hcju/proofwikiqa huggingface dataset:
By Lower Closure of Element is Closed under Directed Suprema: :$x^\preceq$ is closed under directed suprema. By Lower Closure of Singleton: :$\left\{ {x}\right\}^\preceq = x^\preceq$ By Lower Closure is Lower Set: :$x^\preceq$ is a lower set. Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set: :$x^\preceq$ is closed. {{qed}}
By Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set: :$x^\preceq$ is closed. By definition of closed set: :$\relcomp S {x^\preceq} \in \tau$ Thus by definition: :$\relcomp S {x^\preceq}$ is a open set. {{qed}}
=== Sufficient Condition === Let $A$ be open. Thus by definition of Scott topology: :$A$ is an upper set. Let $D$ be a directed subset of $S$ such that :$\sup D \in A$ By definition of Scott topology: :$A$ is inaccessible by directed suprema. By definition of inaccessible by directed suprema: :$A \cap D \ne \varnothing$ By definition of non-empty: :$\exists y: y \in A \cap D$ By definition of intersection: :$y \in A$ and $y \in D$ Thus $y \in D$. Thus by definition of upper set: :$\forall x \in D: y \preceq x \implies x \in A$ {{qed|lemma}} === Necessary Condition === Assume that :$A$ is upper and with property (S). According to the definition of Scott topology it should be proved that :$A$ is upper and inaccessible by directed suprema. Thus by assumption: :$A$ is upper. Let $D$ be a directed subset of $S$ such that :$\sup D \in A$ By definition of property (S): :$\exists y \in D: \forall x \in D: y \preceq x \implies x \in A$ By definition of reflexivity: :$y \preceq y$ Then :$y \in A$ By definition of intersection: :$y \in A \cap D$ Thus by definition of non-empty: :$A \cap D \ne \varnothing$ {{qed}}
Rewrite $J \sqbrk y$ as a sum of two functionals: {{begin-eqn}} {{eqn | l = J \sqbrk y       | r = \int_a^b \map F {x, y, y'} \rd x       | c = }} {{eqn | r = \int_a^c \map F {x, y, y'} \rd x + \int_c^b \map F {x, y, y'} \rd x }} {{eqn | r = J_1 \sqbrk y + J_2 \sqbrk y }} {{end-eqn}} Recall that end points $x = a,x = b$ are fixed. The function $\map y x$ has to be $C^0$ at $x = c$, but otherwise this point can move freely. From general variation of functional, and noting that $y = \map y x$ is an extremal, write down variations for $J_1 \sqbrk y$ and $J_2 \sqbrk y$ separately: $\ds \delta J_1 = \bigvalueat {F_{y'} } {x \to c \mathop - 0} \delta y_1 + \bigvalueat {\paren {F - y' F_{y'} } } {x \to c \mathop - 0} \delta x_1$ $\ds \delta J_2 = \bigvalueat {-F_{y'} } {x \to c \mathop + 0} \delta y_1 - \bigvalueat {\paren {F - y' F_{y'} } } {x \to c \mathop + 0} \delta x_1$ Note that $\delta J_1$ and $\delta J_2$ involve the same increments $\delta x_1$ and $\delta y_1$. {{explain|Why}} Since $y = \map y x$ is an extremum of $J$, we have: {{begin-eqn}} {{eqn | l = \delta J       | r = \delta J_1 + \delta J_2 }} {{eqn | r = \bigvalueat {F_{y'} } {x \mathop = c \mathop - 0} \delta y_1 + \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop - 0} \delta x_1 - \bigvalueat {F_{y'} } {x \mathop = c \mathop + 0} \delta y_1 - \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop + 0} \delta x_1       | c = }} {{eqn | r = \bigvalueat {\paren{F_{y'} } {x \mathop = c \mathop - 0} - \bigvalueat {F_{y'} } {x \mathop = c \mathop + 0} } \delta y_1 + \paren {\bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop - 0} - \bigvalueat {\paren {F - y' F_{y'} } } {x \mathop = c \mathop + 0} } }} {{eqn | r = 0 }} {{end-eqn}} Since $ \delta x_1$ and $ \delta y_1$ are arbitrary, both collections of terms have to vanish independently. {{qed}}
