avewright/proofwiki-math huggingface dataset:
Theorem: Dirichlet's Approximation Theorem

Let $\alpha, x \in \R$.
Then there exist integers $a, q$ such that:
:$\gcd \set {a, q} = 1$
:$1 \le q \le x$
and:
:$\size {\alpha - \dfrac a q} \le \dfrac 1 {q x}$

Proof:
It is sufficient to find $a, q$ not necessarily coprime.
Once such an $a, q$ have been found, then a coprime pair can be found by dividing numerator and denominator of $\dfrac a q$ by the GCD of $a$ and $q$.
Let $X = \floor x$ be the integer part of $x$.
Let $\alpha_q = \alpha q - \floor {\alpha q} \in \hointr 0 1$ for $q = 1, \ldots, X$.
Consider the disjoint real intervals:
:$I_k = \hointr {\dfrac{k-1} {X + 1} } {\dfrac k {X + 1} }, \quad k = 1, \ldots, X + 1$
Suppose there exists $q$ such that:
:$\alpha_q \in \hointr 0 {\dfrac 1 {X + 1} }$
Then:
:$0 \le \alpha q - \floor {\alpha q} < \dfrac 1 {X + 1}$
Hence:
:$\size {\alpha - \dfrac {\floor {\alpha q} } q} < \dfrac 1 {q \paren {X + 1} } < \dfrac 1 {q x}$
Taking $a = \floor {\alpha q}$, the construction is complete.
{{qed|lemma}}
Suppose that not to be the case, but suppose that there exists $q$ such that:
: $\alpha_q \in \hointr {\dfrac X {X + 1} } 1$
Then similarly:
:$\dfrac X {X + 1} < \alpha q - \floor {\alpha q} < 1$
so:
:$-\dfrac 1 {X + 1} < \alpha q - \floor {\alpha q} - 1 < 0$
and:
:$\size {\alpha - \dfrac {\floor {\alpha q} + 1} q} < \dfrac 1 {q \paren {X + 1} } < \dfrac 1 {q x}$
Taking $a = \floor {\alpha q} + 1$, again the construction is complete.
{{qed|lemma}}
If neither of the above cases holds, then the $X - 1$ remaining intervals $I_k$ contain the $X$ values of $\alpha_q$.
Therefore, by the Pigeonhole Principle, there are $k_0 \in \set {2, \ldots, X}$ and $q_1 < q_2$ such that:
:$\alpha_{q_1}, \alpha_{q_2} \in I_{k_0}$
Then for $i = 1, 2$:
:$\dfrac {k_0 - 1} {X + 1} \le \alpha q_i - \floor {\alpha q_i} < \dfrac {k_0} {X + 1}$
Therefore:
:$\size {\alpha q_2 - \alpha q_1 - \floor {\alpha q_2} + \floor {\alpha q_1} } < \dfrac 1 {X + 1}$
and therefore:
:$\size {\alpha - \dfrac{\floor {\alpha q_2} - \floor {\alpha q_1} } {q_2 - q_1} } < \dfrac 1 {\paren {X + 1} \paren {q_2 - q_1} } < \dfrac 1 {x \paren {q_2 - q_1} }$
Taking $a = \floor {\alpha q_2} - \floor {\alpha q_1}$ and $q = q_2 - q_1$ the construction is complete.
{{qed}}
{{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
Theorem: Convergence of Dirichlet Series with Bounded Partial Sums

Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $\C$.
Suppose that there exists $B > 0$ such that for all $n, m \in \N$:
:$\ds \size {\sum_{k \mathop = m}^n a_n} \le B$
Then the Dirichlet series:
:$\ds \map f s = \sum_{n \mathop \ge 1} a_n n^{-s}$
converges locally uniformly to an analytic function on $\map \Re s > 0$.

