nyuuzyou/wikis huggingface dataset:
First Sylow Theorem/Corollary/Proof 2

== Theorem ==
== Proof ==
Let $\order G = m p^n$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.
 
From Cardinality of Set of Subsets, this is given by:
:$N = \dbinom {p^n m} {p^n} = \dfrac {\paren {p^n m} \paren {p^n m - 1} \cdots \paren {p^n m - i} \cdots \paren {p^n m - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$
For $0 < i < p^n$, the highest power of $p$ which divides $m p^n - i$ is the same as that dividing $p^n - i$.
Thus the power of $p$ in the numerator and denominator of $N$ cancel out, except those of $m$.
Thus $m$ and $N$ have the same number of divisors which are power of $p$.
Let $p^r$ be the largest power of $p$ which divides $m$ and $N$.
Now let $G$ act on $\mathbb S$ by the rule:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
That is, $g * S$ is the left coset of $S$ by $g$.
From Group Action on Sets with k Elements, this is a group action.
If every orbit under $*$ were divisible by $p^{r - 1}$, then $p^{r + 1} \divides N$.
But we have stated that $p^r$ is the largest power of $p$ which divides $N$.
Therefore there is at least one orbit under $*$ which can be represented as:
:$\set {S_1, S_2, \ldots, S_k}$
for which $p^{r + 1} \nmid k$.
Let $H = \Stab {S_1}$ denote the stabilizer of $S_1$.
From the Orbit-Stabilizer Theorem, we have that:
:$k = \index G H = \dfrac {\order g} {\order H}$
As $p^r \divides m$, we have:
:$p^{n + r} \divides p^n m$
But:
:$p^n m = \order G = k \order H$
and so:
:$p^{n + r} \divides k \order H$
But we also have that:
:$p^{r + 1} \nmid k$
which means:
:$p^n \divides \order H$
and that:
:$p^n \le \order H$
Now because $H = \Stab {S_1}$, we haveL
:$H g \subseteq S_1$
for all $g \in \S_1$
Thus:
:$\order H = \card {H g} \le \card {S_1} = p^n$
So we have $p^n \le \order H$ and $\order H \le p^n$ which means:
:$\order H = p^n$
and so $H$ is the subgroup of $G$ whose existence we were to demonstrate.
== Sources ==
* 
* : Chapter $2$: The Sylow Theorems: $\S 56$. First Sylow Theorem
Category:First Sylow Theorem
First Sylow Theorem/Examples/Alternating Group on 4 Letters

== Example of Use of First Sylow Theorem ==
The Alternating Group on 4 Letters $A_4$ is of order $12 = 2^2 \times 3$.
Thus the First Sylow Theorem tells us that $A_4$ has:
:at least one subgroup of order $4$
:at least one subgroup of order $2$
:at least one subgroup of order $3$
These it has.
But it has no subgroup of order $6$, although $6 \divides 12$.
== Sources ==
* : $\S 6.5$: Example $121$
Category:First Sylow Theorem
Category:Alternating Group on 4 Letters
First Sylow Theorem/Proof 1

== Theorem ==
== Proof ==
Let $\order G = k p^n$ such that $p \nmid k$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.
 