Find the full differential of Hamiltonian: {{begin-eqn}} {{eqn | l = \rd H       | r = -\rd F + \rd {\sum_{i \mathop = 1}^n y_i' p_i}       | c = {{Defof|Hamiltonian}} }} {{eqn | r = -\rd F + \sum_{i \mathop = 1}^n \paren {\rd {y_i'} p_i + y_i' \rd p_i}       | c = Full differential of a product }} {{eqn | r = -\frac {\partial F} {\partial x} \rd x - \sum_{i \mathop = 1}^n \frac {\partial F} {\partial y_i} \rd y_i - \sum_{i \mathop = 1}^n \frac {\partial F} {\partial y_i'} \rd y_i' + {\sum_{i \mathop = 1}^n \rd y_i' p_i} + \sum_{i \mathop = 1}^n y_i'\rd p_i       | c = {{Defof|Differential of Real-Valued Function}} }} {{eqn | r = -\frac {\partial F} {\partial x} \rd x - \sum_{i \mathop = 1}^n \frac {\partial F} {\partial y_i} \rd y_i - \sum_{i \mathop = 1}^n p_i \rd y_i' + {\sum_{i \mathop = 1}^n \rd y_i' p_i} + \sum_{i \mathop = 1}^n y_i'\rd p_i       | c = {{Defof|Canonical Variable}}: $p_i$ }} {{eqn | r = -\frac {\partial F} {\partial x} \rd x - \sum_{i \mathop = 1}^n \frac {\partial F} {\partial y_i} \rd y_i + \sum_{i \mathop = 1}^n y_i' \rd p_i       | c = Terms with $p_i \rd y_i'$ cancel  }} {{eqn | r = \frac {\partial H} {\partial x} \rd x + \sum_{i \mathop = 1}^n \frac {\partial H} {\partial y_i} \rd y_i + \sum_{i \mathop = 1}^n \frac {\partial H} {\partial p_i} \rd p_i       | c = {{Defof|Differential of Real-Valued Function}} }} {{end-eqn}} By equating coefficients of differentials in last two equations we find that: :$\dfrac {\partial H} {\partial x} = -\dfrac {\partial F} {\partial x}$ :$\dfrac {\partial H} {\partial y_i} = -\dfrac {\partial F} {\partial y_i}$ :$\dfrac {\partial H} {\partial p_i} = y_i'$ From the third identity it follows that: :$\paren {\dfrac {\d y_i} {\d x} = y_i} \implies \paren {\dfrac {\d y_i} {\d x} = \dfrac {\partial H} {\partial p_i} }$ while the second identity together with the definition of $p_i$ assures that: :$\paren {\dfrac {\partial F} {\partial y_i} - \dfrac \d {\d x} \dfrac {\partial F} {\partial y_i} = 0} \implies \paren {\dfrac {\d p_i} {\d x} = -\dfrac {\partial H} {\partial y_i} }$ {{qed}} {{Namedfor|Leonhard Paul Euler|cat = Euler}}
We will prove that :$S$ has property (S). Let $D$ be a directed subset of $S$ such that :$\sup D \in S$ By definition of non-empty set: :$\exists y: y \in D$ Thus $y \in D$. Thus by definition of subset: :$\forall x \in D: y \preceq x \implies x \in S$ {{qed|lemma}} Then $(O3): \quad S \in \tau$ We will prove that :$(O1): \quad \forall F \subseteq \tau: \bigcup F \in \tau$ Let $F \subseteq \tau$. To prove that $\bigcup F \in \tau$ it should be proved that :$\bigcup F$ has property (S). Let $D$ be a directed subset of $S$ such that :$\sup D \in \bigcup F$ By definition of union: :$\exists X \in F: \sup D \in X$ By definition of subset: :$X \in \tau$ By assumption: :$X$ has property (S). By definition of property (S): :$\exists y \in D: \forall x \in D: y \preceq x \implies x \in X$ Thus $y \in D$. Let $x \in D$ such that :$y \preceq x$ Then $x \in X$. Thus by definition of union: :$x \in \bigcup F$ {{qed|lemma}} We will prove that :$(O2): \quad \forall X, Y \in \tau: X \cap Y \in \tau$ Let $X, Y \in \tau$. By assumption: :$X$ and $Y$ have property (S). To prove that $X \cap Y \in \tau$ it should be proved that :$X \cap Y$ has property (S). Let $D$ be a directed subset of $S$ such that :$\sup D \in X \cap Y$ By definition of intersection: :$\sup D \in X$ and $\sup D \in Y$ By definition of property (S): :$\exists x \in D: \forall z \in D: x \preceq z \implies z \in X$ and :$\exists y \in D: \forall z \in D: y \preceq z \implies z \in X$ By definition of directed: :$\exists z \in D: x \preceq z \land y \preceq z$ Thus $z \in D$. Let $u \in D$ such that :$z \preceq u$ By definition of transitivity: :$x \preceq u$ and $y \preceq u$ Then :$u \in X$ and $u \in Y$ Thus by definition of intersection: :$u \in X \cap Y$ {{qed|lemma}} Hence by $(O1)$, $(O2)$, $(O3)$, and definition: :$\left({S, \tau}\right)$ is topological space. {{qed}}
If $G$ is trivial, then $G$ is discrete and cyclic. Let $G$ be non-trivial. Because $x \in G \iff -x \in G$, $G$ has a strictly positive element. Thus $G \cap \R_{>0}$ is non-empty. Let $a = \map \inf {G \cap \R_{>0} }$. === Case 1 === Suppose $a = 0$. We will show that $G$ is dense in $\R$. Let $x \in \R$ and $\epsilon > 0$. Because $a = 0$, there exists $g \in G$ such that $0 < g \le \epsilon$. Then: :$y = g \cdot \floor {\dfrac x g} \in G$, where $\floor {\, \cdot \,}$ denotes the floor function. We have: {{begin-eqn}} {{eqn | l = \size {x - y}       | r = x - g \cdot \floor {\dfrac x g}       | c =  }} {{eqn | r = g \cdot \paren {\frac x g - \floor {\frac x g} }       | c =  }} {{eqn | o = \le       | r = g \cdot 1       | c = Real Number minus Floor }} {{eqn | o = \le       | r = \epsilon       | c =  }} {{end-eqn}} Thus $G$ is dense in $\R$. === Case 2 === Suppose $a > 0$. We show that $G = a \Z$, from which it follows that $G$ is discrete. Let $g \in G$. Then: :$g - a \cdot \floor {\dfrac g a} \in G$ By Real Number minus Floor, we have: {{begin-eqn}} {{eqn | l = 0 \le g - a \cdot \floor {\frac g a}       | r =a \cdot \paren {\dfrac g a - \floor {\dfrac g a} }        | c =  }} {{eqn | o = <       | r = a       | c =  }} {{end-eqn}} Because $a = \map \inf {G \cap \R_{>0} }$, we have: :$g = a \cdot \floor {\dfrac g a}$ Thus: :$g \in a \Z$ {{qed}} Category:Group Theory Category:Topological Groups Category:Real Numbers mytqtos589vo5gt4xsxmpvcd07c3elf
By Subgroup of Real Numbers is Discrete or Dense, there exists $a\in \R$ such that $G=a\Z$. If $a=0$, then $G$ is closed. Let $a>0$. Then: :$\displaystyle\R\setminus G = \bigcup_{z\in \Z}(az,az+a)$ By Union of Open Sets of Metric Space is Open, $\R\setminus G$ is open. Thus $G$ is closed. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\d \Phi} {\d x}       | r = \frac {\partial\Phi} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} \frac {\partial y_i} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial p_i} \frac{\partial p_i} {\partial x} }} {{eqn | r = \frac {\partial \Phi} {\partial x} + \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} \frac {\partial H} {\partial p_i} - \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial p_i} \frac{\partial H} {\partial y_i} }} {{eqn | r = \frac {\partial \Phi} {\partial x} + \sqbrk{\Phi, H} }} {{end-eqn}} For $\Phi$ to be the first integral: :$\dfrac {\d \Phi} {\d x} = 0$ Hence the result. {{qed}} {{finish|get proof for the other condition}}
