hcju/proofwikiqa huggingface dataset:
:400px Let $P = \tuple {x, y}$ be a general point on $C$. Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$. From the above diagram: :$x - X = \pm \rho \sin \psi$ :$Y - y = \pm \rho \cos \psi$ where: :$\rho$ is the radius of curvature of $C$ at $P$ :$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis. Whether the sign is plus or minus depends on whether the curve is convex or concave. By definition of radius of curvature: :$(1): \quad \begin {cases} x - X = \dfrac 1 k \sin \psi \\ Y - y = \dfrac 1 k \cos \psi \end {cases}$ where $k$ is the curvature of $C$ at $P$, given by: :$k = \dfrac {x' y'' - y' x''} {\paren {x'^2 + y'^2}^{3/2} }$ We have that: :$\sin \psi = \dfrac {\d y} {\d s} = \dfrac {y'} {\sqrt {x'^2 + y'^2} }$ :$\cos \psi = \dfrac {\d x} {\d s} = \dfrac {x'} {\sqrt {x'^2 + y'^2} }$ Substituting for $k$ and $\psi$ in $(1)$ gives: {{begin-eqn}} {{eqn | l = x - X       | r = \dfrac {\paren {x'^2 + y'^2}^{3/2} } {x' y'' - y' x''} \dfrac {y'} {\sqrt {x'^2 + y'^2} }       | c =  }} {{eqn | r = \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''}       | c =  }} {{eqn | ll= \leadsto       | l = X       | r = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = Y - y       | r = \dfrac {\paren {x'^2 + y'^2}^{3/2} } {x' y'' - y' x''} \dfrac {x'} {\sqrt {x'^2 + y'^2} }       | c =  }} {{eqn | r = \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''}       | c =  }} {{eqn | ll= \leadsto       | l = Y       | r = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''}       | c =  }} {{end-eqn}} {{qed}}
From Parametric Equations for Evolute: Formulation 1: :$\begin {cases} X = x - \dfrac {y' \paren {1 + y'^2} } {y''} \\ Y = y + \dfrac {1 + y'^2} {y''} \end{cases}$ where: :$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$ :$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$ :$y'$ and $y''$ denote the derivative and second derivative respectively of $y$ {{WRT|Differentiation}} $x$. Thus we have: {{begin-eqn}} {{eqn | l = y'       | r = 2 x       | c =  }} {{eqn | l = y''       | r = 2       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = X       | r = x - \dfrac {2 x \paren {1 + \paren {2 x}^2} } 2       | c =  }} {{eqn | r = x - x - 4 x^3       | c =  }} {{eqn | r = -4 x^3       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = Y       | r = x^2 + \dfrac {1 + \paren {2 x}^2} 2       | c =  }} {{eqn | r = \frac {2 x^2 + 1 + 4 x^2} 2       | c =  }} {{eqn | r = \frac {6 x^2 + 1} 2       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = Y       | r = \frac {6 \paren {-\frac X 4}^{2/3} + 1} 2       | c = substituting for $X$ }} {{eqn | l = Y       | r = 3 \paren {-\frac X 4}^{2/3} + \frac 1 2       | c = substituting for $X$ }} {{eqn | ll= \leadsto       | l = \frac 1 3 \paren {Y - \frac 1 2}       | r = \paren {\frac {X^2} {16} }^{1/3}       | c =  }} {{eqn | ll= \leadsto       | l = 27 X^2       | r = 16 \paren {Y - \dfrac 1 2}^3       | c =  }} {{end-eqn}} The parabola (blue) and its evolute (red) are illustrated below: 400px {{qed}}
