avewright/proofwiki-math huggingface dataset:
Theorem: Either-Or Space is Lindelöf

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a Lindelöf space.

Proof:
We have:
:Either-Or Topology is Compact
:Compact Space is Lindelöf
Hence the result.
{{qed}}
Theorem: Either-Or Space is Lindelöf

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a Lindelöf space.

Proof:
Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$.
So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets.
So $\CC$ has a subcover containing at most three sets.
Hence $T$ is a Lindelöf space by definition.
{{qed}}
Theorem: Subspace of Either-Or Space less Zero is not Lindelöf

Let $T = \struct {S, \tau}$ be the either-or space.
Let $H = S \setminus \set 0$ be the set $S$ without zero.
Then the topological subspace $T_H = \struct {H, \tau_H}$ is not a Lindelöf space.

Proof:
By definition of topological subspace, $U \subseteq H$ is open in $T_H$ {{iff}}:
:$(1): \quad \set 0 \nsubseteq U$
or:
:$(2): \quad \openint {-1} 1 \subseteq U$
But for all $U \subseteq H$, condition $(1)$ holds as $0 \notin H$.
So $T_H$ is by definition a discrete space.
As $T_H$ is uncountable, we have that Uncountable Discrete Space is not Lindelöf holds.
Hence the result.
{{qed}}
Theorem: Either-Or Topology is First-Countable

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a first-countable space.

Proof:
Let $x \in S$ such that $x \ne 0$.
Then $\set x$ is open in $T$ and so on its own forms a local basis of $x$ which is (trivially) countable.
Let $x = 0$.
Let $U \in \tau$ be open in $T$ such that $x \in U$.
Then by definition of the either-or space, $U$ contains the open set $\openint {-1} 1$.
So $\openint {-1} 1$ forms a local basis of $0$ which is (trivially) countable.
Hence the result by definition of first-countable space.
{{qed}}
Theorem: Either-Or Topology is not Separable

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a separable space.

Proof:
From Limit Points of Either-Or Topology, the only limit point of any set of $S$ is $0$.
So the only set whose closure is $S$ are $S \setminus \set 0$ and $S$ itself.
So these two are the only subsets of $S$ which are everywhere dense in $S$.
Both of these are uncountable.
Hence the result, by definition of separable space.
{{qed}}
Theorem: Either-Or Topology is Non-Meager

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is non-meager.

Proof:
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$.
So every non-empty subset of $T$ (apart from $\left\{{0}\right\})$ contains at least one open set of $T$.
So no subset of $T$ is nowhere dense in $T$.
So $T$ is not a countable union of subsets of $S$ which are nowhere dense in $T$.
Hence the result by definition of non-meager.
{{qed}}
Theorem: Either-Or Topology is Non-Meager

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is non-meager.

Proof:
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$.
The result follows directly from Space with Open Point is Non-Meager.
{{qed}}
Theorem: Basis for Either-Or Topology

Let $T = \struct {S, \tau}$ be the either-or space.
Let $\BB$ be the set:
:$\BB := \set {\set x: x \in S, x \ne 0} \cup \set {\openint {-1} 1}$
that is, the set of all singleton subsets of $S$ less $\set 0$ and including the open real interval $\openint {-1} 1$.
Then $\BB$ is a basis for $T$.

Proof:
Let $U \in \tau$ such that $0 \notin U$.
Then:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
where $x \ne 0$.
Hence for all $x \in U$, we have $\set x \in \BB$.
Thus $U$ is the union of elements of $\BB$.
Now suppose $U \in \tau$ such that $0 \in U$.
Then $\openint {-1} 1 \subseteq U$ by definition.
So one of four cases holds:
{{begin-eqn}}
{{eqn | l = U
      | r = \openint {-1} 1
}}
{{eqn | l = U
      | r = \hointr {-1} 1
      | rr= = \openint {-1} 1 \cup \set {-1}
}}
{{eqn | l = U
      | r = \hointl {-1} 1
      | rr= = \openint {-1} 1 \cup \set 1
}}
{{eqn | l = U
      | r = \closedint {-1} 1
      | rr= = \openint {-1} 1 \cup \set {-1} \cup \set 1
}}
{{end-eqn}}
All of these sets are in $\BB$, so that $U$ is the union of elements of $\BB$.
Hence, by definition, $\BB$ is a basis for $T$.
{{qed}}
Category:Either-Or Topology
Category:Examples of Topological Bases
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Theorem: Either-Or Topology is Locally Path-Connected

