hcju/proofwikiqa huggingface dataset:
=== Construction of $G$ === Let $U$ be the class of all sets. {{explain|Perhaps Universal Class can be used here}} Define the mapping $F: \N \to U$ recursively: :$\map F 0 = S$ :$\map F {n + 1} = \bigcup \map F n$ Applying the axiom of union inductively, $\map F n$ is a set for each $n \in \N$. Let $\displaystyle G = \bigcup_{i \mathop = 0}^\infty \map F i$. By the axiom of unions, $G$ is a set. === Transitivity === It is to be proved that $G$ is transitive. That is: :$a \in b, b \in G \implies a \in G$ Let $a \in b$ and $b \in G$. By the definition of $G$, there exists $n \in \N$, $b \in \map F n$. By the definition of $F$: :$\map F {n + 1} = \bigcup \map F n$ Then by the definition of union: :$a \in \map F {n + 1}$ Thus by the definition of $G$: :$a \in G$ {{qed|lemma}} === Minimality === It is to be proved that if $Q$ is a transitive set and $S \subseteq Q$ then $G \subseteq Q$. {{proof wanted|Only an outline so far}} Let $Q$ be transitive and $S \subseteq Q$. Define $F$ as above. Prove by induction that $\map F n \subseteq Q$ for each $n$. {{qed}} p0rijby2lh6rzihk2xa1yutgmiriaxe
{{NotZFC}} Suppose for the sake of contradiction that for each ordinal $i$ the set $G\setminus V_i$ is non-empty.  Let $i$ be any ordinal. Then by the axiom of foundation: $\exists x: x \in G\setminus V_i \text{ and } x \cap \left({G \setminus V_i}\right) = \varnothing$ Since $G$ is transitive, $x \subseteq G$. Since $x \subseteq G$ and $x \cap \left({G \setminus V_i}\right) = \varnothing$: :$x \subseteq V_i$ By the definition of $V$, $x \in V_{i+1}$, so: :$x \notin G \setminus V_{i+1}$ Thus $G \setminus V_{i+1} \subsetneqq G \setminus V_i$ for each ordinal $i$. Let $H: \operatorname{On} \to P \left({G}\right)$ be defined by: :$H \left({i}\right) = G \setminus V_i$ for each ordinal $i$ Since the class of ordinals is well-ordered, applying Strictly Increasing Mapping on Well-Ordered Class proves that for any ordinals $p$ and $q$, :$p < q$ implies $H \left({q}\right) \subsetneqq H \left({p}\right)$. Since the class of ordinals is totally ordered, for any two distinct ordinals $i$ and $j$: : $H \left({i}\right) \ne H \left({j}\right)$ so $H$ is injective. But that contradicts the fact that $P \left({G}\right)$ is a set. {{qed}} Category:Von Neumann Hierarchy m6vhk4yxba1sj9q1ottirbma1w99yrw
Follows directly from External Direct Product Closure. {{finish|Fill in the detail.}} {{qed}} Category:External Direct Products 6fqfvdj53acppdgupdqfawyltkjk0md
Let $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$ be abelian groups. Let $\displaystyle \struct {G, \circ} = \prod_{k \mathop = 1}^n G_k$ be the external direct product of $\struct {G_1, \circ_1}, \struct {G_2, \circ_2}, \ldots, \struct {G_n, \circ_n}$. From External Direct Product of Groups is Group: Finite Product it follows that $\struct {G, \circ}$ is a group. By definition, each of $\circ_1, \circ_2, \ldots, \circ_n$ are commutative operations. From External Direct Product Commutativity: General Result it follows that $\circ$ is commutative. Hence, by definition, $\struct {G, \circ}$ is an abelian group. {{qed}} Category:Abelian Groups Category:Group Direct Products rhjtkdwnymakyv8zp1ejrymhofvjc00