From Cardinality of Set of Subsets, this is given by:
:$N = \dbinom {p^n k} {p^n} = \dfrac {\paren {p^n k} \paren {p^n k - 1} \cdots \paren {p^n k - i} \cdots \paren {p^n k - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$
From Binomial Coefficient involving Power of Prime:
:$\dbinom {p^n k} {p^n} \equiv k \pmod p$
Thus:
:$N \equiv k \pmod p$
Now let $G$ act on $\mathbb S$ by the rule:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
That is, $g * S$ is the left coset of $S$ by $g$.
From Group Action on Sets with k Elements, this is a group action.
Now, let $\mathbb S$ have $r$ orbits under this action.
From Set of Orbits forms Partition, the orbits partition $\mathbb S$.
Let these orbits be represented by $\set {S_1, S_2, \ldots, S_r}$, so that:
If each orbit had length divisible by $p$, then $p \divides N$.
But this can not be the case, because, as we have seen:
:$N \equiv k \pmod p$
So at least one orbit has length which is not divisible by $p$.
Let $S \in \set {S_1, S_2, \ldots, S_r}$ be such that $\size {\Orb S)} = m: p \nmid m$.
Let $s \in S$.
It follows from Group Action on Prime Power Order Subset that:
:$\Stab S s = S$
and so:
:$\size {\Stab S} = \size S = p^n$
From Stabilizer is Subgroup:
:$\Stab S \le G$
Thus $\Stab S$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Theorem $11.4$
Category:First Sylow Theorem
First Sylow Theorem/Proof 2

== Theorem ==
== Proof ==
Let $\Bbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Now let $G$ act on $\Bbb S$ by the rule:
:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
From Set of Orbits forms Partition, the orbits partition $\mathbb S$.
Let these orbits be represented by $\set {S_1, S_2, \ldots, S_n}$, so that:
:$(1): \quad \size {\mathbb S} = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$
Thus each $\Orb {S_i}$ is the orbit under $*$ of some $S_i$ whose order is $p^n$.
By the Orbit-Stabilizer Theorem:
:$(2): \quad \order {\Orb {S_i} } = \dfrac {\order G} {\order {\Stab {S_i} } }$ for all $i \in \set {1, 2, \ldots, n}$
where $\Stab {S_i}$ is the stabilizer of $S_i$ under $*$.
Let $s \in S_i$ and $x \in \Stab {S_i}$.
Then $s x \in S_i$ because $S_i x = S_i$.
Let $s$ be fixed while $x$ runs over $\Stab {S_i}$.
Then:
:$\forall s \in S_i: s \Stab {S_i} \subseteq S_i$
Therefore $S_i$ is the union of all the left cosets $s \Stab {S_i}$.
These left cosets may not all be distinct.
Let $r_i$ be the number of distinct left cosets of $s \Stab {S_i}$.
Then:
:$\card {S_i} = r_i \card {\Stab {S_i} }$
because:
: distinct left cosets are disjoint
: all have the same cardinality $\card {\Stab {S_i} }$.
Thus:
:$(3): \quad \card {\Stab {S_i} } \divides \card {S_i} = p^n$
Because the only divisors of $p^n$ are smaller powers of $p$:
:$\card {\Stab {S_i} } = p^{n_i}$
for some $n_i \le n$, for $i = 1, 2, \ldots, r$.
We can write:
:$g = \order G = k p^n$
where $p$ does not divide $k$.
From $(2)$ and $(3)$:
:$\card {S_i} = k p^{d_i}$
where $d_i = n - n_i$ for $i = 1, 2, \ldots, r$.
From Cardinality of Set of Subsets:
:$\card {\Bbb S} = \dbinom {k p^n} {p_n}$
We have from Binomial Coefficient involving Power of Prime that:
:$\dbinom {k p^n} {p_n} \equiv k \pmod p$
But $p$ does not divide $k$.
Hence $p$ does not divide $\card {\Bbb S}$.
Hence by $(1)$ and $(4)$:
:$p \nmid k \paren {p^{d_1} + p^{d_2} + \cdots p^{d_r} }$
If $d_i > 0$ for all $i = 1, 2, \ldots, r$, then $p$ would divide this integer.
So there must be some $i$ such that $d_i = 0$.
Then, for this $i$:
:$n_i = n$
That is:
:$\card {\Stab {S_i} } = p^n$
Thus $\Stab {S_i}$ is the subgroup of $G$ of order $p_n$ that was asserted.
== Sources ==
* : $\S 6.5$. Orbits: Example $121$
Category:First Sylow Theorem
First Variation Formula