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a locally path-connected space.

Proof:
Consider the set:
:$\BB := \set {\set x: x \in S, x \ne 0} \cup \openint {-1} 1$
Then by Basis for Either-Or Topology, $\BB$ is a basis for $T$.
From Point is Path-Connected to Itself, all $x \in S, x \ne 0$ are path-connected elements of $\BB$.
Finally we consider $\openint {-1} 1 \in \BB$.
Let $p \in \openint {-1} 1$ and consider the mapping $f: \closedint 0 1 \to \openint {-1} 1$:
:<nowiki>$\forall x \in \closedint 0 1: \map f x = \begin{cases}
0 & : x = 0 \\
p & : x \in \hointl 0 1 \end{cases}$</nowiki>
Then $f$ is a continuous mapping and so a path from $0$ to $p$.
So $\openint {-1} 1$ is a path-connected element of $\BB$.
Hence the result by definition of locally path-connected space.
{{qed}}
Theorem: Either-Or Topology is Scattered

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a scattered space.

Proof:
{{Recall|Scattered Space|scattered space}}
{{:Definition:Scattered Space/Definition 1}}
Let $H$ be a non-empty subset of $S$.
Let $x \in H$.
Let $x \ne 0$.
By definition of either-or space, $\set x$ is open in $T$.
So $\set x$ is an open set of $x$ containing only $x$.
Thus by definition $x$ is isolated.
Thus if $H$ is a non-empty subset of $S$ which contains any element other than $0$, $H$ contains at least one isolated point.
The other case is that $H$ contains no $x$ such that $x \ne 0$.
That is: $H = \set 0$.
But from Singleton Point is Isolated, $0$ is an isolated point in $H$.
Thus in all cases $H$ contains at least one point which is isolated in $H$.
So by definition $T$ is a scattered space.
{{qed}}
Theorem: Countable Stability implies Stability for All Infinite Cardinalities

Let $T$ be a complete $\LL$-theory whose language $\LL$ is countable.
If $T$ is $\omega$-stable, then $T$ is $\kappa$-stable for all infinite $\kappa$.

Proof:
We prove the contrapositive.
Let $\kappa$ be an infinite cardinal.
Suppose that $T$ is not $\kappa$-stable.
Then there exists some $\MM \models T$ and $A \subseteq \MM$ with $\card A = \kappa$ such that:
:$\card {\map { {S_n}^\MM} A} > \kappa$
Let $\LL_A$ denote $\LL \cup \set {a: a \in A}$, the language obtained from $\LL$ by adding new constant symbols for each $a \in A$.
For each $\LL_A$-formula $\phi$, let $\sqbrk \phi = \set {p \in \map { {S_n}^\MM} A: \phi \in p}$, the set of complete $n$-types over $A$ which contain $\phi$.
Our goal will be to find a countable set $B$ in $\MM$ such that:
:$\card {\map { {S_n}^\MM} B} = 2^{\aleph_0} \ne \aleph_0$
which will demonstrate non-$\omega$-stability of $T$.
We will do this by constructing a countable binary tree of formulas such that each of the $2^{\aleph_0}$ distinct simple paths from the root of the tree out to infinity correspond to distinct types.
Before we can build the tree, we need the following lemma.
Theorem: Finite Complement Topology is Topology

Let $T = \struct {S, \tau}$ be a finite complement space.
Then $\tau$ is a topology on $T$.