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\Phi_n: \left({\mathcal S_n, \circledcirc_n}\right) \to \left({\mathcal T_n, \circledast_n}\right)$ is a homomorphism. === Basis for the Induction === $P \left({1}\right)$ is true, as this just says: :$\Phi_1 = \phi_1: \left({S_1, \circ_1}\right) \to \left({T_1, \ast_1}\right)$ is a homomorphism which holds by hypothesis. $P \left({2}\right)$ is the case: :$\Phi_2 = \phi_1 \times \phi_2: \left({S_1 \times S_2, \circledcirc_2}\right) \to \left({T_1 \times T_2, \circledast_2}\right)$ is a homomorphism which has been proved in Homomorphism of External Direct Products. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m+1}\right)$ is true. So this is our induction hypothesis: :$\Phi_m: \left({\mathcal S_m, \circledcirc_m}\right) \to \left({\mathcal T_m, \circledast_m}\right)$ is a homomorphism. Then we need to show: :$\Phi_{m+1}: \left({\mathcal S_{m+1}, \circledcirc_{m+1}}\right) \to \left({\mathcal T_{m+1}, \circledast_{m+1}}\right)$ is a homomorphism. === Induction Step === This is our induction step: Let $\left({s_1, \ldots, s_{m+1}}\right)$ and $\left({t_1, \ldots, t_{m+1}}\right)$ be arbitrary elements of $\mathcal S_{m+1}$. {{begin-eqn}} {{eqn | o =        | r = \Phi_{m+1} \left({\left({s_1, \ldots, s_{m+1} }\right) \circledcirc_{m+1} \left({t_1, \ldots, t_{m+1} }\right)}\right) }} {{eqn | r = \Phi_{m+1} \left({\left({\left({s_1, \ldots, s_m}\right), s_{m+1} }\right) \circledcirc_{m+1} \left({\left({t_1, \ldots, t_m}\right), t_{m+1} }\right)}\right)       | c = Definition of Cartesian Product }} {{eqn | r = \left({\Phi_m \left({\left({s_1, \ldots, s_m}\right) \circledcirc_m \left({t_1, \ldots, t_m}\right)}\right), s_{m+1} \circ_{m+1} t_{m+1} }\right)       | c = Definition of External Direct Product }} {{eqn | r = \left({\Phi_m \left({\left({s_1, \ldots, s_m}\right) \circledcirc_m \left({t_1, \ldots, t_m}\right)}\right), \phi_{m+1} \left({s_{m+1} }\right) \ast_{m+1} \phi_{m+1} \left({t_{m+1} }\right) }\right)       | c = $\phi_{m+1}$ is a Homomorphism }} {{eqn | r = \left({\Phi_m \left({s_1, \ldots, s_m}\right) \circledast_m \Phi_m \left({t_1, \ldots, t_m}\right), \phi_{m+1} \left({s_{m+1} }\right) \ast_{m+1} \phi_{m+1} \left({t_{m+1} }\right) }\right)       | c = Induction Hypothesis }} {{eqn | r = \Phi_{m+1} \left({s_1, \ldots, s_m, s_{m+1} }\right) \circledast_{m+1} \Phi_{m+1} \left({t_1, \ldots, t_m, t_{m+1} }\right)       | c = Definition of External Direct Product }} {{end-eqn}} So $P \left({m}\right) \implies P \left({m+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: \Phi_n: \left({\mathcal S_n, \circledcirc_n}\right) \to \left({\mathcal T_n, \circledast_n}\right)$ is a homomorphism. {{qed}} Category:Homomorphisms Category:Cartesian Product 9nsxqrm8m7xr1hby13g8lfa86vf1c0r
{{NotZFC}} By Set has Rank, each element of $C$ has a rank. Let $R$ be the class of ranks of elements of $C$. $R$ is non-empty because $C$ is non-empty. {{explain|This should follow from arbitrary intersections of ordinals being ordinals.}} Since any non-empty class of ordinals has a least element, $R$ has a least element, $q$. By the definition of $R$: :$\exists x \in C: \operatorname{rank}\left({x}\right) = q$ Then $x$ is an element of $C$ of least rank. {{qed}} Category:Von Neumann Hierarchy 9zkt1txf45kfd89za25q4fr3u487f2w