By definition of generator set of filter: :$F = \left({\operatorname{fininfs}\left({G}\right)}\right)^\succeq$ where :$\operatorname{fininfs}\left({G}\right)$ denotes the finite infima set of $G$, :$A^\succeq$ denotes the upper closure of $A$. By Finite Infima Set of Coarser Subset is Coarser than Finite Infima Set: :$\operatorname{fininfs} \left({G}\right)$ is coarser than $\operatorname{fininfs} \left({A}\right)$ Thus by Upper Closure of Coarser Subset is Subset of Upper Closure: :$F \subseteq \left({\operatorname{fininfs} \left({A}\right)}\right)^\succeq$ By filter in ordered set: :$F$ is an upper set. By Set Coarser than Upper Set is Subset: :$A \subseteq F$ Thus by Finite Infima Set and Upper Closure is Smallest Filter: :$\left({\operatorname{fininfs}\left({A}\right)}\right)^\succeq \subseteq F$ Thus by definition of set equality: :$F = \left({\operatorname{fininfs}\left({A}\right)}\right)^\succeq$ Hence $A$ is generator set of $F$. {{qed}}
Let $F: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ be a homotopy between $f_1$ and $f_2$. Let $G: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ be a homotopy between $g_1$ and $g_2$. Define $H: \left[{0 \,.\,.\, 1}\right] \times \left[{0 \,.\,.\, 1}\right] \to S$ by: :$H \left({s, t}\right) = \begin{cases} F \left({2 s, t}\right) & : s \in \left[{0 \,.\,.\, \dfrac 1 2}\right] \\ G \left({2 s - 1, t}\right) & : s \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]\end{cases}$ By Continuous Mapping on Finite Union of Closed Sets, $H$ is continuous. By definition of concatenationof paths, $H$ is a homotopy between $f_1 * g_1$ and $f_2  g_2$. {{qed}}
Let $x \in \operatorname{fininfs} \left({A}\right)$ By definition of finite infima set: :$\exists Y \in \mathit{Fin}\left({A}\right): x = \inf Y$ and $Y$ admits an infimum, where $\mathit{Fin}\left({A}\right)$ denotes the set of all finite subsets of $A$. By definition of coarset subset: :$\forall y \in Y: \exists z \in B: z \preceq y$ By Axiom of Choice: :$\exists f:Y \to B: \forall y \in Y: f\left({y}\right) \preceq y$ We will prove that :$f\left[{Y}\right]$ admits an infimum. In the case when $Y = \varnothing$: By Image of Empty Set is Empty Set/Corollary: :$f\left[{Y}\right] = \varnothing$ Thus :$f\left[{Y}\right]$ admits an infimum. In the case when $Y \ne \varnothing$ By definition of non-empty set: :$\exists y: y \in Y$ By definition of image of set: :$f\left({y}\right) \in f\left[{Y}\right]$ By Image of Mapping from Finite Set is Finite: :$f\left[{Y}\right]$ is a finite set. Thus by Existence of Non-Empty Finite Infima in Meet Semilattice: :$f\left[{Y}\right]$ admits an infimum. {{qed|lemma}} We will prove that :$\inf f\left[{Y}\right]$ is lower bound for $Y$. Let $y \in Y$. By definition of $f$: :$f\left({y}\right) \preceq y$ By definition of image of set: :$f\left({y}\right) \in f\left[{Y}\right]$ By definitions of infimum and lower bound: :$\inf f\left[{Y}\right] \preceq f\left({y}\right)$ Thus by definition of transitivity: :$\inf f\left[{Y}\right] \preceq y$ {{qed|lemma}} By definition of infimum: :$\inf f\left[{Y}\right] \preceq x$ By definition of finite infima set: :$\inf f\left[{Y}\right] \in \operatorname{fininfs} \left({B}\right)$ Thus :$\exists y \in \operatorname{fininfs} \left({B}\right): y \preceq x$ {{qed}}
