hcju/proofwikiqa huggingface dataset:
From the $n$th root test, $S \paren x$ is convergent if $\displaystyle \limsup_{n \mathop \to \infty} \size {a_n \paren {x - \xi}^n}^{1/n} < 1$. Thus: {{begin-eqn}} {{eqn | l = \size {a_n \paren {x - \xi}^n}^{1/n}       | o = <       | r = 1       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \size {a_n}^{1/n} \size {x - \xi}       | o = <       | r = 1       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \size {a_n}^{1/n}       | o = <       | r = \frac 1 {\size {x - \xi} }       | c =  }} {{end-eqn}} The result follows from the definition of radius of convergence. {{qed}} {{Namedfor|Augustin Louis Cauchy|name2 = Jacques Salomon Hadamard}}
{{AimForCont}} $R_A > R_B$. Let $z_0 \in \C$ such that $R_B < \cmod {z_0} < R_A$. Then $A$ is convergent at $z_0$ but $B$ is divergent at $z_0$. But by the Comparison Test, if $A$ is convergent at $z_0$ then $B$ is also convergent at $z_0$. From this contradiction it follows that there can be no such $z_0$. That is: :$R_A \le R_B$ {{qed}}
{{begin-eqn}} {{eqn | l = x^2 + y^2 - 2 x - 2 y - 2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\paren {x - 1}^2 - 1} + \paren {\paren {y - 1}^2 - 1} - 2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x - 1}^2 + \paren {y - 1}^2       | r = 4       | c =  }} {{end-eqn}} This defines a circle whose center is at $1 + i$ and whose radius is $2$. From Modulus of Exponential is Exponential of Real Part: :$\cmod {e^z} = e^x$ If $z \in C$ then from the geometry of the circle $C$: :$-1 \le x \le 3$ Then from Exponential is Strictly Increasing: :$e^{-1} \le e^x \le e^3$ Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \tanh \paren {a + b i}       | r = \frac {\sinh \paren {a + b i} } {\cosh \paren {a + b i} }       | c = {{Defof|Hyperbolic Tangent}} }} {{eqn | r = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}       | c = Hyperbolic Sine of Complex Number and Hyperbolic Cosine of Complex Number }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tanh \paren {a + b i}       | r = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}       | c = Hyperbolic Tangent of Complex Number: Formulation 1 }} {{eqn | r = \dfrac {\tanh a \cos b + i \sin b} {\cos b + i \tanh a \sin b}       | c = multiplying denominator and numerator by $\dfrac 1 {\cosh a}$ }} {{eqn | r = \frac {\tanh a + i \tan b} {1 + i \tanh a \tan b}       | c = multiplying denominator and numerator by $\dfrac 1 {\cos b}$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \paren {a + b i}       | r = \dfrac {\tanh a + i \tan b} {1 + i \tanh a \tan b}       | c = Hyperbolic Tangent of Complex Number: Formulation 2 }} {{eqn | r = \frac {\paren {\tanh a + i \tan b} \paren {1 - i \tanh a \tan b} } {1 + \tanh^2 a \tan^2 b}       | c = multiplying denominator and numerator by $1 - i \tanh a \tan b$ }} {{eqn | r = \frac {\tanh a + i \tan b + i \tanh^2 a \tan b - \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b}       | c =  }} {{eqn | r = \frac {\tanh a - \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b} + \frac {\tan b + \tanh^2 a \tan b} {1 + \tanh^2 a \tan^2 b} i       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \csch {a + b i}       | r = \frac 1 {\map \sinh {a + b i} }       | c = {{Defof|Hyperbolic Cosecant}} }} {{eqn | r = \dfrac 1 {\sinh a \cos b + i \cosh a \sin b}       | c = Hyperbolic Sine of Complex Number }} {{eqn | r = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\paren {\sinh a \cos b + i \cosh a \sin b} \paren {\sinh a \cos b - i \cosh a \sin b} }       | c = multiplying denominator and numerator by $\sinh a \cos b - i \cosh a \sin b$ }} {{eqn | r = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\sinh^2 a \cos^2 b - i^2 \cosh^2 a \sin^2 b}       | c = Difference of Two Squares }} {{eqn | r = \dfrac {\sinh a \cos b - i \cosh a \sin b} {\sinh^2 a \cos^2 b + \cosh^2 a \sin^2 b}       | c = {{Defof|Imaginary Unit}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sech \paren {a + b i}       | r = \frac 1 {\cosh \paren {a + b i} }       | c = {{Defof|Hyperbolic Secant}} }} {{eqn | r = \dfrac 1 {\cosh a \cos b + i \sinh a \sin b}       | c = Hyperbolic Cosine of Complex Number }} {{eqn | r = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\paren {\cosh a \cos b + i \sinh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} }       | c = multiplying denominator and numerator by $\cosh a \cos b - i \sinh a \sin b$ }} {{eqn | r = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b - i^2 \sinh^2 a \sin^2 b}       | c = Difference of Two Squares }} {{eqn | r = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}       | c = {{Defof|Imaginary Unit}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \coth {a + b i}       | r = \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}       | c = Hyperbolic Cotangent of Complex Number: Formulation 1 }} {{eqn | r = \dfrac {\coth a \cos b + i \sin b} {\cos b + i \coth a \sin b}       | c = multiplying denominator and numerator by $\dfrac 1 {\sinh a}$ }} {{eqn | r = \dfrac {\coth a \cot b + i} {\cot b + i \coth a}       | c = multiplying denominator and numerator by $\dfrac 1 {\sin b}$ }} {{eqn | r = \dfrac {1 + i \coth a \cot b} {\coth a - i \cot b}       | c = multiplying denominator and numerator by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \coth {a + b i}       | r = \dfrac {1 + i \coth a \cot b} {\coth a - i \cot b}       | c = Hyperbolic Cotangent of Complex Number: Formulation 2 }} {{eqn | r = \dfrac {\paren {1 + i \coth a \cot b} \paren {\coth a + i \cot b} } {\paren {\coth a - i \cot b} \paren {\coth a + i \cot b} }       | c = multiplying denominator and numerator by $\coth a + i \cot b$ }} {{eqn | r = \dfrac {\paren {1 + i \coth a \cot b} \paren {\coth a + i \cot b} } {\coth^2 a + \cot^2 b}       | c = Difference of Two Squares }} {{eqn | r = \dfrac {\coth a + i \coth b + i \coth^2 a \cot b - \coth a \cot^2 b} {\coth^2 a + \cot^2 b}       | c =  }} {{eqn | r = \dfrac {\coth a - \coth a \cot^2 b} {\coth^2 a + \cot^2 b} + \dfrac {\coth b + \coth^2 a \cot b} {\coth^2 a + \cot^2 b} i       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tanh \paren {a + b i}       | r = \dfrac {\sinh a \cos b + i \cosh a \sin b} {\cosh a \cos b + i \sinh a \sin b}       | c = Hyperbolic Tangent of Complex Number: Formulation 1 }} {{eqn | r = \dfrac {\paren {\sinh a \cos b + i \cosh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} } {\paren {\cosh a \cos b + i \sinh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} }       | c =  }} {{eqn | r = \dfrac {\sinh a \cosh a \cos^2 b + i \cosh^2 a \cos b \sin b - i \sinh^2 a \cos b \sin b + \sinh a \cosh a \sin^2 b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}       | c =  }} {{eqn | r = \dfrac {\sinh a \cosh a \paren {\cos^2 b + \sin^2 b} + i \paren {\cosh^2 a - \sinh^2 a} \cos b \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}       | c =  }} {{eqn | r = \dfrac {\sinh a \cosh a + i \paren {\cosh^2 a - \sinh^2 a} \cos b \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \dfrac {\sinh a \cosh a + i \cos b \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {2 \paren {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b} }       | c = Double Angle Formula for