hcju/proofwikiqa huggingface dataset:
=== Closed Sets Intersect implies Closed Sets are Connected === Let $T = \struct {S, \tau}$ be ultraconnected in the sense that: :no two non-empty closed sets of $T$ are disjoint. Let $F \subseteq S$ be an arbitrary closed set of $T$. {{AimForCont}} $F$ is not connected. Then there exist non-empty closed set $G, H$ in $F$ that are disjoint (and whose union is $F$). By Closed Set in Closed Subspace, $G$ and $H$ are closed in $T$. Because $G \cap H = \O$, $T$ is not ultraconnected. This is a contradiction. Thus $F$ is connected. As $F$ is arbitrary, this applies to all closed set of $T$. Thus $T = \struct {S, \tau}$ is ultraconnected in the sense that: :every closed set of $T$ is connected.. {{qed|lemma}} === Closed Sets are Connected implies Closed Sets Intersect === Let $T = \struct {S, \tau}$ be ultraconnected in the sense that: :every closed set of $T$ is connected. Let $G$ and $H$ be closed sets of $T$. Then their union $G \cup H$ is closed in $T$. By assumption, $G \cup H$ is connected. By Closed Set in Closed Subspace, $G$ and $H$ are closed sets of $G \cup H$. Because $G \cup H$ is connected, $G \cap H$ is non-empty. Because $G$ and $H$ were arbitrary, this applies to intersection of all such closed sets Thus $T = \struct {S, \tau}$ is ultraconnected in the sense that: :no two non-empty closed sets of $T$ are disjoint. {{qed}}
Let $A, B$ be two non-empty closed sets in $\struct {F, \tau}$. By Closed Set in Topological Subspace, $A, B$ are closed in $T$ as well. By {{Defof|Ultraconnected Space}}, $A$ and $B$ are not disjoint. Since $A$, $B$ are arbitrary, no two non-empty closed sets of $\struct {F, \tau}$ are disjoint. Hence the result from {{Defof|Ultraconnected Space}}. {{qed}}
First note that: {{begin-eqn}} {{eqn | l = 18 \, 465 \, 126 \, 257       | r = 4 \times 4 \, 616 \, 281 \, 564 + 1       | c =  }} {{eqn | l = 18 \, 465 \, 126 \, 293       | r = 4 \times 4 \, 616 \, 281 \, 573 + 1       | c =  }} {{end-eqn}} where it can be seen that prime numbers of the form $4 n + 1$ are locally increasing, and: {{begin-eqn}} {{eqn | l = 19 \, 033 \, 524 \, 533       | r = 4 \times 4 \, 758 \, 381 \, 133 + 1       | c =  }} {{eqn | l = 19 \, 033 \, 524 \, 539       | r = 4 \times 4 \, 758 \, 381 \, 135 - 1       | c =  }} {{end-eqn}} where it can be seen at that point prime numbers of the form $4 n - 1$ may now be in front, but this is inconclusive. {{ProofWanted|I'm not actually sure this is correct. If $4 n - 1$ is now in front for the first time since $18 \, 465 \, 126 \, 293$, I would expect the previous prime also to be of the same form -- but it is not.}}
We have that: From $29 \, 912 \, 035 \, 725$ and $34 \, 883 \, 817 \, 075$ are amicable: {{:Odd Amicable Pair/Examples/29,912,035,725-34,883,817,075}} From $31 \, 695 \, 652 \, 275$ and $33 \, 100 \, 200 \, 525$ are amicable: {{:Odd Amicable Pair/Examples/31,695,652,275-33,100,200,525}} From $32 \, 129 \, 958 \, 525$ and $32 \, 665 \, 894 \, 275$ are amicable: {{:Odd Amicable Pair/Examples/32,129,958,525-32,665,894,275}} {{qed}}
{{ProofWanted|Use Polynomial is Linear Combination of Monomials}} Category:Polynomial Theory 2x9lzdnsljh8sd5unxy7zgtw235pu2z
