nyuuzyou/wikis huggingface dataset:
Platonic Solid is Uniform Polyhedron

== Theorem ==
Let $P$ be a Platonic solid.
Then $P$ is also uniform.
== Proof ==
From Platonic Solid is Isogonal, $P$ is isogonal.
That all the faces are regular polygons follows .
Hence the result.
== Sources ==
* 
* 
Category:Platonic Solids
Category:Uniform Polyhedra
Player with Greatest Capital wins Fair Game

== Theorem ==
Let $G$ be a fair game between two players $P_1$ and $P_2$.
Let $P_1$ have greater capital than $P_2$.
Then in a sequence of instances of $G$, $P_1$ has a greater probability of ruining $P_2$ than the other way about.
== Proof ==
== Sources ==
* 
* 
Category:Fair Games
Plug-In Estimator/Examples/Sample Mean

== Example of Plug-In Estimator ==
For a random sample of $n$ observations $x_i$ for $1 = 1, 2, \ldots, n$, the sample mean $\bar x$ is the plug-in estimator of the population mean $\mu$.
== Sources ==
* 
* 
Category:Examples of Plug-In Estimators
Plug-In Estimator/Examples/Sample Variance

== Example of Plug-In Estimator ==
For a random sample of $n$ observations $x_i$ for $1 = 1, 2, \ldots, n$, the sample variance $\ds \sum_i \dfrac {\paren {x_i - \bar x}^2} n$ is the plug-in estimator of the population variance $\sigma^2$.
This is a biased estimator of $\sigma^2$, but for a normal distribution it is the maximum likelihood estimator.
== Sources ==
* 
* 
* 
* 
Category:Examples of Plug-In Estimators
Pluperfect Digital Invariant/Examples/10 Digits

== Examples of $10$-Digit Pluperfect Digital Invariants ==
The only $10$-digit pluperfect digital invariant is:
== Historical Note ==
== Sources ==
* 
* : $4,679,307,774$
* : $4,679,307,774$
* 
Category:Pluperfect Digital Invariants
Pluperfect Digital Invariant/Examples/10 Digits/Historical Note

== Historical Note on $10$-Digit Pluperfect Digital Invariant ==
According to , as reported in his $1985$ work , this $10$-digit pluperfect digital invariant was discovered by .
== Sources ==
* : $4,679,307,774$
* : $4,679,307,774$
Category:Historical Notes
Pluperfect Digital Invariant/Examples/39 Digits

== Examples of $39$-Digit Pluperfect Digital Invariants ==
The $39$-digit pluperfect digital invariants are:
== Sources ==
* : $115,132,219,018,763,992,565,095,597,973,971,522,401$
* : $115,132,219,018,763,992,565,095,597,973,971,522,401$
* 
Category:Pluperfect Digital Invariants
Pluperfect Digital Invariant/Examples/3 Digits/Historical Note

== Historical Note on $3$-Digit Pluperfect Digital Invariants ==
In the (some believe inflammatory and controversial) words of in his of $1940$:
:These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them that appeals to the mathematician. The proofs are neither difficult nor interesting -- merely a little tiresome. The theorems are not serious; and it is plain that one reason ... is the extreme speciality of both the enunciations and the proofs, which are not capable of any significant generalization.
== Sources ==
* : $153$
* : $153$
Category:Historical Notes
Pluperfect Digital Invariant has less than 61 Digits

== Theorem ==
Let $n \in \Z_{>0}$ be a pluperfect digital invariant.
Then $n$ has less than $61$ digits.
== Proof ==
We have that:
:$n \times 9^n < 10^\paren {n - 1}$
when $n > 60$.
So an $n$-digit integer, for $n > 60$, is always greater than the sum of the $n$th powers of its digits.
== Sources ==
* : $61$
Category:Pluperfect Digital Invariants
Poincaré Conjecture

