avewright/proofwiki-math huggingface dataset:
Theorem: Regular Hexagon is composed of Equilateral Triangles

Let $P$ be a regular hexagon.
Let $O$ be the center of $P$.
Then there exists a triangulation of $P$ into six congruent equilateral triangles.
The three vertices of each triangle are $O$ and two adjacent vertices of $P$.

Proof:
Let $V_2$ be a vertex of $P$.
Let $V_1, V_3$ be the two adjacent vertices of $V_1$.
From Regular Polygon is composed of Isosceles Triangles, it follows that there exists a triangulation of $P$ into six congruent isosceles triangles of the type $V_1 V_2 O$.
As $\angle V_1 V_2 O$ and $\angle V_3 V_2 O$ are corresponding angles of the congruent isosceles triangles $\triangle V_1 V_2 O$ and $\triangle V_3 V_2 O$, it follows that $\angle V_1 V_2 O = \angle V_3 V_2 O$.
Let $V_2 O$ denote one side of the triangles $\triangle V_1 V_2 O$ and $\triangle V_1 V_3 O$.
As the angles $\angle V_1 V_2 O$ and $\angle V_3 V_2 O$ are constructed by intersecting the angle $\angle V_1 V_2 V_3$ with $V_2 O$, it follows that:
{{begin-eqn}}
{{eqn | l  = \angle V_1 V_2 V_3
      | r  = \angle V_1 V_2 O + \angle V_3 V_2 O
}}
{{eqn | ll = \leadsto
      | l  = 120 \degrees
      | r  = \angle V_1 V_2 O + \angle V_3 V_2 O
      | c =  by Internal Angles of Regular Polygon, as $\dfrac {\paren {6 - 2} 180 \degrees} 6 = 120 \degrees$
}}
{{eqn | ll = \leadsto
      | l  = 60 \degrees
      | r = \angle V_1 V_2 O
      | c = as $\angle V_1 V_2 O$ = $\angle V_3 V_2 O$
}}
{{end-eqn}}
As $\triangle V_1 V_2 O$ is an isosceles triangle, it follows that:
:$\angle V_2 V_1 O = 60 \degrees$
From Sum of Angles of Triangle equals Two Right Angles used on $\triangle V_1 V_2 O$, it follows that:
:$\angle V_1 O V_2 = 180 \degrees - \angle V_1 V_2 O - \angle V_2 V_1 O = 60 \degrees$
From Equiangular Triangle is Equilateral, it follows that $\triangle V_1 V_2 O$ is an equilateral triangle.
As the triangles in the triangulation of $P$ are congruent, it follows that they are all equilateral.
{{qed}}
Theorem: Morley's Formula

Let $n \in \C$ be a complex number.
Let $\map \Re n < \dfrac 2 3$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty \paren {\dfrac {n^{\overline k} } {k!} }^3
      | r = \dfrac {\map \Gamma {1 - \dfrac {3 n} 2} } {\map {\Gamma^3} {1 - \dfrac n 2} } \map \cos {\dfrac {\pi n} 2} 
      | c = 
}}
{{end-eqn}}