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $I = \closedint a b$ be a closed real interval.
Let $J$ be an open real interval.
Suppose $\gamma : I \to M$ is a unit-speed admissible curve.
Let $\Gamma : J \times I \to M$ be a variation of $\gamma$.
Let $T_p M$ be the tangent space at $p \in M$.
Let $V : M \to T_\gamma M$ be the variation field of $\Gamma$.
Let $L_g$ be the Riemannian length of an admissible curve.
Let $\gamma'$ be the velocity of $\gamma$.
Let $D_t$ be the covariant derivative along $\gamma$.
Let $\tuple {a_0, \ldots, a_k}$ be an admissible subdivision for $\gamma$.
For all $i \in \N_{> 0} : i \le k - 1$ let $\Delta_i \gamma' = \map {\gamma'} {a_i^+} - \map {\gamma'} {a_i^-}$.
Let $\innerprod \cdot \cdot$ be the inner product induced by the Riemannian metric.
Then for all $s \in J$ the mapping $\map {L_g} {\Gamma_s}$ is a smooth function of $s$ and:
:$\ds \valueat {\dfrac d {d s} \map {L_g} {\Gamma_s} } {s \mathop = 0} = - \int_a^b \innerprod V {D_ t \gamma'} \d t - \sum_{i \mathop = 1}^{k - 1} \innerprod {\map V {a_i} } {\Delta_i \gamma'} + \innerprod {\map V b} {\map {\gamma'} b} - \innerprod {\map V a} {\map {\gamma'} a}$
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Minimizing Curves Are Geodesics
Category:Riemannian Manifolds
Category:Curves
First Variation Formula for Proper Variation

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $I = \closedint a b$ be a closed real interval.
Let $J$ be an open real interval.
Suppose $\gamma : I \to M$ is a unit-speed admissible curve.
Let $\Gamma : J \times I \to M$ be a proper variation of $\gamma$.
Let $T_p M$ be the tangent space at $p \in M$.
Let $V : M \to T_\gamma M$ be the variation field of $\Gamma$.
Let $L_g$ be the Riemannian length of an admissible curve.
Let $\gamma'$ be the velocity of $\gamma$.
Let $D_t$ be the covariant derivative along $\gamma$.
Let $\tuple {a_0, \ldots, a_k}$ be an admissible subdivision for $\gamma$.
For all $i \in \N_{> 0} : i \le k - 1$ let $\Delta_i \gamma' = \map {\gamma'} {a_i^+} - \map {\gamma'} {a_i^-}$.
Let $\innerprod \cdot \cdot$ be the inner product induced by the Riemannian metric.
Then for all $s \in J$ the mapping $\map {L_g} {\Gamma_s}$ is a smooth function of $s$ and:
:$\ds \valueat {\dfrac d {d s} \map {L_g} {\Gamma_s} } {s \mathop = 0} = - \int_a^b \innerprod V {D_ t \gamma'} \d t - \sum_{i \mathop = 1}^{k - 1} \innerprod {\map V {a_i} } {\Delta_i \gamma'}$
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Minimizing Curves Are Geodesics
Category:Riemannian Manifolds
Category:Curves
Fisher's Inequality

