hcju/proofwikiqa huggingface dataset:
At a time $t$, let: :the rod be at an angle $\theta$ to the the vertical :the bob be travelling at a speed $v$ :the displacement of the bob from where it is when the rod is vertical, along its line of travel, be $s$. :350px From Motion of Pendulum, the equation of motion of the bob is given by: :$\dfrac {a^2} 2 \paren {\dfrac {\d \theta} {\d t} }^2 = g a \paren {a \cos \theta - a \cos \alpha}$ Differentiating {{WRT|Differentiation}} $t$ and simplifying yields: :$a \dfrac {\d^2 \theta} {\d t^2} = -g \sin \theta$ For small angles, $\sin \theta \approx \theta$ and so: :$\dfrac {\d^2 \theta} {\d t^2} + \dfrac g a \theta = 0$ This is an instance of a $2$nd order linear ODE. From Linear Second Order ODE: $y'' + k^2 y = 0$, this has the solution: :$\theta = c_1 \sin \sqrt {\dfrac g a} t + c_1 \cos \sqrt {\dfrac g a} t$ We have at $t = 0$ that $\theta = \alpha$ and $\dfrac {\d \theta} {\d t} = 0$. This yields: :$\theta = \alpha \cos \sqrt {\dfrac g a} t$ and so the period of this pendulum is: :$T = 2 \pi \sqrt {\dfrac a g}$ {{qed}}
:$T$ is separable {{iff}} :there exists a countable subset of $T$ which is dense by definition of separable space {{iff}} :there exists a subset $A$ of $T$ such that $A$ is dense and exists an injection $A \to \N$ by definition of countable set {{iff}} :there exists a subset $A$ of $T$ such that $A$ is dense and $\left\vert{A}\right\vert \leq \left\vert{\N}\right\vert$ by Injection iff Cardinal Inequality {{iff}} :there exists a subset $A$ of $T$ such that $A$ is dense and $\left\vert{A}\right\vert \leq \aleph_0$ by Aleph Zero equals Cardinality of Naturals {{iff}} :$d \left({T}\right) \leq \aleph_0$ by definition of density where $\left\vert{A}\right\vert$ denotes the cardinality of $A$. {{qed}}
By definition of separated sets: :$(1): \quad A^- \cap B = A \cap B^- = \varnothing$ By Separated Sets are Disjoint: :$A \cap B = \varnothing$ {{begin-eqn}} {{eqn | l = \partial A \cup \partial B       | r = \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)       | c = Union of Boundaries }} {{eqn | r = \partial \left({A \cup B}\right) \cup \varnothing \cup \left({\partial A \cap \partial B}\right)       | c = Boundary of Empty Set is Empty }} {{eqn | r = \partial \left({A \cup B}\right) \cup \left({\partial A \cap \partial B}\right)       | c = Union with Empty Set }} {{end-eqn}} We will proof that :$\partial A \cap \partial B \subseteq \partial \left({A \cup B}\right)$ Let $x \in \partial A \cap \partial B$. Then by definition of intersection: :$x \in \partial A \land x \in \partial B$ Hence by Boundary is Intersection of Closure with Closure of Complement: :$x \in A^- \cap \left({\complement_T \left({A}\right)}\right)^- \land x \in B^- \cap \left({\complement_T \left({B}\right)}\right)^-$ where :$A^-$ denotes the closure of $A$, :$\complement_T \left({A}\right) = T \setminus A$ denotes the relative complement of $A$ in $T$. Then by definition of intersection: :$x \in A^- \land x \in B^-$ Therefore by definition of empty set, intersection, and $(1)$: :$x \notin B \land x \notin A$ Then by definition of union: :$x \notin A \cup B$ By definition of relative complement: :$x \in \complement_T \left({A \cup B}\right)$ By definition of closure: :$\complement_T \left({A \cup B}\right) \subseteq \left({\complement_T \left({A \cup B}\right)}\right)^-$ Then by definition of subset: :$x \in \left({\complement_T \left({A \cup B}\right)}\right)^-$ By Closure of Finite Union equals Union of Closures: :$A^- \cup B^- = \left({A \cup B}\right)^-$ By definition of union: :$x \in \left({A \cup B}\right)^-$ Then by definition of intersection: :$x \in \left({A \cup B}\right)^- \cap \left({\complement_T \left({A \cup B}\right)}\right)^-$ Thus  by Boundary is Intersection of Closure with Closure of Complement: :$x \in \partial \left({A \cup B}\right)$ This ends the proof of inclusion. Thus by Union with Superset is Superset the result: :$\partial A \cup \partial B = \partial \left({A \cup B}\right)$ {{qed}}