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^\infty \paren {\dfrac {n^{\overline k} } {k!} }^3
      | r = \dfrac {6 \map \sin {\dfrac {\pi n} 2} \map \sin {\pi n} } {\pi^2 n^2 \paren {1 + 2 \map \cos {\pi n} } } \times \dfrac {\map {\Gamma^3} {\dfrac n 2 + 1} } {\map \Gamma {\dfrac {3 n} 2 + 1} }
      | c = {{Corollary|Dixon's Hypergeometric Theorem|1}}
}}
{{eqn | r = \dfrac {6 \map \sin {\dfrac {\pi n} 2} \map \sin {\pi n} } {\pi^2 n^2 \paren {1 + 2 \map \cos {\pi n} } } \times \dfrac {\paren {\dfrac n 2}^3 \map {\Gamma^3} {\dfrac n 2} } {\paren {\dfrac {3 n} 2} \map \Gamma {\dfrac {3 n} 2} }
      | c = {{Defof|Gamma Function}}
}}
{{eqn | r = \dfrac {\map \sin {\dfrac {\pi n} 2} \map \sin {\pi n} } {2 \pi^2 \paren {1 + 2 \map \cos {\pi n} } } \times \dfrac {\map {\Gamma^3} {\dfrac n 2} } {\map \Gamma {\dfrac {3 n} 2} } \times \dfrac {\map {\Gamma^3} {1 - \dfrac n 2} } {\map {\Gamma^3} {1 - \dfrac n 2} } \times \dfrac {\map \Gamma {1 - \dfrac {3 n} 2} } {\map \Gamma {1 - \dfrac {3 n} 2} }
      | c = multiplying by $1$ and simplifying
}}
{{eqn | r = \dfrac {\map \sin {\dfrac {\pi n} 2} \map \sin {\pi n} \map \Gamma {1 - \dfrac {3 n} 2} } {2 \pi^2 \paren {1 + 2 \map \cos {\pi n} } \map {\Gamma^3} {1 - \dfrac n 2} } \times \dfrac {\pi^3 } {\paren {\map \sin {\dfrac {\pi n} 2}^3 } } \times \dfrac {\map \sin {\dfrac {3\pi n} 2} } \pi
      | c = Euler's Reflection Formula
}}
{{eqn | r = \dfrac {\map \sin {\dfrac {\pi n} 2} \map \sin {\pi n} \map \Gamma {1 - \dfrac {3 n} 2} } {2 \map \sin {\dfrac {3\pi n} 2} \map {\Gamma^3} {1 - \dfrac n 2} } \times \dfrac 1 {\paren {\map \sin {\dfrac {\pi n} 2}^2} } \times \map \sin {\dfrac {3\pi n} 2} 
      | c = Sine of Integer Multiple of Argument: Formulation 6 and simplifying
}}
{{eqn | r = \dfrac {\map \sin {\pi n} \map \Gamma {1 - \dfrac {3 n} 2} } {2 \map \sin {\dfrac {\pi n} 2} \map {\Gamma^3} {1 - \dfrac n 2} }
      | c = simplifying
}}
{{eqn | r = \dfrac {2 \map \sin {\dfrac {\pi n} 2} \map \cos {\dfrac {\pi n} 2}  \map \Gamma {1 - \dfrac {3 n} 2} } {2 \map \sin {\dfrac {\pi n} 2} \map {\Gamma^3} {1 - \dfrac n 2} }
      | c = Double Angle Formula for Sine
}}
{{eqn | r = \dfrac {\map \Gamma {1 - \dfrac {3 n} 2} } {\map {\Gamma^3} {1 - \dfrac n 2}  } \map \cos {\dfrac {\pi n} 2} 
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Frank Morley|cat = Morley, Frank}}
Theorem: Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 2

The following formulation of the Axiom of Choice:

Proof:
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of sets all of which are non-empty, indexed by $I$ which is also non-empty.
By hypothesis, Formulation 1 holds.
That is, there exists a choice function on every set of non-empty sets.
Let $f$ be a choice function on $\set{X_i}$.
Let $x_i = \map f {X_i}$.
By definition of choice function, each $x_i \in X_i$.
Therefore, $\family {x_i}_{i \mathop \in I}$ satisfies Formulation 2.
Theorem: Real Cosine Function has Zeroes

:$\exists \theta \in \R: \map \cos \theta = 0 \text { and } \map \cos {-\theta} = 0$

