nyuuzyou/wikis huggingface dataset:
Implicit Differentiation/Examples/Arbitrary Example 1

== Example of Implicit Differentiation ==
Consider the implicit function:
:$(1): \quad y^3 + 2 x^2 y + 8 = 0$
Using the technique of implicit differentiation, $(1)$ is differentiated $x$ thus:
:$3 y^2 \dfrac {\d y} {\d x} + 4 x y + 2 x^2 \dfrac {\d y} {\d x} = 0$
from the Product Rule for Derivatives and the Power Rule for Derivatives.
This can then be algebraically manipulated to arrive at:
:$\dfrac {\d y} {\d x} = -\dfrac {4 x y} {3 y^2 + 2 x^2}$
== Sources ==
* 
* 
Category:Examples of Implicit Differentiation
Implicit Function/Examples/x^2 + y^2 + 1 = 0

== Examples of Implicit Functions ==
Consider the equation:
:$(1): \quad x^2 + y^2 + 1 = 0$
where $x, y \in \R$ are real variables.
Solving for $y$, we obtain:
:$y = \pm \sqrt {-1 - x^2}$
and it is seen that no $y \in \R$ can satisfy this equation.
Hence $(1)$ does not define a real function.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function
Category:Examples of Implicit Functions
Implicit Function/Examples/x^2 + y^2 - r^2 = 0

== Example of Implicit Function ==
Consider the equation:
:$(1): \quad x^2 + y^2 - r^2 = 0$
where:
:$x, y \in \R$ are real variables
:$r \in \R_{>0}$ is a strictly positive real constant.
Then $(1)$ defines $y$ as an implicit function of $x$ on the closed interval $\closedint {-r} r$.
== Proof ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function
:::for the special case of $r = 5$
* 
:::for the special case of $r = 1$
* 
:::for the special case of $r = 1$
Category:Examples of Implicit Functions
Implicit Function/Examples/x^2 + y^2 - r^2 = 0/Proof

== Example of Implicit Function ==
== Proof ==
Solving for $y$, we obtain:
:$y = \pm \sqrt {r^2 - x^2}$
As it stands, $(1)$ does not define a real function, because:
:$\text{(a)}: \quad y$ is not defined for values of $x$ outside the range $-r \le x \le r$
:$\text{(b)}: \quad$ For all $x$ within the range $-r \le x \le r$, there are two possible values of $y$ that can be taken.
This can be corrected by:
:$\text{(a)}: \quad$ Specifying the domain to be the closed interval $\closedint {-r} r$ (or a subset thereof)
:$\text{(b)}: \quad$ Specifying which value of $y$ that is to be chosen, for example:
:::$\forall x \in \closedint {-r} r: y = \sqrt {r^2 - x^2}$ (where $\sqrt {}$ in this context specifically means the positive square root)
:::$\forall x \in \closedint {-r} r: y = -\sqrt {r^2 - x^2}$
:::$\forall x \in \closedint {-r} r: y = \begin {cases} \sqrt {r^2 - x^2} & : x \le 0 \\ -\sqrt {r^2 - x^2} & : x > 0 \end {cases}$
Thus we have:
:$y = \map f {x, \map g x}$
where:
:$\map f {x, \map g x} = x^2 + \paren {\sqrt {r^2 - x^2} }^2 - r^2 = 0$
thereby demonstrating that $y$ is an implicit function of $x$.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function: Example $2.84$
:::for the special case of $r = 5$
Category:Examples of Implicit Functions
Implicit Function/Examples/x^3 + y^3 - 3 x y = 0

== Examples of Implicit Functions ==
Consider the equation:
:$(1): \quad x^3 + y^3 - 3 x y = 0$
where $x, y \in \R$ are real variables.
Then $(1)$ defines $y$ as an implicit function of $x$ for all $x \in \R$.
== Proof ==
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function
Category:Examples of Implicit Functions
Implicit Function/Examples/x^3 + y^3 - 3 x y = 0/Proof

