hcju/proofwikiqa huggingface dataset:
Define $g_i: \left[{a_i\,.\,.\,b_i}\right] \to \C$ by $g_i \left({t}\right) = f \left({\gamma_i \left({t}\right) }\right) \gamma_i' \left({t}\right)$ for all $i \in \left\{ {1, \ldots, n}\right\}$. By definition of smooth path, it follows that $\gamma_i$ and $\gamma_i'$ are continuous for all $i \in \left\{ {1, \ldots, n}\right\}$. From Continuity of Composite Mapping/Corollary and Sum Rule for Continuous Functions, it follows that $g_i$ is continuous. From Continuous Complex Function is Complex Riemann Integrable, we find that $\displaystyle \int_{a_i}^{b_i} g_i \left({t}\right) \rd t$ is defined. Similarly, it can be shown that $\displaystyle \int_{c_i}^{d_i} f \left({\sigma_i \left({t}\right) }\right) \sigma_i' \left({t}\right) \rd t$ is defined for all $i \in \left\{ {1, \ldots, n}\right\}$. Hence, all complex Riemann integrals in the theorem are defined. The equality now follows from Contour Integral is Independent of Parameterization. {{qed}}
Let $Y_i = \dfrac {X_i - \mu} {\sigma}$. We have that: : $E \left[{Y_i}\right] = 0$ and: : $E \left[{Y_i^2}\right] = 1$ Then by Taylor's Theorem the characteristic function can be written: :$\phi_{Y_i} = 1 - \dfrac {t^2} 2 + o \left({t^2}\right)$ Now let: {{begin-eqn}} {{eqn | l = U_n        | r = \frac {S_n - n \mu} {\sqrt {n \sigma^2} } }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {X_i - \mu} {\sqrt {n \sigma^2} }  }} {{eqn | r = \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n \left({\frac {X_i - \mu} {\sigma} }\right)  }} {{eqn | r = \frac 1 {\sqrt n} \sum_{i \mathop = 1}^n Y_i  }} {{end-eqn}} Then its characteristic function is given by: {{begin-eqn}} {{eqn | l = \phi_{U_n} \left({t}\right)        | r = E \left[{e^{i t U_n} }\right] }} {{eqn | r = E \left[{\exp \left({\frac {i t} {\sqrt n} \sum_n^{i \mathop = 1} Y_i}\right)}\right]  }} {{eqn | r = \prod_n^{i \mathop = 1} E \left[{\exp \left({\frac{i t} {\sqrt n} Y_i}\right)}\right]       | c = since $Y_i$ are independent identically distributed }} {{eqn | r = \prod_n^{i \mathop = 1} \phi_{Y_i} \left({\frac t {\sqrt n} }\right) }} {{eqn | r = \left({\phi_{Y_i} \left({\frac t {\sqrt n} }\right)}\right)^n       | c = since $Y_i$ are independent identically distributed }} {{eqn | r = \left({1 - \frac {t^2} {2 n} + o \left({t^2}\right)}\right)^n }} {{end-eqn}} {{explain|What is the significance of the limits going from $n$ to $i = 1$ in the summations and products above, or is this a mistake?}} Recall that the characteristic equation of a standard normal is given by: :$e^{-\frac 1 2 t^2}$. Indeed the characteristic equations of the series converges to the standard normal characteristic equation: :$\left({1 - \dfrac {t^2} {2 n} + o \left({t^2}\right)}\right)^n \to e^{-\dfrac 1 2 t^2}$ as $n \to \infty$ Then Lévy's Continuity Theorem applies. In particular, the convergence in distribution of the $U_n$ to some random variable with standard normal distribution is equivalent to continuity of the limiting characteristic equation at $t = 0$. But, $e^{-\frac 1 2 t^2}$ is clearly continuous at $0$. So we have that $\dfrac{S_n - n \mu} {\sqrt{n \sigma^2} }$ converges in distribution to a standard normal random variable. {{qed}} {{Proofread}}