Proof:
By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite.
{{qed|lemma}}
Let $A, B \in \tau$.
Let $H = A \cap B$.
Then:
{{begin-eqn}}
{{eqn | l = H
      | r = A \cap B
      | c =
}}
{{eqn | ll= \leadsto
      | l = \relcomp S H
      | r = \relcomp S {A \cap B}
      | c =
}}
{{eqn | r = \relcomp S A \cup \relcomp S B
      | c = De Morgan's laws: Complement of Intersection
}}
{{end-eqn}}
But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both finite.
Hence their union is also finite.
Thus $\relcomp S H$ is finite.
So $H = A \cap B \in \tau$ as its complement is finite.
{{qed|lemma}}
Let $\UU \subseteq \tau$.
From De Morgan's laws: Complement of Union:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
But as:
:$\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ is finite.
Hence so is their intersection.
So $\ds \relcomp S {\bigcup \UU}$ is finite.
So by definition:
:$\ds \bigcup \UU \in \tau$
{{qed|lemma}}
$\tau$ obeys all of the open set axioms, so it is a topology on $T$.
{{qed}}
Theorem: Limit Points of Infinite Subset of Finite Complement Space

Let $T = \struct {S, \tau}$ be a finite complement space.
Let $H \subseteq S$ be an infinite subset of $S$.
Then every point of $S$ is a limit point of $H$.

Proof:
Let $U \in \tau$ be any open set of $T$.
From Infinite Subset of Finite Complement Space Intersects Open Sets, we have that $U \cap H$ is infinite {{iff}} $H$ is infinite.
Let $x \in S$.
Then every open set $U$ in $T$ such that $x \in U$ also contains an infinite number of points of $H$ other than $x$.
Thus, by definition, $x$ is a limit point of $H$.
{{qed}}
Theorem: Finite Complement Topology is Separable

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a separable space.

Proof:
Let $H$ be a countably infinite subset of $S$.
From Closure of Infinite Subset of Finite Complement Space, the closure of $H$ is $S$.
So by definition $H$ is everywhere dense in $T$.
Hence the result by definition of separable space.
{{qed}}
Theorem: Closure of Infinite Subset of Finite Complement Space

Let $T = \struct {S, \tau}$ be a finite complement space.
Let $H \subseteq S$ be an infinite subset of $S$.
Then $H^- = S$ where $H^-$ is the closure of $S$.

Proof:
Let $H$ be an infinite subset of $S$.
From Limit Points of Infinite Subset of Finite Complement Space, every point of $S$ is a limit point of $H$.
Hence the result from the definition of closure.
{{qed}}
Theorem: Subspace of Finite Complement Topology is Compact

Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then every topological subspace of $T$, including $T$ itself, is a compact space.

Proof:
Let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.
Let $\CC$ be an open cover of $T_H$.
Let $U \in \CC$ be any set in $C$.
$U$ covers all but a finite number of points of $T_H$.
So for each of those points we pick an element of $\CC$ which covers each of those points.
Hence we have a finite subcover of $T_H$.
So by definition $T_H$ is a compact space.
{{qed}}
Theorem: Uncountable Finite Complement Space is not Perfectly T4

Let $T = \struct {S, \tau}$ be a finite complement space on an uncountable set $S$.
Then $T$ is not a perfectly $T_4$ space.

Proof:
Recall the definition of a perfectly $T_4$ space
:Every closed set in $T$ can be written as a countable intersection of open sets of $T$.
Let $V$ be a closed set in $T$.
From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$:
:$V$ is not a $G_\delta$ set.
The result follows by definition of perfectly $T_4$ space.
{{qed}}
Theorem: Kluyver's Formula for Ramanujan's Sum

Let $q \in \N_{>0}$.
Let $n \in \N$.
Let $\map {c_q} n$ be Ramanujan's sum.
Let $\mu$ denote the Möbius function.
Then:
:$\ds \map {c_q} n = \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}$
where $\divides$ denotes divisibility.