== Examples of Implicit Functions ==
== Proof ==
This is the Cartesian form of the equation for the folium of Descartes:
:$x^3 + y^3 - 3 a x y = 0$
for $a = 1$:
:
It is seen that when $x \le 0$, $y$ is uniquely determined by $x$.
Also, from Maximum Abscissa for Loop of Folium of Descartes, $y$ is also uniquely determined by $x$ when $x > 2^{2/3}$.
In between those values, for $y$ to be defined as a function of $x$ it is necessary to choose one of the $3$ possible values that $y$ can take for each $y$.
For example:
: $\quad$ $\quad$ 
Thus we have:
:$y = \map f {x, \map g x}$
where $\map g x$ is not straightforward to define.
This demonstrates that $y$ is an implicit function of $x$.
== Sources ==
* : Chapter $1$: Basic Concepts: Lesson $2 \text D$: Implicit Function: Example $2.85$
Category:Examples of Implicit Functions
Implicit Function Theorem/Real Functions

== Theorem ==
Let $n$ and $k$ be natural numbers.
Let $\Omega \subset \R^n \times \R^k$ be open.
Let $f: \Omega \to \R^k$ be continuous.
Let the partial derivatives of $f$ with respect to $\R^k$ be continuous.
Let $\tuple {a, b} \in \Omega$, with $a\in \R^n$ and $b\in \R^k$.
Let $\map f {a, b} = 0$.
For $\tuple {x_0, y_0} \in \Omega$, let $D_2 \map f {x_0, y_0}$ denote the total derivative of the function $y \mapsto \map f {x_0, y}$ at $y_0$.
Let the linear map $D_2 \map f {a, b}$ be invertible.
Then there exist neighborhoods $U \subset \Omega$ of $a$ and $V \subset \R^k$ of $b$ such that there exists a unique function $g: U \to V$ such that $\map f {x, \map g x} = 0$ for all $x \in U$.
Moreover, $g$ is continuous.
== Proof 1 ==
== Proof Using the Inverse Function Theorem ==
== Examples ==
== Sources ==
* : Appendix $\text C$
Category:Implicit Function Theorem
Implicit Function Theorem/Real Functions/Proof 1

== Theorem ==
== Outline of Proof ==
We apply the Uniform Contraction Mapping Theorem to the function:
:$F: \R^n \times \R^k \to \R^k : \map F {x, y} = y - \map {D_2 \map f {a, b}^{-1} } {\map f {x, y} }$
for appropriate $x \in \R^n$ and $y \in \R^k$.
== Proof ==
==== Reduction to $\tuple {a, b} = \tuple {0, 0}$ ====
We may assume $\tuple {a, b} = \tuple {0, 0}$.
Define:
:$F: \Omega \to \R^k: \map F {x, y} = y - \map {D_2 \map f {a, b}^{-1} } {\map f {x, y} }$
By Linear Function is Continuous, $D_2 \map f {a, b}^{-1}$ is continuous on $\R^k$.
Thus $F$ is continuous on $\Omega$.
==== $F$ is locally a uniform contraction ====
Let $r_1 > 0$ such that the open ball $\map B {0, r_1} \subseteq \Omega$.
We have, for $\tuple {x, y_1}, \tuple {x, y_2} \in \map B {0, r_1}$, by the Mean Value Inequality:
:$\ds \norm {\map F {x, y_2} - \map F {x, y_1} } \le \sup_{y \mathop \in \closedint {y_1} {y_2} } \norm {D_2 \map F {x, y} } \cdot \norm {y_2 - y_1}$
We have, for $\tuple {x, y} \in \Omega$:
:$D_2 \map F {x, y} = I - \paren {D_2 \map F {0, 0} }^{-1} \circ D_2 \map F {x, y}$
where $I$ is the identity mapping.
Thus $D_2 F$ is continuous in $\Omega$.
By definition, $D_2 \map F {0, 0} = 0$.
By continuity, there exists $r_2 > 0$ such that:
:$\norm {D_2 \map F {x, y} } \le \dfrac 1 2$ for $\norm {\tuple {x, y} } \le r_2$
From the above inequality, $F$ is a uniform contraction mapping on the open ball $\map B {0, r_2}$ for the Lipschitz Constant $\dfrac 1 2$.
==== Constructing a stable neighborhood ====
We have, for $x, y \in \map {\overline B} {0, r_2}$:
Because $\map F {0, 0} = 0$ and $F$ is continuous, there exists $r_3>0$ such that $\norm {\map F {x, 0} } \le \dfrac {r_2} 2$.
Then:
:$\norm {\map F {x, y} } \le r_2$ for $\norm y \le r_2$
and:
:$\norm x \le r_3$
Thus the restriction of $F$ to $\map B {0, r_3} \times \map {\overline B} {0, r_2}$ is a uniform contraction:
:$F: \map B {0, r_3} \times \map {\overline B} {0, r_2} \to \map {\overline B} {0, r_2}$
==== Applying the Uniform Contraction Theorem ====
By Subspace of Complete Metric Space is Closed iff Complete, $\map {\overline B} {0, r_2}$ is complete.
By the Uniform Contraction Mapping Theorem, there exists a unique mapping $g: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ such that $\map F {x, \map g x} = \map g x$.
Moreover, $g$ is continuous.
A mapping $h: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ satisfies $\map F {x, \map h x} = \map h x$ it satisfies $\map {D_2 \map f {a, b}^{-1} } {\map f {x, \map h x} } = 0$ it satisfies $\map f {x, \map h x} = 0$, because $D_2 \map f {a, b}^{-1}$ is invertible.
Thus $g$ is the unique mapping $g: \map B {0, r_3} \to \map {\overline B} {0, r_2}$ such that $\map f {x, \map g x} = 0$.
Category:Implicit Function Theorem
Implicit Function Theorem for Differentiable Real Functions