== Theorem ==
Let $\Sigma^m$ be a smooth $m$-manifold.
Let $\Sigma^m$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$.
== Proof ==
The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here. An outline of the $m = 3$ case will be given instead.
=== Dimension $m = 1$ ===
=== Dimension $m = 2$ ===
=== Dimension $m = 3$ ===
=== Dimension $m = 4$ ===
=== Dimension $m = 5$ ===
=== Dimension $m \ge 6$ ===
== Historical Note ==
== Sources ==
* 
* 
* 
* 
* 
* : Appendix $23$: Millennium Prize problems: $7$.
Category:Poincaré Conjecture
Category:Manifolds
Category:Millennium Problems
Poincaré Conjecture/Dimension 1

== Theorem ==
Let $\Sigma^1$ be a smooth $1$-manifold.
Let $\Sigma^1$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^1$ is homeomorphic to the $1$-sphere $\Bbb S^1$.
== Proof ==
Follows from the Classification of Compact One-Manifolds.
Category:Poincaré Conjecture
Poincaré Conjecture/Dimension 2

== Theorem ==
Let $\Sigma^2$ be a smooth $2$-manifold.
Let $\Sigma^2$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_2 \struct {\Sigma; \Z} = \Z$
Then $\Sigma^2$ is homeomorphic to the $2$-sphere $\Bbb S^2$.
== Proof ==
Follows from the Classification of Compact Two-Manifolds.
Category:Poincaré Conjecture
Poincaré Conjecture/Dimension 3

== Theorem ==
Let $\Sigma^3$ be a smooth $3$-manifold.
Let $\Sigma^3$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_3 \struct {\Sigma; \Z} = \Z$
Then $\Sigma^3$ is homeomorphic to the $3$-sphere $\Bbb S^3$.
== Proof ==
Follows from Thurston's Geometrization Conjecture.
== Historical Note ==
Category:Poincaré Conjecture
Poincaré Conjecture/Dimension 4

== Theorem ==
Let $\Sigma^4$ be a smooth $4$-manifold.
Let $\Sigma^4$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_4 \struct {\Sigma; \Z} = \Z$
Then $\Sigma^4$ is homeomorphic to the $4$-sphere $\Bbb S^4$.
== Proof ==
Follows from $4$-dimensional Topological $h$-Cobordism Theorem.
== Historical Note ==
Category:Poincaré Conjecture
Poincaré Conjecture/Dimension 5

== Theorem ==
Let $\Sigma^5$ be a smooth $5$-manifold.
Let $\Sigma^m$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_5 \struct {\Sigma; \Z} = \Z$
Then $\Sigma^5$ is homeomorphic to the $5$-sphere $\Bbb S^5$.
== Proof ==
Summary:
Any $\Sigma^5$ bounds a contractible $6$-manifold $Z$.
Let $\Bbb D^6$ be a $6$-disk (AKA $6$-ball).
Then $Z - \Bbb D^6$ is an $h$-cobordism between $\Sigma$ and $\partial \Bbb D^6 = \Bbb S^5$.
Hence $\Sigma$ is differomorphic to $\Bbb S^5$ by the $h$-Cobordism Theorem.
Category:Poincaré Conjecture
Poincaré Conjecture/Dimension 6 or Greater