Proof:
{{AimForCont}} $\cos x$ is positive everywhere on $\R$.
From Derivative of Cosine Function and Derivative of Sine Function:
:$\map {\dfrac {\d^2} {\d x^2} } {\cos x} = \map {\dfrac \d {\d x} } {-\sin x} = -\cos x$
Thus $\map {\dfrac {\d^2} {\d x^2} } {\cos x} = -\cos x$ would always be negative.
Thus from Second Derivative of Concave Real Function is Non-Positive, $\cos x$ would be concave everywhere on $\R$.
But from Real Cosine Function is Bounded, $\cos x$ is bounded on $\R$.
By Differentiable Bounded Concave Real Function is Constant, $\cos x$ would then be a constant function.
This contradicts the fact that $\cos x$ is not a constant function.
Thus, by Proof by Contradiction, $\cos x$ cannot be positive everywhere on $\R$.
Therefore:
:$\exists \phi \in \R: \cos \phi < 0$
From Cosine of Zero is One:
:$\cos 0 = 1 > 0$
Since the Cosine Function is Continuous, by the Intermediate Value Theorem:
:$\exists \theta \in \R: \cos \theta = 0$
Hence from Cosine Function is Even:
:$\map \cos {-\theta} = 0$
{{qed}}
Category:Cosine Function
evcqzk890vks74af4nswybaiytn89n2
Theorem: Sine and Cosine are Periodic on Reals/Sine/Proof 1

The sine function is periodic with the same period as the cosine function.
:820px

Proof:
Since Real Cosine Function is Periodic, let $K$ be its period.
Then:
:$\cos K = \map \cos {0 + K} = \cos 0$
Because Cosine of Zero is One:
:$\cos K = 1$
Furthermore:
{{begin-eqn}}
{{eqn | l = \cos^2 K + \sin^2 K
      | r = 1
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | l = \sin^2 K
      | r = 0
      | c = $\cos K = 1$
}}
{{eqn | l = \sin K
      | r = 0
}}
{{end-eqn}}
Then, the following holds:
{{begin-eqn}}
{{eqn | l = \map \sin {x + K}
      | r = \sin x \cos K + \cos x \sin K
      | c = Sine of Sum
}}
{{eqn | r = \sin x \cdot 1 + \cos x \cdot 0
      | c = $\cos K = 1$; $\sin K = 0$
}}
{{eqn | r = \sin x}}
{{end-eqn}}
Thus $\sin$ is periodic with some period $L \leq K$.
{{qed|lemma}}
The following also hold:
{{begin-eqn}}
{{eqn | l = \sin L
      | r = \map \sin {0 + L}
}}
{{eqn | r = \sin 0
      | c = Period of Periodic Real Function
}}
{{eqn | r = 0
      | c = Sine of Zero is Zero
}}
{{eqn | l = \cos L
      | r = \bigvalueat { \dfrac{\d \sin x} {\d x} } {x \mathop = L}
      | c = Derivative of Sine Function
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac{\map \sin {L + h} - \map \sin L}{h}
      | c = Derivative of Real Function at Point
}}
{{eqn | r = \lim_{h \mathop \to 0} \dfrac{\map \sin {0 + h} - \map \sin 0}{h}
      | c = Period of Periodic Real Function
}}
{{eqn | r = \bigvalueat { \dfrac{\d \sin x} {\d x} } {x \mathop = 0}
      | c = Derivative of Real Function at Point
}}
{{eqn | r = \cos 0
      | c = Derivative of Sine Function
}}
{{eqn | r = 1
      | c = Cosine of Zero is One
}}
{{end-eqn}}
So we may conclude:
{{begin-eqn}}
{{eqn | l = \map \cos {x + L}
      | r = \cos x \cos L - \sin x \sin L
      | c = Cosine of Sum
}}
{{eqn | r = \cos x \cdot 1 - \sin x \cdot 0
      | c = $\cos L = 1$ and $\sin L = 0$
}}
{{eqn | r = \cos x}}
{{end-eqn}}
Therefore the period of cosine is at most that of sine:
:$K \leq L$
But if $K \leq L \leq K$ then:
:$K = L$
{{qed}}
Theorem: Constant Loop is Loop

Let $\struct {T, \tau}$ be a topological space.
Let $p \in T$.
Let $c_p : \closedint 0 1 \to T$ be the constant mapping defined by:
:$\forall t \in \closedint 0 1 : \map {c_p} t = p$
Then $c_p$ is a loop in $T$.