First, we prove that $\rho$ is a smooth path: {{begin-eqn}} {{eqn | l = \rho' \left({t}\right)       | r = \gamma' \left({\psi\left({t}\right) }\right) \psi' \left({t}\right)       | c = Derivative of Complex Composite Function }} {{eqn | r = -\gamma' \left({\psi\left({t}\right) }\right)       | c = Derivatives of Function of $a x + b$: $\psi' \left({t}\right) = -1$ }} {{eqn | o = \ne       | r = 0       | c = {{Defof|Smooth Path (Complex Analysis)|Smooth Path}}: $\gamma' \left({\psi\left({t}\right)}\right) \ne 0$ }} {{end-eqn}} As $\gamma'$ is continuously differentiable, and $\psi$ is continuous, it follows that $\rho'$ is continuous. Then $\rho$ is a parameterization of a directed smooth curve $-C$.  {{qed|lemma}} Next, we prove that the definition of $-C$ is independent of the parameterization $\gamma$. Suppose $\sigma$ is another parameterization of $C$: : $\sigma = \gamma \circ \phi$ Here $\phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ is a bijective differentiable strictly increasing function. Define $\tilde \psi: \left[{c \,.\,.\, d}\right] \to \left[{c \,.\,.\, d}\right]$ by $\tilde \psi \left({t}\right) = c + d - t$. Define $\tilde \rho: \left[{c \,.\,.\, d}\right] \to \C$ by $\tilde \rho = \sigma \circ \tilde{\psi}$. We now prove that $\rho$ and $\tilde \rho$ both are parameterizations of the same directed smooth curve $-C$. Both $\psi$ and $\tilde \psi$ are bijective with $\psi^{-1} = \psi$ and $\tilde \psi^{-1} = \tilde \psi$. Define $\tilde \phi: \left[{c \,.\,.\, d}\right] \to \left[{a \,.\,.\, b}\right]$ by $\tilde \phi = \psi^{-1} \circ \phi \circ \tilde \psi$. From Composite of Bijections is Bijection, it follows that $\tilde \phi$ is bijective. From Derivative of Composite Function, it follows that $\tilde \phi$ is differentiable with {{begin-eqn}} {{eqn | l = \tilde \phi' \left({t}\right)       | r = \dfrac 1 {\psi' \left({\psi^{-1} \left({\phi \left({\tilde \psi \left({t}\right) }\right) }\right) }\right) } \phi' \left({\tilde \psi \left({t}\right) }\right) \tilde \psi' \left({t}\right)       | c = Derivative of Inverse Function }} {{eqn | r = \phi' \left({\tilde \psi \left({t}\right)}\right)       | c = Derivatives of Function of $a x + b$ }} {{eqn | o = >       | r = 0       | c = Derivative of Monotone Function }} {{end-eqn}} From Derivative of Monotone Function, it follows that $\tilde \phi$ is strictly increasing. As: :$\rho \circ \tilde \phi = \gamma \circ \psi \circ \psi^{-1} \circ \phi \circ \tilde \psi = \gamma \circ \phi \circ \tilde \psi = \sigma \circ \tilde \phi = \tilde \rho$ it follows that $\rho$ and $\tilde{\rho}$ are parameterizations of $-C$. {{qed}}
Let $C_i$ be parameterized by the smooth path $\gamma_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$ for all $i \in \left\{ {1, \ldots, n}\right\}$. From Reversed Directed Smooth Curve is Directed Smooth Curve, it follows that $-C_i$ is independent of the paraterization $\gamma_i$ of $C_i$. We now prove that the end point of $-C_i$ is equal to the start point of $-C_{i-1}$ for all $i \in \left\{ {2, \ldots, n}\right\}$. By definition of reversed directed smooth curve, $-C_i$ is parameterized by $\rho_i: \left[{a_i \,.\,.\, b_i}\right] \to \C$. Here, $\rho_i \left({t}\right) = \gamma_i \left({a_i + b_i - t}\right)$. From Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints, it follows that the endpoints $\rho_i \left({a_i}\right)$ and $\rho_i \left({b_i}\right)$ are independent of the parameterization $\rho_i$. Then: {{begin-eqn}} {{eqn | l = \rho_i \left({b_i}\right)       | r = \phi_i \left({a_i + b_i - b_i}\right) }} {{eqn | r = \phi_i \left({a_i}\right) }} {{eqn | r = \phi_{i - 1} \left({b_{i - 1} }\right)       | c = {{Defof|Contour (Complex Plane)}} }} {{eqn | r = \rho_{i - 1} \left({a_{i - 1} }\right) }} {{end-eqn}} By definition, it follows that $-C_n, \ldots, -C_1$ can be concatenated to form a contour. {{qed}}