Proof:
By Exponential is Entire, the partial sums:
:$\ds \map {f_N} s = \sum_{n \mathop = 1}^N a_n n^{-s}$
are analytic.
So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence.
For $0 < A < \pi / 2$, $\delta > 0$ we let:
:$D_{A, \delta} = \set {s \in \C : -A < \map \arg s < A, \map \Re s > \delta}$
Then for any $s \in \C$ such that $\map \Re s > 0$, we can choose $A$, $\delta$ such that $s \in D_{A, \delta}$.
So it is sufficient to prove locally uniform convergence in this region.
Also note that if $\lambda_n = \log n$, then:
:$\ds \map f s = \sum_{n \mathop \ge 1} a_n e^{-\lambda_n s}$
By Abel's Lemma: Formulation 2, for $N, M \in \N$ we have:
:$\ds \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} = \sum_{n \mathop = M}^{N - 1} \paren {\sum_{k \mathop = M}^n a_n} \paren {e^{-\lambda_n s} - e^{-\lambda_{n + 1} s} } + e^{-\lambda_N s} \paren {\sum_{n \mathop = M}^N a_n}$
Let $\epsilon > 0$ be arbitrary, and choose $N_0 \in \N$ such that:
:$\ds \forall n \ge N_0: \size {e^{-\lambda_n s} } < \epsilon$
Now for $N, M \ge N_0$, by the above we obtain:
:$(1): \quad \ds \size {\sum_{n \mathop = M}^N a_n e^{-\lambda_n s} } \le B \sum_{n \mathop = M}^{N - 1} \size {e^{-\lambda_n s} - e^{-\lambda_{n + 1} s} } + B \epsilon$
For any $\alpha, \beta \in \R$ we have:
{{begin-eqn}}
{{eqn | l = \size {e^{-\alpha s} - e^{-\beta s} }
      | r = \size {s \int_\alpha^\beta e^{-x s} \rd x}
      | c = 
}}
{{eqn | o = \le
      | r = \size s \int_\alpha^\beta e^{-x \sigma} \rd x
      | c = where $\sigma = \map \Re s$
}}
{{eqn | r = \frac {\size s} \sigma \paren {e^{-\alpha \sigma} - e^{-\beta \sigma} }
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \size {\sum_{n \mathop = M}^N a_n e^{-\lambda_n s} }
      | o = \le
      | r = \frac {B \size s} \sigma \sum_{n \mathop = M}^{N - 1} \paren {e^{-\lambda_n \sigma} - e^{-\lambda_{n + 1} \sigma} } + B \epsilon
      | c = by $(1)$
}}
{{eqn | r = \frac {B \size s} \sigma \paren {e^{-\lambda_N \sigma} - e^{-\lambda_M \sigma} } + B \epsilon
      | c = 
}}
{{eqn | o = \le
      | r = \frac {B \epsilon \size s} \sigma + B \epsilon
      | c = 
}}
{{end-eqn}}
Finally letting $\map \Im s = t$:
:$\dfrac {\size s} \sigma = \dfrac {\sqrt {\sigma^2 + t^2} } \sigma = \sqrt {1 + \dfrac {t^2} {\sigma^2} }$
and:
:$\dfrac {\pi} 2 > A \ge \map \arg s = \map \arctan {\dfrac t \sigma}$
so $\dfrac t \sigma < \tan A$ is bounded uniformly in $D_{A, \delta}$.
Thus letting $N \to \infty$, we have shown that:
:$\ds \size {\sum_{n \mathop = M}^\infty a_n e^{-\lambda_n s} } \to 0$
as $M \to \infty$ uniformly in $D_{A, \delta}$.
{{qed}}
Category:Analytic Number Theory
9h5i288vrya0q824l02fx4b7kczrv03
Theorem: Convergence of Dirichlet Series with Bounded Coefficients

Let $\sequence {a_n}_{n \mathop \in \N}$ be a bounded sequence in $\C$.
Then the Dirichlet series:
:$\ds \map f s = \sum_{n \mathop \ge 1} a_n n^{-s}$
converges absolutely and locally uniformly to an analytic function on $\map \Re s > 1$.