== Theorem ==
Let $\Sigma^m$ be a smooth $m$-manifold where $m \ge 6$.
Let $\Sigma^m$ satisfy:
:$H_0 \struct {\Sigma; \Z} = 0$
and:
:$H_m \struct {\Sigma; \Z} = \Z$
Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$.
== Proof ==
We can cut two small $m$-disks $D', D' '$ from $\Sigma$.
The remaining manifold, $\Sigma \setminus \paren {D' \cup D' '}$ is an h-cobordism between $\partial D'$ and $\partial D' '$.
These are just two copies of $\Bbb S^{m-1}$.
By the $h$-cobordism theorem, there exists a diffeomorphism:
:$\phi: \Sigma \setminus \paren {D' \cup D' '} \to \Bbb S^{m - 1} \times \closedint 0 1$
which can be chosen to restrict to the identity on one of the $\Bbb S^{m - 1}$.
Let $\Xi$ denote this $\Bbb S^{m - 1}$ such that $\phi$ restricts to the identity.
Since $\psi \vert_\Xi = Id$, we can extend $\psi$ across $D' '$, the interior of $\Xi$ to obtain a diffeomorphism $\phi': \Sigma \setminus D' ' \to \Bbb S^{m - 1} \cup D'$.
Let $\Bbb D^m$ denote this latter manifold, which is merely an $m$-disk.
Our diffeomorphism $\phi': \Sigma \setminus D' ' \to \Bbb D^m$ induces a diffeomorphism on the boundary spheres $\Bbb S^{m - 1}$.
Any diffeomorphism of the boundary sphere $\Bbb S^{m - 1}$ can be extended radially to the whole disk:
:$\map {\operatorname {int} } {\Bbb S^{m - 1} } = D' '$
but only as a homeomorphism of $D' '$. 
Hence the extended function $\phi' ': \Sigma \to \Bbb S^m$ is a homeomorphism.
Category:Poincaré Conjecture
Poincaré Conjecture/Historical Note

== Theorem ==
The Poincaré Conjecture was first posed in $1904$ by .
For $n = 1$ and $n = 2$ the result was long known to be true.
For $n \ge 5$ it was proved by in $1960$.
The case for $n = 4$ was solved by in $1982$.
The remaining case for $n = 3$ was finally resolved by the work of , who solved Thurston's Geometrization Conjecture in $2003$ (although some sources say $2004$).
He did this by using the Ricci flow method.
== Sources ==
* 
* 
* 
Category:Historical Notes
Poincaré Duality Theorem

== Theorem ==
Let $M$ be an $n$-manifold.
Let its $n$th homology group $\map {H_n} M$ be infinite and cyclic (that is, that $M$ is an orientable manifold).
Then $\map {H_r} M$ is homeomorphic to the $\paren {n - r}$th cohomology group $\map {H^{n - r} } M$ for all $r$.
=== Corollary ===
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:Poincaré Duality Theorem
Category:Riemannian Geometry
Poincaré Duality Theorem/Corollary

== Corollary to Poincaré Duality Theorem ==
Let $V$ be an $n$-space.
Let $S$ and $T$ be subspaces of $M$ of dimension $r$ and $n - r$.
Then $S$ and $T$ usually meet at a single point.
== Proof ==
== Sources ==
* 
* 
Category:Poincaré Duality Theorem
Poincaré Plane is Abstract Geometry

== Theorem ==
The Poincaré plane $\struct {\H, L_H}$ is an abstract geometry.
== Proof ==
We will show that the axioms for an abstract geometry hold.
=== Axiom 1 ===
Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be two points in $\struct {\H, L_H}$.
If $x_1 = x_2 = a$ then $P, Q \in {}_a L$.
If $x_1 \ne x_2$ then let:
:$c = \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {2 \paren {x_2 - x_1} }$
:$r = \sqrt{\paren{x_1 - c}^2 + y_1^2}$
From definition of ${}_c L_r$:
:${}_c L_r := \set {\tuple {x, y} \in \H : \paren {x - c}^2 + y^2 = r^2}$
We have:
Thus $P, Q \in {}_c L_r$.
So any two points in $\H$ lie on a line in $L_H$.
=== Axiom 2 ===
For any $a \in \R$:
:$\tuple {a, 1}, \tuple {a, 2} \in {}_a L$
Also for any $c \in \R$ and $r \in \R_{>0}$:
:$\tuple {c + \dfrac 1 2 r, \dfrac {\sqrt 3} 2 r}, \tuple {c - \dfrac 1 2 r, \dfrac {\sqrt 3} 2 r} \in {}_c L_r$
So every line in $L_H$ has at least two points.
Hence $\struct {\H, L_H}$ is an abstract geometry.
== Sources ==
* : $\S 2.1$
Category:Poincaré Plane
Category:Abstract Geometries
Poincaré Recurrence Theorem