== Theorem ==
For a BIBD $\struct {v, k, \lambda}$, the number of blocks $b$ must be greater than or equal to the number of points $v$:
:$ b \ge v$
== Proof ==
Let $A$ be the incidence matrix.
By Product of Incidence Matrix of BIBD with its Transpose, we have that:
:$A^\intercal \cdot A = \begin{bmatrix} 
r & \lambda & \cdots & \lambda \\
\lambda & r & \cdots & \lambda \\
\vdots & \vdots & \ddots & \vdots \\
\lambda & \lambda & \cdots & r \\
\end{bmatrix}$
From Necessary Condition for Existence of BIBD:
:$r > \lambda$
So we can write $r = \lambda + \mu$ for some $\mu > 0$ and so:
:$A^\intercal \cdot A = \begin{bmatrix} 
\lambda + \mu & \lambda & \cdots & \lambda \\
\lambda & \lambda + \mu & \cdots & \lambda \\
\vdots & \vdots & \ddots & \vdots \\
\lambda & \lambda & \cdots & \lambda + \mu \\
\end{bmatrix}$
This is a combinatorial matrix of order $v$.
So:
Now, since $k < v$ (this is obvious) and using the properties of a BIBD, we get that $r > \lambda$. 
So $\map \det {A^\intercal A} \ne 0$.
Since $A^\intercal A$ is a $v \times v$ matrix, then the rank, $\rho$, of $A^\intercal A = v$.
Using the facts that $\map \rho {A^\intercal A} \le \map \rho A$, and $\map \rho A \le b$ (since $A$ only has $b$ columns), we have that:
:$v \le \map \rho A \le b$
Category:Balanced Incomplete Block Designs
Fisher-Tippett-Gnedenko Theorem

== Theorem ==
In general, for iid $\set {X_i} _{i\mathop = 1}^n$ with cdf $\map F x$ we have $\mathbb P \sqbrk {\max_{i \mathop = 1}^n \set {X_i} > x} = \map {F^n} x$.
Let $x_* := \sup \set {x: \map F x < 1}$.
Then as $n \to \infty$:
:$\ds \max_{i \mathop = 1, \mathop \ldots, n} \set {X_i} \stackrel {prob} {\to} x_*$
We normalize it by constants $a_n > 0, b_n$ such that:
:$\dfrac {\ds \max_{i \mathop = 1, \mathop \ldots, n} \set {X_i} - b_n} {a_n} \stackrel {dis} {\to} X_*$
for some non-degenerate $X_*$ with distribution $\map G x$.
We have the following three possible limits
* Frechet: If $x > 0$, $\map G x = \map \exp {-x^{-a} }$ for some $a > 0$. If $x \le 0$, $\map G x = 0$.
* Gumbel: If $x \in \R$, $\map G x = \map \exp {-e^{-\frac {x - \mu} {\sigma} } }$ for $\mu\in \R, \sigma>0$. 
* Weibull: If $x \ge 0$, $\map G x = 1 - \map \exp {-\paren {\dfrac x \lambda}^k}$ for $\lambda, k > 0$. If $x < 0$ , $\map G x = 0$.
Fisher-Tippet-Gnedenko theorem
:$(1): \quad \map G x$ will be Frechet $\map F x < 1$ and $\ds \lim_{t \mathop \to +\infty} \dfrac {1 - \map F {t x} } {1 - \map F t} = x^{-\theta}$ for $x > 0$ and $\theta > 0$. Here we can set $b_n := 0$ and $a_n := F^{-1} \paren {1 - \dfrac 1 n}$.
:$(2): \quad \map G x$ will be Weibull $x_* : = \sup \set {x: \map F x < 1} < \infty$ and $\ds \lim_{t \mathop \to +\infty} \dfrac {1 - \map F {x_* + t x} } {1 - \map F {x_* - t} } = \paren {-x}^{-\theta}$ for $x < 0$ and $\theta > 0$. Here we can set $b_n := x_*$ and $a_n := x_* - \map {F^{-1} } {1 - \dfrac 1 n}$.
:$(3): \quad G$ will be Gumbel if the density $\map f x := \dfrac \d {\d x} \map F x > 0$ and it is differentiable in $\openint {x_1} {x_*}$ for some $x_1$, and $\ds \lim_{x \mathop \to x_*} \map {\dfrac \d {\d x} } {\dfrac {1 - \map F x} {\map f x} } = 0$. Here we can set $b_n := \map {F^{-1} } {1 - \dfrac 1 n}$ and $a_n := \dfrac 1 {n \map f {b_n} }$.
== Proof ==
Proof under construction
Category:Probability Theory
Fitting Chord Into Circle