:400px Let $P = \left({x, y}\right)$ be a general point on $C$. Let $Q = \left({X, Y}\right)$ be the center of curvature of $C$ at $P$. From the above diagram: :$(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\ Y - y = \pm \rho \cos \psi \end{cases}$ where: :$\rho$ is the radius of curvature of $C$ at $P$ :$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis. Whether the sign is plus or minus depends on whether the curve is convex or concave. For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive. The case for $k < 0$ can then be treated similarly. Thus we have $k > 0$ and so $(1)$ can be written: :$(2): \quad \begin{cases} X = x - \rho \sin \psi \\ Y = y +\rho \cos \psi \end{cases}$ By definition of curvature: :$k = \dfrac {\mathrm d \psi} {\mathrm d s}$ and: :$\rho = \dfrac 1 k = \dfrac {\mathrm d s} {\mathrm d \psi}$ Hence: {{begin-eqn}} {{eqn | l = \rho \sin \psi       | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d y} {\mathrm d s}       | c =  }} {{eqn | n = 3       | r = \dfrac {\mathrm d y} {\mathrm d \psi}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \rho \cos \psi       | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d x} {\mathrm d s}       | c =  }} {{eqn | n = 4       | r = \dfrac {\mathrm d x} {\mathrm d \psi}       | c =  }} {{end-eqn}} Differentiating $(2)$ {{WRT|Differentiation}} $\psi$: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d X} {\mathrm d \psi}       | r = \dfrac {\mathrm d x} {\mathrm d \psi} - \rho \cos \psi - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi       | c =  }} {{eqn | r = - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi       | c = from $(3)$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d Y} {\mathrm d \psi}       | r = \dfrac {\mathrm d y} {\mathrm d \psi} - \rho \sin \psi + \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi       | c =  }} {{eqn | r = \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi       | c = from $(4)$ }} {{end-eqn}} By assumption, $\dfrac {\mathrm d \rho} {\mathrm d \psi} \ne 0$ on $C$. Hence we have: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d Y} {\mathrm d X}       | r = -\dfrac {\cos \psi} {\sin \psi}       | c =  }} {{eqn | r = -\dfrac 1 {\tan \psi}       | c =  }} {{end-eqn}} We have Slope of Normal is Minus Reciprocal of Tangent. Thus the slope of the tangent to $E$ equals the slope of the normal. The result follows. Note the case when $\dfrac {\mathrm d \rho} {\mathrm d \psi} \ne 0$ on $C$. In this case $C$ is a circle. By Evolute of Circle is its Center its evolute is a single point and so has no tangent. {{qed}}
From Length of Arc of Evolute equals Difference in Radii of Curvature: :the length of arc of the evolute $E$ of $C$ between any two points $Q_1$ and $Q_2$ of $C$ is equal to the difference between the radii of curvature at the corresponding points $P_1$ and $P_2$ of $C$. Thus $C$ exhibits precisely the property of the involute of $E$. {{qed}}
Let $C$ be the cycloid defined by the equations: :$\begin {cases} x = a \paren {\theta - \sin \theta} \\ y = a \paren {1 - \cos \theta} \end {cases}$ From Parametric Equations for Evolute: Formulation 2: :$(1): \quad \begin{cases} X = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''} \\ Y = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''} \end{cases}$ where: :$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$ :$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$ :$x'$ and $x''$ denote the derivative and second derivative respectively of $x$ {{WRT|Differentiation}} $\theta$ :$y'$ and $y''$ denote the derivative and second derivative respectively of $y$ {{WRT|Differentiation}} $\theta$. Thus we have: {{begin-eqn}} {{eqn | l = x'       | r = a \paren {1 - \cos \theta}       | c =  }} {{eqn | l = x''       | r = a \sin \theta       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y'       | r = a \sin \theta       | c =  }} {{eqn | l = y''       | r = a \cos \theta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \dfrac {x'^2 + y'^2} {x' y'' - y' x''}       | r = \dfrac {a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta} {a \paren {1 - \cos \theta} a \cos \theta - a \sin \theta \, a \sin \theta}       | c = substituting for $x'$, $x''$, $y'$, $y''$ }} {{eqn | r = \dfrac {\paren {1 - 2 \cos \theta + \cos^2 \theta} + \sin^2 \theta} {\cos \theta - \cos^2 \theta - \sin^2 \theta}       | c = multiplying out, cancelling out $a^2$ }} {{eqn | r = \dfrac {2 - 2 \cos \theta} {\cos \theta - 1}       | c = Sum of Squares of Sine and Cosine }} {{eqn | n = 2       | r = -2       | c = dividing top and bottom by $1 - \cos \theta$ }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = X       | r = a \paren {\theta - \sin \theta} - a \sin \theta \paren {-2}       | c = substituting for $x$ and $y'$ and from $(2)$ in $(1)$ }} {{eqn | r = a \paren {\theta + \sin \theta}       | c = simplifying }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = Y       | r = a \paren {1 - \cos \theta} + a \paren {1 - \cos \theta} \paren {-2}       | c = substituting for $y$ and $x'$ and from $(2)$ in $(1)$ }} {{eqn | r = -a \paren {1 - \cos \theta}       | c =  }} {{end-eqn}} The cycloid $C$ (blue) and its evolute (red) are illustrated below: 600px {{qed}}
Let $a, b \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$. Then :$\exists x \in D_1, y \in D_2: a = x \wedge y$ and :$\exists z \in D_1, t \in D_2: b = z \wedge t$ By definition of directed subset: :$\exists g \in D_1: x \preceq g \land z \preceq g$ and :$\exists h \in D_2: y \preceq h \land t \preceq h$ By Meet Semilattice is Ordered Structure: :$x \wedge y \preceq g \wedge h$ and $z \wedge t \preceq g \wedge h$ :$g \wedge h \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$ Thus :$\exists c \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}: a \preceq c \land b \preceq c$ Hence by definition :$\left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$ is directed. {{qed}}
Let $x, y \in f^\to\left({D}\right)$. By definition of image of set: :$\exists a \in D: x = f \left({a}\right)$ and :$\exists b \in D: y = f \left({b}\right)$ By definition of directed subset: :$\exists c \in D: a \preceq c \land b \preceq c$ By definition of image of set: :$f\left({c}\right) \in f^\to\left({D}\right)$ By definition of increasing mapping: :$x \precsim f\left({c}\right)$ and $y \precsim f\left({c}\right)$ Thus by definition :$f^\to\left({D}\right)$ is a directed subset of $T$. {{qed}}
From Kepler's Third Law of Planetary Motion: :$T^2 = \left({\dfrac {4 \pi^2} {G M} }\right) a^3$ where: :$T$ is the orbital period of the planet in question (in this case, the Earth) :$a$ is the distance from the planet (in this case, the Earth) to the sun :$M$ is the mass of the sun :$G$ is the gravitational constant In MKS units: :$T = 60 \times 60 \times 24 \times 365.24219 \, \mathrm s$ by definition of year :$a = 149 \, 597 \, 870 \, 700 \, \mathrm m$ by definition of astronomical unit :$G = 6.674 \times 10^{-11} \, \mathrm N \, \mathrm m^2 \, \mathrm{kg}^{-2}$ by measurement, as by hypothesis. :$M$ will be the result in kilograms. Thus: {{begin-eqn}} {{eqn | l = M       | r = \dfrac {4 \pi^2 a^3} {T^2 G}       | c =  }} {{eqn | r = \frac {4 \times \left({3.14159}\right)^2 \times \left({149.60 \times 10^9}\right)^3} {\left({31.557 \times 10^6}\right) \times 6.674 \times 10^{-11} }       | c =  }} {{end-eqn}} The calculation has been left as an exercise for anyone who has the patience. {{qed}}