== Theorem ==
Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Then for each $A \in \BB$:
:$\ds \map \mu {A \setminus \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A} = 0$
That is, for $\mu$-almost all $x\in A$ there are integers $0 < n_1 < n_2 < \cdots$ such that $\map {T^{n_i} }x \in A$.
== Proof ==
Let:
:$\ds A_\infty := \bigcap_{N \mathop = 1}^\infty \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A$
For $N \in \N$, let:
:$\ds A_N := \bigcup_{n \mathop = N}^\infty T^{-n} \sqbrk A$
so that:
:$\ds A_\infty = \bigcap _{N \mathop = 1} ^\infty A_N$
Now, we need to show:
:$\ds \map \mu {A \setminus A_\infty} = 0$
First, for all $N \in \N$:
On the other hand, for all $N \in \N$:
Hence, because $T$ is a $\mu$-preserving transformation:
:$\map \mu {A_N} = \map \mu {T^{-1} \sqbrk {A_N} } = \map \mu {A_{N + 1} }$
Therefore:
== Sources ==
* Theorem $2.11$
* Theorem $1.4$
Category:Measures
Category:Ergodic Theory
Point Evaluations are Continuous Linear Functionals on Space of Complex-Valued Continuous Functions on Compact Hausdorff Space

== Theorem ==
Let $K$ be a compact Hausdorff space. 
Let $\map \CC K$ be the space of complex-valued continuous functions of compact Hausdorff space.
For $k \in K$, define $\delta_k : \map \CC K \to \C$ by: 
:$\map {\delta_k} f = \map f k$ 
for each $f \in \map \CC K$.
Then $\delta_k$ is a continuous linear functional on $\map \CC K$ and further: 
:$\norm {\delta_k}_{\map \CC {K}^\ast} = 1$ 
== Proof ==
Let $k \in K$. 
$\delta_k$ is clearly linear, so we need to show that it is continuous.
From Continuity of Linear Functionals, it is enough to show that $\delta_k$ is bounded.
For each $f \in \map \CC K$ we have: 
:$\ds \cmod {\map f k} \le \sup_{x \in K} \cmod {\map f x}$ 
That is: 
:$\cmod {\map {\delta_k} f} \le \norm f_\infty$
If $K$ is a singleton, then we have $\cmod {\map {\delta_k} f} = \norm f_\infty$ for each $f \in \map \CC K$ and hence $\norm {\delta_k}_{\map \CC {K}^\ast} = 1$. 
If $K$ is not a singleton, pick distinct points $k, \ell \in K$.
From Urysohn's Lemma, there exists a continuous function $f : K \to \closedint 0 1 \subseteq \C$ with $\map f \ell = 0$ and $\map f k = 1$.
For this $f \in \map \CC K$, we have: 
:$\cmod {\map {\delta_k} f} = \cmod {\map f k} = 1 = \sup_{x \in K} \cmod {\map f x} = \norm f_\infty$ 
Hence $\norm {\delta_k}_{\map \CC K^\ast} = 1$ in this case as well.
Category:Space of Continuous Functions on Compact Hausdorff Space
Point Finite Set of Open Sets in Separable Space is Countable