Assume $y \in \R$ where $-\dfrac \pi 2 \le y \le \dfrac \pi 2$. {{begin-eqn}} {{eqn | l = y       | r = \arcsin x }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \sin y }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \frac 1 {2 i} \paren {e^{i y} - e^{-i y} }       | c = Sine Exponential Formulation }} {{eqn | ll= \leadstoandfrom       | l = 2 i x       | r = e^{i y} - e^{-i y}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 2 i x e^{i y}       | r = e^{2 i y} - 1 }} {{eqn | ll= \leadstoandfrom       | l = e^{2 i y} -2 i x e^{i y}       | r = -1 }} {{eqn | ll= \leadstoandfrom       | l = e^{2 i y} - 2 i x e^{i y} - x^2       | r = 1 - x^2 }} {{eqn | ll= \leadstoandfrom       | l = \paren {e^{i y} - i x}^2       | r = 1 - x^2 }} {{eqn | ll= \leadstoandfrom       | l = e^{i y} - i x       | r = \sqrt {1 - x^2} }} {{eqn | ll= \leadstoandfrom       | l = i y       | r = \map \ln {i x + \sqrt {1 - x^2} } }} {{eqn | ll= \leadstoandfrom       | l = y       | r = \dfrac 1 i \map \ln {i x + \sqrt {1 - x^2} } }} {{end-eqn}} {{qed}}
Assume $y \in \R$ such that $0 \le y \le \pi$. {{begin-eqn}} {{eqn | l = y       | r = \arccos x }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \cos y }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \frac 1 2 \paren {e^{-i y} + e^{i y} }       | c = Cosine Exponential Formulation }} {{eqn | ll= \leadstoandfrom       | l = 2 x       | r = e^{-i y} + e^{i y} }} {{eqn | ll= \leadstoandfrom       | l = 2 x e^{i y}       | r = 1 + e^{2 i y} }} {{eqn | ll= \leadstoandfrom       | l = e^{2 i y} - 2 x e^{i y}       | r = -1 }} {{eqn | ll= \leadstoandfrom       | l = e^{2 i y} - 2 x e^{i y} + x^2       | r = x^2 - 1 }} {{eqn | ll= \leadstoandfrom       | l = \paren {e^{i y} - x}^2       | r = x^2 - 1 }} {{eqn | ll= \leadstoandfrom       | l = e^{i y} - x       | r = \sqrt {x^2 - 1} }} {{eqn | ll= \leadstoandfrom       | l = i y       | r = \map \ln {x + \sqrt {x^2 - 1} } }} {{eqn | ll= \leadstoandfrom       | l = y       | r = \dfrac 1 i \map \ln {x + \sqrt {x^2 - 1} } }} {{end-eqn}} {{qed}}
Assume $y \in \R$, $ -\dfrac \pi 2 \le y \le \dfrac \pi 2 $. {{begin-eqn}} {{eqn | l = y       | r = \arctan x }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \tan y }} {{eqn | ll= \leadstoandfrom       | l = x       | r = i \frac {1 - e^{2 i y} } {1 + e^{2 i y} }       | c = Tangent Exponential Formulation }} {{eqn | ll= \leadstoandfrom       | l = i x       | r = \frac {e^{2 i y} - 1} {e^{2 i y} + 1}       | c = $ i^2 = -1 $ }} {{eqn | ll= \leadstoandfrom       | l = i x \paren {e^{2 i y} + 1}       | r = e^{2 i y} - 1 }} {{eqn | ll= \leadstoandfrom       | l = i x e^{2 i y} + i x       | r = e^{2 i y} - 1 }} {{eqn | ll= \leadstoandfrom       | l = e^{2 i y} - i x e^{2 i y}       | r = 1 + i x }} {{eqn | ll= \leadstoandfrom       | l = e^{2 i y}       | r = \frac {1 + i x} {1 - i x} }} {{eqn | ll= \leadstoandfrom       | l = e^{-2 i y}       | r = \frac {1 - i x} {1 + i x} }} {{eqn | ll= \leadstoandfrom       | l = -2 i y       | r = \map \ln {\frac {1 - i x} {1 + i x} } }} {{eqn | ll= \leadstoandfrom       | l = y       | r = \frac 1 2 i \map \ln {\frac {1 - i x} {1 + i x} } }} {{end-eqn}} {{qed}}