First, suppose that $C$ is a directed smooth curve in $\C$. Let $C$ be parameterized by the smooth path $\gamma: \closedint a b \to \C$. By definition of reversed directed smooth curve, $-C$ is parameterized by a smooth path $\rho: \closedint a b \to \C$ with $\rho = \gamma \circ \psi$. Here, $\psi: \closedint a b \to \closedint a b$ is defined by $\map \psi t = a + b - t$. From Derivatives of Function of $a x + b$: :$\map {\psi'} t = -1$ for all $t \in \closedint a b$ Then: {{begin-eqn}} {{eqn | l = \displaystyle \int_C \map f z \rd z        | r = \int_a^b \map f {\map \rho t} \map {\rho'} t \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = \int_a^b \map f {\map \gamma {\map \psi t} } \map {\gamma'} {\map \psi t} \map {\psi'} t \rd t       | c = Derivative of Complex Composite Function }} {{eqn | r = -\int_a^b \map f {\map \gamma u} \map {\gamma'} u \rd u       | c = substitution with $u = \map \psi t$, where $\map \psi a > \map \psi b$ }} {{eqn | r = -\int_C \map f z \rd z }} {{end-eqn}} {{qed|lemma}} Next, suppose that $C$ is a contour. Then $C$ is a concatenation of a finite sequence $C_1, \ldots, C_n$ of directed smooth curves. By definition of reversed contour, $-C$ is a concatenation of the finite sequence $-C_n, \ldots, -C_1$ of directed smooth curves. Then: {{begin-eqn}} {{eqn | l = \int_C \map f z \rd z        | r = \sum_{i \mathop = 1}^n \int_{-C_i} \map f z \rd z        | c = Contour Integral of Concatenation of Contours }} {{eqn | r = -\sum_{i \mathop = 1}^n \int_{C_i} \map f z \rd z        | c = from the result above, as each $C_i$ is a directed smooth curve }} {{eqn | r = -\int_C \map f z \rd z }} {{end-eqn}} {{qed}}
{{NotZFC}} The general strategy of the proof is as follows: We will recursively define a certain subset, $a$, of $A$. We will use the fact that $\prec$ is a foundational relation to choose a minimal element, $m$, of $a$. Then we will prove that $m$ is in fact a minimal element of $A$. For each $x \in A$, let $\map {\prec^{-1} } x$ denote the preimage under $\prec$ of $x$ in $A$. For each class $C$, let $\map R C$ denote the set of elements of $C$ of minimal rank, and let $\map R \O = \O$. That is: For a given class $C$, let $\alpha_C$ be the smallest ordinal such that: :$C \cap \map V {\alpha_C} \ne \O$ where $V$ is the von Neumann hierarchy. Then let $\map R C$ denote the set $C \cap \map V {\alpha_C}$. Let $F$ be a function defined recursively: :$\map F 0 = \map R A$ :$\displaystyle \map F {n + 1} = \bigcup_{y \mathop \in \map F n} \map R {\map {\prec^{-1} } y}$ {{refactor|Extract this lemma into its own page.