Let $\map f x$ be a strictly convex real function. Let $p = \map {f'} x$. By definition of the Legendre transform, the transformed real function is of the form: :$\map {f^*} p = - \map f {\map x p} + p \map x p$ By Legendre Transform of Strictly Convex Real Function is Strictly Convex, $f^*$ is strictly convex. Let $t = \map { {f^*}'} p$. Let $\map {\paren {f^*}^*} t = - \map {f^*} {\map p t} + t \map p t$. Then: {{begin-eqn}} {{eqn | l = \paren {f^*}^*       | r = -f^* + t p  }} {{eqn | r = -\paren {-f + p t} + t p  }} {{eqn | r = f }} {{end-eqn}} $t$ is an arbitrary independent variable, hence can be renamed. Set $t = x$. Then: :$\tuple {t,f^{**} } = \tuple {x, f}$ which is the original pair of function and its variable. Hence, by definition, the Legendre transform is an involution. {{qed}}
Let $x \in A^\succeq$ By definition of upper closure of subset: :$\exists y \in A: y \preceq x$ By definition of coarser subset: :$\exists z \in B: z \preceq y$ By definition of transitivity: :$z \preceq x$ Thus by definition of upper closure of subset: :$x \in B^\succeq$ {{qed}}
Let $x \in A$. By definition of coarser subset: :$\exists y \in B: y \preceq x$ Thus by definition of upper set: :$x \in B$ {{qed}}
Function $g$ acquires an extremum along $p$, when its first derivative along $p$ vanishes: {{begin-eqn}} {{eqn | l = \frac {\partial g} {\partial p}         | r = -\frac {\partial f^*} {\partial p} + x }} {{eqn | r = 0       | c = Extremum condition }} {{eqn | ll= \leadsto       | l = \map { {f^*}'} p       | r = x }} {{end-eqn}} To check if the extremum is a global maximum, consider the second derivative: :$\dfrac {\partial^2 g} {\partial p^2} = - \dfrac {\partial^2 f^*} {\partial p^2}$ By Legendre Transform of Strictly Convex Real Function is Strictly Convex and Real Function is Strictly Convex iff Derivative is Strictly Increasing, it holds that: {{begin-eqn}} {{eqn | l = \frac {\partial^2 f^*} {\partial p^2}       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \frac {\partial^2 g} {\partial p^2}       | o = <       | r = 0 }} {{eqn | ll= \leadsto       | l = \valueat {\frac {\partial^2 g} {\partial p^2} } {\frac {\partial g} {\partial p} \mathop = 0}       | o = <       | r = 0 }} {{end-eqn}} Negative second derivative at the extremum implies extremum being a global maximum. Therefore: :$\displaystyle \max_p \paren{-\map {f^*} p + x p} = \bigvalueat {\paren {-\map {f^*} p + x p} } {x \mathop = \map { {f^*}'} p}$ The {{RHS}} is the Legendre Transform of $f^*$. However, $f^*$ is a Legendre transformed $f$. By Legendre Transform is Involution, this equals $f$. {{qed}} {{Finish|clean up local/global extrema ambiguity}}
Let $x \in f\left[{\N}\right]$. By definition of image of mapping: :$\exists n \in \N: x = f\left({n}\right)$ By definition of $g$: :$g\left({n}\right) = \inf \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$ By definition of reflexivity: :$n \le n$ Then :$f\left({n}\right) \in \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$ By definitions of infimum and lower bound: :$g\left({n}\right) \preceq x$ By definition of image of mapping: :$g\left({n}\right) \in g\left[{\N}\right]$ Thus :$\exists y \in g\left[{\N}\right]: y \preceq x$ {{qed}}
