hcju/proofwikiqa huggingface dataset:
From Multiplicative Group of Rationals is Subgroup of Reals, $\struct {\Q, \times} \lhd \struct {\R, \times}$. From Multiplicative Group of Reals is Subgroup of Complex, $\struct {\R, \times} \lhd \left({\C, \times}\right)$. Thus $\struct {\Q, \times} \le \struct {\C, \times}$. From Non-Zero Complex Numbers under Multiplication form Abelian Group, $\struct {\C, \times}$ is abelian. From Subgroup of Abelian Group is Normal it follows that $\struct {\Q, \times} \lhd \struct {\C, \times}$. {{qed}} Category:Multiplicative Group of Rational Numbers Category:Multiplicative Group of Complex Numbers Category:Normal Subgroups r2dngx9z4talkqh8mt674djbe9q9wyu
From Rational Numbers form Subfield of Real Numbers, $\struct {\Q, +, \times}$ is a subfield of $\struct {\R, + \times}$. From Real Numbers form Subfield of Complex Numbers, $\struct {\R, +, \times}$ is a subfield of $\struct {\C, + \times}$. Thus from Subfield of Subfield is Subfield $\struct {\Q, +, \times}$ is a subfield of $\struct {\C, + \times}$. {{qed}}
Let $K_1$ be a subfield of $R$ and $K_2$ be a subfield of $K_1$. Then by definition: :$K_1 \subseteq R$ :$K_2 \subseteq K_1$ From Subset Relation is Transitive it follows that $K_2 \subseteq R$ So by definition $K_2$ is a subfield of $R$. {{qed}} Category:Subfields g8qj1gxkqsgvlury7jpgaahfpiooiz9
Recall the definition of Jacobson radical of $A$: :$\map {\operatorname {Jac} } A$ is the intersection of all maximal ideals of $R$. === $\subset$ Direction === {{AimForCont}} that $x \in \map {\operatorname {Jac} } A$ such that there exists $y \in A$ such that $1_A - x y \notin A^\times$.  By definition $x$ is contained in all maximal ideals of $A$. Let $\mathfrak m \subseteq A$ be one particular such maximal ideal of $A$. Then $x \in \map {\operatorname {Jac} } A \subseteq \mathfrak m$ implies that $xy \in \mathfrak m$ and therefore $1_A \in \mathfrak m$. But if $1_A \in \mathfrak m$ then from Ideal of Unit is Whole Ring: Corollary: :$\mathfrak m = R$ This contradicts the definition of $\mathfrak m$ as a maximal ideal of $A$. Hence by Proof by Contradiction: :$\map {\operatorname {Jac} } A \subseteq \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$ {{qed|lemma}} === $\supset$ Direction === Now suppose that $x \notin \map {\operatorname {Jac} } A$. Then $x \notin \mathfrak m$ for some maximal ideal $\mathfrak m$ of $A$. Because $\mathfrak m$ is maximal, $x$ and $\mathfrak m$ generate $A$. Therefore there exist $w \in \mathfrak m$ and $y \in A$ such that $w + x y = 1_A$. Thus $1_A - x y \in \mathfrak m$, and $1_A - x y \notin A^\times$. Hence: :$\map {\operatorname {Jac} } A \supseteq \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$ {{qed|lemma}} The result follows by definition of set equality. {{qed}} Category:Commutative Algebra sr2jx31q78xi1evsr3ym5z0zl65mp6a
Since $J \subset R$, it follows from: :Quotient Ring of Commutative Ring is Commutative and: :Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a commutative ring with unity.  Let $0_{R / J}$ be the zero of $R / J$. === Sufficient Condition === Let $J$ be a prime ideal. We need to show that if: :$x + J, \ y + J \in \struct {R / J, +, \circ}$ such that: :$\paren {x + J} \circ \paren {y + J} = \paren {x \circ y} + J = 0_{R / J}$ then: :$x + J = 0_{R/J}$ or $y + J = 0_{R/J}$. The zero of $R / J$ is $0_{R / J} = J$. Therefore: :$\paren {x \circ y} + J = 0_{R / J} \implies x \circ y \in J$ Because $J$ is a prime ideal, it follows that either $x \in J$ or $y \in J$. {{WLOG}} we assume that $x \in J$. But then: :$x + J = J = 0_{R / J}$ {{qed|lemma}} === Necessary Condition === Let $R / J$ be an integral domain. Let $x, y \in R$ be such that $x \circ y \in J$. Then: :$0_{R / J} = J = x \circ y + J = \paren {x + J} \circ \paren {y + J}$ Because $R / J$ is an integral domain it follows that: :$x + J = 0_{R / J}$ or: :$y + J = 0_{R / J}$ If $x + J = 0_{R / J}$, then $x \in J$. If $y + J = 0_{R / J}$, then $y \in J$. Thus either $x \in J$ or $y \in J$, and so by definition $J$ is a prime ideal of $R$. {{qed}}