Proof:
By Exponential is Entire, the partial sums:
:$\ds \map {f_N} s = \sum_{n \mathop = 1}^N a_n n^{-s}$
are analytic.
So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence.
Let $B$ be a bound for the $a_n$:
:$\forall n \in \N: \size {a_n} \le B$
Let $D$ be any open subset of $\map \Re s > 1$.
So for some $\kappa > 0$:
:$\forall s \in D: \map \Re s \ge 1 + \kappa$
Now:
{{begin-eqn}}
{{eqn | l = \size {\map {f_N} s}
      | o = \le
      | r = \sum_{n \mathop = 1}^N \size {a_n} \size {n^{-s} }
      | c = 
}}
{{eqn | o = \le
      | r = B \sum_{n \mathop = 1}^N \frac 1 {n^{1 + \kappa} }
      | c = 
}}
{{end-eqn}}
which we know to be finite (!).
{{MissingLinks|I would imagine the statement (!) is proved somewhere, but can't think what it'd be called; Here Convergence of P-Series/Real}}
Category:Analytic Number Theory
k5pvajlvhvvwv5niunt3k5i1wwyhoc4
Theorem: Dirichlet L-Function from Trivial Character

Let $\chi_0$ be the trivial Dirichlet character modulo $q$.
{{explain|Trivial character we got, Dirichlet character we got, we still need a page for trivial Dirichlet character. There exists on another page a link to Definition:Dirichlet Character/Trivial Character which ought to be straightforward to construct.}}
Let $\map L {s, \chi_0}$ be the Dirichlet $L$-function associated to $\chi_0$.
Let $\map \zeta s$ be the Riemann zeta function.
Then:
:$\ds \map L {s, \chi_0} = \map \zeta s \cdot \prod_{p \mathop \divides q} \paren {1 - p^{-s} }$
where $\divides$ denotes divisibility.
{{explain|For which $s$? The proof seemingly requires a stronger condition like $\map \Re s > 1$.}}

Proof:
By definition:
:$\map {\chi_0} a = \begin{cases} 1 & : \gcd \set {a, q} = 1 \\ 0 & : \text{otherwise} \end{cases}$
{{explain|Worth specifying the domain of $a$ for added clarity here.}}
Therefore:
{{begin-eqn}}
{{eqn | l = \map L {s, \chi_0}
      | r = \prod_p \paren {1 - \map \chi p p^{-s} }^{-1}
      | c = {{Defof|Euler Product}}: $p$ ranges over the primes
}}
{{eqn | r = \prod_{p \mathop \nmid q} \paren {1 - p^{-s} }^{-1}
      | c = Fundamental Theorem of Arithmetic: $\gcd \set {p, q} > 1$ {{iff}} $p \divides q$
}}
{{eqn | r = \prod_{p \mathop \divides q} \paren {1 - p^{-s} } \prod_p \paren {1 - p^{-s} }^{-1}
      | c =
}}
{{eqn | r = \map \zeta s \prod_{p \mathop \divides q} \paren {1 - p^{-s} }
      | c = {{Defof|Euler Product}}
}}
{{end-eqn}}
{{explain|Not sure whether Fundamental Theorem of Arithmetic is where you need to go for $\gcd \set {p, q} > 1$ {{iff}} $p \divides q$ -- I'm fairly sure there's a result on {{ProofWiki}} demonstrating it more directly}}
Hence the result.
{{Qed}}
Category:Analytic Number Theory
9bec69i297aqwlihkm385l1b7n2rmk1
Theorem: Separable Metacompact Space is Lindelöf

Let $T = \struct {S, \tau}$ be a separable topological space which is also metacompact.
Then $T$ is a Lindelöf space.

Proof:
{{Recall|Separable Space|separable space}}
{{:Definition:Separable Space}}
{{Recall|Metacompact Space|metacompact space}}
{{:Definition:Metacompact Space}}
{{Recall|Lindelöf Space|Lindelöf space}}
{{:Definition:Lindelöf Space}}
Having established the definitions, we proceed.
Let $T = \struct {S, \tau}$ be separable and metacompact.
{{AimForCont}} there exists an open cover $\UU$ of $S$ which has no countable subcover.
As $T$ is metacompact, $\UU$ has an open refinement $\VV$ which is point finite.
By nature of $\UU$, which has no countable subcover, $\VV$ is uncountable.
{{explain|How would $\VV$ being countable lead to a countable subcover of $\UU$? ACC is helpful for this.}}
We have {{hypothesis}} that $T$ is separable.
Let $\SS$ be a countable subset of $S$ which is everywhere dense.
Then each $V \in \VV$ contains some $s \in \SS$.
{{explain|why? Also, this is not true if $\VV$ contains the empty set.}}
So some $s \in \SS$ is contained in an uncountable number of elements of $\VV$.
Thus, by definition, $\VV$ is not point finite.
Thus no uncountable open refinement $\VV$ of $\UU$ exists which is point finite.
It follows that $\VV$ must be countable.
Thus $\UU$ has a countable subcover.
That is, by definition, $T$ is a Lindelöf space.
{{qed}}
Theorem: Separable Metacompact Space is Lindelöf

Let $T = \struct {S, \tau}$ be a separable topological space which is also metacompact.
Then $T$ is a Lindelöf space.