Proof:
Let $\alpha \in \R$.
Let $e: \R \to \C$ be the mapping defined as:
:$\map e \alpha := \map \exp {2 \pi i \alpha}$
Let $\zeta_q$ be a primitive $q$th root of unity.
Let:
:$\ds \map {\eta_q} n := \sum_{1 \mathop \le a \mathop \le q} \map e {\frac {a n} q}$
By Complex Roots of Unity in Exponential Form this is the sum of $n$th powers of all $q$th roots of unity.
Therefore:
:$\ds \map {\eta_q} n = \sum_{d \mathop \divides q} \map {c_d} n$
By the Möbius Inversion Formula, this gives:
:$\ds \map {c_q} n = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}$
Now by Sum of Powers of Primitive Complex Roots of Unity, we have:
:$\ds \map {\eta_d} n = \begin{cases} d & : d \divides n \\ 0 & : d \nmid n \end{cases}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {c_q} n
      | r = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}
}}
{{eqn | r = \sum_{\substack {d \mathop \divides q \\ d \mathop \nmid n} } \map {\eta_d} n \map \mu {\frac q d} + \sum_{\substack {d \mathop \divides q \\ d \mathop \divides n} } \map {\eta_d} n \map \mu {\frac q d}
}}
{{eqn | r = 0 + \sum_{\substack {d \mathop \divides q \\ d \mathop \divides n} } d \map \mu {\frac q d}
}}
{{eqn | r = \sum_{d \mathop \divides \gcd \set {q, n} } d \map \mu {\frac q d}
      | c = Common Divisor Divides GCD
}}
{{end-eqn}}
as required.
{{qed}}
{{Namedfor|Jan Cornelis Kluyver|cat = Kluyver}}
Category:Analytic Number Theory
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Theorem: Condition for Complex Root of Unity to be Primitive

Let $n, k \in \N$.
Then $\alpha_k = \map \exp {\dfrac {2 \pi i k} n}$ is a primitive $n$th root of unity {{iff}} $\gcd \set {n, k} = 1$.

Proof:
Let $U_n = \set {\map \exp {\dfrac {2 \pi i k} n}: 0 \le k \le n - 1}$.
Let $V = \set {1, \dotsc, {\alpha_k}^{n - 1} }$.
By Complex Roots of Unity in Exponential Form it is sufficient to show that $U_n = V$ {{iff}} $\gcd \set {n, k} = 1$.
Let $\gcd \set {n, k} = d > 1$.
Then there are $n', k' \in \N$ such that:
:$n' = d n$
and:
:$k' = d k$
Then we have:
:$\alpha_k = \map \exp {\dfrac {2 \pi i k'} {n'} }$
and:
:$\alpha_k^{n'} = \map \exp {2 \pi i k'} = 1$
Therefore:
:$V = \set {1, \dotsc, \alpha^{n' - 1} }$
such that $n' < n$.
So:
:$\cmod V = n' < n = \cmod {U_n}$
and $U_n \ne V$.
Let $\gcd \set {n, k} = 1$.
Let:
:$\map \exp {\dfrac {2 \pi i k} n}^d = \map \exp {\dfrac {2 \pi i k} n} = 1$
Then it must be the case that $\dfrac {k d} n \in \Z$.
Since $\gcd \set {n, k}\ = 1$ it follows that:
:$n \divides d$
and so:
:$d \ge n$
Therefore $\set {1, \dotsc, \alpha^{n - 1} }$ are distinct
Hence $\card V = \card {U_n}$.
Moreover each element of $V$ can be written in the form:
:$\map \exp {\dfrac {2 \pi i k} n}$
with $0 \le k \le n - 1$.
It follows that $V = U_n$.
{{qed}}
Theorem: Relation of Boubaker Polynomials to Chebyshev Polynomials

The Boubaker polynomials are related to Chebyshev polynomials through the equations:
:$\map {B_n} {2 x} = \dfrac {4 x} n \dfrac \d {\d x} \map {T_n} x - 2 \map {T_n} x$
:$\map {B_n} {2 x} = -2 \map {T_n} x + 4 x \map {U_{n - 1} } x$
where:
:$T_n$ denotes the Chebyshev polynomials of the first kind
:$U_n$ denotes the Chebyshev polynomials of the second kind.