Hyperbolic Sine, Double Angle Formula for Sine }} {{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh^2 a \paren {1 + \cos 2 b} +  \sinh^2 a \paren {1 - \cos 2 b} }       | c = Double Angle Formula for Cosine }} {{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh^2 a + \cosh^2 a \cos 2 b + \sinh^2 a - \sinh^2 a \cos 2 b}       | c =  }} {{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh 2 a + \paren {\cosh^2 a - \sinh^2 a} \cos 2 b}       | c = Double Angle Formula for Hyperbolic Cosine }} {{eqn | r = \dfrac {\sinh 2 a + i \sin 2 b} {\cosh 2 a + \cos 2 b}       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \paren {a + b i}       | r = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}       | c = Tangent of Complex Number: Formulation 1 }} {{eqn | r = \dfrac {\paren {\sin a \cosh b + i \cos a \sinh b} \paren {\cos a \cosh b + i \sin a \sinh b} } {\paren {\cos a \cosh b - i \sin a \sinh b} \paren {\cos a \cosh b + i \sin a \sinh b} }       | c =  }} {{eqn | r = \dfrac {\sin a \cos a \cosh^2 b + i \cos^2 a \cosh b \sinh b + i \sin^2 a \cosh b \sinh b - \sin a \cos b \sinh^2 b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}       | c =  }} {{eqn | r = \dfrac {\sin a \cos a \paren {\cosh^2 b - \sinh^2 b} + i \paren {\cos^2 a + \sin^2 a} \cosh b \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}       | c =  }} {{eqn | r = \dfrac {\sin a \cos a + i \paren {\cos^2 a + \sin^2 a} \cosh b \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{eqn | r = \dfrac {\sin a \cos a + i \cosh b \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b} {2 \paren {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b} }       | c = Double Angle Formula for Sine, Double Angle Formula for Hyperbolic Sine }} {{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\paren {1 + \cos 2 a} \cosh^2 b + \paren {1 - \cos 2 a} \sinh^2 b}       | c = Double Angle Formula for Cosine }} {{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\cosh^2 b + \cos 2 a \cosh^2 b + \sinh^2 b - \cos 2 a \sinh^2 b}       | c =  }} {{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\cos 2 a \paren {\cosh^2 b - \sinh^2 b} + \cosh 2 b}       | c = Double Angle Formula for Hyperbolic Cosine }} {{eqn | r = \dfrac {\sin 2 a + i \sinh 2 b } {\cos 2 a + \cosh 2 b}       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{end-eqn}} {{qed}}
=== Necessary Condition === {{:Characterisation of Cauchy Sequence in Non-Archimedean Norm/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Characterisation of Cauchy Sequence in Non-Archimedean Norm/Sufficient Condition}}{{qed}}
Let $\epsilon > 0$ be given. By the definition of a Cauchy sequence: :$\exists N: \forall n, m > N: \norm {x_n - x_m} < \epsilon$ So :$\exists N: \forall n > N: \norm {x_{n + 1} - x_n} < \epsilon$ Hence the result follows: :$\lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n} = 0$.
Let $\epsilon > 0$ be given. By assumption $\exists N \in \N$ such that: :$(1) \quad  \forall n > N: \norm {x_{n + 1} - x_n} < 0$ Suppose $n, m > N$, and $m = n + r > n$. Then: {{begin-eqn}} {{eqn | l = \norm {x_m - x_n}       | r = \norm {x_{n + r} - x_{n + r - 1} + x_{n + r - 1} - x_{n + r - 2} + \dotsb + x_{n + 1} - x_n} }} {{eqn | r = \max \set {\norm {x_{n + r} - x_{n + r - 1} }, \norm {x_{n + r - 1} - x_{n + r - 2} }, \dotsc, \norm {x_{n + 1} - x_n} }       | c = as $\norm {\,\cdot\,}$ is non-Archimedean }} {{eqn | r = \norm {x_{n + s} - x_{n + s - 1} }       | c = for some $s$: $0 < s \le r$ }} {{eqn | o = <       | r = \epsilon       | c = by $(1)$ }} {{end-eqn}} It follows that: ::$\sequence {x_n}$ is a Cauchy sequence.