== Theorem ==
Let $\Omega \subset \R^{n+k}$ be open.
Let $f : \Omega \to \R^k$ be differentiable.
Let the $i$th partial derivatives of $f$ be continuous in $\Omega$ for $n+1 \leq i \leq n+k$.
Let $(a,b) \in \Omega$, with $a\in \R^n$ and $b\in \R^k$.
Let $f(a,b) = 0$.
For $(x_0,y_0)\in\Omega$, let $D_2 f(x_0,y_0)$ denote the differential of the function $y\mapsto f(x_0, y)$ at $y_0$.
Let the linear map $D_2 f(a,b)$ be invertible.
Then there exist neighborhoods $U\subset\Omega$ of $a$ and $V\subset\R^k$ of $b$ such that there exists a unique function $g : U \to V$ such that $f(x, g(x)) = 0$ for all $x\in U$.
Moreover, $g$ is differentiable, and its differential satisfies:
:$dg (x) = - \left( (D_2f)(x, g(x)) \right)^{-1} \circ (D_1 f)(x, g(x))$ for all $x\in U$.
== Proof ==
Category:Implicit Functions
Implicit Function Theorem for Lipschitz Contraction at Point

== Theorem ==
Let $M$ and $N$ be metric spaces.
Let $M$ be complete.
Let $f: M \times N \to M$ be a uniform contraction.
Then for all $t \in N$ there exists a unique $\map g t \in M$ such that $\map f {\map g t, t} = \map g t$, and if $f$ is Lipschitz continuous at a point $\tuple {\map g t, t}$, then $g$ is Lipschitz continuous at $t$.
== Proof ==
For every $t \in N$, the mapping:
:$f_t : M \to M : x \mapsto \map f {x, t}$ is a contraction mapping.
By the Banach Fixed-Point Theorem, there exists a unique $\map g t \in M$ such that $\map {f_t} {\map g t} = \map g t$.
Let $f$ be Lipschitz continuous at $\tuple {\map g t, t}$.
We show that $g$ is Lipschitz continuous at $t$.
Let $K < 1$ be a uniform Lipschitz constant for $f$.
Let $L$ be a Lipschitz constant for $f$ at $a$.
Let $s\in N$.
Then
and thus:
Thus $g$ is Lipschitz continuous at $t$.
== Also see ==
* Implicit Function Theorem for Lipschitz Contractions
Category:Implicit Functions
Implicit Function Theorem for Lipschitz Contractions