By Boundary is Intersection of Closure with Closure of Complement: :$\partial_T \O = \O^- \cap \relcomp T \O^-$ where $\O^-$ denotes the closure of $\O$. By Closure of Empty Set is Empty Set: :$\O^- = \O$ Thus the result follows by Intersection with Empty Set. {{qed}}
From Open and Closed Sets in Topological Space, $\O$ is clopen in $T$. The result follows from Set is Clopen iff Boundary is Empty. {{qed}}
Let $B$ start from rest. The differential equation governing the motion of $B$ is given by: :$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \paren {\dfrac {\d \mathbf s} {\d t} }^2$ Dividing through by $m$ and setting $c = \dfrac k m$ gives: :$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \paren {\dfrac {\d \mathbf s} {\d t} }^2$ By definition of velocity: :$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v^2$ for some constant $c$. and so taking magnitudes of the vector quantities: {{begin-eqn}} {{eqn | l = \int \dfrac {\d v} {g - c v^2}       | r = \int \rd t       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 {2 c} \sqrt {\frac c g} \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }       | r = t + c_1       | c = Primitive of $\dfrac 1 {a^2 - x^2}$: Logarithm Form }} {{end-eqn}} We have $v = 0$ when $t = 0$ which leads to $\ln 1 = 0 + c_1$ and thus $c_1 = 0$. Hence: {{begin-eqn}} {{eqn | l = \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }       | r = 2 t \sqrt {c g}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v}       | r = e^{2 t \sqrt {c g} }       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {\frac c g} + v       | r = \paren {\sqrt {\frac c g} - v} e^{2 t \sqrt {c g} }       | c =  }} {{eqn | ll= \leadsto       | l = v \paren {e^{2 t \sqrt {c g} } + 1}       | r = v \paren {e^{2 t \sqrt {c g} } - 1} \sqrt {\frac g c}       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \sqrt {\frac g c} \paren {\frac {e^{2 t \sqrt {c g} } - 1} {e^{2 t \sqrt {c g} } + 1} }       | c =  }} {{eqn | r = \sqrt {\frac g c} \paren {\frac {1 - e^{-2 t \sqrt {c g} } } {1 + e^{-2 t \sqrt {c g} } } }       | c =  }} {{end-eqn}} Since $c > 0$ it follows that $v \to \sqrt {\dfrac g c}$ as $t \to \infty$. Thus in the limit: :$v = \sqrt {\dfrac g c} =  \sqrt {\dfrac {g m} k}$ {{qed}}
Let a projectile $B$ of mass $m$ be fired vertically upwards from the surface of $P$ at such a speed that it escapes the gravitational field of $P$ completely. $F$ be the force exerted on the projectile by the gravitational field of $P$. Let $x$ be the distance of $B$ from the surface of $P$ at time $t$. We have: :$F = -\dfrac k {x^2}$ where $k$ can be calculated at the surface of $P$ as: {{begin-eqn}} {{eqn | l = F       | r = m g       | c =  }} {{eqn | r = \dfrac k{R^2}       | c =  }} {{eqn | ll= \leadsto       | l = k       | r = R^2 m g       | c =  }} {{end-eqn}} Thus: :$F = -\dfrac {m g R^2} {x^2}$ Let the $B$ be travelling away from $P$ with speed $v$ at time $t$. Thus: {{begin-eqn}} {{eqn | l = F       | r = -m \frac {\d v} {\d t}       | c =  }} {{eqn | ll= \leadsto       | l = -\frac {m g R^2} {x^2}       | r = m v \frac {\d v} {\d x}       | c =  }} {{end-eqn}} So: :$\displaystyle -\int \frac {R^2 g} {x^2} \rd x = \int v \rd v$ This gives: :$\dfrac {R^2 g} x = \dfrac {v^2} 2 + C$ Now when $x = R$, $v = v_0$, the launch velocity, which leads to: :$C = R g - \dfrac {v_0^2} 2$ and so: :$\dfrac {v^2} 2 = \dfrac {R^2 g} x - R g + \dfrac {v_0^2} 2$ For $\dfrac {v^2} 2 > 0$ we need: :$\dfrac {v_0^2} 2 > R g \paren {1 - \dfrac R x}$ whatever $x$ may be. So when $x \to \infty$ we have: :$\dfrac {v_0^2} 2 > R g$ Hence: :$v_0 > \sqrt {2 R g}$ {{qed}}