Let $C$ denote the sequence in question. We have that $554 \, 688 \, 278 \, 429$ is prime. First note that: :$\dfrac {554 \, 688 \, 278 \, 429 - 1} 2 = 277 \, 344 \, 139 \, 214 = 2 \times 138 \, 672 \, 069 \, 607$ and so is not prime. Thus $554 \, 688 \, 278 \, 429$ is not a safe prime, and thus fulfils the requirement for $C$ to be a Cunningham chain of the first kind. Then: {{begin-eqn}} {{eqn | n = 1       | l = 2 \times 554 \, 688 \, 278 \, 429 + 1       | r = 1 \, 109 \, 376 \, 556 \, 859       | c = which is prime }} {{eqn | n = 2       | l = 2 \times 1 \, 109 \, 376 \, 556 \, 859 + 1       | r = 2 \, 218 \, 753 \, 113 \, 719       | c = which is prime }} {{eqn | n = 3       | l = 2 \times 2 \, 218 \, 753 \, 113 \, 719 + 1       | r = 4 \, 437 \, 506 \, 227 \, 439       | c = which is prime }} {{eqn | n = 4       | l = 2 \times 4 \, 437 \, 506 \, 227 \, 439 + 1       | r = 8 \, 875 \, 012 \, 454 \, 879       | c = which is prime }} {{eqn | n = 5       | l = 2 \times 8 \, 875 \, 012 \, 454 \, 879 + 1       | r = 17 \, 750 \, 024 \, 909 \, 759       | c = which is prime }} {{eqn | n = 6       | l = 2 \times 17 \, 750 \, 024 \, 909 \, 759 + 1       | r = 35 \, 500 \, 049 \, 819 \, 519       | c = which is prime }} {{eqn | n = 7       | l = 2 \times 35 \, 500 \, 049 \, 819 \, 519 + 1       | r = 71 \, 000 \, 099 \, 639 \, 039       | c = which is prime }} {{eqn | n = 8       | l = 2 \times 71 \, 000 \, 099 \, 639 \, 039 + 1       | r = 142 \, 000 \, 199 \, 278 \, 079       | c = which is prime }} {{eqn | n = 9       | l = 2 \times 142 \, 000 \, 199 \, 278 \, 079 + 1       | r = 284 \, 000 \, 398 \, 556 \, 159       | c = which is prime }} {{eqn | n = 10       | l = 2 \times 284 \, 000 \, 398 \, 556 \, 159 + 1       | r = 568 \, 000 \, 797 \, 112 \, 319       | c = which is prime }} {{eqn | n = 11       | l = 2 \times 568 \, 000 \, 797 \, 112 \, 319 + 1       | r = 1 \, 136 \, 001 \, 594 \, 224 \, 639       | c = which is prime }} {{eqn | n = 12       | l = 2 \times 1 \, 136 \, 001 \, 594 \, 224 \, 639 + 1       | r = 2 \, 272 \, 003 \, 188 \, 449 \, 279       | c = which is $19 \times 119 \, 579 \, 115 \, 181 \, 541$ prime }} {{end-eqn}} Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $12$ can be done by a computer search. {{qed}}
Let $S$ be bounded above by $x \in \R$. By the Archimedean Principle, there exists an integer $n \ge x$. Then $S$ is bounded above by $n$. By Set of Integers Bounded Above by Integer has Greatest Element, $S$ has a greatest element. {{qed}}
Let $S$ be bounded below by $x \in \R$. By the Archimedean Principle, there exists an integer $n \le x$. Then $S$ is bounded below by $n$. By Set of Integers Bounded Below by Integer has Smallest Element, $S$ has a smallest element. {{qed}}
=== Existence === By Set of Integers Bounded Above by Real Number has Greatest Element, the set: :$S = \left\{{m \in \Z: m \le x}\right\}$ has a greatest element, say $n$. Because $n+1>n$, $n+1\notin S$. Thus $n+1> x$. Thus $n\leq x < n+1$. {{qed|lemma}} === Uniqueness === Let $n\in\Z$ be such that: :$n \leq x < n + 1$ We show that $n$ is a greatest element of the set: :$S = \left\{{m \in \Z: m \le x}\right\}$ so that the uniqueness follows from Greatest Element is Unique. Because $n\leq x$, we have $n \in S$. Let $m \in S$. Because $m \leq x < n+1$, $n+1 > m$. By Weak Inequality of Integers iff Strict Inequality with Integer plus One: :$n \geq m$. Because $m$ was arbitrary, $n$ is a greatest element of $S$. {{qed}}
By Supremum of Set of Integers equals Greatest Element, $S$ has a greatest element $n \in \Z$, that is equals to the supremum of $S$. {{qed}}