== Theorem ==
Let $\struct {X, \tau}$ be a separable space.
Let $\FF$ be a point finite set of open sets of $X$.
Then $\FF$ is countable.
== Proof ==
Since $\struct {X, \tau}$ is separable, $X$ has a countable everywhere dense subset $S$.
, assume that $\O \notin \FF$.
By the definition of point finite, $\set {V \in \FF: x \in V}$ is finite for each $x \in S$.
From Open Set Characterization of Denseness, each element of $\FF$ contains an element of $S$.
From Union of Set of Sets when a Set Intersects All:
:$\FF = \ds \bigcup_{x \mathop \in S} \set {V \in \FF: x \in V}$
Thus by Countable Union of Finite Sets is Countable, $\FF$ is countable.
Category:Topology
Point Spectrum of Densely-Defined Linear Operator consists of its Eigenvalues

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a densely-defined linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Then $\lambda \in \map {\sigma_p} T$ $\lambda$ is an eigenvalue of $T$.
== Proof ==
We have that $\lambda \in \map {\sigma_p} T$ :
:$T - \lambda I$ is not injective.
That is, there exists $x \in \map D T \setminus \set 0$ such that: 
:$\paren {T - \lambda I} x = \map {\paren {T - \lambda I} } 0 = 0$
So $\lambda \in \map {\sigma_p} T$ there exists $x \in \map D T \setminus \set 0$ such that: 
:$T x = \lambda I x = \lambda x$
That is, $\lambda \in \map {\sigma_p} T$ $\lambda$ is an eigenvalue of $T$.
Category:Point Spectra (Densely-Defined Linear Operators)
Point Spectrum of Symmetric Densely-Defined Linear Operator is Real

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\struct {\map D T, T}$ be a symmetric densely-defined linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $\struct {\map D T, T}$.
Then: 
:$\map {\sigma_p} T \subseteq \R$
== Proof ==
If $\map {\sigma_p} T = \O$, the result is immediate.
Let $\lambda \in \map {\sigma_p} T$. 
Then, from Point Spectrum of Densely-Defined Linear Operator consists of its Eigenvalues, there exists $x \in \HH \setminus \set 0$ such that: 
:$T x = \lambda x$
Then, we have: 
:$\innerprod {T x} x = \innerprod {\lambda x} x = \lambda \innerprod x x$
and: 
So, we have: 
:$\paren {\lambda - \overline \lambda} \innerprod x x = 0$
Since $x \ne 0$, we have $\innerprod x x \ne 0$, so:
:$\lambda = \overline \lambda$ 
From Complex Number equals Conjugate iff Wholly Real, we then have: 
:$\lambda \in \R$
So:
:$\map {\sigma_p} T \subseteq \R$
from the definition of set inclusion.
== Sources ==
* $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators
Category:Symmetric Densely-Defined Linear Operators
Category:Point Spectra (Densely-Defined Linear Operators)
Point at Distance Zero from Closed Set is Element

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be arbitrary.
Let $\map d {x, H}$ denote the distance between $x$ and $H$:
:$\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$
Let $H$ be closed in $M$.
Then:
:$\map d {x, H} = 0$ $x \in H$.
== Proof ==
=== Necessary Condition ===
Let $x \in H$.
Then from Distance from Subset to Element:
:$\map d {x, H} = 0$
whether $H$ is closed or not.
=== Sufficient Condition ===
Let $\map d {x, H} = 0$.
From Point at Zero Distance from Subset of Metric Space is Limit Point or Element, either:
:$x \in H$
or:
:$x$ is a limit point of $H$.
As $H$ is closed in $M$, it follows by definition that:
:$H$ contains all its limit points.
Hence if $x$ is a limit point of $H$ it follows that $x \in H$.
In either case it follows that $x \in H$.
== Also see ==
* Subset of Metric Space is Closed iff contains all Zero Distance Points, which presents this result as:
$H$ is closed in $M$ :
:$\forall x \in A: \map d {x, H} = 0 \implies x \in H$
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 5 \ \text {(a)}$
Category:Metric Spaces
Category:Closed Sets
Point at Infinity of Intersection of Parallel Lines