Let $L$ be the length of one arch of $y = \sin x$. Then: {{begin-eqn}} {{eqn | l = L       | r = 2 \int_0^{\pi/2} \sqrt {1 + \paren {\map {\frac \d {\d x} } {\sin x} }^2} \rd x       | c = {{Defof|Length of Curve}} }} {{eqn | r = 2 \int_0^{\pi/2} \sqrt {1 + \cos^2 x} \rd x       | c = Derivative of Sine Function }} {{eqn | r = 2 \int_0^{\pi/2} \sqrt {2 - \sin^2 x} \rd x       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 2 \sqrt 2 \int_0^{\pi/2} \sqrt {1 - \paren {\frac {\sqrt 2} 2} \sin^2 x} \rd x       | c = extracting $\sqrt 2$ }} {{eqn | r = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}       | c = {{Defof|Complete Elliptic Integral of the Second Kind}} }} {{end-eqn}} {{qed}}
The arc length of a small length increment $\d s$ is given in polar co-ordinates by: :$\paren {\d s}^2 = \paren {r \d \theta}^2 + \paren {\d r}^2$ from which: :$\dfrac {\d s} {\d \theta} = \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2}$ Half of one lobe of the lemniscate is achieved when $\theta$ goes from $0$ to $\pi / 4$. Therefore the total length of the lemniscate of Bernoulli is given by: :$\displaystyle L = 4 \int_0^{\pi/4} \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$ First we show: {{begin-eqn}} {{eqn | l = r^2       | r = a^2 \cos 2 \theta       | c =  }} {{eqn | ll= \leadsto       | l = 2 r \frac {\d r} {\d \theta}       | r = -2 a^2 \sin 2 \theta       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d r} {\d \theta}       | r = -\frac {a^2 \sin 2 \theta} {a \sqrt {\cos 2 \theta} }       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\dfrac {\d r} {\d \theta} }^2       | r = \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = r^2 + \paren {\dfrac {\d r} {\d \theta} }^2       | r = a^2 \cos 2 \theta + \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}       | c =  }} {{eqn | r = a^2 \frac {\cos^2 2 \theta + \sin^2 2 \theta} {\cos 2 \theta}       | c =  }} {{eqn | r = \frac {a^2} {\cos 2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = L       | r = 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {\cos 2 \theta} }       | c =  }} {{eqn | r = 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {1 - 2 \sin^2 \theta} }       | c =  }} {{eqn | r = 4 a \map F {\sqrt 2, \dfrac \pi 4}       | c = {{Defof|Incomplete Elliptic Integral of the First Kind}} }} {{end-eqn}} {{qed}}
By Intersection is Subset: :$\complement_S \left({A}\right) \cap \complement_S \left({B}\right) \subseteq \complement_S \left({A}\right) \land \complement_S \left({A}\right) \cap \complement_S \left({B}\right) \subseteq \complement_S \left({B}\right)$ Then by Topological Closure of Subset is Subset of Topological Closure: :$\left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \subseteq \left({\complement_S \left({A}\right)}\right)^- \land \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \subseteq \left({\complement_S \left({B}\right)}\right)^-$ Hence by Boundary is Intersection of Closure with Closure of Complement: :$\left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap A^- \subseteq \partial A \land \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap B^- \subseteq \partial B$ Thus {{begin-eqn}} {{eqn | l = \partial \left({A \cup B}\right)       | r = \left({\complement_S \left({A \cup B}\right)}\right)^- \cap \left({A \cup B}\right)^-       | c = Boundary is Intersection of Closure with Closure of Complement }} {{eqn | r = \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap \left({A \cup B}\right)^-       | c = Complement of Union }} {{eqn | r = \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap \left({A^- \cup B^-}\right)       | c = Closure of Finite Union equals Union of Closures }} {{eqn | r = \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap A^- \cup \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap B^-       | c = Intersection Distributes over Union }} {{eqn | o = \subseteq       | r = \partial A \cup \partial B       | c = Set Union Preserves Subsets }} {{end-eqn}} {{qed}}