Proof:
{{Recall|Metacompact Space|metacompact space}}
{{:Definition:Metacompact Space}}
{{Recall|Lindelöf Space|Lindelöf space}}
{{:Definition:Lindelöf Space}}
Having established the definitions, we proceed.
Let $\UU$ be an open cover of $S$.
Let $\VV$ be a point finite open refinement of $\UU$.
By Point Finite Set of Open Sets in Separable Space is Countable, $\VV$ is countable.
Define a mapping $H$ on $\VV$ thus:
:$\forall V \in \VV: \map H V = \set {U \in \UU: V \subseteq U}$
By Image of Countable Set under Mapping is Countable, the image of $H$ is countable.
Call this image $I$.
Since $\VV$ is a refinement of $\UU$, $\O \notin I$.
By the {{Axiom-link|Countable Choice}}, $I$ has a choice function $c$.
Then $G = c \circ H: \VV \to \UU$ is a mapping such that:
:$\forall V \in \VV: V \subseteq \map G V$
Then $\QQ = \map G \VV$ is countable by Image of Countable Set under Mapping is Countable.
Each element of $\QQ$ is an element of $\UU$ by the definition of $G$.
Let $x \in S$.
Then since $\VV$ is a cover for $S$:
: $\exists V \in \VV: x \in V$
Then $x \in V \subseteq \map G V \in \QQ$.
Thus $\QQ$ is a countable subcover of $\UU$.
Thus each open cover of $S$ has a countable subcover, so $T$ is a Lindelöf space.
{{qed}}
Theorem: Functional Equation for Dirichlet L-Functions

Let $\chi$ be a primitive Dirichlet character to the modulus $q \geq 1$.
Let $\map \Lambda {s, \chi}$ be the completed $L$-function for $\chi$.
Let $\map \tau \chi$ denote the Gaussian sum.
Then for all $s \in \C$:
:$\map \Lambda {s, \chi} = i^{-\kappa} \dfrac {\map \tau \chi} {\sqrt q} \map \Lambda {1 - s, \overline \chi}$
where $\kappa = \dfrac 1 2 \paren {1 - \map \chi {-1} }$.

Proof:
{{proof wanted}}
Category:Analytic Number Theory
6b4oru2c2fsvpefnpvjo41xwl64m5up
Theorem: Orthogonality Relations for Characters

Let $G$ be a finite abelian group with identity $e$.
Let $G^*$ be the dual group of characters $\chi : G \to \C_{\ne 0}$.
Let $\chi_0$ be the trivial character on $G$.
Let $\psi: G \to \C_{\ne 0}$ be any character.
Let $y \in G$ be arbitrary.
Then:
:$\ds \sum_{x \mathop \in G} \map \psi x = \begin {cases}
\order G & : \psi = \chi_0 \\
0 & : \psi \ne \chi_0 \end {cases}$
and:
:$\ds \sum_{\chi \mathop \in G^*} \map \chi y = \begin {cases}
\order {G^*} & : y = e \\
0 & : y \ne e \end {cases}$