From the definition of directed smooth curve, it follows that  $\sigma_i = \gamma_i \circ \phi_i$ for all $i \in \set {1, \ldots, n}$. Here, $\phi_i: \closedint {c_i} {d_i} \to \closedint {a_i} {b_i}$ is a bijective differentiable strictly increasing function.  For all $i \in set {1, \ldots, n}$, $\gamma_i$ and $\sigma_i$ are continuous. From Complex Modulus Function is Continuous and Continuity of Composite Mapping: Corollary, it follows that $\size {\gamma_i'}$ and $\size {\sigma_i'}$ are continuous. From Continuous Real Function is Darboux Integrable, we find that $\displaystyle \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \size {\map {\gamma_i'} t} \rd t$ and $\displaystyle \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \size {\map {\sigma_i'} t} \rd t$ are defined. Hence, all real integrals in the theorem are defined. Then: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \int_{a_i }^{b_i} \size {\map {\gamma_i'} t} \rd t       | r = \sum_{i \mathop = 1}^n \int_{\map {\phi_i^{-1} } {a_i} }^{\map {\phi_i^{-1} } {b_i} } \size {\map {\gamma_i'} {\map \phi u} } \map {\phi'} u \rd u       | c = substitution with $t = \map \phi u$ }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{\map {\phi_i^{-1} } {a_i} }^{\map {\phi_i^{-1} } {b_i} } \size {\map {\gamma_i'} {\map \phi u} \map {\phi'} u}\rd u       | c = as $\map {\phi'} u > 0$ by Derivative of Monotone Function }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{\map {\phi_i^{-1} } {a_i} }^{\map {\phi_i^{-1} } {b_i} } \size {\map {\sigma'} u} \rd u       | c = Derivative of Complex Composite Function }} {{eqn | r = \sum_{i \mathop = 1}^n \int_{c_i}^{d_i} \size {\map {\sigma'} u} \rd u       | c = Reparameterization of Directed Smooth Curve Maps Endpoints To Endpoints }} {{end-eqn}} {{qed}} Category:Contour Integration ijjqjztb5xqpzf8c0vs04d5aqltuys9
From Ring of Integers Modulo m is Ring, $2$ is a ring with unity. Furthermore, the identities: :$0 \cdot 0 = 0$ :$1 \cdot 1 = 1$ show that $2$ is also an idempotent ring. Hence the result, by definition of Boolean ring. {{qed}}
Let $x, y \in R$. Then: {{begin-eqn}} {{eqn | l = x + y       | r = \paren {x + y}^2       | c = {{Defof|Idempotent Ring}} }} {{eqn | r = x^2 + x \circ y + y \circ x + y^2       | c = Binomial Theorem: Ring Theory }} {{eqn | r = x + x \circ y + y \circ x + y       | c = {{Defof|Idempotent Ring}} }} {{end-eqn}} Subtracting $x + y$ from both sides yields: {{begin-eqn}} {{eqn | l = x \circ y + y \circ x       | r = 0_R }} {{eqn | r = y \circ x + y \circ x       | c = Idempotent Ring has Characteristic Two }} {{end-eqn}} Finally, subtracting $y \circ x$ from both sides, we obtain: :$x \circ y = y \circ x$ and conclude $R$ is a commutative ring. {{qed}}
Let $x \in R$. Then: {{begin-eqn}} {{eqn | l = x + x       | r = \paren {x + x}^2       | c = {{Defof|Idempotent Ring}} }} {{eqn | r = \paren {x + x} \paren {x + x} }} {{eqn | r = x \paren {x + x} + x \paren {x + x}       | c = {{Ring-axiom|D}} }} {{eqn | r = x^2 + x^2 + x^2 + x^2       | c = {{Ring-axiom|D}} again }} {{eqn | r = x + x + x + x       | c = {{Defof|Idempotent Ring}} }} {{eqn | ll= \leadsto       | l = 0_R       | r = x + x       | c = $\struct {\R, +}$ is a group, so Cancellation Laws apply }} {{end-eqn}} Hence the result. {{qed}}
By Structure Induced by Ring Operations is Ring, $\struct {2^S, +, \cdot}$ is a ring. By Unity of Induced Structure, $\struct {2^S, +, \cdot}$ also has a unity. By Induced Structure is Idempotent, $\cdot$ is an idempotent operation. Hence $\struct {2^S, +, \cdot}$ is a Boolean ring. {{qed}}