By Up-Complete Product: :$\struct {S \times S, \precsim}$ is up-complete. By Up-Complete Product/Lemma 1: :$D_1 \times D_2$ is directed subsets of $S \times S$ By definition of mapping preserves directed suprema: :$f$ preserves the supremum of $D_1 \times D_2$ By definition of up-complete: :$D_1 \times D_2$ admits a supremum and $D_1$ and $D_2$ admit suprema. Thus {{begin-eqn}} {{eqn | l = \paren {\sup D_1} \wedge \paren {\sup D_2}       | r = \map f {\sup D_1, \sup D_2}       | c = Definition of $f$ }} {{eqn | r = \map f {\map \sup {D_1 \times D_2} }       | c = Supremum of Cartesian Product }} {{eqn | r = \map \sup {\map {f^\to} {D_1 \times D_2} }       | c = Mapping Preserves Supremum }} {{eqn | r = \sup \set {\map f {x, y}: \tuple {x, y} \in D_1 \times D_2}       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | r = \sup \set {x \wedge y: x \in D_1, y \in D_2}       | c = Definition of $f$ }} {{end-eqn}} {{qed}}
Let $I, J \in \map {\it Ids} {\mathscr S}$ such that :$\set {I, J}$ admits an infimum in $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$. By definition of image of set: :$\map {f^\to} {\set {I, J} } = \set {\map f I, \map f J}$ Thus by definition of meet semilattice: :$\map {f^\to} {\set {I, J} }$ admits an infimum in $\mathscr S$. Thus: {{begin-eqn}} {{eqn | l = \map \inf {\map {f^\to} {\set {I, J} } }       | r = \map f I \wedge \map f J       | c = {{Defof|Meet (Order Theory)}} }} {{eqn | r = \paren {\sup I} \wedge \paren {\sup J}       | c = Definition of $f$ }} {{eqn | r = \sup \set {x \wedge y: x \in I, y \in J}       | c = Assumption }} {{eqn | r = \map \sup {I \wedge J}       | c = Meet in Set of Ideals }} {{eqn | r = \map f {I \wedge J}       | c = definition of $f$ }} {{eqn | r = \map f {\inf \set {I, J} }       | c = {{Defof|Meet (Order Theory)}} }} {{end-eqn}} {{qed}}
By definition of multplication: {{begin-eqn}} {{eqn | l = \mathbf i \times \mathbf j       | r = \mathbf k       | c =  }} {{eqn | l = \mathbf j \times \mathbf i       | r = -\mathbf k       | c =  }} {{end-eqn}} {{qed}}
Each of the norm axioms is examined in turn: === $N1$: Positive Definiteness === This is proved in Field Norm of Quaternion is Positive Definite. {{qed|lemma}} === $N2$: Multiplicativity === This is proved in Field Norm of Quaternion is Multiplicative. {{qed|lemma}} === $N3$: Triangle Inequality === For example: :$\map n {1 + 1} = 4 > 2 = \map n 1  + \map n 1$ and so $N3$ is not satisfied. {{qed|lemma}} Not all the norm axioms are fulfilled. Hence the result. {{qed}} Category:Quaternions Category:Norm Theory q26q9bmf8pl0w3600m2bv534c6iiqun
{{begin-eqn}} {{eqn | l = \map n {\mathbf x}       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \cmod {\mathbf x \overline {\mathbf x} }       | r = 0       | c ={{Defof|Field Norm of Quaternion}} }} {{eqn | ll= \leadstoandfrom       | l = a^2 + b^2 + c^2 + d^2       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a = 0, b = 0, c = 0, d = 0       | o =        | c =  }} {{eqn | ll= \leadstoandfrom       | l = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k       | r = \mathbf 0       | c =  }} {{end-eqn}} Hence $n$ is positive definite by definition. {{qed}}
{{begin-eqn}} {{eqn | l = \map n {\mathbf x \mathbf y}       | r = \mathbf x \, \mathbf y \ \overline {\mathbf x \, \mathbf y}       | c = {{Defof|Field Norm of Quaternion}} }} {{eqn | r = \mathbf x \, \mathbf y \paren {\overline {\mathbf y} \, \overline {\mathbf x} }       | c = Product of Quaternion Conjugates }} {{eqn | r = \mathbf x \paren {\mathbf y \paren {\overline {\mathbf y} \, \overline {\mathbf x} } }       | c =  }} {{eqn | r = \mathbf x \paren {\paren {\mathbf y \, \overline {\mathbf y} } \overline {\mathbf x} }       | c =  }} {{eqn | r = \mathbf x \paren {\map n {\mathbf y} \overline {\mathbf x} }       | c =  }} {{eqn | r = \map n {\mathbf y} \paren {\mathbf x \, \overline {\mathbf x} }       | c =  }} {{eqn | r = \map n {\mathbf y} \, \map n {\mathbf x}       | c =  }} {{eqn | r = \map n {\mathbf x} \, \map n {\mathbf y}       | c =  }} {{end-eqn}} {{qed}}