Proof:
If $\psi = \chi_0$, then it is straightforward that:
{{begin-eqn}}
{{eqn | l = \sum_{x \mathop \in G} \map \psi x
      | r = \sum_{x \mathop \in G} \map {\chi_0} x
      | c = {{hypothesis}}
}}
{{eqn | r = \sum_{x \mathop \in G} 1
      | c = {{Defof|Trivial Character}}
}}
{{eqn | r = \order G
      | c = 
}}
{{end-eqn}}
If $\psi \ne \chi_0$, then $\exists y \in G$ such that $\map \psi y \ne 1$.
As $x$ runs through $G$ in the summation, $y x$ also runs through $G$.
So:
{{begin-eqn}}
{{eqn | l = \sum_{x \mathop \in G} \map \psi x
      | r = \sum_{x \mathop \in G} \map \psi {y x}
      | c = by the above claim {{explain|which claim?}}
}}
{{eqn | r = \sum_{x \mathop \in G} \map \psi y \map \psi x
      | c = $\psi$ as a character is by definition a homomorphism
}}
{{eqn | r = \map \psi y \sum_{x \mathop \in G} \map \psi x
      | c = 
}}
{{end-eqn}}
Since by assumption $\map \psi y \ne 1$, it must be true that:
:$\ds \sum_{x \mathop \in G} \map \psi x = 0$
{{qed}}
{{refactor|Suggestion: Move the dual space version into a corollary, as it is a direct application of the main theorem on the dual space, with identifying $x \in G$ as a character: $x: \chi \mapsto \map \chi x$|level = medium}}
Category:Analytic Number Theory
42hc8d8kfe3m3ykc1uekwzwyo2ioshx
Theorem: Analytic Continuation of Dirichlet L-Function

Let $\chi : G := \paren {\Z / q \Z}^\times \to \C^\times$ be a Dirichlet character modulo $q$.
{{explain|$\C^\times$}}
Let $\map L {s, \chi}$ be the Dirichlet $L$-function for $\chi$.
Let $\chi$ be the trivial character.
Then $\map L {s, \chi}$ has an analytic continuation to $\C$ except for a simple pole at $s = 1$.
Let $\chi$ be non-trivial.
Then $\map L {s, \chi}$ is analytic on $\map \Re s > 0$.

Proof:
Let $\chi$ be the trivial character.
Then by Dirichlet L-Function from Trivial Character:
:$\ds \map L {s, \chi} = \map \zeta s \cdot \prod_{p \mathop \divides q} \paren {1 - p^{-s} }$
where $\divides$ denotes divisibility.
Also, by Poles of Riemann Zeta Function, $\zeta$ is analytic on $\C$ except for a simple pole at $s = 1$.
Since $\map L {s, \chi}$ is just $\zeta$ times some constant, the same holds for this function.
If $\chi$ is non-trivial, then by the Orthogonality Relations for Characters:
:$\ds \sum_{x \mathop \in G} \map \chi x = 0$
By definition, $\chi$ is $q$-periodic, and zero on integers not coprime to $q$.
So for any $M \in \N$:
:$\ds \sum_{n \mathop = M + 1}^{M + Q} \map \chi n = 0$
Let $M, N \in \N$ be arbitrary.
Let $d$ be such that $M + q d \le N \le M + q \paren {d + 1}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = M}^N \map \chi n
      | r = \sum_{k \mathop = 0}^{d - 1} \sum_{n \mathop = 0}^{q - 1} \map \chi {M + k q + n} + \sum_{n \mathop = M + q d}^N \map \chi n
      | c =
}}
{{eqn | r = \sum_{n \mathop = M \mathop + q d}^N \map \chi n
      | c = because $\chi$ is $q$-periodic, and zero on integers not coprime to $q$
}}
{{eqn | o = \le
      | r = q
      | c = because $\size {N - M + q d} \le q$
}}
{{end-eqn}}
So the coefficients $\map \chi n$ of $\map L {s, \chi}$ have bounded partial sums.
Therefore, by Convergence of Dirichlet Series with Bounded Partial Sums, $\map L {s, \chi}$ converges locally uniformly to an analytic function on $\map \Re s > 0$.
{{qed}}
{{expand|Should be extended to all of $\C$, put the above down to get to Dirichlet's theorem}}
Category:Analytic Number Theory
63tfl9bx1torcdtibt79ngoufyt0ur2
Theorem: L-Function does not Vanish at One

Let $\psi$ be a non-trivial Dirichlet character modulo $q$.
Let $\map L {s, \chi}$ be the Dirichlet $L$-function associated to $\chi$.
Then $\map L {1, \chi} \ne 0$.