== Theorem ==
Let $M$ and $N$ be metric spaces.
Let $M$ be complete.
Let $f : M \times N \to M$ be a Lipschitz continuous uniform contraction.
Then for all $t\in N$ there exists a unique $\map g t \in M$ such that $\map f {\map g t, t} = \map g t$, and the mapping $g : N \to M$ is Lipschitz continuous.
== Proof ==
For every $t\in N$, the mapping:
:$f_t : M \to M : x \mapsto \map f {x, t}$ is a contraction mapping.
By the Banach Fixed-Point Theorem, there exists a unique $\map g t \in M$ such that $\map {f_t} {\map g t} = \map g t$.
We show that $g$ is Lipschitz continuous.
Let $K<1$ be a uniform Lipschitz constant for $f$.
Let $L$ be a Lipschitz constant for $f$.
Let $s,t\in N$.
Then
and thus:
Thus $g$ is Lipschitz continuous.
Category:Implicit Functions
Implicit Function Theorem for Smooth Real Functions

== Theorem ==
Let $\Omega \subset \R^{n+k}$ be open.
Let $f : \Omega \to \R^k$ be smooth.
Let $(a,b) \in \Omega$, with $a\in \R^n$ and $b\in \R^k$.
Let $f(a,b) = 0$.
For $(x_0,y_0)\in\Omega$, let $D_2 f(x_0,y_0)$ denote the differential of the function $y\mapsto f(x_0, y)$ at $y_0$.
Let the linear map $D_2 f(a,b)$ be invertible.
Then there exist neighborhoods $U\subset\Omega$ of $a$ and $V\subset\R^k$ of $b$ such that there exists a unique function $g : U \to V$ such that $f(x, g(x)) = 0$ for all $x\in U$.
Moreover, $g$ is smooth, and its differential satisfies:
:$dg (x) = - \left( (D_2f)(x, g(x)) \right)^{-1} \circ (D_1 f)(x, g(x))$ for all $x\in U$.
== Proof ==
Category:Implicit Functions
Implicit Set Definition/Examples/Letters of the Alphabet

== Example of Implicit Set Definition ==
An example in natural language of implicit set definition is:
:$S := \set {A, B, C, D, \dotsc, Z}$
That is, $S$ is the set of capital letters of the alphabet.
This definition is actually ambiguous, as it is not made clear exactly which alphabet is under consideration here.
The English one is assumed by context.
== Sources ==
* : Chapter $1$ Numbers: Sets
Category:Examples of Sets
Implicitly Defined Real-Valued Function

== Theorem ==
Let $F: \struct {\mathbf X' \subseteq \R^{n + 1} } \to \struct {\mathbb I' \subseteq \R}$ have continuous partial derivatives. 
Let $\tuple {\mathbf x, z}$ denote an element of $\R^{n + 1}$, where $\mathbf x \in \R^n$ and $z \in \R$.
Suppose $\exists \tuple {\mathbf x_0, z_0} \in \mathbf X'$ such that:
:$\map F {\mathbf x_0, z_0} = 0$
:$\dfrac \partial {\partial z} \map F {\mathbf x_0, z_0} \ne 0$
Then there exists a unique mapping of the form:
:$g: \mathbf X \to \mathbb I$
where $\mathbf X \subseteq \R^n$ contains $\mathbf x_0$ and $\mathbb I$ is an open real interval containing $z_0$, such that:
:$\forall \mathbf x \in \mathbf X, z \in \mathbb I: \map F {\mathbf x, z} = 0 \iff z = \map g {\mathbf x}$
and $g$ itself has continuous partial derivatives.
== Proof ==
This is a special case of the Implicit Function Theorem.
== Sources ==
* http://www.math.dartmouth.edu/archive/m14f04/public_html/math_14_lagrange.pdf, Theorem $2$
Category:Calculus
Impossible Proposition Strictly Implies Every Proposition/Examples/2+2=5