== Theorem ==
Into a given circle, it is possible to fit a chord equal to a given line segment which is not greater than the diameter of the circle.
== Construction ==
:
Let $ABC$ be the given circle and $D$ be the given line segment.
Let $BC$ be a diameter of circle $ABC$.
If $D = BC$ then the task is complete, as $BC$ is already fitted into $ABC$.
Otherwise, let $CE$ be cut off from $BC$ equal to $D$.
With center $C$ and radius $CE$, draw circle $EFG$.
Let $G$ be one of the points at which $EFG$ meets $ABC$.
Then $CG$ is the chord required.
== Proof ==
As $C$ is the center of $EFG$, $CG = CE$ and so $CG = D$.
But $CG$ is a chord of $ABC$.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Five Color Theorem/Historical Note

== Historical Note on Five Color Theorem ==
The Five Color Theorem is not the strongest result possible.
It was proved by in $1890$.
In the same year he showed a flaw in 's supposed $1879$ proof of the Four Color Theorem which was not mended for almost another $100$ years.
It was not until $1976$ that and demonstrated that four colors suffice.
Their proof relies heavily on computers and for the moment is not to be found on .
Category:Historical Notes
Five Cube Theorem

== Theorem ==
Every integer can be expressed as a sum of five cube numbers.
== Proof ==
Let $r$ be an integer.
Then $r$ can be expressed in exactly one the following forms:
:$\paren {6 m + 0}$
:$\paren {6 m + 1}$
:$\paren {6 m + 2}$
:$\paren {6 m + 3}$
:$\paren {6 m + 4}$
:$\paren {6 m + 5}$
for some $m \in \Z$.
:$\paren {6 m + 0} = \paren {m + 1}^3 + \paren {m - 1}^3 + \paren {- m}^3 + \paren {- m}^3 + 0^3$
:$\paren {6 m + 1} = \paren {m + 1}^3 + \paren {m - 1}^3 + \paren {- m}^3 + \paren {- m}^3 + 1^3$
:$\paren {6 m + 2} = \paren {m}^3 + \paren {m - 2}^3 + \paren {1 - m}^3 + \paren {1 - m}^3 + 2^3$
:$\paren {6 m + 3} = \paren {m - 3}^3 + \paren {m - 5}^3 + \paren {- m + 4}^3 + \paren {- m + 4}^3 + 3^3$
:$\paren {6 m + 4} = \paren {m + 3}^3 + \paren {m + 1}^3 + \paren {- m - 2}^3 + \paren {- m - 2}^3 + \paren {-2}^3$
:$\paren {6 m + 5} = \paren {m + 2}^3 + \paren {m}^3 + \paren {- m - 1}^3 + \paren {- m - 1}^3 + \paren {-1}^3$
== Also see ==
* Number is Sum of Five Cubes, which shows that there is in fact an infinite number of ways to represent any integer as a sum of five cubes.
Category:Cube Numbers
Category:Sums of Cubes
Five Platonic Solids

== Theorem ==
There exist exactly five platonic solids:
:$\paren 1: \quad$ the regular tetrahedron
:$\paren 2: \quad$ the cube
:$\paren 3: \quad$ the regular octahedron
:$\paren 4: \quad$ the regular dodecahedron
:$\paren 5: \quad$ the regular icosahedron.
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : $5$
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
* : $5$
* 
* 
* : Chapter $2$: The Logic of Shape: The golden mean
Category:Five Platonic Solids
Category:Platonic Solids
Category:5
Five Platonic Solids/Proof 1