Proof:
From Constant Mapping is Continuous, it follows that $c_p$ is continuous.
By definition of path, it follows that $c_p$ is a path in $T$.
We have:
:$\map {c_p} 0 = \map {c_p} 1 = p$
Hence, $c_p$ is a loop in $T$.
{{qed}}
Category:Loops (Topology)
cmezlz5elde4qtdq5udjhkpwekca3ug
Theorem: Lasota-Yorke Inequality/One-Sided Shift Space of Finite Type

Let $\struct {X_\mathbf A ^+, \sigma}$ be a one-sided shift of finite type.
Let $F_\theta ^+$ be the space of Lipschitz functions on $X_\mathbf A ^+$.
Let $\norm \cdot_\theta$ be the Lipschitz norm on $F_\theta ^+$.
Let $\norm \cdot_\infty$ be the supremum norm on $F_\theta ^+$.
Let $f \in F_\theta ^+$.
Let $u := \map \Re f$ be the real part of $f$.
Let $\LL_f$ and $\LL_u$ denote the transfer operators.
If $\LL_u$ is normalized, then there is a $C > 0$ such that:
:$\norm {\LL_f ^n w}_\theta \le C \norm w_\infty + \theta ^n \norm w_\theta$
for all $w \in F_\theta ^+$ and $n \in \N$.

Proof:
Recall the basic inequality:
:There exists a $C_0 > 0$ so that we have:
{{begin-eqn}}
{{eqn | n = 1
      | l = \size {\LL_f ^n w}_\theta
      | o = \le
      | r = C_0 \norm w_\infty + \theta ^n \size w_\theta
}}
{{end-eqn}}
:for all $w \in F_\theta ^+$ and $n \in \N$.
On the other hand, we have:
{{begin-eqn}}
{{eqn | n = 2
      | l = \norm {\LL_f w}_\infty
      | o = \le
      | r = \norm w_\infty
}}
{{end-eqn}}
since for all $x \in X_\mathbf A ^+$:
{{begin-eqn}}
{{eqn | l = \cmod {\map {\LL_f w } x}
      | r = \cmod {\sum_{y \mathop \in \map {\sigma^{-1} } x} e^{\map f y} \map w y}
}}
{{eqn | o = \le
      | r = \sum_{y \mathop \in \map {\sigma^{-1} } x} \cmod {e^{\map f y} \map w y}
}}
{{eqn | r = \sum_{y \mathop \in \map {\sigma^{-1} } x} e^{\map u y} \cmod {\map w y}
}}
{{eqn | o = \le
      | r = \norm w_\infty \sum_{y \mathop \in \map {\sigma^{-1} } x} e^{\map u y}
}}
{{eqn | r = \norm w_\infty \map {\LL _u 1} x
}}
{{eqn | r = \norm w_\infty
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \norm {\LL_f ^n w}_\theta
      | r = \norm {\LL_f ^n w}_\infty + \size {\LL_f ^n w}_\theta
      | c = {{Defof|Lipschitz Norm}}
}}
{{eqn | o = \le
      | r = \norm w_\infty + \size {\LL_f ^n w}_\theta
      | c = by $(2)$
}}
{{eqn | o = \le
      | r = \norm w_\infty + C_0 \norm w_\infty + \theta ^n \size w_\theta
      | c = by $(1)$
}}
{{eqn | r = \paren {C_0 + 1} \norm w_\infty + \theta ^n \size w_\theta
}}
{{eqn | o = \le
      | r = \paren {C_0 + 1} \norm w_\infty + \theta ^n \size w_\theta + \theta ^n \norm w_\infty
}}
{{eqn | r = \paren {C_0 + 1} \norm w_\infty + \theta ^n \norm w_\theta
      | c = {{Defof|Lipschitz Norm}}
}}
{{end-eqn}}
{{finish|Details}}
{{Namedfor|Andrzej Lasota|cat = Lasota|name2 = James Alan Yorke|cat2 = Yorke}}
Category:Ergodic Theory
Category:Functional Analysis
sn0uu05rwytxxkv3rweaufj1x9ahotb
Theorem: Locally Euclidean Space has Countable Local Basis Homeomorphic to Open Balls