== Example of the Paradoxes of Strict Implication ==
The proposition:
:$2 + 2 = 5$ strictly implies that grass is blue
is an example of the Paradoxes of Strict Implication:
:Impossible Proposition Strictly Implies Every Proposition.
== Sources ==
* 
* 
Category:Paradoxes of Strict Implication
Improper Fraction/Examples

== Examples of Improper Fractions ==
=== Example: $\frac 4 3$ ===
=== Example: $\frac 8 7$ ===
=== Example: $\frac {-5} 4$ ===
=== Example: $\frac {16} {10}$ ===
=== Example: $1 \frac 1 2$ ===
Category:Examples of Improper Fractions
Improper Integral of Partial Derivative

==Theorem==
Let $I$ and $U$ be open intervals of $\R$, and $f: U \times I \to \R$ be a function such that:
# for all $u \in U$, $f(\cdot,u) : I \to \R$ is integrable;
# for all $t \in I$, $f(t,\cdot) : U \to \R$ is $\CC^1$;
# there is an integrable function $\phi : I \to \R$ such that for all $(t,u) \ in I \times U$,
:$\ds \size {\partial_u f(t,u)} \leq \phi(t)$
Then the function $F : U\to \R$ defined by
:$\ds F(u) = \int_I f(t,u) \rd t$
is $\CC^1$ and
:$\ds F'(u) = \int_I \partial_u f (t,u) \rd t$
In other words
:$\ds \frac \d {\d u} \int_I f(t,u) \rd t = \int_I \partial_u f(t,u) \rd u$
for all $u \in U$.
This is a straightforward corollary of the following measure theoretic theorem.
Let $X \subset \R$ be an open interval, and $\Omega$ a measure space.
If $f : X \times \Omega \to \R$ is a function satisfying:
#for all $x \in X$, the function $f(x,\cdot) : \Omega \to \R$ is Lebesque integrable;
#for almost all $\omega \in \Omega$, the partial derivative $\partial_x f$ along $X$ of $x$ is defined
#there is an integrable function $\phi: X \to \R$ such that for all $x \in X$, and for almost all $\omega \in \Omega$,
:$\ds \size {\partial_x f(x,\omega)} \leq \phi(x)$
Then the function $F : X \to \R$ defined as
:$\ds F(x) = \int_\Omega f(x,\omega) \rd \omega$
has a derivative at all point $x\in I$ given by the identity
:$\ds F'(x) = \int_\Omega \partial_x f(x,\omega) \rd \omega $
More concisely:
:$\ds \frac \d {\d x} \int_\Omega f(x,\omega) \rd \omega = \int_\Omega \partial_x f(x,\omega) \rd \omega$
for all $x\in X$.
Improved Euler Method

== Proof Technique ==
Consider the first order ODE:
:$(1): \quad y' = \map f {x, y}$ subject to the initial condition $\map y {x_0} = y_0$
where $\map f {x, y}$ is continuous.
Let $\map y x$ be the particular solution of $(1)$.
For all $n \in \N_{>0}$, we define:
:$x_n = x_{n - 1} + h$
where $h \in \R_{>0}$.
Then for all $n \in \N_{>0}$ such that $x_n$ is in the domain of $y$:
:$y_{n + 1} = y_n + \dfrac h 2 \paren {\map f {x_n, y_n} + \map f {x_{n + 1}, z_{n + 1} } }$
where:
:$z_{n + 1} = y_n + h \map f {x_n, y_n}$
is an approximation to $\map y {x_{n + 1} }$.
== Proof ==
== Sources ==
* : Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods
Category:Proof Techniques
Category:Differential Equations
Category:Numerical Analysis
Impulse Imparted by Constant Force