From Maximal Ideal iff Quotient Ring is Field: :the quotient ring $R / M$ is a field. It follows from Field is Integral Domain that $R / M$ is an integral domain. Finally it follows from Prime Ideal iff Quotient Ring is Integral Domain that $M$ is a prime ideal. {{qed}}
Since $J \subset R$, it follows from: :Quotient Ring of Commutative Ring is Commutative and: : Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a commutative ring with unity. Let $0_{R / J}$ be the zero of $R / J$. === Sufficient Condition === Let $J$ be radical. We need to show that: :if $x + J \in R / J$ such that $\left({x + J}\right)^n = 0_{R / J}$ for some positive integer $n$ :then $x + J = 0_{R / J}$. Let $\left({x + J}\right)^n = 0_{R / J}$. Then: :$x^n + J = 0_{R / J}$ and therefore: :$x^n \in J$ Because $J$ is radical it follows that: :$x \in J$ That is: :$x + J = 0_{R / J}$ Therefore $R / J$ is reduced. {{qed|lemma}} === Necessary Condition === Let $R / J$ be reduced. Let $x \in R$ be such that $x^n \in J$. Then: :$0_{R / J} = J = x^n + J = \left({x + J}\right)^n$ Because $A / J$ is reduced, this implies that: :$x + J = 0_{R / J}$ That is: :$x \in J$ This shows that $J$ is radical. {{qed}} Category:Quotient Rings c8zunqlot5mf4vpehyl9wn5288fmfge
Let $\ds m_1 = \prod_{j \mathop \in J} X_j^{k_j}, m_2 = \prod_{j \mathop \in J} X_j^{k_j'}$ be two monomials. Their product is: :$\ds m_1 \circ m_2 = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$ If $k_j + k_j' \ne 0$ then either $k_j \ne 0$ or $k_j' \ne 0$ (or both are nonzero). Therefore if $k_j + k_j' \ne 0$ for infinitely many $j$, then either $m_1$ or $m_2$ is not a monomial. {{qed}} Category:Monomials h2lb3nxhe53agesh51ic1k8xmovh20p
Let $M$ be the set of all monomials on the indexed set $\family {X_j: j \in J}$. We are required to show that the following properties hold: {{begin-axiom}} {{axiom | q = \forall m_1, m_2 \in M         | m = m_1 \circ m_2 \in M         | t = Closure }} {{axiom | q = \forall m_1, m_2, m)3 \in M         | m = \paren {m_1 \circ m_2} \circ m_3 = m_1 \circ \paren {m_2 \circ m_3}         | t = Associativity }} {{axiom | q = \forall m_1, m_2 \in M         | m = m_1 \circ m_2 = m_2 \circ m_1         | t = Commutativity }} {{axiom | q = \exists e_m \in M: \forall m_1 \in M         | m = m_1 \circ e_M = m_1         | t = Identity }} {{end-axiom}} First note that using the multiindex notation described in the definition of monomials, for $r \in \N$, $m_i = \mathbf X^{k^i} \in M$, $i = 1, \ldots, r$, the product of the $m_i$ is given by: :$m_1 \circ \cdots \circ m_r = \mathbf X^{k^1 + \cdots + k^r}$ Here the superscripts enumerate the multiindices, and do not indicate raising to a power. So to show the closure, associativity and commutativity of monomials under $\circ$, it is sufficient to show the corresponding properties for multiindices under addition defined by: :$\family {k^1 + k^2}_j := k^1_j + k^2_j$ In the following $k^1, k^2, k^3$ are multiindices, that is, families of non-negative integers indexed by $J$ such that only finitely many entries are non-zero. === Proof of Closure === Let $\family {k^1 + k^2}_j = k^1_j + k^2_j \ne 0$. By definition