Let $M$ be a locally Euclidean space of some dimension $d$.
Let $m \in M$.
Then: 
:there exists a countable local basis $\family{U_n}_{n \in \N}$ of $m$ where each $U_n$ is the homeomorphic image of an open ball of $\R^d$

Proof:
By definition of a locally Euclidean space:
:there exists an open neighbourhood $U$ of $m$ which is homeomorphic to an open subset $V$ of Euclidean space $\R^d$.
Let $\phi: U \to V$ be a homeomorphism.
By definition of the Euclidean space $\R^d$ the topology on $\R^d$ is the topology induced by the metric:
:$\ds \map {d_2} {x, y} := \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{1 / 2}$
where $x = \tuple {x_1, x_2, \ldots, x_d}, y = \tuple {y_1, y_2, \ldots, y_d} \in \R^d$.
By definition of the induced topology:
:$\exists \epsilon > 0 : \map {B_\epsilon} {\map \phi m} \subseteq V$
where $\map {B_\epsilon} {\map \phi m}$ is the open ball of radius $\epsilon$ and center $\map \phi m$
Consider the set of open balls:
:$\BB_m = \set{\map {B_{\epsilon / n}} {\map \phi m} : n \in \N_{>0}}$
From Sequence of Reciprocals is Null Sequence:
:the sequence $\sequence{\dfrac \epsilon n}_{n \in \N_{>0}}$ is a null sequence
From Null Sequence induces Local Basis in Metric Space:
:$\BB_m$ is a countable local basis of $\map \phi m$ in $\R^d$
From Local Basis of Open Subspace iff Local Basis:
:$\BB_m$ is a countable local basis of $\map \phi m$ in $V$
For each $n \in \N_{>0}$, let:
:$U_n = \phi^{-1} \sqbrk {\map {B_{\epsilon / n}} {\map \phi m}}$
Consider the set:
:$\BB'_m = \set{U_n : n \in \N_{>0}}$
From Inverse of Homeomorphism is Homeomorphism:
:$\phi^{-1} : U \to V$ is a homeomorphism
From Homeomorphic Image of Local Basis is Local Basis:
:$\BB'_m$ is a local basis of $m$ in $U$
From Local Basis of Open Subspace iff Local Basis:
:$\BB'_m$ is a local basis of $m$ in $M$
From Restriction of Homeomorphism is Homeomorphism:
:$\forall n \in \N_{>0}: U_n = \phi^{-1} \sqbrk {\map {B_{\epsilon / n}} {\map \phi m}}$ is homeomorphic to the open ball $\map {B_{\epsilon / n}} {\map \phi m}$
The result follows.
{{qed}}
Category:Locally Euclidean Spaces
l1tjvq105ch7awymr2xti4uwtf1yybe
Theorem: Homeomorphic Image of Local Basis is Local Basis

Let $T_\alpha = \struct{S_\alpha, \tau_\alpha}$ and $T_\beta = \struct{S_\beta, \tau_\beta}$ be topological spaces.
Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.
Let $s \in S_\alpha$.
Let $\BB$ be a local basis of $s$ in $T_\alpha$.
Then:
:$\BB' = \set{ \phi \sqbrk B : B \in \BB}$ is a local basis of $\map \phi s$ in $T_\beta$

Proof:
By definition of homeomorphism:
:$\forall U \in \tau_\alpha : \phi \sqbrk U \in \tau_\beta$
Hence:
:$\BB'$ is a set of open sets in $T_\beta$ containing $\map \phi s$
Let $U \in \tau_\beta$ containing $\map \phi s$.
By definition of homeomorphism:
:$\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$
By definition of local basis:
:$\exists V \in \BB: s \in V \subseteq \phi^{-1} \sqbrk U$
By definition of image of subset:
:$\map \phi s \in \phi \sqbrk V$
From Subset Maps to Subset:
:$\phi \sqbrk V \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$
From Image of Preimage under Mapping:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$
From Subset Relation is Transitive]:
:$\phi \sqbrk V \subseteq U$
Hence:
:$\exists \phi \sqbrk V \in \BB': \map \phi s \in \phi \sqbrk V \subseteq U$
{{qed}}
Theorem: Local Basis of Open Subspace iff Local Basis

Let $T = \struct {S, \tau}$ be a topological space.
Let $U \subseteq S$ be an open subset
Let $\tau_U$ denote the subspace topology on $U$.
Let $s \in U$.
Let $\BB \subseteq \powerset U$.
Then:
:$\BB$ is a local basis of $s$ in $\struct {U, \tau_U}$
{{iff}}:
:$\BB$ is a local basis of $s$ in $\struct {S, \tau}$.