== Theorem ==
Let a constant force $\mathbf F$ be applied to a particle $P$ from time $t_1$ to time $t_2$.
Then the impulse $\mathbf J$ imparted to $P$ is:
:$\mathbf J = \mathbf F \paren {t_2 - t_1}$
== Proof ==
== Sources ==
* 
* 
Category:Impulse
Impulse equals Change in Momentum

== Theorem ==
Let a force $\mathbf F$ be applied to a particle $P$.
Then the impulse $\mathbf J$ imparted to $P$ is equal to the change of momentum of $P$.
== Proof ==
Hence the impulse $\mathbf J$ equals the change of (linear) momentum between $t_1$ and $t_2$, as required.
== Sources ==
* 
* 
Category:Impulse
In Connected Smooth Manifold Any Two Points can be Joined by Admissible Curve

== Theorem ==
Let $M$ be a connected smooth manifold with or without a boundary.
Let $p, q \in M$ be points.
Let $\gamma : \closedint a b \to M$ be an admissible curve.
Then:
:$\forall p, q \in M : \exists \gamma \subset M : \paren {\map \gamma a = p} \land \paren {\map \gamma b = q}$
== Proof ==
For $p, q \in M$, we write:
:$p \sim q$
 there exists an admissible curve $\gamma : \closedint a b \to M$ such that $\map \gamma a = p$ and $\map \gamma b = q$
=== Lemma 1 ===
The $\sim$ is a equivalent relation on $M$.
=== Proof of Lemma 1 ===
For each $p \in M$, we define:
:$N_p := \set {q \in M : p \sim q}$
Now we need to show that:
:$\forall p \in M : M = N_p$
=== Lemma 2 ===
For each $p \in N$, $N_p$ is a non-empty open set.
=== Proof of Lemma 2 ===
Let $p \in M$.
Let $q \in N_p$.
Let $\struct {U, \phi}$ be a chart such that $q \in U$.
As $U$ is open, there is an open ball such that:
:$\map {B_\epsilon} {\map \phi q} \subseteq \phi \sqbrk U$
Let:
:$V := {\phi ^{-1} } \sqbrk {\map {B_\epsilon} {\map \phi q} }$
We shall show that:
:$V \subseteq N_p$
Let $r \in V \setminus \set q$.
Define a regular curve segment:
:$\ell_r : \closedint 0 1 \to M$
by:
:$\ds \map {\ell_r} t := \map {\phi^{-1} } { \map \phi q + t \paren {\map \phi r - \map \phi q} }$
In particular, $\ell_r$ is an admissible curve such that:
:$\map {\ell_r} a = q$
:$\map {\ell_r} b = r$
Therefore:
:$q \sim r$
As $p \sim q$, we have by Lemma 2:
:$p \sim r$
That is:
:$r \in N_p$
=== Lemma 3 ===
If $N_p \cap N_q \ne \O$, then $N_p = N_q$.
=== Proof of Lemma 3 ===
Let $r \in N_p \cap N_q$.
That is, $r \sim p$ and $r \sim q$.
By Lemma 1, then $p \sim q$.
Let $p \in M$.
Recall $N_p$ is an non-empty open set.
Observe:
:$\ds M \setminus N_p = \bigcup_{q \mathop \in M \setminus N_p} N_q$
is also an open set.
Thus $N_p$ is an non-empty clopen set.
Since $M$ is connected, we have:
:$M = N_p$
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Riemannian Geometry
Incidence of Double Letters in Fibonacci String