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Idempotence can be written as: : $\neg \left({p \lor p}\right) \lor p$ This evaluates as follows: :$\begin{array}{|cccc|c|c|} \hline \neg & (p & \lor & p) & \lor & p \\ \hline 2 & 2 & 1 & 1 & 2 & 1 \\ 1 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics r07evkbzlhssv5d3yrb3yass8oq8v9l
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Addition can be written as: : $\neg q \lor \left({p \lor q}\right)$ This evaluates as follows: :$\begin{array}{|cc|c|ccc|} \hline \neg & q & \lor & (p & \lor & q) \\ \hline 2 & 1 & 2 & 1 & 1 & 1 \\ 1 & 2 & 2 & 1 & 2 & 2 \\ 2 & 1 & 2 & 2 & 2 & 1 \\ 1 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics rk497e2li883lwe4k6qna1aco309ty8
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Rule of Commutation can be written as: :$\neg \left({p \lor q}\right) \lor \left({q \lor p}\right)$ This evaluates as follows: :$\begin{array}{|cccc|c|ccc|} \hline \neg & (p & \lor & q) & \lor & (q & \lor & p) \\ \hline 2 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 2 & 2 & 2 & 2 & 2 & 1 \\ 1 & 2 & 2 & 1 & 2 & 1 & 2 & 2 \\ 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics nhvy61sitluk232bj2i72se5qmw6xrs
By Set of Integers Bounded Above by Real Number has Greatest Element, $S$ has a greatest element, say $n \in S$. By Greatest Element is Supremum, $n$ is the supremum of $S$. {{qed}}
By the definitional abbreviation for the conditional: :$\mathbf A \implies \mathbf B =_{\text{def}} \neg \mathbf A \lor \mathbf B$ the Factor Principle can be written as: :$\neg \left({\neg p \lor q}\right) \lor \left({\neg \left({r \lor p}\right) \lor \left ({r \lor q}\right)}\right)$ This evaluates as follows: :$\begin{array}{|ccccc|c|cccccccc|} \hline \neg & (\neg & p & \lor & q) & \lor & (\neg & (r & \lor & p) & \lor & (r & \lor & q)) \\ \hline 1 & 2 & 1 & 2 & 1 & 2 & 2 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 2 & 1 & 2 & 2 & 2 & 1 \\ 1 & 2 & 1 & 2 & 2 & 2 & 2 & 1 & 1 & 1 & 2 & 1 & 2 & 2 \\ 1 & 2 & 1 & 2 & 2 & 2 & 1 & 2 & 2 & 1 & 2 & 2 & 2 & 2 \\ 2 & 1 & 2 & 1 & 1 & 2 & 1 & 1 & 2 & 2 & 1 & 1 & 1 & 1 \\ 2 & 1 & 2 & 1 & 1 & 2 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 1 \\ 1 & 1 & 2 & 2 & 2 & 2 & 1 & 1 & 2 & 2 & 2 & 1 & 2 & 2 \\ 1 & 1 & 2 & 2 & 2 & 2 & 1 & 2 & 2 & 2 & 2 & 2 & 2 & 2 \\ \hline \end{array}$ {{qed}} Category:Formal Semantics g378broae9uk77f3lxwet48mpsu2f3y
We must show that $H^{- \circ - \circ} = H^{- \circ}$. First we show that $H^{- \circ - \circ} \subseteq H^{- \circ}$. By the definition of interior: :$H^{- \circ} \subseteq H^-$ By Topological Closure of Subset is Subset of Topological Closure: :$H^{- \circ -} \subseteq H^{--} = H^-$ By Interior of Subset: :$H^{- \circ - \circ } \subseteq H^{- \circ}$ {{qed|lemma}} Next we show $H^{- \circ - \circ} \supseteq H^{- \circ}$. Again by the definition of interior: : $H^{- \circ -} \supseteq H^{- \circ}$ By Interior of Subset: : $H^{- \circ - \circ}  \supseteq H^{- \circ \circ} = H^{- \circ}$ {{qed|lemma}} Thus by definition of set equality: : $H^{- \circ - \circ} = H^{- \circ}$ {{qed}}