By Intersection is Subset: :$A \cap B \subseteq A \land A \cap B \subseteq B$ Then by Topological Closure of Subset is Subset of Topological Closure: :$\paren {A \cap B}^- \subseteq A^- \land \paren {A \cap B}^- \subseteq B^-$ Hence by Boundary is Intersection of Closure with Closure of Complement: :$\paren {A \cap B}^- \cap \paren {\relcomp S A}^- \subseteq \partial A \land \paren {A \cap B}^- \cap \paren {\relcomp S B}^- \subseteq \partial B$ Thus {{begin-eqn}} {{eqn | l = \partial \paren {A \cap B}       | r = \paren {A \cap B}^- \cap \paren {\relcomp S {A \cap B} }^-       | c = Boundary is Intersection of Closure with Closure of Complement }} {{eqn | r = \paren {A \cap B}^- \cap \paren {\relcomp S A \cup \relcomp S B}^-       | c = Complement of Intersection }} {{eqn | r = \paren {A \cap B}^- \cap \paren {\paren {\relcomp S A}^- \cup \paren {\relcomp S B}^-}       | c = Closure of Finite Union equals Union of Closures }} {{eqn | r = \paren {A \cap B}^- \cap \paren {\relcomp S A}^- \cup \paren {A \cap B}^- \cap \paren {\relcomp S B}^-       | c = Intersection Distributes over Union }} {{eqn | o = \subseteq       | r = \partial A \cup \partial B       | c = Set Union Preserves Subsets }} {{end-eqn}} {{qed}}
From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$: {{begin-eqn}} {{eqn | l = \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi       | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2} }} {{eqn | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi       | c = $\sin 0 = \cos \dfrac \pi 2 = 0$ }} {{eqn | r = \cdots }} {{eqn | r = \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi }} {{eqn | r = \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map K k       | r = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt{1 - k^2 \sin^2 \phi} } }} {{eqn | r = \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {- \frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi       | c = General Binomial Theorem }} {{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 1 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi       | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}} }} {{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {1 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }       | c = from above }} {{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}       | c =  }} {{end-eqn}} {{qed}}
From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$: {{begin-eqn}} {{eqn | l = \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi       | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2} }} {{eqn | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi       | c = $\sin 0 = \cos \dfrac \pi 2 = 0$ }} {{eqn | r = \cdots }} {{eqn | r = \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi }} {{eqn | r = \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map E k       | r = \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \phi} \, \rd \phi }} {{eqn | r = \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {\frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi       | c = General Binomial Theorem }} {{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 3 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi       | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}} }} {{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {3 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }       | c = From above }} {{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 