Proof:
('''Using Riordan Matrix''')
The ordinary generating function of the Boubaker Polynomials can be expressed in terms of Riordan matrices:
:$\ds \sum_{n \geqslant 0} \map {B_n} t x^n = \frac {1 + 3 x^2} {1 - x t + x^2} = \paren {1 + 3 x^2 \mid 1 + x^2} \paren {\dfrac 1 {1 - x t} }$       
By considering the Riordan matrices of Chebyshev polynomials of the first and second kind $T_n$ and $U_n$, respectively:
:$\map T {\dfrac 1 4 - \dfrac {x^2} 4 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$
and:
:$\map T {\dfrac 1 2 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$
and by setting:
:$\map T {1 + 3 x^2 \mid 1 + x^2} = \map T {1 + 3 x^2 \mid 1} \map T {\dfrac 1 2 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$
we obtain finally:
:$\map {B_n} x = \map {U_n} {\dfrac x 2} + 3 \map {U_{n - 2} } {\dfrac x 2}$
which gives, due the relation between Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$ and their derivatives:
{{explain|What is that relation between the CPFK and CPSK?}}
:$\map {B_n} {2 x} = -2 \map {T_n} x + 4 x \map {U_{n - 1} } x$
:$\map {B_n} {2 x} = \dfrac {4 x} n \dfrac \d {\d x} \map {T_n} x - 2 \map {T_n} x$
{{qed}}
Theorem: Relation of Boubaker Polynomials to Dickson Polynomials

The Boubaker polynomials $B_n$ are linked to the  Dickson polynomials by the relations:
:$\map {B_{n + 1} } x \map {B_{n + j} } x - \map {B_{n + j + 1} } x \map {B_n} x = \paren {3 x^2 + 4} \map {D_{n + 1} } {x, \dfrac 1 4}$
:$\map {B_n} x = \map {D_n} {2 x, \dfrac 1 4} + 4 \map {D_{n - 1} } {2 x, \dfrac 1 4}$

Proof:
{{ProofWanted|The material on this page was originally raised by User:Yohji kinomoto.}}
Theorem: Relation of Boubaker Polynomials to Fermat Polynomials

The Boubaker polynomials are related to Fermat polynomials by:
:$\forall n \in \N: \map {B_n} x = \dfrac 1 {\paren {\sqrt 2}^n} \map {F_n} {\dfrac {2 \sqrt 2 x} 3} + \dfrac 3 {\paren {\sqrt 2}^{n - 2} } \map {F_{n - 2} } {\dfrac {2 \sqrt 2 x} 3}$

Proof:
('''Using Riordan Matrix''')
Since the ordinary generating function of the Boubaker Polynomials can be expressed in terms of Riordan matrices:
:$\ds \sum_{n \mathop \geqslant 0}^{} \map {B_n} t x^n = \frac {1 + 3x^2 }{1 - x t + x^2} = \paren {1 + 3 x^2 \biggl\lvert 1 + x^2} \paren {\frac 1 {1 - x t} }$     
then, by writing differently:
:$\ds \map T {1 + 3x^2 \biggl\lvert 1 + x^2} = \map T {1 + 3x^2 \biggl \lvert 1} \map T {\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2} \map T {2 \biggl\lvert 2}$
and:
:$\ds \map T {1 + 3x^2 \biggl\lvert 1 + x^2} = \map T {1 + 3x^2 \biggl\lvert 1} \map T {1 \biggl\lvert \sqrt 2} \map T {\frac 1 3 \biggl\lvert \frac 1 3 + \frac 2 3 x^2} \map T {3 \biggl\lvert \frac 3 {\sqrt 2} }$
and taking into account the Riordan matrix of Fermat polynomials:
:$\ds \map T {\frac 1 3 \biggl\lvert \frac 1 3 + \frac 1 3 x^2}$
we obtain finally:
:$\forall n \in \N: \map {B_n} x = \dfrac 1 {\paren {\sqrt 2}^n} \map {F_n} {\dfrac {2 \sqrt 2 x} 3} + \dfrac 3 {\paren {\sqrt 2}^{n - 2} } \map {F_{n - 2} } {\dfrac {2 \sqrt 2 x} 3}$
{{qed}}
Theorem: F-Sigma and G-Delta Subsets of Uncountable Finite Complement Space

Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then countably infinite subsets of $S$ are $F_\sigma$ sets and are neither open nor closed sets.
Their relative complements in $S$ are $G_\delta$ sets, and are also neither open nor closed sets.

Proof:
Let $U$ be a countably infinite subset of $S$.
Then it can be written as:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
But as $\set x$ is a finite subset of $S$, it is by definition the complement of an open set in $T$.
So by definition $\set x$ is closed in $T$.
So:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
is a countable union of closed sets of $T$.
That is, by definition, $U$ is an $F_\sigma$ set.
As $U$ is countably infinite, its complement is uncountable, not finite, and so $U$ is not open in $T$.
But as $U$ is also not finite, it is not the complement of an open set and so is not closed in $T$ either.
From Complement of $F_\sigma$ Set is $G_\delta$ Set it follows that the complement of $U$ is a $G_\delta$ set, and similarly neither open nor closed.
{{qed}}
Theorem: Skolem's Paradox

Let $\LL$ be a countable first-order language.
Let $T$ be an $\LL$-theory which axiomatizes some version of set theory (for example, ZFC).
There is a countable model of $T$.

Proof:
This is a straightforward application of the downward Löwenheim-Skolem Theorem.
{{qed}}
{{refactor|From here on down, decide where it belongs and how it is to be presented. At the moment it is too much like an encyclopedia article to be compatible with {{ProofWiki}}'s dictionary style.|level = advanced}}
Theorem: Uncountable Finite Complement Space is not First-Countable

Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then $T$ is not first-countable.

Proof:
{{AimForCont}} some $x \in S$ has a countable local basis.
That means:
:there exists a countable set $\BB_x \subseteq \tau$
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every open neighborhood of $x$ contains some $B \in \BB_x$.
Let $y \in S$ with $y \ne x$.
Because its complement relative to $S$ is finite, we have that $S \setminus \set y$ is an open set containing $x$.
Because  $\BB_x$ is a neighborhood basis at $x$, there exists some $B \in \BB_x$ such that:
:$B \subseteq S \setminus \set y$
whence $y \notin B$.
So:
:$y \notin \bigcap \BB_x$
So:
{{begin-eqn}}
{{eqn | l = \bigcap \BB_x
      | r = \set x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = S \setminus \set x
      | r = S \setminus \bigcap \BB_x
      | c = 
}}
{{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}
      | c = De Morgan's Laws: Difference with Intersection
}}
{{end-eqn}}
By definition, each of $S \setminus B$ is finite.
From Countable Union of Countable Sets is Countable it follows that $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is countable.
So $S \setminus \set x$ and therefore $S$ is also countable.
From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false.
Hence by definition $T$ can not be first-countable.
{{qed}}
Theorem: Separable Space is not necessarily First-Countable

Let $T = \struct {S, \tau}$ be a topological space which is separable.
Then it is not necessarily the case that $T$ is first-countable.