{{begin-eqn}} {{eqn	| l = \left\vert{\int_{C_r} f\left({z}\right) \rd z}\right\vert 	| r = \left\vert{\int_{C_r} e^{i a z} g\left({z}\right) \rd z}\right\vert  }} {{eqn	| r = \left\vert{i r}\right\vert \left\vert{\int_0^\pi e^{i a r e^{i \theta} } g \left({r e^{i \theta} }\right) e^{i \theta} \rd \theta}\right\vert 	| c = {{Defof|Complex Contour Integral}} }} {{eqn	| r = \left\vert{i r}\right\vert \left\vert{\int_0^\pi e^{i a r \left({i \sin \theta + \cos \theta}\right)} g \left({r e^{i \theta} }\right) e^{i \theta} \rd \theta}\right\vert 	| c = Euler's Formula }} {{eqn	| r = r \left\vert{\int_0^\pi e^{a r \left({i \cos \theta - \sin \theta}\right)} g \left({r e^{i \theta} }\right) e^{i \theta} \rd \theta}\right\vert 	| c = $i^2 = -1$, $\left\vert{i}\right\vert = 1$ }} {{eqn	| o = \le 	| r = r \int_0^\pi \left\vert{e^{i a r \cos \theta} }\right\vert \left\vert{e^{-a r \sin \theta} }\right\vert \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \left\vert{e^{i \theta} }\right\vert \rd \theta 	| c = Modulus of Complex Integral }} {{eqn	| r = r \int_0^\pi e^{-a r \sin \theta} \left\vert{g \left({r e^{i\theta} }\right)}\right\vert \rd \theta 	| c = $\left\vert{e^{i \theta} }\right\vert = 1$ for real $\theta$ }} {{eqn	| o = \le  	| r = r \int_0^\pi e^{-a r \sin \theta} \max_{0 \le \theta \le \pi} \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \rd \theta 	| c = {{Defof|Supremum of Real-Valued Function}} }} {{eqn  	| o = \le 	| r = r \max_{0 \le \theta \le \pi} \left\vert{g \left({r e^{i \theta} }\right)}\right\vert \int_0^\pi e^{-\left({2 a r \theta}\right) / \pi} \rd \theta 	| c = Shape of Sine Function }} {{eqn	| r = r \max_{0 \le \theta \le \pi} \left\vert{g\left({r e^{i \theta} }\right)}\right\vert \left[{- \frac{\pi e^{\left({2 a r \theta}\right) / \pi} } {2 a r} }\right]_0^\pi 	| c = Primitive of Exponential of a x, Fundamental Theorem of Calculus  }} {{eqn	| r = \max_{0 \le \theta \le \pi} \left\vert{g\left({r e^{i \theta} }\right)}\right\vert \frac{\pi \left({1 - e^{-2 a r} }\right)} {2 a} }} {{eqn	| o = \le 	| r = \frac \pi {2 a} \max_{0 \le \theta \le \pi} \left\vert{g\left({r e^{i \theta} }\right)}\right\vert 	| c = $e^{-ar} < 1$ for positive $a, r$ }} {{eqn	| o = \le 	| r = \frac \pi a \max_{0 \le \theta \le \pi} \left\vert{g\left( {r e^{i \theta} }\right)}\right\vert }} {{end-eqn}} {{qed}}
Follows from Dirichlet Series of Convolution of Arithmetic Functions {{ProofWanted}} Category:Dirichlet Series e03r49gdkg8ry56gqsgk754xuk3dgql
{{ProofWanted}} Using Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function: The functional equation: :$\map \xi s = \map \xi {1 - s}$ follows upon observing that this integral is invariant under $s \mapsto 1 - s$. {{ProofWanted|using the formula for $\zeta\cdot\Gamma$ and contour integration}}
We have: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty n^{-s}       | r = \sum_{n \mathop = 1}^\infty n \left({ n^{-s} - \left({n + 1}\right)^{-s} }\right)       | c = Abel's Lemma: Formulation 2 }} {{eqn | r = s \sum_{n \mathop = 1}^\infty n \int_n^{n+1} x^{-s - 1} \rd x       | c = }} {{eqn | r = s \int_1^\infty \left\lfloor{x}\right\rfloor x^{-s - 1} \rd x       | c = where $\left\lfloor{x}\right\rfloor$ denotes the floor function of $x$ }} {{eqn | l =       | r = \frac s {s - 1} - s \int_1^\infty \left\{ {x}\right\} x^{-s - 1} \rd x       | c = where $\left\{ {x}\right\}$ denotes the fractional part of $x$ }} {{end-eqn}} {{qed}}
Use Riemann Zeta Function in terms of Dirichlet Eta Function and Binomial Theorem. {{ProofWanted}} Category:Riemann Zeta Function 9ik01re4fpmavul7jfabh52lrh1ne3z