Consider the numbers of the form $2^m - 1$ which are triangular: {{begin-eqn}} {{eqn | l = 2^m - 1       | r = \frac {r \left({r + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | ll= \iff       | l = 8 \left({2^m - 1}\right)       | r = 4 r \left({r + 1}\right)       | c =  }} {{eqn | ll= \iff       | l = 2^{m + 3} - 8       | r = 4 r^2 + 4 r       | c =  }} {{eqn | n = 1       | ll= \iff       | l = 2^{m + 3} - 7       | r = 4 r^2 + 4 r + 1       | c =  }} {{eqn | r = \left({2 r + 1}\right)^2       | c =  }} {{end-eqn}} Let: :$n = m - 3$ :$x = 2 r + 1$ and it can be seen that $(1)$ is equivalent to: :$x^2 + 7 = 2^n$ From Solutions of Ramanujan-Nagell Equation: :$x = 1, 3, 5, 11, 181$ Setting $r = \dfrac {x - 1} 2$ it is seen that the corresponding triangular numbers are: :$\dfrac{\left({x - 1}\right) \left({x + 1}\right)} 8$ Thus the corresponding Ramanujan-Nagell numbers are: :$0, 1, 3, 15, 4095$ {{qed}} Category:Ramanujan-Nagell Equation Category:5 2rxxepfalugnsewvv9inswphac8pxvi
Euler's equation for $J \sqbrk {y, p}$: {{begin-eqn}} {{eqn | l = \frac {\delta J \sqbrk{y, p} } {\delta p}       | r = \frac {\partial} {\partial p} \paren {-\map H {x, y, p} + p y'}       | c = Depends only on $p$ and not its derivatives }} {{eqn | r = -\frac {\partial H} {\partial p} + y' }} {{eqn | r = 0 }} {{eqn | ll= \leadsto       | l = y'       | r = \frac {\partial H} {\partial p} }} {{end-eqn}} Substitute this result back into the functional $J \sqbrk {y, p}$: {{begin-eqn}} {{eqn | l = \bigvalueat {J \sqbrk {y, p} } {\frac {\delta J \sqbrk {y, p} } {\delta p} \mathop = 0}       | r = \int_a^b \paren {-H + p \frac{\partial H}{\partial p} } \rd x }} {{eqn | r = \int_a^b \paren {\map F {x, y, y'} - p y' + p \frac {\partial \paren {-\map F {x, y, y'} + p y'} } {\partial p} } \rd x       | c = {{Defof|Hamiltonian}} }} {{eqn | r = \int_a^b \paren {\map F {x, y, y'} - p y' + p y'} \rd x }} {{eqn | r = \int_a^b \map F {x, y, y'} \rd x }} {{end-eqn}} {{qed}}
By Set is Coarser than Image of Mapping of Infima: :$f\left[{\N}\right]$ is coarser than $g\left[{\N}\right]$. By definition of generator set of filter: :$F = \left({\operatorname{fininfs}\left({f\left[{\N}\right]}\right)}\right)^\succeq$ where :$\operatorname{fininfs}\left({f\left[{\N}\right]}\right)$ denotes the finite infima set of $f\left[{\N}\right]$, :for subset $A$ of $S$: $A^\succeq$ denotes upper closure of $A$. By Finite Infima Set and Upper Closure is Smallest Filter: :$f\left[{\N}\right] \subseteq F$ We will prove that :$g\left[{\N}\right] \subseteq F$ Let $x \in g\left[{\N}\right]$. By definition of image of mapping: :$\exists n \in \N: x = g\left({n}\right)$ By definition of $g$: :$x = \inf \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$ By definition: :$\left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$ is a finite set. By definition of subset: :$\left\{ {f\left({m}\right): m \in \N \land m \le n}\right\} \subseteq F$ By definition of reflexivity: :$f\left({n}\right) \in \left\{ {f\left({m}\right): m \in \N \land m \le n}\right\}$ Thus by Filtered in Meet Semilattice with Finite Infima: :$x \in F$ {{qed|lemma}} By Subset of Set is Coarser than Set :$g\left[{\N}\right]$ is coarser than $F$. Thus by Coarser Between Generator Set and Filter is Generator Set of Filter: :$g\left[{\N}\right]$ is generator set of $F$. {{qed}}