Let $\operatorname{Re}$ and $\operatorname{Im}$ denote real parts and imaginary parts respectively. Let $\phi \left({a}\right) \le \phi \left({b}\right)$. Then: {{begin-eqn}} {{eqn | l = \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \, \mathrm d t       | r = \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t       | c = Definition of Complex Riemann Integral }} {{eqn | r = \int_a^b \operatorname{Re} \left({ f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u + i \int_a^b \operatorname{Im} \left({ f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u       | c = Integration by Substitution }} {{eqn | r = \int_a^b \operatorname{Re} \left({ f \left({ \phi \left({u}\right) }\right) \phi' \left({u}\right) }\right) \, \mathrm d u + i \int_a^b \operatorname{Im} \left({ f \left({ \phi \left({u}\right) }\right) \phi' \left({u}\right) }\right) \, \mathrm d u       | c = Multiplication of Real and Imaginary Parts }} {{eqn | r = \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u }} {{end-eqn}} Let $\phi \left({a}\right) > \phi \left({b}\right)$. Then: {{begin-eqn}} {{eqn | l = \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} f \left({t}\right) \, \mathrm d t       | r = \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} \operatorname{Re} \left({ f \left({t}\right) }\right) \, \mathrm d t + i \int_{\phi \left({b}\right)}^{\phi \left({a}\right)} \operatorname{Im} \left({ f \left({t}\right) }\right) \, \mathrm d t       | c = Definition of Complex Riemann Integral }} {{eqn | r = \int_b^a \operatorname{Re} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u + i \int_b^a \operatorname{Im} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u       | c = Integration by Substitution }} {{eqn | r = -\int_a^b \operatorname{Re} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u - i \int_a^b \operatorname{Im} \left({f \left({ \phi \left({u}\right) }\right) }\right) \phi' \left({u}\right) \, \mathrm d u       | c = Definition of Definite Integral }} {{eqn | r = -\int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \, \mathrm d u }} {{end-eqn}} {{qed}}
Suppose that $f$ is not strictly monotone. That is, there exist $x, y, z \in I$ with $x < y < z$ such that either $f \left({x}\right) \le f \left({y}\right)$ and $f \left({y}\right) \ge f \left({z}\right)$, or $f \left({x}\right) \ge f \left({y}\right)$ and $f \left({y}\right) \le f \left({z}\right)$. Suppose $f \left({x}\right) \le f \left({y}\right)$ and $f \left({y}\right) \ge f \left({z}\right)$. If $f \left({x}\right) = f \left({y}\right)$, or $f \left({y}\right) = f \left({z}\right)$, or $f \left({x}\right) = f \left({z}\right)$, $f$ is not injective, which is a contradiction. Thus, $f \left({x}\right) < f \left({y}\right)$ and $f \left({y}\right) > f \left({z}\right)$. Suppose $f \left({x}\right) < f \left({z}\right)$. That is: : $f \left({x}\right) < f \left({z}\right) < f \left({y}\right)$ As $f$ is continuous on $I$, the Intermediate Value Theorem can be applied. Hence there exists $c \in \left({x \,.\,.\, y}\right)$ such that $f \left({c}\right) = f \left({z}\right)$. As $z \notin \left({x \,.\,.\, y}\right)$, we have $c \ne z$. So $f$ is not injective, which is a contradiction. Suppose instead $f \left({x}\right) > f \left({z}\right)$. That is: : $f \left({z}\right) < f \left({x}\right) < f \left({y}\right)$ Again, as $f$ is continuous on $I$, the Intermediate Value Theorem can be applied. Then, there exists $c \in \left({y \,.\,.\, z}\right)$ such that $f \left({c}\right) = f \left({x}\right)$. So $f$ is not injective, which is a contradiction. If we suppose $f \left({x}\right) \ge f \left({y}\right)$ and $f \left({y}\right) \le f \left({z}\right)$, we reach a similar contradiction. By Proof by Contradiction, $f$ is strictly monotone. {{qed}} Category:Continuous Functions rq33x2iks7zqezdi3sp7eax9o3txvzb