== Theorem ==
Let $S_n$ denote the $n$th Fibonacci string.
Then:
:$(1):\quad$ There are no instances of $2$ $\text a$'s together
:$(2):\quad$ There are no instances of $3$ $\text b$'s together
in $S_n$.
== Proof ==
The proof proceeds by strong induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_n$.
$\map P 1$ is the case:
:$S_1 = \text a$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$S_2 = \text b$
$\map P 3$ is the case:
:$S_3 = \text {ba}$
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $2 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_k$.
and:
:There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_{k - 1}$.
from which it is to be shown that:
:There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_{k + 1}$.
=== Induction Step ===
This is the induction step:
By definition of Fibonacci string:
:$S_{k + 1} = S_k S_{k - 1}$
concatenated.
By the induction hypothesis, neither $S_k$ nor $S_{k - 1}$ have any double $\text a$'s or triple $\text b$'s.
The only way for $S_{k + 1}$ to have a double $\text a$ is for $S_n$ to end with $\text a$ and for $S_{k - 1}$ to begin with $\text a$.
But from Fibonacci String Begins with ba, for all $k > 2$, $S_{k - 1}$ begins with $\text b$.
When $k = 2$, $S_{k - 1}$ does begin with $\text a$, but then $S_{k + 1} = \text {ba}$, which has no double $\text a$.
Similarly, the only way for $S_{k + 1}$ to have a triple $\text b$, either $S_k$ must end with double $\text b$ or $S_{k - 1}$ must begin with double $\text b$.
From Fibonacci String Begins with ba, $S_{k - 1}$ does not begin with double $\text b$.
From Fibonacci String Ends with ab or ba, $S_k$ does not begin with double $\text b$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:For all $n \in \Z_{>0}$, there are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_n$.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $36$
Category:Fibonacci Strings
Included Set Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau_H}$ be an included set space.
Then $\tau_H$ is a topology on $S$, and $T$ is a topological space.
== Proof ==
=== ===
Let $U_1, U_2 \in \tau_H$.
By definition of included set topology:
:$H \subseteq U_1$ and $H \subseteq U_2$
From Intersection is Largest Subset:
:$H \subseteq U_1 \cap U_2$
By definition of included set topology:
:$U_1 \cap U_2 \in \tau_H$
=== ===
Let $\UU \subseteq \tau_H$.
By definition of included set topology:
:$\forall U \in \UU: H \subseteq U$
Hence from Subset of Union:
:$H \subseteq \bigcup \UU$
=== ===
By definition of included set topology:
:$S \in \tau_H$
So all the properties are fulfilled for $\tau_H$ to be a topology on $S$.
Category:Included Set Topology
Included Set Topology on Finite Intersection

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space on a set $S$.
Let $A_1, A_2, \ldots, A_n$ be a finite set of subsets of $S$:
:$\forall i \in \closedint 1 n: A_i \subseteq S$
Let $\forall i \in \closedint 1 n: \map T {A_i} = \struct {S, \tau_{A_i} }$ be the included set spaces on $S$ by $A_i$.
Let:
:$\forall i \in \closedint 1 n: \map T {A_i} \le T$
where $\map T {A_i} \le T$ denotes that $\map T {A_i}$ is coarser than $T$.
Then:
:$\ds \map T {\bigcap A_i} \le T$
where $\ds \map T {\bigcap A_i}$ is the included set space on $S$ by $\ds \bigcap_{i \mathop = 1}^n A_i$.
== Proof ==
For ease of notation, define:
:$A := \ds \bigcap_{i \mathop = 1}^n A_i$
and let $\tau_A$ denote the included set topology on $S$ by $A$.
Let $U \in \tau_A$ be nonempty.
Then by definition, $A \subseteq U$.
Hence there is a subset $Z \subseteq S$ of $S$, such that $U = A \cup Z$; that is:
:$U = \ds \paren {\bigcap_{i \mathop = 1}^n A_i} \cup Z = \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$
where the latter equality follows from Union Distributes over Intersection.
For every $i \in \closedint 1 n$, one has $A_i \cup Z \in \tau_{A_i}$ by definition of included set topology.
Further, by assumption, $\map T {A_i} \le T$, that is $\tau_{A_i} \subseteq \tau$.
It follows that $A_i \cup Z \in \tau$ for all $i \in \closedint 1 n$.
Hence $U = \ds \bigcap_{i \mathop = 1}^n \paren {A_i \cup Z}$ is in $\tau$, because $\tau$ is a topology.
This comes down to:
:$\tau_A \subseteq \tau$
Hence by definition of coarser topology:
:$\map T {\bigcap A_i} \le T$
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets: Exercise $1 \ \text{(b)}$
Category:Included Set Topology