of multiindex, at least one of $k^1_j$ and $k^2_j$ must be non-zero, and this can only be true for a finite number of entries. Furthermore, since $k^1_j,\ k^2_j \ge 0$, we have $k^1_j + k^2_j \ge 0$. Therefore $k^1 + k^2$ has finitely many non-zero entries, and these are all positive, and multiindices are closed under addition. === Proof of Associativity === Using associativity of integer addition, we have:  :$\family {\paren {k^1 + k^2} + k^3}_j = \paren {k^1_j + k^2_j} + k^3_j = k^1_j + \paren {k^2_j + k^3_j} = \family {k^1 + \paren {k^2 + k^3} }_j$ So addition of multiindices is associative. === Proof of Commutativity === Using commutativity of integer addition, we have  :$\family {k^1 + k^2}_j = k^1_j + k^2_j = k^2_j + k^1_j = \family {k^2 + k^1}_j$ So addition of multiindices is commutative. === Proof of Existence of Identity === Let $e_M$ be the multiindex such that $\family {e_M}_j = 0$ for all $j \in J$. Then: :$\family {e_M + k^1}_j = \family {e_M}_j + k^1_j = \family {k^1}_j$ so $e_M$ is an identity for the set of monomials. {{qed}} Category:Examples of Monoids Category:Polynomial Theory igkuc9y7o4iylu6pr4yx1afrbjflqrh
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity. Let $\left\{{X_j: j \in J}\right\}$ be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$  :$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$  :$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$  be arbitrary polynomials in the indeterminates $\left\{{X_j: j \in J}\right\}$ over $R$. Then: {{begin-eqn}} {{eqn|l = \left({ f + g }\right) + h      |r = \sum_{k \mathop \in Z} \left({ \left({ a_k + b_k }\right) + c_k }\right) \mathbf X^k      |c = applying the definition of polynomial addition twice }} {{eqn|r = \sum_{k \mathop \in Z} \left({ a_k + \left({ b_k + c_k }\right) }\right) \mathbf X^k      |c = because $+$ in $R$ is associative }}  {{eqn|r = f + \left({ g + h }\right)      |c = applying the definition of polynomial addition twice }}  {{end-eqn}} {{qed}} Category:Polynomial Theory khbinhw56oq156ek6gvmmv21h544jhg
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity with zero $0_R$. Let $\left\{{X_j: j  \in J}\right\}$  be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$  be an arbitrary polynomial form in the indeterminates $\left\{{X_j: j \in J}\right\}$ over $R$. Let: :$\displaystyle N = \sum_{k \mathop \in Z} 0_R \mathbf X^k$  be the null polynomial. Then: {{begin-eqn}} {{eqn | l = f + N       | r = \sum_{k \mathop \in Z} \left({ a_k + 0_R }\right) \mathbf X^k       | c = Definition of Polynomial Addition }} {{eqn | r = \sum_{k \mathop \in Z} a_k \mathbf X^k       | c =  }}  {{eqn | r = f       | c = Definition of Polynomial Addition }}  {{end-eqn}} Therefore, $N + f = f$ for all polynomial forms $f$. Therefore, $N$ is an additive identity for the set of polynomial forms. {{qed}} Category:Polynomial Theory owfhks8vopppzlob0sxveyfurzp2jir
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity with zero $0_R$. Let $\left\{{X_j: j  \in J}\right\}$ be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ :$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ be arbitrary polynomials in the indeterminates $\left\{{X_j: j \in J}\right\}$ over $R$. Then: {{begin-eqn}} {{eqn | l = f + g       | r = \sum_{k \mathop \in Z} \left({ a_k + b_k }\right) \mathbf X^k       | c = by definition of polynomial addition }} {{eqn | r = \sum_{k \mathop \in Z} \left({ b_k + a_k }\right) \mathbf X^k       | c = because addition in $R$ is commutative }}  {{eqn | r = g + f       | c = by definition of polynomial addition }}  {{end-eqn}} So $f + g = g + f$ for all polynomials $f$ and $g$. Therefore polynomial addition is commutative. {{qed}} Category:Polynomial Theory 9es01ie7qkr8hwqtnlzft4e1kqkmze8