Proof:
Let $\map \BB s$ denote the set of open sets containing $s$ in $\struct {S, \tau}$
Let $\map \CC s$ denote the set of open sets containing $s$ in $\struct {U, \tau_U}$
From Open Set in Open Subspace:
:$\BB \subseteq \map \BB s$ {{iff}} $\BB \subseteq \map \CC s$
Theorem: Distance between Points in Regular Hexagon

Let $H$ be a regular hexagon embedded in the Euclidean plane $\R^2$.
Let $s \in \R_{>0}$ be the side length of $H$.
Let $\mathbf x, \mathbf y \in \R^2$ such that $\mathbf x$ and $\mathbf y$ lie in the interior of $H$, or on the circumference of $H$.
Then:
:$\map d {\mathbf x, \mathbf y} \le 2s$
where $\map d {\mathbf x, \mathbf y}$ denotes the Euclidean distance between $\mathbf x$ and $\mathbf y$.

Proof:
From Regular Polygon is Cyclic, it follows that $H$ can be inscribed in a circle with center $\mathbf c$.
The circumcircle intersects all vertices of $H$.
From Regular Hexagon is composed of Equilateral Triangles, it follows that the side length $s$ is equal to the distance from $\mathbf c$ to any vertex of $H$.
It follows that the radius of the circumcircle is equal to $s$.
Hence:
{{begin-eqn}}
{{eqn | o = \le
      | l = \map d {\mathbf x, \mathbf y} 
      | r = \map d {\mathbf x, \mathbf c} + \map d {\mathbf c, \mathbf y}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | o = \le
      | r = s + s
      | c = as $\mathbf x, \mathbf y$ lie in $H$, they also lie in the circumcircle 
}}
{{eqn | r = 2s
}}
{{end-eqn}}
{{qed}}
Category:Hexagons
fjzli9wmmnvdk0wg1v4cfekuo39g91y
Theorem: Homeomorphic Image of Neighborhood Basis is Neighborhood Basis

Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.
Let $s \in S_\alpha$.
Let $\NN$ be a neighborhood basis of $s$ in $T_\alpha$.
Then:
:$\NN' = \set {\phi \sqbrk N : N \in \NN}$ is a neighborhood basis of $\map \phi s$ in $T_\beta$.

Proof:
Let $N$ be a neighborhood of $s$ in $T_\alpha$.
By definition of neighborhood:
:$\exists V \in \tau_\alpha : s \in V \subseteq N$
By definition of image of subset:
:$\map \phi s \in \phi \sqbrk V$
From Subset Maps to Subset:
:$\phi \sqbrk V \subseteq \phi \sqbrk N$
By definition of homeomorphism:
:$\phi \sqbrk V \in \tau_\beta$
It follows that $\phi \sqbrk N$ is a neighborhood of $\map \phi s$ in $T_\beta$.
Hence:
:$\NN' = \set{ \phi \sqbrk N : N \in \NN}$ is a set of neighborhoods of $\map \phi s$ in $T_\beta$
Let $M$ be a neighborhood of $\map \phi s$ in $T_\beta$.
By definition of neighborhood:
:$\exists U \in \tau_\beta : \map \phi s \in U \subseteq M$
By definition of homeomorphism:
:$\phi^{-1} \sqbrk U \in \tau_\alpha$ containing $s$
By definition of neighborhood basis:
:$\exists N \in \NN: s \in N \subseteq \phi^{-1} \sqbrk U$
By definition of image of subset:
:$\map \phi s \in \phi \sqbrk N$
From Subset Maps to Subset:
:$\phi \sqbrk N \subseteq \phi \sqbrk {\phi^{-1} \sqbrk U}$
From Image of Preimage under Mapping:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} \subseteq U$
From Subset Relation is Transitive:
:$\phi \sqbrk N \subseteq M$
Hence:
:$\exists \phi \sqbrk N \in \BB': \map \phi s \in \phi \sqbrk N \subseteq M$
The result follows.
{{qed}}
Theorem: Neighborhood Basis of Open Subspace iff Neighborhood Basis