== Theorem ==
== Proof ==
A solid angle cannot be constructed from only two planes.
Therefore at least three faces need to come together to form a vertex.
Let $P$ be a platonic solid.
Let the polygon which forms each face of $P$ be a equilateral triangles.
We have that:
:each vertex of a regular tetrahedron is composed of $3$ equilateral triangles
:each vertex of a regular octahedron is composed of $4$ equilateral triangles
:each vertex of a regular icosahedron is composed of $5$ equilateral triangles.
$6$ equilateral triangles, placed together at a vertex, form $4$ right angles.
From :
:a solid angle is contained by plane angles which total less than $4$ right angles.
Thus it is not possible to form $P$ such that its vertices are formed by $6$ equilateral triangles.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $6$ equilateral triangles.
Hence there are only $3$ possible platonic solids whose faces are equilateral triangles.
We have that each vertex of a cube is composed of $3$ squares.
$4$ squares, placed together at a vertex, form $4$ right angles.
Thus, again from :
:it is not possible to form $P$ such that its vertices are formed by $4$ squares.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ squares.
Hence there is only $1$ possible platonic solid whose faces are squares.
We have that each vertex of a regular dodecahedron is composed of $3$ regular pentagons.
From :
:the vertices of a regular pentagon equal $1 \dfrac 1 5$ right angles.
$4$ regular pentagons, placed together at a vertex, form $4 \dfrac 4 5$ right angles.
Thus, again from :
:it is not possible to form $P$ such that its vertices are formed by $4$ regular pentagons.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ regular pentagons.
Hence there is only $1$ possible platonic solid whose faces are regular pentagons.
$3$ regular hexagons, placed together at a vertex, form $4$ right angles.
Thus, again from :
:it is not possible to form $P$ such that its vertices are formed by $3$ or more regular hexagons.
Regular polygons with more than $6$ sides have vertices which are greater than those of a regular hexagon.
Therefore $3$ such regular polygons, placed together at a vertex, form more than $4$ right angles.
Thus, again from :
:it is not possible to form $P$ such that its vertices are formed by $3$ or more regular polygons with more than $6$ sides.
Hence the $5$ possible platonic solids have been enumerated and described.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Five Platonic Solids
Five Platonic Solids/Proof 2

== Theorem ==
== Proof ==
Consider a convex regular polyhedron $P$.
Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$.
Let $n$ be the number of those polygons which meet at each vertex of $P$.
From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$.
The sum of the internal angles must be less than $360^\circ$.
So:
But $m$ and $n$ are both greater than $2$.
So:
:if $m = 3$, $n$ can only be $3$, $4$ or $5$
:if $m = 4$, $n$ can only be $3$
:if $m = 5$, $n$ can only be $3$
and $m$ cannot be greater than $3$.
There are $5$ possibilities in all.
Therefore all platonic solids have been accounted for.
== Sources ==
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Five Platonic Solids
Five Ramanujan-Nagell Numbers

== Theorem ==
There exist exactly $5$ Ramanujan-Nagell numbers: positive integers of the form $2^m - 1$ which are triangular:
:$0, 1, 3, 15, 4095$
== Proof ==
Consider the numbers of the form $2^m - 1$ which are triangular:
Let:
:$n = m - 3$
:$x = 2 r + 1$
and it can be seen that $(1)$ is equivalent to:
:$x^2 + 7 = 2^n$
From Solutions of Ramanujan-Nagell Equation:
:$x = 1, 3, 5, 11, 181$
Setting $r = \dfrac {x - 1} 2$ it is seen that the corresponding triangular numbers are:
:$\dfrac {\paren {x - 1} \paren {x + 1} } 8$
Thus the corresponding Ramanujan-Nagell numbers are:
:$0, 1, 3, 15, 4095$
Category:Ramanujan-Nagell Equation
Category:5
Fixed-Point Property is Topological

== Theorem ==
Let $T \sim T'$ be homeomorphic topological spaces.
Suppose every continuous $f : T \to T$ has a fixed point.
Then, every continuous $g : T' \to T'$ also has a fixed point.
== Proof ==
Let $\phi : T \to T'$ be a homeomorphism.
Let $g : T' \to T'$ be an arbitrary continuous mapping.
Consider $f_g : T \to T$ defined as:
:$\map {f_g} x = \map {\phi^{-1}} {\map g {\map \phi x}}$
By definition of homeomorphism and Composite of Continuous Mappings is Continuous:
:$f_g$ is continuous
Therefore, , there is a fixed point $x_0 \in T$ of $f_g$.
We then have:
Therefore, $\map \phi {x_0}$ is a fixed point of $g$.
As $g$ was arbitrary, the result holds.
Category:Fixed Point Theorems
Category:Topological Properties
Fixed Elements form 1-Cycles