}} === Lemma === $\map F n$ is a set for each $n \in \omega$. === Proof === Proceed by induction: $\map R A$ is a set, so $\map F 0$ is a set. Suppose that $\map R {\map F n}$ is a set. We know that for each $y \in \map F n$, $\map R {\map {\prec^{-1} } y}$ is a set, so by the Axiom of Unions, $\map F {n + 1}$ is a set. {{qed|lemma}} Let $\displaystyle a = \bigcup_{n \mathop \in \omega} \map F n$. By the Axiom of Unions, $a$ is a set. Since $\map F n \subseteq A$ for each $n \in \omega$, $a \subseteq A$. By Non-Empty Class has Element of Least Rank, $\map F 0 \ne \O$, so $a \ne \O$. Since $\prec$ is foundational, $a$ has a $\prec$-minimal element $m$. {{AimForCont}} that $m$ is not a $\prec$-minimal element of $A$. Then, by Characterization of Minimal Element,  :$\map {\prec^{-1} } m \ne \O$ By Non-Empty Class has Element of Least Rank, $\map {\prec^{-1} } m$ has an element $w$ of least rank. {{begin-eqn}} {{eqn | ll= \exists n \in \N:       | l = m       | o = \in       | r = \map F n       | c = Definition of $a$ }} {{eqn | l = w       | o = \in       | r = \map F {n + 1}       | c = Definition of $F$ }} {{eqn | l = w       | o = \in       | r = a       | c = Definition of $a$ }} {{end-eqn}} Therefore, $a \cap \map {\prec^{-1} } m \ne \O$, contradicting the fact that $m$ is $\prec$-minimal in $a$. Thus we conclude that $m$ is a $\prec$-minimal element of $A$. {{qed}}
{{explain|generally, why the following purports to be a proof. Suggest the specific basis axioms are invoked.}} $U$ is non-empty by the Ultrafilter Lemma. The topology of the Stone space is defined as the topology generated by the basis $Q$ consisting of all sets of the form :$\set {x \in U: b \in x}$ for some $b \in B$. We must show that $Q$ is in fact a basis. $Q$ is trivially a subset of the power set of $U$. Suppose that $x \in U$. Then by the definition of ultrafilter, $x \ne \O$. Thus $x$ has some element $b$. Then $x \in \set {x \in U: b \in x} \in Q$. So $Q$ covers $U$. Let $P, Q \in Q$. Then for some $b, c \in B$, :$P = \set {x \in U: b \in x}$ and :$Q = \set {y \in U: c \in y}$. Let $z \in P \cap Q$. Then by the definitions of $P$ and $Q$, :$b \in z$ and $c \in z$. So $z$ is an ultrafilter on $B$ containing $b$ and $c$. Since $z$ is a filter containing $b$ and $c$, it also contains $b \wedge c$. Let $R = \set {w \in U: b \wedge c \in w}$. $R$ is clearly an element of $Q$ containing $z$. Suppose that $v \in R$. Then $b \wedge c \in v$. Since $v$ is a filter, $b \in v$ and $c \in v$. Thus $v \in P \cap Q$. Since this holds for each $v \in R$, $R \subseteq P \cap Q$ {{qed}} Category:Stone Spaces h9iz9f3u9277lm6yi1nmkt1wtydxlaq
Consider the external direct product of rings $\Z \oplus 2 \Z$. From Integer Multiples form Commutative Ring, $2 \Z$ does not admit a unity. By Unity of External Direct Sum of Rings, neither does $\Z \oplus 2 \Z$. Now consider the ideal $\set 0 \times 2 \Z$ of $\Z \oplus 2 \Z$. We have for all $a \in \Z$ and $b, c \in 2 \Z$ that: :$\tuple {0, b} - \tuple {0, c} = \tuple {0, b - c}$ :$\tuple {a, b} \cdot \tuple {0, c} = \tuple {0, b \cdot c}$ so by the Test for Ideal, indeed $\set 0 \times 2 \Z$ is an ideal in $\Z \oplus 2 \Z$. By Quotient Ring of External Direct Sum of Rings, we have: :$\paren {\Z \oplus 2 \Z} / \paren {\set 0 \times 2 \Z} \cong \paren {Z / \set 0} \oplus \paren {2 \Z / 2 \Z}$ By Quotient Ring by Null Ideal and Quotient Ring Defined by Ring Itself is Null Ring, this last ring is isomorphic to $\Z \times \set 0 \cong \Z$. Since $\Z$ has a unity, this construction provides an example of the required kind. {{qed}}