Proof:
Let $G^*$ be the group of characters modulo $q$.
{{explain|"group of characters modulo $q$" - we have the definition for character, and we have (somewhere) the definition of the ring of integers modulo $q$, but not this entity.}}
Let $\ds \map {\zeta_q} s = \prod_{\chi \mathop \in G^*} \map L {s, \chi} R$.
Among the factors of $\zeta_q$ is the the $L$-function associated to the trivial character, which by Analytic Continuation of Dirichlet L-Function we know to have a simple pole at $s = 1$.
{{AimForCont}} $\map L {1, \psi} = 0$. 
Then the zero of this factor kills the pole of the principal $L$-function.
So by Analytic Continuation of Dirichlet L-Function $\zeta_q$ is analytic on $\map \Re s > 0$.
For $\map \Re s > 1$ we have:
{{begin-eqn}}
{{eqn | l = \map {\zeta_q} s
      | r = \prod_{\chi \mathop \in G^*} \prod_p \frac 1 {1 - \map \chi p p^{-s} }
      | c = {{Defof|Euler Product}}, where $p$ ranges over the primes
}}
{{eqn | r = \prod_{p \mathop \nmid q} \prod_{\chi \mathop \in G^*}  \frac 1 {1 - \map \chi p p^{-s} }
      | c = because the Euler product is absolutely convergent
}}
{{end-eqn}}
For any prime $p$, let $f_p$ be the order of $p$ mod $q$.
{{explain|order}}
Then:
:$\map \chi p^{f_p} = \map \chi {p^{f_p} } = \map \chi 1 = 1$
So $\map \chi p$ is an $f_p$th root of unity.
Moreover by the Orthogonality Relations for Characters each distinct such root occurs $\map \phi q / f_p$ times among the numbers $\map \chi p$ where $\chi \in G^*$.
Also, letting $\xi$ be a primitive $f_p$th root of unity we find that for any $u \in \C$:
:$\ds \prod_{i \mathop = 0}^{f_p} \paren {1 - \xi^i u} = 1 - u^\xi$
Putting these facts together:
:$\ds \prod_{\chi \mathop \in G^*} \frac 1 {1 - \map \chi p p^{-s} } = \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}$
Therefore:
{{begin-eqn}}
{{eqn | l = \map {\zeta_q} s
      | r = \prod_{p \mathop \divides q} \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}
      | c =
}}
{{eqn | r = \prod_{p \mathop \divides q} \paren {1 + p^{-f_p s} + p^{-2 f_p s} + \cdots}^{\map \phi q / f_p}
      | c =
}}
{{end-eqn}}
Also, if $\chi_0$ is the trivial character modulo $q$, by definition of Euler product we have:
:$\ds \map L {\map \phi q s, \chi_0} = \prod_{p \mathop \nmid q} \paren {1 + p^{-\map \phi q s} + p^{-2 \map \phi q s} + \cdots}$
from which we see that for $s \in \R_{\ge 0}$:
:$\map {\zeta_q} s \ge \map L {\map \phi q s, \chi_0}$
However, by Analytic Continuation of Dirichlet L-Function, $\map L {\map \phi q s, \chi_0}$ diverges for $s = \map \phi q^{-1}$, and therefore so does $\map {\zeta_q} s$.
But we showed above that $\map {\zeta_q} s$ converges for $\map \Re s > 0$, a contradiction.
{{qed}}
Category:Analytic Number Theory
leidf5r00y4ufq6n3z73ajohijn7tw6
Theorem: Logarithm of Dirichlet L-Functions

Let $\chi$ be a Dirichlet character modulo $q$.
The Dirichlet series:
:$\map f s = \ds \sum_{n \mathop \ge 1} \sum_p \frac {\map \chi p^n} {n p^{n s} }$
converges absolutely to an analytic function, where $p$ ranges over the primes.
Moreover, $\map f s$ defines a branch of $\ln \map L {s, \chi}$.