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$.
Let $\Fix \pi$ be the set of elements fixed by $\pi$.
For any $\pi \in S_n$, all the elements of $\Fix \pi$ form $1$-cycles.
== Proof ==
Let $\pi$ be a permutation, and let $x \in \Fix \pi$.
From the definition of a fixed element:
:$\map \pi x = x$
From the definition of a $k$-cycle, we see that $1$ is the smallest $k \in \Z: k > 0$ such that:
:$\map {\pi^k} x = x$
The result follows.
Category:Fixed Elements under Permutations
Category:Symmetric Groups
Fixed Point Formulation of Explicit ODE

== Theorem ==
Let $x' = \map f {t, x}$ with $\map x {t_0} = x_0$ be an explicit ODE of dimension $n$.
For $a, b \in \R$, let $\XX = \map {\CC} {\closedint a b; \R^n}$ be the space of continuous functions on the closed interval $\closedint a b$.
Let $T: \XX \to \XX$ be the map defined by:
:$\ds \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$
Then a fixed point of $T$ in $\XX$ is a solution to the above ODE.
== Proof ==
Let $\map y t$ be a fixed point of the map $T$.
That is:
:$\ds \map y t = x_0 + \int_{t_0}^t \map f {s, \map y s} \rd s$
Then:
:$\ds \map y {t_0} = x_0 + \int_{t_0}^{t_0} \map f {s, \map y s} \rd s = x_0$
By the fundamental theorem of calculus we have that $y$ is differentiable, and for $t \in \closedint a b$:
This shows that $y$ is a solution to the ODE as claimed.
Category:Ordinary Differential Equations
Fixed Point Set of Continuous Self-Map on Hausdorff Space is Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $f: T \to T$ be a continuous mapping on $T$.
Let $W$ be the set defined as:
:$W = \set {x \in T: \map f x = x}$
Then $W$ is closed in $T$.
== Proof ==
Let $g: T \to T$ be the identity mapping on $T$:
:$\forall x \in T: \map g x = x$
From Identity Mapping is Continuous, $g$ is a continuous mapping on $T$.
From Equal Images of Mappings to Hausdorff Space form Closed Set:
:$\set {x \in T: \map f x = \map g x}$ is closed in $T$.
and the result follows.
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 6$
Category:Hausdorff Spaces
Fixed Point of Composition of Inflationary Mappings

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be inflationary mappings.
Let $x \in S$.
Then:
:$x$ is a fixed point of $f \circ g$
:
:$x$ is a fixed point of both $f$ and $g$.
== Proof ==
=== Necessary Condition ===
Follows from Fixed Point of Mappings is Fixed Point of Composition.
=== Sufficient Condition ===
Let $h = f \circ g$.
Let $x$ be a fixed point of $h$.
Then by definition of composition:
:$\map f {\map g x} = x$
Since $f$ is inflationary:
:$x \preceq \map g x$
 $x \ne \map g x$.
Then $x \prec \map g x$.
Since $f$ is also inflationary:
:$\map g x \preceq \map f {\map g x}$
Thus by Extended Transitivity:
:$x \prec \map f {\map g x}$
But this contradicts the assumption that $x$ is a fixed point of $f \circ g$.
Therefore, $x = \map g x$, and $x$ is a fixed point of $g$.
 $\map f x \ne x$.
Then $x \prec \map f x$.
As we have shown that $x = \map g x$, it follows that:
:$x \prec \map f {\map g x}$
But this contradicts assumption that $x$ is a fixed point of $f \circ g$.
Hence, $x$ is also a fixed point of $f$.
Category:Order Theory