This proof assumes that $\mathbb K = \R$. === Necessary Condition === Assume that $f$ is differentiable in $z$. By definition of open set, there exists $r \in \R_{>0}$ such that  $\left({z - r \,.\,.\, z + r}\right) \subseteq D$. We have the open ball $B_r \left({0}\right) = \left({-r \,.\,.\, r}\right)$. Define $\epsilon: \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\} \to \R$ by: :$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$ If $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$, then $z+h \in \left({z - r \,.\,.\, z + r}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined. From Continuity of Composite Mapping/Corollary, it follows that $\epsilon$ is continuous in $\left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$. As $f$ is differentiable at $z$, it follows that: :$\displaystyle \lim_{h \to 0} \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac {f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$ If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in \left({-r \,.\,.\, r}\right) \setminus \left\{ {0}\right\}$: :$f \left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$ {{qed|lemma}} === Sufficient condition === Rewrite the equation of the assumption to get: :$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$ From Sum Rule for Limits of Functions, it follows that: :$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({\alpha + \epsilon \left({h}\right)}\right) = \alpha$ By definition of differentiability, $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$. {{qed}}
This proof assumes that $\mathbb K = \C$. === Necessary Condition === Assume that $f$ is differentiable in $z$. By definition of open set, there exists $r \in \R_{>0}$ such that the open ball $B_r \left({z}\right) \subseteq D$. Define $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ by: :$\epsilon \left({h}\right) = \dfrac{f \left({z + h}\right) - f \left({z}\right)} h - f' \left({z}\right)$ If $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$, then $z+h \in B_r \left({z}\right) \setminus \left\{ {z}\right\} \subseteq D$, so $\epsilon$ is well-defined. As $f$ is differentiable in $z$, it follows that: :$\displaystyle \lim_{h \to 0}  \ \epsilon \left({h}\right) = \lim_{h \to 0} \dfrac{ f \left({z + h}\right) - f \left({z}\right) } h - f' \left({z}\right) = f' \left({z}\right) - f' \left({z}\right) = 0$ If we put $\alpha = f' \left({z}\right)$, it follows that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$: :$f\left({z + h}\right) = f \left({z}\right) + h \left({\alpha + \epsilon \left({h}\right) }\right)$ {{qed|lemma}} === Sufficient condition === Rewrite the equation of the assumption as: :$\dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \alpha + \epsilon \left({h}\right)$ From Sum Rule for Limits of Functions: :$\displaystyle \lim_{h \to 0} \dfrac{f \left({z + h}\right) - f \left({z}\right)} h = \lim_{h \to 0} \left({ \alpha + \epsilon \left({h}\right) }\right) = \alpha$ By definition of differentiability, $f$ is differentiable at $z$ with $f' \left({z}\right) = \alpha$. {{qed}}
By Von Neumann Hierarchy Comparison, $V(x) \in V(y)$. (1) By (1) and the Axiom:Axiom of Foundation, $V(x) \ne V(y)$. Furthermore, by (1) and  Von Neumann Hierarchy is Supertransitive, {{ProofWanted}} aqsb7mcpwjsa3idzrn0l74m6ckk2kad