Suppose, for the sake of contradiction, that $M$ is not a prime ideal. Then by Prime Ideal in Lattice, there are elements $a$ and $b$ of $L$ such that :$a \wedge b \in M$ :$a \notin M$ :$b \notin M$ === Lemma === There do not exist $m$ and $n$ in $M$ such that $m \vee a \in F$ and $n \vee b \in F$. === Proof === Suppose such exist. Since: :$m \vee \left({n \vee b}\right) \ge n \vee b$ :$n \vee b \in F$ :$F$ is a filter it follows that: :$m \vee \left({n \vee b}\right) \in F$ Applying associativity yields: :$\left({m \vee n}\right) \vee b \in F$ By the same argument: :$\left({m \vee n}\right) \vee a \in F$ By the definition of a filter: :$\left({\left({m \vee n}\right) \vee b}\right) \wedge \left({\left({n \vee m}\right) \vee a}\right) \in F$ Distributing: :$\left({m \vee n}\right) \wedge \left({b \vee a}\right) \in F$ But by assumption: :$b \vee a \in M$ and by the definition of an ideal: :$m \vee n \in M$ so again by the definition of an ideal: :$\left({m \vee n}\right) \wedge \left({b \vee a}\right) \in M$ contradicting the supposition that $M$ is disjoint from $F$. {{qed|lemma}} {{WLOG}}, we can thus suppose that: :$m \vee a \notin F$ for all $m \in M$ Let $N = \left\{ {x \in L: x \le m \vee a \text{ for some } m \in M}\right\}$. === Lemma === $N$ is an ideal in $L$. === Proof === Let: :$x \in N$ :$y \in L$ :$y \le x$ Then by the definition of $N$ there exists an $m \in M$ such that: :$x \le m \vee a$ Since $y \le x$, it follows that: :$y \le m \vee a$ so $y \in N$. Let: :$x \in N$ :$y \in N$ Then there exist $m_x$ and $m_y$ in $M$ such that: :$x \le m_x \vee a$ :$y \le m_y \vee a$ Then: : $x \vee y \le \left({m_x \vee a}\right) \vee \left({m_y \vee a}\right) = \left({m_x \vee m_y}\right) \vee a$ But $m_x \vee m_y \in M$, so: :$x \vee y \in N$ {{qed|lemma}} === Lemma === $M \subsetneq N$ === Proof === Let $m \in M$. Then: :$m \le \left({m \vee a}\right)$ so $m \in N$. Thus $M \subseteq N$. We have: :$a \le \left({m \vee a}\right)$ so: :$a \in N$ but: :$a \notin M$ Thus $M \subsetneq N$. {{qed|lemma}} === Lemma === $N \cap F = \varnothing$ === Proof === Suppose $x \in N \cap F$. Then: :$x \in N$ so for some $m \in M$: :$x \le m \vee a$ Furthermore, $x \in F$. So by the definition of a filter: :$m \vee a \in F$ But this contradicts our assumption that $m \vee a \notin F$ for all $m \in M$. {{qed|lemma}} By assuming that $M$ is not a prime ideal, we have constructed an ideal $N$ properly containing $M$ that is disjoint from $F$, contradicting the maximality of $M$. Thus $M$ is a prime ideal. {{qed}} Category:Order Theory Category:Lattice Theory ii74u4scsol96px9414dsbisl09sn58
=== Necessary Condition === Let $I$ be a prime ideal. Let $a, b \in L$ such that $a, b \notin I$. Then $a, b \in L \setminus I$. By the definition of prime ideal $L \setminus I$ is a filter. By the definition of a filter: :$\exists c \in L \setminus I: c \le a, c \le b$ By the definition of meet: :$c \le a \wedge b$ Since $c \in L \setminus I$, the definition of a filter implies that : $a \wedge b \in L \setminus I$ {{refactor|The above argument should be extracted into a separate page which may or may not be instantiated as a separate definition of a filter. As it is, we haven't even got the definition of "meet" written yet - although whether there is any point in doing so as $a \wedge b :{{=}} \inf \left\{ {a, b}\right\}$ should be sufficient.