Let $T = \struct{S, \tau}$ be a topological space.
Let $U \subseteq S$ be an open subset
Let $\tau_U$ denote the subspace topology on $U$.
Let $s \in U$.
Let $\NN \subseteq \powerset U$.
Then:
:$\NN$ is a neighborhood basis of $s$ in $\struct{U, \tau_U}$ {{iff}} $\NN$ is a neighborhood basis of $s$ in $\struct{S, \tau}$

Proof:
Let $\map \NN s$ denote the set of neighborhoods of $s$ in $\struct{S, \tau}$
Let $\map \MM s$ denote the set of neighborhoods of $s$ in $\struct{U, \tau_U}$
From Neighborhood in Open Subspace:
:$\NN \subseteq \map \NN s$ {{iff}} $\NN \subseteq \map \MM s$
Theorem: Locally Euclidean Space has Countable Neighborhood Basis Homeomorphic to Closed Balls

Let $M$ be a locally Euclidean space of some dimension $d$.
Let $m \in M$.
Then: 
:there exists a countable neighborhood basis $\family{N_n}_{n \in \N}$ of $m$ where each $N_n$ is the homeomorphic image of a closed ball of $\R^d$

Proof:
By definition of a locally Euclidean space:
:there exists an open neighbourhood $U$ of $m$ which is homeomorphic to an open subset $V$ of Euclidean space $\R^d$.
Let $\phi: U \to V$ be a homeomorphism.
By definition of the Euclidean space $\R^d$ the topology on $\R^d$ is the topology induced by the metric:
:$\ds \map {d_2} {x, y} := \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{1 / 2}$
where $x = \tuple {x_1, x_2, \ldots, x_d}, y = \tuple {y_1, y_2, \ldots, y_d} \in \R^d$.
By definition of the induced topology:
:$\exists \epsilon > 0 : \map {B_\epsilon} {\map \phi m} \subseteq V$
where $\map {B_\epsilon} {\map \phi m}$ is the open ball of radius $\epsilon$ and center $\map \phi m$
Consider the set of closed balls:
:$\NN_m = \set{\map {B^-_{\epsilon / 2 n}} {\map \phi m} : n \in \N_{>0}}$
From Sequence of Reciprocals is Null Sequence:
:the sequence $\sequence{\dfrac \epsilon {2 n}}_{n \in \N_{>0}}$ is a null sequence
From Null Sequence induces Neighborhood Basis of Closed Sets in Metric Space:
:$\NN_m$ is a countable neighborhood basis of $\map \phi m$ in $\R^d$
From Neighborhood Basis of Open Subspace iff Neighborhood Basis:
:$\NN_m$ is a countable local basis of $\map \phi m$ in $V$
For each $n \in \N_{>0}$, let:
:$U_n = \phi^{-1} \sqbrk {\map {B_{\epsilon / n}} {\map \phi m}}$
Consider the set:
:$\NN'_m = \set{U_n : n \in \N_{>0}}$
From Inverse of Homeomorphism is Homeomorphism:
:$\phi^{-1} : U \to V$ is a homeomorphism
From Homeomorphic Image of Neighborhood Basis is Neighborhood Basis and Continuity Defined from Closed Sets:
:$\NN'_m$ is a Neighborhood basis of $m$ in $U$
From Neighborhood Basis of Open Subspace iff Neighborhood Basis:
:$\NN'_m$ is a local basis of $m$ in $M$
From Restriction of Homeomorphism is Homeomorphism:
:$\forall n \in \N_{>0}: N_n = \phi^{-1} \sqbrk {\map {B_{\epsilon / n}^-} {\map \phi m}}$ is homeomorphic to the closed ball $\map {B_{\epsilon / n}^-} {\map \phi m}$
The result follows.
{{qed}}
Category:Locally Euclidean Spaces
tn62fb3302s2sdkt630g4kpvw9rm6q5
Theorem: Generated Submodule may not equal Set of Linear Combinations