Proof:
;Proof by Counterexample
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
We have that a Finite Complement Topology is Separable.
But we also have that an Uncountable Finite Complement Space is not First-Countable.
Hence the result.
{{qed}}
Theorem: Ramanujan Sum is Multiplicative

Let $q \in \N_{>0}$, $n \in \N$.
Let $\map {c_q} n$ be the Ramanujan sum.
Then $\map {c_q} n$ is multiplicative in $q$.

Proof:
Let $q, r \in \N$ such that:
:$\gcd \set {q, r} = 1$
where $\gcd \set {q, r}$ denotes the greatest common divisor of $q$ and $r$.
Then:
{{begin-eqn}}
{{eqn | l = \map {c_q} n \, \map {c_r} n
      | r = \sum_{d_1 \mathop \divides \gcd \set {n, q} } d_1 \, \map \mu {\frac q {d_1} } \sum_{d_2 \mathop \divides \gcd \set {n, r} } d_2 \, \map \mu {\frac r {d_2} }
      | c = Kluyver's Formula for Ramanujan's Sum
}}
{{eqn | r = \sum_{\substack {d_1 \mathop \divides \gcd \set {n, q} \\ d_2 \mathop \divides \gcd \set {n, r} } } d_1 d_2 \, \map \mu {\frac q {d_1} } \, \map \mu {\frac r {d_2} }
      | c = 
}}
{{eqn | r = \sum_{d \mathop \divides \gcd \set {n, q} \gcd \set {n, r} } d \, \map \mu {\frac {q r} d}
      | c = Möbius Function is Multiplicative, and setting $d = d_1 d_2$
}}
{{eqn | r = \sum_{d \mathop \divides \gcd \set {n, q r} } d \, \map \mu {\frac {q r} d}
      | c = GCD with One Fixed Argument is Multiplicative Function
}}
{{eqn | r = \map {c_{q r} } n
      | c = Kluyver's Formula for Ramanujan's Sum
}}
{{end-eqn}}
This completes the proof.
{{qed}}
Category:Analytic Number Theory
92a5u5ragspb52bp1qa1wgrtfjbbata
Theorem: Asymptotic Growth of Euler Phi Function

Let $\phi$ be the Euler $\phi$ function.
For any $\epsilon > 0$ and sufficiently large $n$:
:$n^{1 - \epsilon} < \map \phi n < n$

Proof:
It is clear that $\map \phi n < n$ for all $n$, so it is sufficient to prove that:
:$\ds \lim_{n \mathop \to \infty} \frac {n^{1 - \epsilon} } {\map \phi n} = 0$
By Multiplicative Function that Converges to Zero on Prime Powers it is sufficient to prove that:
:$\ds \lim_{p^k \mathop \to \infty} \frac {p^{k \paren {1 - \epsilon} } } {\map \phi {p^k} } = 0$
as $p^k$ ranges through all prime powers.
By Euler Phi Function of Prime Power we have:
:$\map \phi {p^k} = p^k - p^{k - 1}$
for a prime power $p^k$.
Therefore:
{{begin-eqn}}
{{eqn | l = \frac {p^{k \paren {1 - \epsilon} } } {\map \phi {p^k} }
      | r = \frac {p^{k \paren {1 - \epsilon} } } {p^k - p^{k - 1} }
      | c = 
}}
{{eqn | r = \frac p {p - 1} \frac {p^{k \paren {1 - \epsilon} } } {p^k}
      | c = 
}}
{{eqn | r = \frac p {p - 1} \frac 1 {p^{k \epsilon} }
      | c = 
}}
{{eqn | o = \le
      | r = \frac 2 {p^{k \epsilon} }
      | c = since $p / \paren {p - 1} \le 2$ for all primes $p$
}}
{{end-eqn}}
Therefore:
:$\ds \lim_{p^k \mathop \to \infty} \frac {p^{k \paren {1 - \epsilon} } } {\map \phi {p^k} } \le \lim_{p^k \mathop \to \infty} \frac 2 {p^{k \epsilon} } = 0$
{{qed}}
Category:Analytic Number Theory
Category:Euler Phi Function
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