}} Thus $a \wedge b \notin I$. By contraposition it follows that $a \wedge b \in I$ implies that $a \in I$ or $b \in I$. {{qed|lemma}} === Sufficient Condition === For all $a$ and $b$ in $L$, let $a \wedge b \in I$ imply that $a \in I$ or $b \in I$. Let $x, y \in L \setminus I$. Then $x, y \notin I$. By supposition, $x \wedge y \notin I$, so: :$x \wedge y \in L \setminus I$ Let $p \in L \setminus I, q \in L$ such that $p \le q$. Let $q \in I$. Then by definition of ideal, $p \in I$. This contradicts $p \in L \setminus I$. Thus $q \notin I$. Thus $q \in L \setminus I$. Therefore, $L \setminus I$ is a filter. {{qed}} Category:Order Theory nbzym84t11zdrzd9hov1qc77usdkscr
Proof by Counterexample: Let $S := \mathcal P \left({\left\{{a, b}\right\}}\right)$. Let $T := \left\{{1, 2, 3, 4}\right\}$. From Subset Relation on Power Set is Partial Ordering, we have that $\left({S, \subseteq}\right)$ is an ordered set. Clearly so is $\left({T, \le}\right)$ (although its ordering is in fact total, it is still technically an ordered set). Let $\phi: S \to T$ be defined as: : $\phi \left({\varnothing}\right) = 1$ : $\phi \left({\left\{{a}\right\}}\right) = 2$ : $\phi \left({\left\{{b}\right\}}\right) = 3$ : $\phi \left({\left\{{a, b}\right\}}\right) = 4$ By inspection it can be seen that $\phi$ is a bijection. Also by inspection it can be seen that $\phi$ is increasing. But note that while $2 \le 3$, it is not the case that $\left\{{a}\right\} \subseteq \left\{{b}\right\}$. That is, $\phi^{-1}$ is ''not'' increasing. {{qed}}
By definition, an order isomorphism is a bijection $\phi$ such that: : $\phi: S \to T$ is order-preserving : $\phi^{-1}: T \to S$ is order-preserving. So, by definition, there exists a bijection between $S$ and $T$. The result follows by definition of set equivalence. {{qed}}
By definition of contour, $C$ is a finite sequence $C_1, \ldots, C_n$ of directed smooth curves. Let $C_i$ be parameterized by the smooth path $\gamma_i: \closedint {a_i} {b_i} \to \C$ for all $i \in \set {1, \ldots, n}$. Then: {{begin-eqn}} {{eqn | l = \int_C \paren {\lambda \map f z + \mu \map g z} \rd z       | r = \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \paren {\lambda \map f {\map {\gamma_i} t} + \mu \map g {\map {\gamma_i} t} } \map {\gamma_i'} t \rd t       | c = {{Defof|Complex Contour Integral}} }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {\lambda \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t + \mu \int_{a_i}^{b_i} \map g {\map {\gamma_i} t} \map {\gamma_i'} t \rd t}       | c = Linear Combination of Complex Integrals }} {{eqn | r = \lambda \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map f {\map {\gamma_i} t} \map {\gamma_i'} t \rd t + \mu \sum_{i \mathop = 1}^n \int_{a_i}^{b_i} \map g {\map {\gamma_i} t} \map {\gamma_i'} t \rd t }} {{eqn | r = \lambda \int_C \map f z \rd z + \mu \int_C \map g z \rd z  }} {{end-eqn}} {{qed}}