{{AimForCont}} that $\set {1, i, j}$ forms a basis for an algebra of $3$ dimensions with real scalars. Let $1$ and $i$ have their usual properties as they do as complex numbers: :$\forall a: 1 a = a 1 = a$ :$i \cdot i = -1$ Then: :$i j = a_1 + a_2 i + a_3 j$ for some $a_1, a_2, a_3 \in \R$. Multiplying through by $i$: :$(1): \quad i \paren {i j} = \paren {i i} j = -j$ and: {{begin-eqn}} {{eqn | l = i \paren {a_1 + a_2 i + a_3 j}       | r = a_1 i - a_2 + a_3 i j       | c =  }} {{eqn | r = a_1 i - a_2 + a_3 \paren {a_1 + a_2 i + a_3 j}       | c =  }} {{eqn | r = a_1 i - a_2 + a_1 a_3 + a_2 a_3 i + {a_3}^2 j       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = \paren {a_1 a_3 - a_2} + \paren {a_1 + a_2 a_3} i + \paren { {a_3}^2 + 1} j       | c = from $(1)$ }} {{end-eqn}} But this implies that ${a_3}^2 = -1$, which contradicts our supposition that $a_3 \in \R$. Hence the result by Proof by Contradiction. {{qed}}
We will prove that :$\left({\sup X_1, \sup X_2}\right)$ is upper bound for $X_1 \times X_2$ Let $\left({a, b}\right) \in X_1 \times X_2$. By definition of Cartesian product: :$a \in X_1$ and $b \in X_2$ By definitions of supremum and upper bound: :$a \preceq_1 \sup X_1$ and $b \preceq_2 \sup X_2$ Thus by definition Cartesian product of ordered sets: :$\left({a, b}\right) \precsim \left({\sup X_1, \sup X_2}\right)$ {{qed|lemma}} We will prove that :$\forall \left({a, b}\right) \in S1 \times S_2: \left({a, b}\right)$ is upper bound for $X_1 \times X_2 \implies \left({\sup X_1, \sup X_2}\right) \precsim \left({a, b}\right)$ Let $\left({a, b}\right) \in S1 \times S_2$ such that :$\left({a, b}\right)$ is upper bound for $X_1 \times X_2$ We will prove as sublemma that :$a$ is upper bound for $X_1$ Let $c \in X_1$. By definition of non-empty set: :$\exists d: d \in X_2$ By definition of Cartesian product: :$\left({c, d}\right) in X_1 \times X_2$ By definition of upper bound: :$\left({c, d}\right) \precsim \left({a, b}\right)$ Thus be definition of Cartesian product of ordered sets: :$c \preceq_1 a$ This ends the proof of sublemma. Analogically we have that :$b$ is upper bound for $X_2$ By definition of supremum; :$\sup X_1 \preceq_1 a$ and $\sup X_2 \preceq_2 b$ Thus by definition of Cartesian product of ordered sets: :$\left({\sup X_1, \sup X_2}\right) \precsim \left({a, b}\right)$ {{qed|lemma}} Thus by definition :$X_1 \times X_2$ admits a supremum and :$\sup \left({X_1 \times X_2}\right) = \left({\sup X_1, \sup X_2}\right)$ {{qed}}
By definition of Lie algebra, it suffices to prove two properties: :$(1): \forall a \in S: a \times a = 0$ :$(2): \forall a, b, c \in S: a \times \paren {b \times c} + b \times \paren {c \times a} + c \times \paren {a \times b} = 0$ === Proof of $(1)$ === Cross Product of Vector with Itself is Zero {{Qed|lemma}} === Proof of $(2)$ === Vector Cross Product satisfies Jacobi Identity {{Qed|lemma}} Both properties hold, and the result follows. {{qed}}