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$ such that $\LL_1$ and $\LL_2$ are parallel.
By Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ can be expressed as the general equations:
The point at infinity of $\LL_1$ and $\LL_2$ can thence be expressed in homogeneous Cartesian coordinates as $\tuple {-m, l, 0}$.
== Proof ==
Let $\LL_1$ be expressed in the form:
:$l x + m y + n = 0$
Hence let $\LL_2$ be expressed in the form:
:$l x + m y + k n = 0$
where $k \ne 1$.
Let their point of intersection be expressed in homogeneous Cartesian coordinates as $\tuple {X, Y, Z}$
Then:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Point at Infinity
Category:Parallel Lines
Point at Zero Distance from Subset of Metric Space is Limit Point or Element

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be arbitrary.
Let $\map d {x, H}$ denote the distance between $x$ and $H$:
:$\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$
Then:
:$\map d {x, H} = 0$
:
:either $x \in H$ or $x$ is a limit point of $H$.
== Proof ==
=== Necessary Condition ===
Let $x$ be such that either $x \in H$ or $x$ is a limit point of $H$.
If $x \in H$ then:
:$\map d {x, H} = 0$
from Distance from Subset to Element.
Otherwise $x$ is a limit point of $H$.
Then from Limit Point of Subset of Metric Space is at Zero Distance‎:
:$\map d {x, H} = 0$
=== Sufficient Condition ===
Let $\map d {x, H} = 0$.
 neither $x \in H$ nor is $x$ a limit point of $H$.
Because $x \notin H$ it follows from Point not in Subset of Metric Space iff Distance from Elements is Greater than Zero that:
:$\forall y \in H: \map d {x, y} > 0$
By definition of limit point, it is not the case that:
:$\forall \epsilon \in \R_{>0}: \set {y \in H: 0 < \map d {y, x} < \epsilon} \ne \O$
So:
:$\exists \epsilon \in \R_{>0}: \set {y \in H: 0 < \map d {y, x} < \epsilon} = \O$
which means either that $x = y$ or:
:$\exists \epsilon \in \R_{>0}: \forall y \in H: \map d {y, x} \ge \epsilon$
But we know that $x \notin H$ so it cannot be the case that $x = y$.
Hence it follows that:
:$\ds \inf_{y \mathop \in H} \paren {\map d {x, y} } \ge \epsilon$
By definition of the distance between $x$ and $H$:
:$\map d {x, H} \ge \epsilon > 0$
From this contradiction it follows that it cannot be the case that neither $x \in H$ nor is $x$ a limit point of $H$.
That is:
:either $x \in H$ or $x$ is a limit point of $H$.
Category:Limit Points
Category:Distance Function
Point at which Prime-Counting Function becomes less than Eulerian Logarithmic Integral

== Theorem ==
Let $\map \pi n$ denote the prime-counting function.
Let $a \uparrow b$ be interpreted as Knuth notation for $a^b$.
=== Using Riemann Hypothesis ===
=== Not using Riemann Hypothesis ===
Category:Point at which Prime-Counting Function becomes less than Eulerian Logarithmic Integral
Category:Prime Number Theorem
Point at which Prime-Counting Function becomes less than Eulerian Logarithmic Integral/Not using Riemann Hypothesis

== Theorem ==
Let $\map \pi n$ denote the prime-counting function.
Let $a \uparrow b$ be interpreted as Knuth notation for $a^b$.
Suppose the Riemann Hypothesis does not hold.
Then:
:$\exists n < 10 \uparrow \paren {10 \uparrow \paren {10 \uparrow 964} }: \map \pi n - \ds \int_2^n \frac {\d x} {\ln x} < 0$
== Sources ==
* 
Category:Point at which Prime-Counting Function becomes less than Eulerian Logarithmic Integral
Point at which Prime-Counting Function becomes less than Eulerian Logarithmic Integral/Using Riemann Hypothesis