Note that by Riemann Zeta Function is Analytic, $\zeta(s)$ is indeed analytic for $\Re(s)>1$. By Complex Plane minus Point is Connected, $\C\setminus\{1\}$ is connected. By Uniqueness of Analytic Continuation, there is at most one analytic continuation of $\zeta$ to $\C\setminus\{1\}$. {{finish|link to proofs of various analytic continuations}} * Functional Equation for Riemann Zeta Function, analyticity is shown at Analytic Continuation of Riemann Zeta Function * The stepwise method as in Riemann Zeta Function in terms of Dirichlet Eta Function
Note that by Riemann Zeta Function is Analytic, $\zeta(s)$ is indeed analytic for $\Re(s)>1$. By Complex Half-Plane minus Point is Connected, $\{\sigma>0\}\setminus\{1\}$ is connected. By Uniqueness of Analytic Continuation, there is at most one analytic continuation of $\zeta$ to $\{\sigma>0\}\setminus\{1\}$. By either: * Analytic Continuation of Riemann Zeta Function using Dirichlet Eta Function * Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part * Analytic Continuation of Riemann Zeta Function using Jacobi Theta Function there exists one. {{qed}}
{{ProofWanted|Use Functional Equation for Riemann Zeta Function.}} Category:Riemann Zeta Function 0wb6hf9c3s8vwlyid6qewmaur6ei6zt
Let $\varepsilon$ be the identity arithmetic function. By Dirichlet Series of Identity Arithmetic Function, $\varepsilon$ has Dirichlet series $E(s) = 1$. By Dirichlet Series of Convolution of Arithmetic Functions, $F(s)G(s) = 1$. {{qed}} Category:Dirichlet Series b8l8wup13ftuuhffklh4wu1yb17bdtn
Let $*$ denote Dirichlet convolution. Let $\varepsilon$ denote the identity arithmetic function. === Sufficient Condition === Let $f$ have a Dirichlet inverse $g$: :$f * g = \varepsilon$ Then: {{begin-eqn}} {{eqn | l = 1       | r = \map \varepsilon 1       | c = {{Defof|Identity Arithmetic Function}} }} {{eqn | r = \map {\paren {f * g} } 1 }} {{eqn | r = \sum_{d \mathop \divides 1} \map f d \, \map g {d^{-1} }       | c = {{Defof|Dirichlet Convolution}} }} {{eqn | r = \map f 1 \, \map g 1       | c = Divisors of One }} {{end-eqn}} Thus $\map f 1 \ne 0$. {{qed|lemma}} === Necessary Condition === Suppose that $\map f 1 \ne 0$. We want to find an arithmetic function $g$ such that: :$(1): \quad \map {\paren {f * g} } 1 = 1$ :$(2): \quad \map {\paren {f * g} } n = 0$ for all $n > 1$ We have: :$\map {\paren {f * g} } 1 = \map f 1 \, \map g 1$ So we have no choice but to define: :$\map g 1 = \paren {\map f 1}^{-1}$ and condition $(1)$ is satisfied. Condition $(2)$ can be written as: {{begin-eqn}} {{eqn | l = 0       | r = \sum_{d \mathop \divides n} \map f d \, \map g {\frac n d} }} {{eqn | r = \map f 1 \, \map g n + \sum_{\substack {d \mathop \divides n \\ d \mathop > 1} } \map f d \, \map g {\frac n d} }} {{end-eqn}} That is: :$\map g n = -\dfrac 1 {\map f 1} \displaystyle \sum_{\substack {d \mathop \divides n \\ d \mathop > 1} } \map f d \, \map g {\frac n d}$ We can recursively define $g$ by this formula, starting with $\map g 1$ as above. By definition, $g$ then satisfies: :$g * f = \varepsilon$ {{qed}}