Let $(R, +, \circ)$ be a commutative ring with unity with zero $0_R$. To improve readability of the expressions used, we will write the ring product $\circ$ in multiplicative notation. Let $\left\{{X_j: j \in J}\right\}$ be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$  :$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ :$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$  be arbitrary polynomials in the indeterminates $\left\{{X_j: j  \in J}\right\}$ over $R$. Then it follows from Polynomials Closed under Ring Product that: :$\displaystyle f \circ \left({ g \circ h }\right) = \sum_{k \mathop \in Z} m_k \mathbf X^k$ for some $m_k \in R$, and: :$\displaystyle \left({ f \circ g }\right) \circ h = \sum_{k \mathop \in Z} n_k \mathbf X^k$ for some $n_k \in R$. To establish associativity of $\circ$ we compute $m_k$ and $n_k$, and check that they are equal.  We have: {{begin-eqn}} {{eqn|l = m_k      |r = \sum_{r + s \mathop = k} a_r \sum_{p + q \mathop = s} b_p c_q      |c = by definition of polynomial multiplication }} {{eqn|r = \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = s} a_r b_p c_q      |c = by the properties of finite sums }}  {{eqn|r = \sum_{p + q + r \mathop = k} a_r b_p c_q      |c = by the properties of finite sums }}  {{end-eqn}} {{explain|what properties?}} Similarly we compute: {{begin-eqn}} {{eqn|l = n_k      |r = \sum_{r + s \mathop = k} \left({ \sum_{p + q \mathop = r} a_p b_q }\right) c_s      |c = by definition of polynomial multiplication }} {{eqn|r = \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = r} a_p b_q c_s      |c = by the properties of finite sums }}  {{eqn|r = \sum_{p + q + s \mathop = k} a_p b_q c_s      |c = by the properties of finite sums }}  {{end-eqn}} {{explain|again, what properties?}} Since $p$, $q$, $r$ and $s$ are all dummy variables, it follows that $m_k = n_k$ for all $k \in Z$. Therefore, $f \circ \left({ g \circ h }\right) = \left({ f \circ g }\right) \circ h$ for all polynomials $f$, $g$ and $h$. Hence multiplication of polynomials is associative. {{qed}} Category:Polynomial Theory hx02dcqwubyh5v5javvpj1i7mldfrmq
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity with multiplicative identity $1_R$ and additive identity $0_R$. Let $\left\{{X_j: j \in J}\right\}$ be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$  be an arbitrary polynomial in the indeterminates $\left\{{X_j: j \in J}\right\}$ over $R$. Let: :$\displaystyle N = 1_R \mathbf X^0 = \sum_{k \mathop \in Z} b_k \mathbf X^k$  where $b_k = 0_R$ if $k \ne 0$ and $b_0 = 1_R$. Then: {{begin-eqn}} {{eqn|l = f \circ N      |r = \sum_{k \mathop \in Z} \left({\sum_{p \mathop + q \mathop = k} a_p b_q}\right) \mathbf X^k      |c = applying the definition of polynomial multiplication }} {{eqn|r = \sum_{k \mathop \in Z} \left({\sum_{p \mathop + 0 \mathop = k} a_p 1_R}\right) \mathbf X^k      |c = by the definition of $b_k$ }}  {{eqn|r = \sum_{k \mathop \in Z} a_p \mathbf X^k      |c = because $1_R$ is a multiplicative identity }}  {{eqn|r = f      |c =  }}  {{end-eqn}} Therefore, $f \circ N = f$ for all polynomials $f$. Therefore, $N$ is a multiplicative identity for the set of polynomials. {{qed}} Category:Polynomial Theory 74q1ml9so06jni8k2c3xa3zhy3b7g4k