Let $\struct { R, +_R, \times_R }$ be a ring.
Let $\struct { M, +_M, \circ }_R$ be an $R$-module.
Let $S$ be a subset of $M$.
Let $H_1$ be the submodule generated by $S$.
Let $H_2$ be the set of all linear combinations of elements of $S$. 
Then it is possible to select $\struct { R, +_R, \times_R }$, $\struct { M, +_M, \circ }_R$, and $S$ such that:
:$H_1 \ne H_2$

Proof:
Let $R = \set {2 k : k \in \Z}$ be the set of all even integers.
From Integers form Commutative Ring, it follows that $\Z$ is a ring.
From Ideal of Ring/Examples/Set of Even Integers, it follows that $R$ is an ideal of $\Z$.
From Ideal is Subring, it follows that $\struct {R, +, \times}$ is a ring, where $+$ denotes integer addition, and $\times$ denotes integer multiplication.
Set $M := R$.
It follows that $\struct {R, +, \times }_R$ is a module over $R$, where the scalar multiplication $\times$ is defined as the integer multiplication $\times$ in $R$.
That is, $R$ is considered as a module over itself.
Set $S := \set 2$.
Let $H_1$ be the submodule generated by $S$, which means that:
:$\ds H_1 := \bigcap \set { M' \subseteq R : \set 2 \subseteq M', \textrm {$M'$ is a submodule of $R$} } $
Let $M'$ be a submodule of $R$ such that $2 \in M'$.
We show that $M' = R$.
From Congruence Modulo Integer is Equivalence Relation, it follows that all $x \in \Z$ is of the form $x := 4 k' + r$ for some $k \in \Z, r \in \set{ 0,1,2,3 }$.
Suppose $x \in R$ such that $x = 4 k'$ for some $k' \in \Z$
Then:
:$4 k' = 2 k' \times 2$
As $M'$ is closed for scalar product, it follows that $4 k' \in M'$.
Suppose $x \in R$ such that $x = 4 k' + 2$ for some $k' \in \Z$.
As $M'$ is closed for addition, and $2 \in M'$, it follows that $4 k' + 2 \in M'$.
If instead $x = 4 k' + 1$ or $x = 4 k' + 3$, then $x$ cannot be an even integer, so $x \notin R$.
It follows that $R \subseteq M'$.
As $M'$ is a submodule of $R$, it follows that $M' \subseteq R$.
As we have $M' = R$, it follows that:
:$\bigcap \set { M' \subseteq R : \set {2} \subseteq M', \textrm {$M'$ is a submodule of $R$} } = \bigcap \set {R} = R$
Let $H_2$ be the set of all linear combinations of elements of $\set {2}$, which means that:
:$H_2 = \ds \set {\sum_{i \mathop = 1}^n \lambda_i \times 2 : n \in \N_{\ge 1}, \lambda_1, \ldots, \lambda_n \in R}$
With $\ds \sum_{i \mathop= 1}^n \lambda_i \times 2 \in H_2$, it follows that:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop= 1}^n \lambda_i \times 2
      | r = \sum_{i \mathop= 1}^n \paren{ 2 k_i } \times 2
      | c = for some $k_1, \ldots, k_n \in \Z$
}}
{{eqn | r = 4 \sum_{i \mathop= 1}^n k_i
}}
{{end-eqn}}
As $\ds \sum_{i \mathop= 1}^n k_i \in \Z$, it follows that:
:$H_2 = \set { 4 k : k \in Z}$
Hence, $H_1 \ne H_2$.
{{qed}}