Let $s = \sigma + i t$. By Integral Representation of Riemann Zeta Function in terms of Fractional Part, the above integral coincides with $\map \zeta s$ for $\sigma > 1$. We show that it is analytic for $0 \le \sigma \le 1$. For $n \ge 1$, let: {{begin-eqn}} {{eqn | l = \cmod {a_n}       | r = \cmod {\int_n^{n + 1} \fractpart x x^{-s - 1} \rd x}       | c =  }} {{eqn | r = \cmod s \int_n^{n + 1} \cmod {\fractpart x x^{-s - 1} } \rd x       | o = \le       | c = Modulus of Complex Integral }} {{eqn | r = \cmod s \int_n^{n + 1} \fractpart x x^{-\sigma - 1} \rd x }} {{eqn | o = \le       | r = \cmod s \int_n^{n + 1} x^{-\sigma-1} \rd x       | c = as $\fractpart x \le 1$ }} {{eqn | r = -\intlimits {\dfrac {\cmod s} \sigma x^{-\sigma} } n {n + 1}       | c =  }} {{eqn | r = \dfrac {\cmod s} \sigma \paren {\frac 1 {n^\sigma} - \frac 1 {\paren {n + 1}^\sigma} }       | c = }} {{eqn | r = \dfrac {\cmod s} \sigma \paren {\frac {\paren {n + 1}^\sigma - n^\sigma} {n^\sigma \paren {n + 1}^\sigma} }       | c = }} {{end-eqn}} By the Mean Value Theorem, for some $n \le \theta \le n + 1$: :$\paren {n + 1}^\sigma - n^\sigma = \sigma \theta^{\sigma - 1}$ and for $\sigma \le 1$ we have: :$\sigma \theta^{\sigma - 1} \le \sigma \paren {n + 1}^{\sigma - 1}$ Thus we have: {{begin-eqn}} {{eqn | l = \cmod {a_n}       | o = \le       | r = \dfrac {\cmod s} \sigma \paren {\frac {\paren {n + 1}^\sigma - n^\sigma} {n^\sigma \paren {n + 1}^\sigma} } }} {{eqn | o = \le       | r = \dfrac {\cmod s} \sigma \paren {\frac {\sigma \paren {n + 1}^{\sigma - 1} } {n^\sigma \paren {n + 1}^\sigma} } }} {{eqn | r = \cmod s \frac 1 {n^\sigma \paren {n + 1} } }} {{eqn | o = \le       | r = \frac {\cmod s} {n^{\sigma + 1} } }} {{end-eqn}} Thus on any compact subset of the half-plane $\sigma > 0$ which is bounded away from $0$, we have that the series: :$\displaystyle \frac s {s - 1} - \sum_{n \mathop = 1}^\infty a_n = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$ converges uniformly as: {{begin-eqn}} {{eqn | l = \cmod {\sum_{n \mathop = M}^\infty a_n}       | o = \le       | r = \sum_{n \mathop = M}^\infty \cmod {a_n} }} {{eqn | o = \le       | r = \sum_{n \mathop = M}^\infty \frac {\cmod s} {n^{\sigma + 1} } }} {{end-eqn}} Since it is a compact region, $\cmod s$ is bounded by some constant $C$. So: :$\cmod {\displaystyle \sum_{n \mathop = M}^\infty a_n} \le C \sum_{n \mathop = M}^\infty \frac 1 {n^{\sigma + 1} }$ As Dirichlet series are uniformly convergent on compact regions bounded away from their abscissa of convergence, we may find an $M$ large enough so that the above is smaller than $\epsilon$ for all $\sigma$. Therefore the series is uniformly convergent. Thus by Uniform Limit of Analytic Functions is Analytic the series defines an analytic function on the strip $0 < \sigma \le 1$ and $s \ne 1$. This is also equal to $\zeta$ on the halfplane $\sigma>1$. Thus the series defines the unique analytic continuation of $\zeta$ onto the halfplane $\sigma > 0$. {{qed}}
Let $S$ be a set. === Equivalence of Definitions 1, 2 and 3 === {{:Equivalence of Definitions of Ultrafilter on Set/Equivalence of Definitions 1, 2 and 3}}: {{qed|lemma}} === Definition 1 implies Definition 4 === Let $\FF \subseteq \powerset S$ a filter on $S$ which fulfills the condition: :whenever $\GG$ is a filter on $S$ and $\FF \subseteq \GG$ holds, then $\FF = \GG$ By definition of filter, $\FF$ satisfies the Finite Intersection Property and is a non-empty set of subsets of $S$. Let $U \subseteq S$. {{AimForCont}} $U \notin \FF$. Let $V = U^c$, where $U^c$ denotes the relative complement of $U$ in $S$. Suppose there exists $C \in \FF$ such that: :$C \cap V = \O$ By Empty Intersection iff Subset of Complement, it follows that: :$C \subset U$ Thus $U \in \FF$, which contradicts our hypothesis that $U \notin \FF$. Hence the