From Euler's Formula: :$\map \exp {i \theta} = \cos \theta + i \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^\infty \map \sin {2 k + 1} \theta r^k       | r = \map \Im {\sum_{k \mathop = 0}^\infty r^k \map \exp {\paren {2 k + 1} i \theta} }       | c =  }} {{eqn | r = \map \Im {\map \exp {i \theta} \sum_{k \mathop = 0}^\infty \paren {r \map \exp {2 i \theta} }^k}       | c = as $\map \Im {\exp \paren {i \times 0 \times x} } = \map \Im 1 = 0$ }} {{eqn | r = \map \Im {\frac {\map \exp {i \theta} } {1 - r \map \exp {2 i \theta} } }       | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$ }} {{eqn | r = \map \Im {\frac {\map \exp {i \theta} \paren {1 - r \map \exp {-2 i \theta} } } {\paren {1 - r \map \exp {-2 i \theta} } \paren {1 - r \map \exp {2 i \theta} } } }       | c =  }} {{eqn | r = \map \Im {\frac {\map \exp {i \theta} \paren {1 - r \map \exp {-2 i \theta} } } {1 - r \paren {\map \exp {2 i \theta} + \map \exp {- 2 i \theta} } + r^2} }       | c =  }} {{eqn | r = \map \Im {\frac {\map \exp {i \theta} \paren {1 - r \map \exp {-2 i \theta} } } {1 - 2 r \cos 2 \theta + r^2} }       | c = Cosine Exponential Formulation }} {{eqn | r = \map \Im {\frac {\paren {\cos \theta + i \sin \theta} \paren {1 - r \paren {\cos 2 \theta - i \sin 2 \theta} } } {1 - 2 r \cos 2 \theta + r^2} }       | c = Euler's Formula, Corollaryto Euler's Formula }} {{eqn | r = \map \Im {\frac {\cos \theta + i \sin \theta - r \cos \theta \cos 2 \theta + i r \cos \theta \sin 2 \theta - i r \sin \theta \cos 2 \theta - r \sin \theta \sin 2 \theta} {1 - 2 r \cos 2 \theta + r^2} }       | c = expanding }} {{eqn | r = \frac {\sin \theta + r \cos \theta \sin 2 \theta - r \sin \theta \cos 2 \theta} {1 - 2 r \cos 2 \theta + r^2}       | c = taking imaginary part }} {{eqn | r = \frac {\sin \theta + r \paren {\sin 2 \theta \cos \theta - \cos 2 \theta \sin \theta} } {1 - 2 r \cos 2 \theta + r^2}       | c =  }} {{eqn | r = \frac {\sin \theta + r \map \sin {2 \theta - \theta} } {1 - 2 r \cos 2 \theta + r^2}       | c = Sine of Difference }} {{eqn | r = \dfrac {1 + r \sin \theta} {1 - 2 r \cos 2 \theta + r^2}       | c = after simplification }} {{end-eqn}} It is noted that when $\theta$ is a multiple of $\pi$ then: :$1 - 2 r \cos 2 \theta + r^2 = 1 - 2 + 1 = 0$ leaving the {{RHS}} undefined. {{qed}}
Let $\theta \ne \dfrac {m \pi} 2$ for any $m \in \Z$. Then $\size {\cos 2 \theta} < 1$. {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} \sin \paren {2 k + 1} \theta r^k       | r = \dfrac {\paren {1 + r} \sin \theta} {1 - 2 r \cos 2 \theta + r^2}       | c = Power Series of Sine of Odd Theta: $\size r < 1$ }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 0}^\infty \paren {\cos 2 \theta}^k \sin \paren {2 k + 1} \theta       | r = \dfrac {\paren {1 + \cos 2 \theta} \sin \theta} {1 - 2 \cos 2 \theta \cos 2 \theta + \paren {\cos 2 \theta}^2}       | c = setting $r = \cos 2 \theta$ }} {{eqn | r = \dfrac {\paren {1 + \cos 2 \theta} \sin \theta} {1 - \paren {\cos 2 \theta}^2}       | c = simplifying }} {{eqn | r = \dfrac {\paren {1 + 2 \cos^2 \theta - 1} \sin \theta} {1 - \paren {2 \cos^2 \theta - 1}^2}       | c = Double Angle Formula for Cosine }} {{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {1 - \paren {4 \cos^4 \theta - 4 \cos^2 \theta + 1} }       | c = simplifying and expanding }} {{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {4 \cos^2 \theta \paren {1 - \cos^2 \theta} }       | c = simplifying }} {{eqn | r = \dfrac {2 \cos^2 \theta \sin \theta} {4 \cos^2 \theta \sin^2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \dfrac 1 {2 \sin \theta}       | c = simplifying }} {{eqn | r = \dfrac {\csc \theta} 2       | c = {{Defof|Real Cosecant Function}} }} {{end-eqn}} {{qed}}
By the Reverse Triangle Inequality, we have: :$\cmod {\norm {x_n} - \norm l} \le \norm {x_n - l}$ Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\norm {x_n} \to \norm l$ as $n \to \infty$. {{Qed}}
Either: :$\map d {x, z} = \map d {y, z}$ or: :$\map d {x, z} \ne \map d {y, z}$ By Three Points in Ultrametric Space have Two Equal Distances: :$\map d {x, z} = \map d {y, z}$ or $\map d {x, y} = \max \set {\map d {x, z}, \map d {y, z} }$ In either case two of the distances are equal. {{Qed}}
=== Necessary Condition === {{:Non-Archimedean Norm iff Non-Archimedean Metric/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Non-Archimedean Norm iff Non-Archimedean Metric/Sufficient Condition}}{{qed}}
{{AimForCont}} $\dfrac x y = \dfrac y z = \dfrac z x$. {{begin-eqn}} {{eqn | o =        | r = \dfrac x y = \dfrac y z = \dfrac z x       | c =  }} {{eqn | ll= \leadsto       | o =        | r = x^2 z = y^2 x = z^2 y       | c = multiplying top and bottom by $x y z$ }} {{eqn | ll= \leadsto       | o =        | r = x z = y^2, x y = z^2, y z = x^2       | c =  }} {{eqn | ll= \leadsto       | o =        | r = x y z = y^3, x y z = z^3, z y z = x^3       | c =  }} {{eqn | ll= \leadsto       | o =        | r = x^3 = y^3 = z^3       | c =  }} {{eqn | ll= \leadsto       | o =        | r = x = y = z       | c =  }} {{end-eqn}} This contradicts the assertion that $x, y, z$ are all unequal. Hence the result by Proof by Contradiction. {{qed}}
Proof by Counterexample: Let $S = \set {1, 2, 3}, T = \set {2, 3, 4}, X = \set {1, 2, 3, 4}$. {{begin-eqn}} {{eqn | l = S \cup T       | r = \set {1, 2, 3, 4}       | c =  }} {{eqn | ll= \leadsto       | l = X       | o = \subseteq       | r = S \cup T       | c =  }} {{eqn | ll= \leadsto       | l = X       | o = \in       | r = \powerset {S \cup T}       | c =  }} {{end-eqn}} But note that $X \nsubseteq S \land X \nsubseteq T$. Thus: {{begin-eqn}} {{eqn | l = X       | o = \nsubseteq       | r = S \land X \nsubseteq T       | c =  }} {{eqn | ll= \leadsto       | l = X       | o = \notin       | r = \powerset S \land X \notin \powerset T       | c =  }} {{eqn | ll= \leadsto       | l = \neg (X       | o = \in       | r = \powerset S \lor X \in \powerset T)       | c =  }} {{eqn | ll= \leadsto       | l = X       | o = \notin       | r = \powerset S \cup \powerset T       | c =  }} {{eqn | ll= \leadsto       | l = \powerset {S \cup T}       | o = \nsubseteq       | r = \powerset S \cup \powerset T       | c =  }} {{end-eqn}} So: :$\powerset {S \cup T} \ne \powerset S \cup \powerset T$ {{qed}}