By definition of continuity: :$\forall \epsilon > 0: \exists \delta > 0: \left\vert{z - z_0}\right\vert < \delta \implies \left\vert{f \left({z}\right) - f \left({z_0}\right)}\right\vert < \epsilon$ Given $\epsilon > 0$, we find $\delta > 0$ so for all $z \in \C$ with $\left\vert{z - z_0}\right\vert < \delta$: {{begin-eqn}} {{eqn |l= \epsilon       |o= >       |r= \left\vert{f \left({z}\right) - f \left({z_0}\right) }\right\vert }} {{eqn |o= \ge       |r= \left\vert{ \operatorname{Re} \left({f \left({z}\right) - f \left({z_0}\right) }\right) }\right\vert       |c= by Modulus Larger than Real Part }} {{eqn |r= \left\vert{ \operatorname{Re} \left({f \left({z}\right) }\right) - \operatorname{Re} \left({f \left({z_0}\right) }\right) }\right\vert       |c= by Addition of Real and Imaginary Parts }} {{end-eqn}} It follows that: :$\forall \epsilon > 0: \exists \delta > 0: \left\vert{z - z_0}\right\vert < \delta \implies \left\vert{ \operatorname{Re} \left({ f \left({z}\right) }\right) - \operatorname{Re} \left({f \left({z_0}\right)} \right)} \right\vert < \epsilon$ Then equation $(1)$ is proven by: {{begin-eqn}} {{eqn |l= \lim_{z \to z_o} \operatorname{Re} \left({ f \left({z}\right) }\right)       |r= \operatorname{Re} \left({ f \left({z_0}\right) }\right)       |c= by definition of limit }} {{eqn |r= \operatorname{Re} \left({ \lim_{z \to z_o} f \left({z}\right) }\right)       |c= by definition of continuity }} {{end-eqn}} The proof for equation $(2)$ with imaginary parts follows when $\operatorname{Re}$ is replaced by $\operatorname{Im}$ in the equations above. {{qed}} Category:Complex Analysis pp0l9bv7yjwersw3h80slmzhqm50w6x
{{begin-eqn}} {{eqn | l = S_n       | r = \sum_{j \mathop = 1}^n \paren {2 j - 1}       | c = Odd Number Theorem }} {{eqn | r = \sum_{j \mathop = 1}^{n - 1} \paren {2 j - 1} + \paren {2 n - 1}       | c = {{Defof|Summation}} }} {{eqn | r = S_{n - 1} + \paren {2 n - 1}       | c = Odd Number Theorem }} {{end-eqn}} {{qed}} Category:Odd Number Theorem e5imnpnznujgeroh2tc2wl2ur5udrfg
Let $T = \left({S, \tau}\right)$ be such that each filter on $S$ has a limit point in $S$. Let $\mathcal F$ be an ultrafilter on $S$. By hypothesis, $\mathcal F$ has a limit point $x \in S$. By Limit Point iff Superfilter Converges, there exists a filter $\mathcal F'$ on $S$ which converges to $x$ satisfying $\mathcal F \subseteq \mathcal F'$. Because $\mathcal F$ is an ultrafilter, $\mathcal F = \mathcal F'$. {{explain}} Thus $\mathcal F$ converges to $x$.
Let $\mathcal F$ be a  filter on $S$. By the Ultrafilter Lemma, there exists an ultrafilter $\mathcal F'$ such that $\mathcal F \subseteq \mathcal F'$. By hypothesis, $\mathcal F'$ converges to some $x \in S$. This, by Limit Point iff Superfilter Converges, implies that $x$ is a limit point of $\mathcal F$. {{qed}}
Let $a, b, c \in S$. Then {{begin-eqn}} {{eqn | l = \paren {a * b} \circ c       | r = c \circ \paren {a * b}       | c = $\circ$ is commutative }} {{eqn | r = \paren {c \circ a} * \paren {c \circ b}       | c = $\circ$ is left distributive over $*$ }} {{eqn | r = \paren {a \circ c} * \paren {b \circ c}       | c = $\circ$ is commutative }} {{end-eqn}} So $\circ$ is right distributive over $*$. Since $\circ$ is both left distributive and right distributive over $*$, it is distributive over $*$. {{qed}} Category:Commutativity Category:Distributive Operations fjs1jy5xpe42hhx0jeubt3pigqlnxfq