set $\Omega = \set {C \cap V: C \in \FF}$ is a non-empty set of non-empty set. Note that $\Omega$ is downward directed, that is, $\Omega$ is a filter basis. From Filter Basis Generates Filter, $\Omega$ generates a filter $\GG$ on $S$ that contains $\FF$ and $\set V$. By hypothesis, we have: :$\FF =\GG$ Hence $U^c = V \in \FF$. Thus $\FF$ is an ultrafilter by Definition 2. {{qed|lemma}} === Definition 4 implies Definition 1 === Let $S$ be a non-empty set and $\FF$ a non-empty set of subsets on $S$ which fulfills the conditions: :$\FF$ has the finite intersection property :For all $U \subseteq S$, either $U \in \FF$ or $U^c \in \FF$. Because $\FF$ has the finite intersection property, it follows that $\O \notin \FF$. As $\O \notin \FF$, it follows that $\O^c = S \in \FF$. Let $U, V \in \FF$. Then either $U \cap V \in \FF$ or $\paren {U \cap V}^c \in \FF$. However we observe that: :$\paren {U \cap V}^c \cap U \cap V = \O$ By finite intersection property, it follows that $U \cap V \in \FF$. Now let $U \in \FF$ and $V \in S$ such that $U \subseteq V \subseteq S$. Then either $V \in \FF$ or $V^c \in \FF$. However we observe that: :$V^c \cap U \subseteq V^c \cap V = \O$ By finite intersection property, it follows that $V \in \FF$. Therefore $\FF$ is a filter. Let $\GG$ be a filter on $S$ such that $\FF \subseteq \GG$. Assume that $\FF \subsetneq \GG$. Then there exists $A \in \GG \setminus \FF$. Since $\O \notin \GG$ this implies that $A^c \notin \GG$. As $\FF \subsetneq \GG$, it follows that $A^c \notin \FF$. Therefore neither $A \in \FF$ nor $A^c \in \FF$, a contradiction to our assumption. Thus $\FF = \GG$, which implies that $\FF$ is an ultrafilter by definition 1. {{qed}} Category:Filter Theory 4ovq5r6ysetl91ke0xuuy9zzqs54szi
By Riemann Zeta Function in terms of Dirichlet Eta Function, it coincides with $\zeta$ for $\map \Re s > 1$. By Dirichlet Eta Function is Analytic, it is analytic for $\map \Re s > 0$, except at $s = 1$. {{qed}}
By Integral Representation of Riemann Zeta Function in terms of Jacobi Theta Function, it coincides with $\zeta(s)$ for $\Re(s)>1$. Interchanging integral and derivative, one shows that the integral is analytic for $\Re(s)>0$. {{finish}}
=== Definition 1 equals Definition 2 === Follows from Infimum of Set of Integers equals Smallest Element. {{qed|lemma}} === Definition 1 equals Definition 3 === Let $S$ be the set: :$S = \left\{ {m \in \Z: m \ge x}\right\}$ Let $n = \inf S$. By Infimum of Set of Integers is Integer, $n \in \Z$. By Infimum of Set of Integers equals Smallest Element, $n\in S$. Because $n\in S$, we have $n \geq x$. Because $n-1 < n$, we have by definition of supremum: :$n-1 \notin S$ Thus $n-1 < x$. Thus $n$ is an integer such that: :$n-1 < x \leq n$ So $n$ is the ceiling function by definition 3. {{qed|lemma}} === Definition 3 equals Definition 2 === Let $n$ be an integer such that: :$n-1 < x \leq n$ We show that $n$ is the smallest element of the set: :$S = \left\{ {m \in \Z: m \ge x}\right\}$ Let $m \in \Z$ such that $m \ge x$. We show that $n\leq m$. {{AimForCont}} $m < n$. By Weak Inequality of Integers iff Strict Inequality with Integer minus One: :$m \le n - 1$ and so from the definition of $g$ it follows that $m < x$. By Proof by Contradiction it follows that $m \ge n$. Because $m \in S$ was arbitrary, $n$ is the smallest element of $S$. Thus $n$ is the ceiling function by definition 2. {{qed}}