lcm \set {a_1, a_2} \Z + x = \bigcup \AA$
where:
:$\AA := \set {\lcm \set {a_1, a_2} \Z + x} \subseteq \BB$
== Also see ==
* Definition:Furstenberg Topology
Category:Furstenberg Topology
Futurama Theorem

== Theorem ==
Let $A_{n - 2} \subset A_n$ be a subgroup of the alternating group on $n$ letters $A_n$.
For any element $x \in A_{n - 2}$, let $x = x_1 x_2 \dots x_k$, where $x_i \in H$ is a transposition.
Then there exists $y$ which can be represented as a series of transpositions $y_1 y_2 \dots y_j \in A_n$ such that:
:$(1): \quad y x = z$, where $z$ contains no transpositions from $H$
:$(2): \quad y_a \ne x_b$ for all $a, b$.
== Proof ==
Let $w = (n [n - 1])$, that is, the transposition of the $n^{th}$ and $n - 1^{th}$ letters that we consider $A_n$ acting on.
Then the permutation $x^{-1} w$ is the $y$ of the theorem.
== Historical Note ==
Category:Permutations
Category:Named Theorems