Proof:
By Convergence of Dirichlet Series with Bounded Coefficients, $\map f s$ converges absolutely on $\map \Re s > 1$ to an analytic function.
For fixed $s \in \set {\map \Re s > 1}$:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop \ge 1} \sum_p \frac {\map \chi p^n} {n p^{n s} }
      | r = \sum_p \paren {\frac {\map \chi p} {p^s} + \frac {\map \chi p^2} {2 p^{2 s} } + \cdots}
      | c = Manipulation of Absolutely Convergent Series
}}
{{eqn | r = \sum_p \map \ln {\frac 1 {1 - \map \chi p p^{-s} } }
      | c = Taylor Series of Logarithm; this is the branch of $\ln$ with $\map \ln {1 + x} = x - \dfrac {x^2} 2 + \cdots$ for small $x$
}}
{{eqn | r = \ln \prod_p \paren {\frac 1 {1 - \map \chi p p^{-s} } }
      | c = Logarithm is Continuous; this is not necessarily the same branch of $\ln$
}}
{{eqn | r = \ln \map L {s, \chi}
      | c = {{Defof|Euler Product}}
}}
{{end-eqn}}
Hence $\map f s$ is an analytic branch of $\ln$ on $\map \Re s > 1$.
{{MissingLinks|"analytic branch" -- we have "analytic function" and "branch", but not a definition for what an "analytic branch" is}}
{{qed}}
Category:Analytic Number Theory
gria3cahizmvqobo58eobm66m492u5i
Theorem: Dirichlet's Theorem on Arithmetic Sequences

Let $a, q$ be coprime integers.
Let $\PP_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$.
Then $\PP_{a, q}$ has Dirichlet density:
:$\map \phi q^{-1}$
where $\phi$ is Euler's phi function.
In particular, $\PP_{a, q}$ is infinite.

Proof:
=== Lemma 1 ===
{{:Dirichlet's Theorem on Arithmetic Sequences/Lemma 1}}{{qed|lemma}}
Define:
:$\eta_{a, q} : n \mapsto \begin {cases} 1 & : n \equiv a \pmod q \\ 0 & : \text {otherwise} \end {cases}$
Theorem: Discrete Fourier Transform on Abelian Group

Let $G$ be a finite abelian group.
Let $G^*$ be the dual group of characters $G \to \C^\times$.
Let $\eta: G \to \C$ be a mapping from $G$ to the set of complex numbers.
Then for all $x \in G$:
:$\ds \map \eta x = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \innerprod \eta \chi_G \map \chi x$
where:
:$\ds \innerprod \eta \chi_G = \sum_{x \mathop \in G} \map \eta x \map {\overline \chi} x$

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \innerprod \eta \chi_G \map \chi y
      | r = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \sum_{x \mathop \in G} \map \eta x \map {\overline \chi} x \map \chi y
      | c =
}}
{{eqn | r = \frac 1 {\map \phi q} \sum_{x \mathop \in G} \map \eta x \sum_{\chi \mathop \in G^*} \map {\overline \chi} x \map \chi y
      | c =
}}
{{eqn | r = \frac 1 {\map \phi q} \map \eta y \map \phi y
      | c = Orthogonality Relations for Characters
}}
{{eqn | r = \map \eta y
      | c =
}}
{{end-eqn}}
{{qed}}
Category:Fourier Analysis
Category:Representation Theory
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Theorem: Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods

Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $V_1$ and $V_2$ be compact sets in $T$ which are disjoint:
:$V_1 \cap V_2 = \O$
Then $V_1$ and $V_2$ have disjoint neighborhoods.

Proof:
=== Lemma ===
{{:Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma}}
Let $\FF$ be the set of all ordered pairs $\tuple {Z, W}$ such that:
:$Z, W \in \tau$
:$V_1 \subseteq Z$
:$Z \cap W = \O$
By the {{Lemma|Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods}}, $\Img \FF$ covers $V_2$.
{{explain|It is not clear what mapping or relation $\Img \FF$ refers to.}}
By the definition of compact space, there exists a finite subset $K$ of $\Img \FF$ which also covers $V_2$.
By the definition of topology, $\ds \bigcup K$ is open.
By the Principle of Finite Choice, there exists a bijection $\GG \subseteq \FF$ such that $\Img \GG = K$.
{{explain}}
Then $\GG$, and hence its preimage, will be finite.
Let $\ds J = \bigcap \Preimg \GG$.
By Intersection is Largest Subset, $V_1 \subseteq J$.
By the definition of a topology, $J$ is open.
Then $\ds \bigcup K$ and $J$ are disjoint open sets such that $\ds V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$.
{{qed}}