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\set {X_j: j \in J}$ be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ :$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$. Then: {{begin-eqn}} {{eqn | l = f \circ g       | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k       | c = {{Defof|Polynomial Multiplication}} }} {{eqn | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_q b_p} \mathbf X^k       | c = Addition of multiindices is commutative, and $p$ and $q$ are dummy variables }}  {{eqn | r = g \circ f       | c = {{Defof|Polynomial Multiplication}} }}  {{end-eqn}} Therefore, $f \circ g = g \circ f$ for all polynomials $f$ and $g$. Therefore, polynomial multiplication is commutative. {{qed}} Category:Polynomial Theory 19n11usqlh1gwbqe8bbn35tlbqa9asg
Let $\struct {R, +, \circ}$ be a commutative ring with unity. Let $\set {X_j: j \in J}$ be a set of indeterminates. Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$. Let :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$  :$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$  :$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$  be  arbitrary polynomials in the indeterminates $\set{X_j: j \in J}$ over $R$. By Multiplication of Polynomials is Commutative, it is sufficient to prove that $\circ$ is left distributive over addition only. Then {{begin-eqn}} {{eqn | l = f \circ \paren {g + h}       | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} a_p \paren {b_q + c_q} \mathbf X^k       | c = by the definitions of polynomial multiplication and addition }} {{eqn | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} \paren {a_p b_q + a_p c_q} \mathbf X^k       | c = by the properties of finite sums }}  {{eqn | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} a_p b_q \mathbf  X^k + \sum_{k \mathop \in Z} \sum_{p + q = k} a_p c_q \mathbf X^k       | c = by the properties of finite sums }}  {{eqn | r = f \circ g + f \circ h       | c = by the definitions of polynomial multiplication and addition }}  {{end-eqn}} Therefore, $f \circ \paren {g + h} = f \circ g + f \circ h$ for all polynomials $f, g, h$. Therefore, polynomial multiplication is distributive over addition. {{qed}} Category:Polynomial Theory Category:Distributive Operations tqn5xtor14xks6rbwjlg4k61mt8a8uh
Let $\struct {T, \oplus, \otimes}$ be an algebraic structure in which $\otimes$ distributes over $\oplus$. Let $f, g, h \in T^S$.  Then: {{begin-eqn}} {{eqn | l = \map {\paren {f \otimes \paren {g \oplus h} } } x       | r = \map f x \otimes \paren {\map g x \oplus \map h x}       | c = {{Defof|Pointwise Operation}} }} {{eqn | r = \paren {\map f x \otimes \map g x} \oplus \paren {\map f x \otimes \map h x}       | c = because $\otimes$ distributes over $\oplus$ in $T$ }} {{eqn | r = \map {\paren {\paren {f \otimes g} \oplus \paren {f \otimes h} } } x       | c = {{Defof|Pointwise Operation}} }} {{end-eqn}} This shows left distributivity. The proof for right distributivity is identical. {{qed}} Category:Pointwise Operations Category:Distributive Operations c2unl9bb8a9qwficzc2xs3madjanz72
Suppose that $f$ converges absolutely at $\sigma_0 + i t_0$. If $\sigma \ge \sigma_0$, then: {{begin-eqn}} {{eqn | l = \size {\frac {a_n} {n^s} }       | r = \frac {\size {a_n} } {n^\sigma} }} {{eqn | o = \le       | r = \frac {\size {a_n} } {n^{\sigma_0} } }} {{eqn | r = \size {\frac {a_n} { n^{s_0} } } }} {{end-eqn}} Therefore absolute convergence of $\map f {s_0}$ directly implies absolute convergence of $\map f s$ for all $s = \sigma + i t$ with $\sigma > \sigma_0$. {{qed}}