3} {2 j} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} } k^{2 i}       | c =  }} {{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 3} {2 j} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} } k^{2 i}       | c =  }} {{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\frac {2 - 3} {2 i - 1} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 k^{2 i}       | c =  }} {{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}       | c =  }} {{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}       | c = Product equals $1$ when $i = 0$ }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Immediate from Separable Discrete Space is Countable. {{qed|lemma}} === Necessary Condition === Immediate from Countable Space is Separable. {{qed}}
:$X$ is countable {{iff}} :there exists an injection $X \to \N$ by definition of countable set {{iff}} :$\size X \le \size \N$ by Injection iff Cardinal Inequality {{iff}} :$\size X \le \aleph_0$ by Aleph Zero equals Cardinality of Naturals. {{qed}} Category:Countable Sets odwikgixvfndjekh3c1vm2d04dez9q6
Let the random variable $k$ have the binomial distribution with parameters $n$ and $p$, that is: :$k \sim \Binomial n p$ where $k$ denotes the number of successes of the $n$ independent trials of the event with probability $p$. From Expectation of Binomial Distribution: :$\expect k = n p \leadsto \dfrac 1 n \expect k = p$ Linearity of Expectation Function gives: :$ \expect {\dfrac k n} = p =: \mu$ Similarly, from Variance of Binomial Distribution: :$\var k = n p \paren {1 - p} \leadsto \dfrac 1 {n^2} \var k = \dfrac {p \paren {1 - p} } n$ From Variance of Linear Combination of Random Variables: :$\var {\dfrac k n} = \dfrac {p \paren {1 - p} } n =: \sigma^2$ By applying Chebyshev's Inequality to $\dfrac {k} {n}$, we have for any $l>0$: :$\map \Pr {\size {\dfrac k m - \mu} \ge l \sigma} \le \dfrac 1 {l^2}$ Now, let $\epsilon > 0$ and choose $l = \dfrac \epsilon \sigma$, to get: :$\map \Pr {\size {\dfrac k m - \mu} \ge \dfrac \epsilon \sigma \cdot \sigma} \le \dfrac {\sigma^2} {\epsilon^2}$ Simplifying and plugging in the values of $\mu$ and $\sigma^2$ defined above yields: :$\map \Pr {\size {\dfrac k n - p} \ge \epsilon} \le \dfrac {p \paren {1 - p} } {n \epsilon^2}$ Scaling both sides by $-1$ and adding $1$ to both sides yields: :$1 - \map \Pr {\size {\dfrac k n - p} \ge \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n \epsilon^2}$ Applying Union of Event with Complement is Certainty to the left hand side: :$\map \Pr {\size {\dfrac k n - p} \le \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n\epsilon^2}$ Taking the limit as $n$ approaches infinity on both sides, we have: :$\displaystyle \lim_{n \mathop \to \infty} \map \Pr {\size {\frac k n - p} < \epsilon} = 1$ {{qed}}
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid. Let $A$ be located at the origin of a cartesian plane. We have the equations of the cycloid: {{begin-eqn}} {{eqn | l = x       | r = a \paren {\theta - \sin \theta} }} {{eqn | l = y       | r = a \paren {1 - \cos \theta} }} {{end-eqn}} Differentiating with respect to theta: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = \paren {1 - \cos \theta}       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = a \sin \theta }} {{end-eqn}} Let $s$ be the distance along $AB$ from the origin. Then: {{begin-eqn}} {{eqn | l = \paren {\frac {\d s} {\d \theta} }^2       | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2       | c =  }} {{eqn | r = a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta       | c =  }} {{eqn | r = a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}       | c =  }} {{eqn | r = 2 a^2 \paren {1 - \cos \theta}       | c =  }} {{eqn | r = 4 a^2 \map {\sin^2} {\frac \theta 2}       | c =  }} {{end-eqn}} Thus: :$\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$ Now: {{begin-eqn}} {{eqn | l = v       | r = \dfrac {\d s} {\d t}       | c =  }} {{eqn | ll= \leadsto       | l = \int \rd t       | r = t       | c =  }} {{eqn | r = \int \frac {\d s} v       | c =  }} {{end-eqn}} We also have, from the Principle of Conservation of Energy, that: {{begin-eqn}} {{eqn | l = m g y       | r = \frac {m v^2} 2       | c =  }} {{eqn | ll= \leadsto       | l = v       | r = \dfrac {\d s} {\d t}       | c =  }} {{eqn | r = \sqrt {2 g y}       | c =  }} {{eqn | r = \sqrt {2 g a \paren {1 - \cos \theta} }       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \frac {\d t} {\d \theta}       | r = \frac {\d t} {\d s} \theta       | c =  }} {{eqn | r = \frac {a \sqrt {2 \paren {1 - \cos \theta} } } {\sqrt {2 g a \paren {1 - \cos \theta} } }       | c =  }} {{eqn | r = \sqrt {\frac a g}       | c =  }} {{end-eqn}} Let $T$ be the time taken to slide down the brachistochrone. At the top: :$\theta = 0$ and at the bottom: :$\theta = \pi$ Then: {{begin-eqn}} {{eqn | l = T       | r = \int \rd t       | c =  }} {{eqn | r = \sqrt {\frac a g} \int_0^\pi \rd \theta       | c =  }} {{eqn | r = \pi \sqrt {\frac a g}       | c =  }} {{end-eqn}} So the time to slide down this brachistochrone from the top to the bottom is: :$T = \pi \sqrt {\dfrac a g}$ {{qed}}
That the curve $AB$ is indeed a cycloid is demonstrated in Brachistochrone is Cycloid. Let $A$ be located at the origin of a cartesian plane. Let the bead slide from an intermediate point $\theta_0$. We have: :$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$ which leads us, via the same route as for Time of Travel down Brachistochrone, to: {{begin-eqn}} {{eqn | l = T       | r = \int_{\theta_0}^\pi \sqrt {\frac {2 a^2 \paren {1 - \cos \theta} } {2 g a \paren {\cos \theta_0 - \cos \theta} } } \rd \theta       | c =  }} {{eqn | r = \sqrt {\frac a g} \int_{\theta_0}^\pi \sqrt {\frac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta       | c =  }} {{end-eqn}} Using the Half Angle Formula for Cosine and Half Angle Formula for Sine, this gives: :$\displaystyle T = \sqrt{\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$ Now we make the substitution: {{begin-eqn}} {{eqn | l = u       | r = \frac {\map \cos {\theta / 2} } {\map \cos {\theta_0 / 2} }       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d \theta}       | r = -\frac {\map \sin {\theta / 2} } {2 \map \cos {\theta_0 / 2} }       | c =  }} {{end-eqn}} Recalculating the limits: :when $\theta = \theta_0$ we have $u = 1$ :when $\theta = \pi$ we have $u = 0$. So: {{begin-eqn}} {{eqn | l = T       | r = -2 \sqrt {\frac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }       | c =  }} {{eqn | r = \intlimits {2 \sqrt {\frac a g} \sin^{-1} u} 0 1       | c =  }} {{eqn | r = \pi \sqrt {\frac a g}       | c =  }} {{end-eqn}} Thus the time to slide down a brachistochrone from any arbitrary point $\theta_0$ is: :$T = \pi \sqrt {\dfrac a g}$ {{qed}}
Define the set of subsets of $\R$ as: :$\mathcal F := \left\{{\left\{{x}\right\}: x \in \Q}\right\}$ By Closed Real Interval is Closed Set: :$\forall x \in \Q: \left[{x \,.\,.\, x}\right] = \left\{{x}\right\}$ is closed (in topological sense) Then: :$\forall A \in \mathcal F: A$ is closed By Cardinality of Set of Singletons: :$\left\vert{\mathcal F}\right\vert = \left\vert{\Q}\right\vert$ where $\left\vert{\mathcal F}\right\vert$ denotes the cardinality of $\mathcal F$. By Rational Numbers are Countably Infinite: :$\Q$ is countable. Therefore by Set is Countable if Cardinality equals Cardinality of Countable Set: :$\mathcal F$ is countable. By Union of Set of Singletons: :$\bigcup \mathcal F = \Q$ Thus, by definition, $\Q$ is an $F_\sigma$ set. {{qed}}