By $p$-Product Metrics are Lipschitz Equivalent and Lipschitz Equivalent Metrics are Topologically Equivalent, it suffices to consider the case $p = \infty$. Let $\struct{A \times B, \tau'}$ be the product space of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$. By the definition of $d$, it follows that an open ball in $M$ is the (Cartesian) product of an open ball in $M_A$ and an open ball in $M_B$. Since an open ball is an open set, it follows from Equivalence of Definitions of Analytic Basis that $\tau \subseteq \tau'$. Let $W \in \tau'$, $\tuple{x, y} \in W$. By the definition of $\tau'$, it follows from Equivalence of Definitions of Analytic Basis that there exists a $U$ open in $M_A$ and a $V$ open in $M_B$ such that $\tuple{x, y} \in U \times V \subseteq W$. By the definition of an open set: :There exists a strictly positive real number $\alpha \in \R_{>0}$ such that the open $\alpha$-ball of $x$ in $M_A$ is contained in $U$ :There exists a strictly positive real number $\beta \in \R_{>0}$ such that the open $\beta$-ball of $y$ in $M_B$ is contained in $V$ Let $\epsilon = \min \set{\alpha, \beta}$. Then $\epsilon \in \R_{>0}$, and the open $\epsilon$-ball of $\tuple{x, y}$ in $M$ is contained in $U \times V$. Hence, $W$ is open in $M$. That is, $\tau' \subseteq \tau$. By definition of set equality: : $\tau = \tau'$ as desired. {{qed}} Category:Metric Spaces Category:P-Product Metrics lqibvwe21ff2d5d9bz5iuk99rslkxwi
From $p$-Product Metric is Metric, $M_p$ is a metric space. For all real numbers $p \ge 1$, it follows from the definition of $d_p$ that: :$\forall x, y \in A \times B: d_{\infty} \left({x, y}\right) \le d_p \left({x, y}\right) \le 2^{1/p} d_{\infty} \left({x, y}\right)$ Hence, $d_p$ and $d_{\infty}$ are Lipschitz equivalent. The result follows from Lipschitz Equivalence is Equivalence Relation. {{qed}} Category:Lipschitz Equivalence Category:P-Product Metrics e5q0v5e89hcpltsp1iq4vys99oipf9s
From Ordered Sum of Tosets is Totally Ordered Set, $S_1 + S_2$ is a toset. Suppose $T_{n-1}$ is a toset. Given that $S_n$ is a toset, it follows from Ordered Sum of Tosets is Totally Ordered Set that $T_{n-1} + S_n$ is also a toset. The result follows by induction. {{qed}} Category:Total Orderings Category:Ordered Sums 422v3esqiz5ug0v76ygkcvqvi1lkwuq
Suppose that $s = \sup \bigcup \mathbb T \in S$. By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$. Hence by Supremum of Subset: :$\forall T \in \mathbb T: \sup T \preceq s$ Suppose now that $a \in S$ satisfies: :$\forall T \in \mathbb T: \sup T \preceq a$ Then by transitivity of $\preceq$: :$\forall t \in T: t \preceq a$ Since this holds for any $T \in \mathbb T$, also: :$\forall t \in \bigcup \mathbb T: t \preceq a$ Hence $s \preceq a$, by definition of supremum. That is, $s = \sup {\set {\sup T: T \in \mathbb T} }$. {{qed|lemma}} Suppose now that $r = \sup {\set {\sup T: T \in \mathbb T} } \in S$. By definition of supremum, for all $T \in \mathbb T$ and $t \in T$: :$t \preceq \sup T$ By transitivity of $\preceq$: :$\forall T \in \mathbb T: \forall t \in T: t \preceq r$ Hence for all $t \in \bigcup \mathbb T$: :$t \preceq r$ Suppose that $a \in S$ satisfies: :$\forall t \in \bigcup \mathbb T: t \preceq a$ In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$: :$\sup T \preceq a$ and therefore by definition of supremum, also: :$r \preceq a$ That is, $r = \sup \bigcup \mathbb T$. {{qed}} Category:Order Theory 2fhnbjumyyvbt8a4ykp6kutfy4l6h0w
From Ordered Product of Tosets is Totally Ordered Set, $S_1 \cdot S_2$ is a toset. Suppose $T_{n-1}$ is a toset. Given that $S_n$ is a toset, it follows from Ordered Product of Tosets is Totally Ordered Set that $T_{n-1} \cdot S_n$ is also a toset. The result follows by induction. {{qed}} Category:Total Orderings Category:Ordered Products 1902bsmwyduw1aka6zdb5jqe3xfbpig