{{AimForCont}} there exists some compact subset $K$ of $U$ such that $f_n$ does not converge uniformly to $f$ on $K$. Consider $K^* := K \cup \set \sigma$. From Subsets Inherit Uniform Convergence, $f_n$ does not converge uniformly to $f$ on $K^*$. From Uniformly Convergent iff Difference Under Supremum Norm Vanishes, the above is equivalent to: :$\exists \epsilon > 0 : \forall N \in \N : \exists n \ge N : \norm {f_n - f}_{K^*} \ge \epsilon$ where $\norm {\cdot}_{K^*}$ denotes the supremum norm over $K^*$. From Finite Union of Compact Sets is Compact, $K^{*}$ is compact. Since $\sequence {f_n}$ is a  normal family, there is some subsequence $\sequence {f_{n_r} }$ of $\sequence {f_n}$ and some mapping $g \in \map {\mathcal H} U$ such that: :$\sequence {f_{n_r} }$ converges uniformly to $g$ on $K^*$. Further: {{begin-eqn}} {{eqn | l = \map f \sigma       | r = \lim_{n \mathop \to \infty} \map {f_n} \sigma       | c = Definition of $f$ at $\sigma$ }} {{eqn | r = \lim_{r \mathop \to \infty} \map {f_{n_r} } \sigma       | c = Limit of Subsequence equals Limit of Sequence }} {{eqn | r = \map g \sigma       | c = Definition of $g$ at $\sigma$ }} {{end-eqn}} From the Identity Theorem, $f$ and $g$ agree on $U$. From Uniformly Convergent iff Difference Under Supremum Norm Vanishes: :$\exists N \in \N: r \ge N \implies \norm {f_{n_r} - f}_{K^*} < \epsilon$ This contradicts the result that: :$\forall N \in \N: \exists n \ge N: \norm {f_n - f}_{K^*} \ge \epsilon$ Hence the result, by Proof by Contradiction. {{qed}} {{Namedfor|Giuseppe Vitali|cat = Vitali}}
We want to show that: :$\size {\map {f_n} {a_n} - \map f a} \to 0$ as $n \to \infty$ From the Triangle Inequality: :$\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a}$ Now fix $\epsilon \in \R^{>0}$. Since $f$ is continuous and $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that: :$\exists M \in \N: n \ge M \implies \size {\map {f_n} {a_n} - \map f {a_n} } < \dfrac \epsilon 2$ Since $\sequence {f_n}$ is uniformly convergent on $K$: :$\exists N \in \N: n \ge N \implies \size {\map {f_n} {a_n} \to \map f {a_n} } < \dfrac \epsilon 2$ So if $n \ge \max \set {M, N}$: :$\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a} < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$ This completes the proof. {{qed}} Category:Metric Spaces gdg71i7b3w7cjce1vw37x9kwe8x2sr3
=== Sufficient Condition === Let $\mathscr S$ be meet-continuous. Define $\II$, the set of all ideals in $\mathscr S$ Define a mapping $f: \II \to S$ such that :$\forall I \in \II: \map f I = \sup I$ By Meet-Continuous iff Ideal Supremum is Meet Preserving: :$f$ preserves meet. Let $I, J \in \II$. By definition of mapping preserves meet: :$f$ preserves the infimum of $\set {I, J}$ Thus {{begin-eqn}} {{eqn | l = \paren {\sup I} \wedge \paren {\sup J}       | r = \map f I \wedge \map f J       | c = Definition of $f$ }} {{eqn | r = \inf \set {\map f I, \map f J}       | c = {{Defof|Meet (Order Theory)}} }} {{eqn | r = \map \inf {\map {f^\to} {\set {I, J} } }       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | r = \map f {\inf \set {I, J} }       | c = {{Defof|Mapping Preserves Infimum|subdef = Subset}} }} {{eqn | r = \map f {I \wedge J}       | c = {{Defof|Meet (Order Theory)}} }} {{eqn | r = \map \sup {I \wedge J}       | c = Definition of $f$ }} {{eqn | r = \sup \set {i \wedge j: i \in I, j \in J}       | c = Meet in Set of Ideals }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Assume that for every ideals $I, J$ in $\mathscr S$: :$\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$ By Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving: :$f$ preserves meet. Thus by Meet-Continuous iff Ideal Supremum is Meet Preserving: :$\mathscr S$ is meet-continuous. {{qed}}