First it is noted that: :$\forall x \in \R_{<0}: x \notin A$ and that by definition of closed real interval: :$\forall n \in \Z_{>0}: 0 \in A_n$ and so by definition of intersection: :$0 \in A$ It remains to demonstrate that: :$\forall x \in \R_{>0}: x \notin A$ {{AimForCont}} $\exists x \in \R_{>0}: x \in A$. By the Archimedean Principle: :$\exists N \in \Z: N > \dfrac 1 x$ and so from Reciprocal Function is Strictly Decreasing: :$\exists N \in \Z: \dfrac 1 N < x$ Thus: :$x \notin A_N$ and so by definition of intersection: :$x \notin A$ This contradicts our supposition that $x \in A$. Hence the only element of $A$ is $0$, and so: :$A = \set 0$ {{qed}}
{{begin-eqn}} {{eqn | l = P       | r = A \cap B \cap \paren {C \cup D}       | c =  }} {{eqn | r = \paren {A \cap B \cap C} \cup \paren {A \cap B \cap D}       | c = Intersection Distributes over Union }} {{end-eqn}} Let $P$ be expressed as: :$P = X \cup Y$ where: :$X = A \cap B \cap D$ :$Y = A \cap B \cap C$ Then: {{begin-eqn}} {{eqn | l = X       | r = A \cap B \cap D       | c =  }} {{eqn | o = \subseteq       | r = A \cap D       | c = Intersection is Subset }} {{eqn | o = \subseteq       | r = Q       | c = Set is Subset of Union }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = Y       | r = A \cap B \cap C       | c =  }} {{eqn | o = \subseteq       | r = B \cap C       | c = Intersection is Subset }} {{eqn | o = \subseteq       | r = Q       | c = Set is Subset of Union }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = X \cup Y       | o = \subseteq       | r = Q       | c = Union is Smallest Superset }} {{eqn | ll= \leadsto       | l = P       | o = \subseteq       | r = Q       | c =  }} {{end-eqn}} {{qed}}
The number of different partitions of $S$ is '''defined''' as $B_n$. From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind: :$B_n = \displaystyle \sum_{k \mathop = 0}^n {n \brace k}$ where $\displaystyle {n \brace k}$ denotes a Stirling number of the second kind. From Number of Set Partitions by Number of Components, $\displaystyle {n \brace k}$ is the number of partitions of $S$ into $k$ components. Hence the result. {{qed}} === Examples === {{:Total Number of Set Partitions/Examples}} Category:Set Partitions lqzeox4gck17yui3dpj6j295y84c1ia
Since $\sequence {x_n}$ converges to $l \ne 0$, by Sequence Converges to Within Half Limit then: :$\exists k \in \N: \forall n \in \N: \dfrac {\norm l} 2 < \norm {x_{k + n} }$ By {{NormAxiom|1}}: :$\forall n \in \N : x_{k + n} \ne 0$ Let $\sequence {y_n}$ be the subsequence of $\sequence {x_n}$ where $y_n = x_{k + n}$.  By Limit of Subsequence equals Limit of Sequence, $\sequence {y_n}$ is convergent with: :$\displaystyle \lim_{n \mathop \to \infty} y_n = l$ It also follows that:  :$\forall n \in \N : y_n \ne 0$ So $\sequence { {y_n}^{-1} }$ is well-defined and: :$\sequence { {y_n}^{-1} } = \sequence { {x_{k + n} }^{-1} }$ === Lemma === {{:Combination Theorem for Sequences/Normed Division Ring/Inverse Rule/Lemma}}{{qed}}