== Theorem ==
Let $\map \pi n$ denote the prime-counting function.
Let $a \uparrow b$ be interpreted as Knuth notation for $a^b$.
Suppose the Riemann Hypothesis holds.
Then:
:$\exists n < 10 \uparrow \paren {10 \uparrow \paren {10 \uparrow 34} }: \map \pi n - \ds \int_2^n \frac {\d x} {\ln x} < 0$
Here, $10 \uparrow \paren {10 \uparrow \paren {10 \uparrow 34} }$ is Skewes' number.
== Proof ==
== Sources ==
* 
* : $10^{10^{10^{34}}}$
* : $10^{10^{10^{34}}}$
* 
Category:Point at which Prime-Counting Function becomes less than Eulerian Logarithmic Integral
Category:Skewes' Number
Point dividing Line Segment between Two Points in Given Ratio

== Theorem ==
Let $A$ and $B$ be points whose position vectors relative to an origin $O$ of a Euclidean space are $\mathbf a$ and $\mathbf b$.
Let $\mathbf r$ be the position vector of a point $R$ on $AB$ which divides $AB$ in the ratio $m : n$.
:
Then:
:$\mathbf r = \dfrac {n \mathbf a + m \mathbf b} {m + n}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Point dividing Line Segment between Two Points in Given Ratio
Category:Division in Ratio
Category:Vector Algebra
Category:Straight Lines
Point dividing Line Segment between Two Points in Given Ratio/Proof 1

== Theorem ==
== Proof ==
We have that:
:$\vec {A B} = \mathbf b - \mathbf a$
and so:
:$\vec {A R} = \dfrac m {m + n} \paren {\mathbf b - \mathbf a}$
Hence the position vector $\mathbf r$ of $R$ is given by:
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Centroids: $8$. Division of a line in a given ratio
Category:Point dividing Line Segment between Two Points in Given Ratio
Point dividing Line Segment between Two Points in Given Ratio/Proof 2

== Theorem ==
== Proof ==
Let the coordinates of $A$ be $\tuple {x_1, y_1}$.
Let the coordinates of $B$ be $\tuple {x_2, y_2}$.
Let the coordinates of $R$ be $\tuple {X, Y}$.
Then we have:
:$\dfrac {x_2 - X} {X - x_1} = \dfrac n m$
and so:
:$X \paren {m + n} = m x_2 + n x_1$
Similarly for $Y$, giving $R$ as:
:$\tuple {\dfrac {m x_2 + n x_1} {m + n}, \dfrac {m y_2 + n y_1} {m + n} }$
The result follows.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: To find the coordinates of the point dividing the join of $A$ and $B$ in the ratio $\lambda : \mu$
Category:Point dividing Line Segment between Two Points in Given Ratio
Point in Closure of Subset of Metric Space iff Limit of Sequence

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H^-$ denote the closure of $H$.
Let $a \in A$.
Then $a \in H^-$ there exists a sequence $\sequence {x_n}$ of points of $H$ which converges to the limit $a$.
== Proof ==
From definition of closure, $H^- = H' \cup H^i$.
Suppose that $a \in H^-$.
If $a \in H^i$, then $a \in H$ and so $\sequence {a, a, \ldots}$ is a sequence in $H$ that converges to $a$.
If $a \in H'$, then by there exists a sequence in $H$ that converges to $a$.
For the converse, let $\sequence {x_n}$ be a sequence in $H$ that converges to a point $a \in A$.
So let $U$ be an open set that contains $a$.
By the definition of an open set, it follows that there exists an open ball centered at $a$ such that $\map {B_\epsilon} a \subseteq U$.
But then because $\sequence {x_n}$ converges to $a$ there exists $m \in \N_{\gt 0}$ such that:
Because from Set Intersection Preserves Subsets it is seen that $H \cap \map {B_\epsilon} a \subseteq H \cap U$.
It has been shown that every open set that contains $a$ also contains a point in $H$.
So by Condition for Point being in Closure, it is seen that $a \in H^-$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$
Category:Set Closures