Let $S$ be the set of all complex numbers $s$ such that $f \left({s}\right)$ converges absolutely. By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $f \left({s_0}\right)$ converges absolutely, so $S$ is not empty. Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Absolute Convergence Lemma that $f \left({s}\right)$ converges absolutely for all $s \in \C$, a contradiction of our assumptions. Therefore the infimum: :$\sigma_a = \inf \left\{{\sigma: s = \sigma + i t \in S}\right\} \in \R$ is well defined. Now if $s = \sigma + it$ with $\sigma > \sigma_a$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $f \left({s'}\right)$ is absolutely convergent. Then it follows from Dirichlet Series Absolute Convergence Lemma that $f \left({s}\right)$ is absolutely convergent. If $s = \sigma + it$ with $\sigma < \sigma_a$, and $f \left({s}\right)$ is absolutely convergent then $s$ contradicts the definition of $\sigma_a$. Therefore, $\sigma_a$ has the claimed properties. {{qed}}
We begin with a lemma. === Lemma === {{:Dirichlet Series Convergence Lemma/Lemma}} === Proof of Theorem === Suppose that $f \left({s}\right)$ converges at $s_0 = \sigma_0 + it_0$. Then by Convergent Sequence in Metric Space is Bounded, $\left\vert \displaystyle \sum_{k\mathop=1}^n \frac {a_k} {k^{s_0}} \right\vert$ is bounded. Thus the results of the lemma hold. Choose any $s = \sigma + it$ with $\sigma > \sigma_0$. By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$, there exists an $N>0$ such that for all $m>n>N$ :$\displaystyle \left\vert \sum_{k \mathop = 1}^m \frac {a_k} {k^s} - \sum_{k \mathop = 1}^n \frac {a_k} {k^s} \right\vert = \left\vert \sum_{k \mathop = n+1}^m \frac {a_k} {k^s} \right\vert < \epsilon$ The lemma shows that for a given $s$ there exists a constant $C$ independent of $N$ such that: :$\displaystyle \left\vert{\sum_{k \mathop = n+1}^n a_n n^{-s} }\right\vert \le C \left({n+1}\right)^{\sigma_0 - \sigma}$ Since $\sigma_0 - \sigma <0$, the right hand side tends to zero as $n \to \infty$. Thus we may choose $N$ large enough so that for $n>N$ :$\displaystyle \left({n+1}\right)^{\sigma_0 - \sigma} < \dfrac \epsilon C$ Which gives us $\displaystyle \left\vert{\sum_{k \mathop = n+1}^n a_n n^{-s} }\right\vert < \epsilon$ {{qed}}
If there does not exist an $s_0$ such that $\map f {s_0}$ converges, then $\sigma_0 = \infty$ and the theorem is vacuously true If there does exist such an $s_0$, let $\sigma_0$ be the infimum of the real part of all such $s_0$, where $\sigma_0= -\infty$ if the set is not bounded from below. *It is clear that $\map f s$ diverges if $\sigma < \sigma_0$ by construction of $\sigma_0$ as the infimum of all convergent $s$ *As $\sigma_0$ is the infimum of the the real part of the set all s such that $\map f s$ converges, for any $\epsilon > 0$, there exists an $s$ such that $\map f s$ converges and $\sigma_0 \le \sigma < \sigma_0+ \epsilon$ as otherwise $\sigma_0 + \epsilon$ would be the infimum of the set.  :Therefore, for any $s$ such that $\sigma > \sigma_0$, we can pick $\epsilon = \sigma - \sigma_0$ and find a $q$ such that $\map f q$ converges and $\map \Re q < \sigma$. :By Dirichlet Series Convergence Lemma, We therefore have that $\map f s$ converges. {{qed}} Category:General Dirichlet Series 1gzj420mgixxuwfb1mzgc4zjamfldt3
It is trivial that $\sigma_a \ge \sigma_c$. Suppose $s_0 = \sigma_0 + i t_0 \in \C$ such that $\map f {s_0}$ converges. It is sufficient to show that $\map f s$ converges absolutely for all $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$. Pick an upper bound $M$ for the real numbers $\size {a_n n^{-s_0} }$. Then for $s = \sigma + i t$ with $\sigma > \sigma_0 + 1$: :$\size {\dfrac {a_n} {n^s} } = \size {\dfrac {a_n} {n^{s_0} n^{s - s_0} } } = \dfrac M {n^{\sigma - \sigma_0} }$ Then by the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n n^{-s} }$ converges. {{qed}}