By definition of Bell numbers: :$B_n$ is the number of partitions of a (finite) set whose cardinality is $n$. First consider the case where $n > 0$. From Number of Set Partitions by Number of Components, the number of partitions of $S$ into $k$ components is $\displaystyle {n \brace k}$. Thus the total number of all partitions of $S$ is the sum of $\displaystyle {n \brace k}$ where $k$ runs from $1$ to $n$. Hence: :$B_n = \displaystyle \sum_{k \mathop = 1}^n {n \brace k}$ But $\displaystyle {n \brace 0} = 0$ for all $n \in \Z_{> 0}$, and so: :$B_n = \displaystyle \sum_{k \mathop = 0}^n {n \brace k}$ Now consider the edge case where $n = 0$, that is, $S = \O = \set{}$. There is by conventional definition one partition of $\O$, that is: $\set \O$ In this case, we have: :$\displaystyle {0 \brace 0} = 1$ and so: :$B_0 = \displaystyle \sum_{k \mathop = 0}^0 {n \brace k} = {0 \brace 0} = 1$ {{qed}} Category:Bell Numbers Category:Stirling Numbers e0es4327xn9bkjt34d3zg9aenijqz3u
From Bell Number as Summation over Lower Index of Stirling Numbers of the Second Kind, we have that: :$B_n = \displaystyle \sum_{k \mathop = 0}^n {n \brace k}$ But when $n > 0$: :$\displaystyle {n \brace 0} = 0$ Hence the result. {{qed}} Category:Bell Numbers Category:Stirling Numbers ldgte9oswzxmfqn4olwei3ptsrjkkxs
Let us assume that the '''PFI''' is true. Let $S \subseteq \N$ which satisfy: :$(A): \quad 0 \in S$ :$(B): \quad \set {0, 1, \ldots, n} \subseteq S \implies n + 1 \in S$. We want to show that $S = \N$, that is, the '''PCI''' is true. Let $P \paren n$ be the propositional function: :$P \paren n \iff \set {0, 1, \ldots, n} \subseteq S$ We define the set $S'$ as: :$S' = \set {n \in \N: P \paren n \text { is true} }$ $P \paren 0$ is true by $(A)$, so $0 \in S'$. Assume $P \paren k$ is true where $k \ge 0$. So $k \in S'$, and by hypothesis: :$\set {0, 1, \ldots, k} \subseteq S$ So by $(B)$: :$k + 1 \in S$ Thus: :$\set {0, 1, \ldots, k, k + 1} \subseteq S$. That last statement means $P \paren {k + 1}$ is true. This means $k + 1 \in S'$. Thus we have satisfied the conditions: :$0 \in S'$ :$n \in S' \implies n + 1 \in S'$ That is, $S' = \N$, and $P \paren n$ holds for all $n \in \N$. Hence, by definition: :$S = \N$ So '''PFI''' gives that $S = \N$. Therefore '''PFI''' implies '''PCI'''.
Let us assume that the '''PCI''' is true. Let $\O \subset S \subseteq \N$. We need to show that $S$ has a minimal element, and so demonstrate that the '''WOP''' holds. {{AimForCont}} that: :$(C): \quad S$ has no minimal element. Let $\map P n$ be the propositional function: :$n \notin S$ Suppose $0 \in S$. We have that $0$ is a lower bound for $\N$. Hence by Lower Bound for Subset, $0$ is also a lower bound for $S$. $0 \notin S$, otherwise $0$ would be the minimal element of $S$. This contradicts our supposition $(C)$, namely, that $S$ does not have a minimal element. So $0 \notin S$ and so $\map P 0$ holds. Suppose $\map P j$ for $0 \le j \le k$. That is: :$\forall j \in \closedint 0 k: j \notin S$ where $\closedint 0 k$ denotes the closed interval between $0$ and $k$. Now if $k + 1 \in S$ it follows that $k + 1$ would then be the minimal element of $S$. So then $k + 1 \notin S$ and so $\map P {k + 1}$. Thus we have proved that: :$(1): \quad \map P 0$ holds :$(2): \quad \paren {\forall j \in \closedint 0 k: \map P j} \implies \map P {k + 1}$ So we see that '''PCI''' implies that $\map P n$ holds for all $n \in \N$. But this means that $S = \O$, which is a contradiction of the fact that $S$ is non-empty. So, by Proof by Contradiction, $S$ must have a minimal element. That is, $\N$ satisfies the Well-Ordering Principle. Thus '''PCI''' implies '''WOP'''.