Let $x \in A$. By definition of subset: :$x \in B$ By definition of reflexivity: :$x \preceq x$ Thus :$\exists y \in B: y \preceq x$ {{qed}}
Let $16 \le n \le 29$. Then: :$n < 30 < 2 n$ and we have: :$30 = 2 \times 3 \times 5$ which is a product of $3$ distinct primes. Hence the result holds for $n$ in that range. Let $n \ge 30$. Then by the Division Theorem: :$\exists q, r \in \N: n = 6 q + r$, $0 \le r < 6$, $q \ge 5$ By Bertrand-Chebyshev Theorem, there is a prime $p$ where $5 \le q < p < 2 q$. Hence $p$ is not $2$ or $3$, and: {{begin-eqn}} {{eqn | l = n       | o = <       | r = 6 \paren {q + 1}       | c =  }} {{eqn | o = \le       | r = 2 \times 3 \times p }} {{eqn | o = \le       | r = 12 q }} {{eqn | o = \le       | r = 2 n }} {{end-eqn}} This proves the result. {{qed}}
Recall Pascal's triangle: {{:Definition:Pascal's Triangle}} By definition, the entry in row $n$ and column $m$ contains the binomial coefficient $\dbinom n m$. Thus the $2$nd column contains all the elements of the form $\dbinom n 2$. The $m$th diagonal consists of the elements in column $n - m$. Thus the $m$th diagonal contains the binomial coefficients $\dbinom n {n - m}$. By Symmetry Rule for Binomial Coefficients: :$\dbinom n {n - m} = \dbinom n m$ Thus the $2$nd diagonal also contains the binomial coefficients $\dbinom n 2$. By Binomial Coefficient with Two: Corollary, the triangular numbers are precisely those numbers of the form $\dbinom n 2$. Hence the result. {{qed}}
Assume that :$\complement_S\left({A}\right)$ is irreducible. Let $x, y \in \tau$ such that :$A = x \wedge y$ By definition of topological space: :$x \cap y \in \tau$ By Meet in Inclusion Ordered Set: :$x \wedge y = x \cap y$ By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection: :$\complement_S\left({A}\right) = \complement_S\left({x}\right) \cup \complement_S\left({y}\right)$ By definition: :$\complement_S\left({x}\right)$ and $\complement_S\left({y}\right)$ are closed. By definition of irreducible: :$\complement_S\left({A}\right) = \complement_S\left({x}\right)$ or $\complement_S\left({A}\right) = \complement_S\left({y}\right)$ Thus by Relative Complement of Relative Complement: :$A = x$ or $A = y$ {{qed}}
{{begin-eqn}} {{eqn | l = T_n + 2 T_{n - 1}       | r = \frac {n \paren {n + 1} } 2 + 2 \frac {\paren {n - 1} n} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {n^2 + n + 2 \paren {n^2 - n} } 2       | c =  }} {{eqn | r = \frac {3 n^2 - n} 2       | c =  }} {{eqn | r = \frac {3 n \paren {n - 1} } 2       | c =  }} {{eqn | r = P_n       | c = Closed Form for Pentagonal Numbers }} {{end-eqn}} {{qed}}
Since $n^2 < n \paren {n + 1} < \paren {n + 1}^2$, $n \paren {n + 1}$ cannot be a square number. Thus there are infinitely many distinct integer solutions to the Pell's Equation: :$x^2 - n \paren {n + 1} y^2 = 1$ and for each solution: {{begin-eqn}} {{eqn | l = T_n T_{x^2 - 1}       | r = \frac {n \paren {n + 1} } 2 \times \frac {x^2 \paren {x^2 - 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {n \paren {n + 1} } 2 \times \frac {x^2 n \paren {n + 1} y^2 } 2 }} {{eqn | r = \paren {\frac {x y n \paren {n + 1} } 2}^2 }} {{end-eqn}} Hence the result. {{qed}}