hcju/proofwikiqa huggingface dataset:
From Right Module over Commutative Ring induces Left Module, $\struct{G, +_G, \circ’}$ is a left module. Let $\lambda, \mu \in R$ and $x \in G$. Then: {{begin-eqn}} {{eqn | l = \lambda \circ’ \paren{x \circ \mu}       | r = \paren{x \circ \mu} \circ \lambda       | c = Definition of $\circ’$ }} {{eqn | r = x \circ \paren {\mu \circ \lambda}       | c = Right module axiom $(RM \,3)$ : Associativity of Scalar Multiplication }} {{eqn | r = x \circ \paren {\lambda \circ \mu}       | c = Ring product $\circ$ is commutative }} {{eqn | r = \paren{x \circ \lambda} \circ \mu       | c = Right module axiom $(RM \,3)$ : Associativity of Scalar Multiplication }} {{eqn | r = \paren{\lambda \circ’ x} \circ \mu       | c = Definition of $\circ’$ }} {{end-eqn}} Hence $\struct{G, +_G, \circ’, \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition. {{qed}}
:450px Let $AC$ and $BD$ meet at $E$. Consider the triangles $\triangle ABD$ and $\triangle CBD$. We have that: :$AB = CB$ :$AD = CD$ :$BD$ is common. Hence by Triangle Side-Side-Side Equality, $\triangle ABD$ and $\triangle CBD$ are congruent. Consider the triangles $\triangle ABE$ and $\triangle CBE$. We have from the congruence of $\triangle ABD$ and $\triangle CBD$ that: :$\angle ABE = \angle CBE$ :$AB = CB$ and $BE$ is common. Hence by Triangle Side-Angle-Side Equality, $\triangle ABE$ and $\triangle CBE$ are congruent. We have that $AC$ is a straight line. We have from the congruence of $\triangle ABE$ and $\triangle CBE$ that: :$\angle BEC = \angle BEA$ From Two Angles on Straight Line make Two Right Angles, $\angle BEC + \angle BEA$ make two right angles. Thus: :$2 \angle BEC = 2 \angle BEA = 2$  right angles and so: :$\angle BEC = \angle BEA$ are both right angles. That is, $AC$ and $BD$ are perpendicular. {{qed}}
:450px Let $AC$ and $BD$ meet at $E$. From Diagonals of Kite are Perpendicular, $AC$ and $BD$ are perpendicular. That is: :$\angle AEB = \angle CEB$ both being right angles. Consider the triangles $\triangle ABE$ and $\triangle CBE$. We have that: :$\angle AEB = \angle CEB$ are both right angles. :$AB = BC$ :$BE$ is common Hence by Triangle Right-Angle-Hypotenuse-Side Equality, $\triangle ABE$ and $\triangle CBE$ are congruent. Thus: :$AE = CE$ and it has already been established that $BD$ is perpendicular to $AC$. Hence the result by definition of perpendicular bisector. {{qed}}
By definition, a topological group is a topological semigroup. Hence $\struct {G, *, \tau_{_G} }$ is a topological semigroup. From Product Rule for Continuous Mappings to Topological Semigroup: :$f * g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is a continuous mapping. {{qed}}
By definition, a topological group is a topological semigroup. Hence $\struct {G, *, \tau_{_G}}$ is a topological semigroup. From Multiple Rule for Continuous Mappings to Topological Semigroup: :$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ are continuous mappings. {{qed}}
By definition of a topological group: :$\phi: \struct {G, \tau_{_G} } \to \struct {G, \tau_{_G} }$ such that $\forall x \in G: \map \phi x = x^{-1}$ is a continuous mapping From Composite of Continuous Mappings is Continuous: :the composition $\phi \circ g: \struct {S, \tau_{_S} } \to \struct {G, \tau_{_G} }$ is continuous. Now: {{begin-eqn}} {{eqn | lo= \forall x \in S :       | l = \map {\paren {g^{-1} } } x       | r = \paren {\map g x}^{-1}       | c = Definition of $g^{-1}$ }} {{eqn | r = \map \phi {\map g x}       | c = Definition of $\phi$ }} {{eqn | r = \map {\paren {\phi \circ g} } x       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} From Equality of Mappings: :$g^{-1} = \phi \circ g$ The result follows. {{qed}} Category:Combination Theorem for Continuous Mappings to Topological Group gp5er6ofunk1qidomwbeyoortiade0q
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors. From Continuous Mapping to Topological Product, $f \times g$ is continuous {{iff}} $\operatorname{pr}_1 \circ \paren {f \times g}$ and $\operatorname{pr}_2 \circ \paren {f \times g}$ are continuous. Now: {{begin-eqn}} {{eqn | lo = \forall x \in T’ :        | l = \map {\pr_1 \circ \paren {f \times g} } x       | r = \map {\pr_1} {\map {\paren {f \times g} } x}       | c = {{Defof|Composite Mapping|composite mapping}} }} {{eqn | r = \map {\pr_1} {\tuple {\map f x, \map g x} }       | c = Definition of the mapping $f \times g$ }} {{eqn | r = \map f x       | c = first projection from $T$ }} {{end-eqn}} From Equality of Mappings, $\pr_1 \circ \paren {f \times g} = f$. Similarly, $\pr_2 \circ \paren {f \times g} = g$. The result follows. {{qed}} Category:Product Spaces Category:Continuous Mappings pvgtezt85lqkazptqaqj2ks9z6ot0vf
Let $f \times g: S \to G \times G$ be the mapping defined by: :$\forall x \in S : \map {\paren {f \times g} } x = \tuple {\map f x, \map g x}$ From Pointwise Operation is Composite of Operation with Mapping to Cartesian Product: :$f * g = * \circ \paren {f \times g}$ Let $\tau$ be the product topology on $G \times G$. From Corollary to Continuous Mapping to Topological Product: :$f \times g : \struct{S, \tau_{_S}} \to \struct {G \times G, \tau}$ is a continuous mapping. Now $* : \struct {G \times G, \tau} \to \struct {G, \tau_{_G} }$ is continuous by definition of a topological semigroup. From Composite of Continuous Mappings is Continuous: :$* \circ \paren {f \times g}$ is continuous. The result follows. {{qed}} Category:Combination Theorem for Continuous Mappings to Topological Semigroup 9r2rotf3hy5lypz6h49yg6b4upaoe4i
Let $c_\lambda : S \to G$ be the constant mapping defined by: :$\forall x \in S: \map {c_\lambda} x = \lambda$ From Constant Mapping is Continuous, $c_\lambda$ is continuous. From Product Rule for Continuous Mappings to Topological Semigroup: :$c_\lambda * f$ and $f * c_\lambda$ are continuous. Now: {{begin-eqn}} {{eqn | lo= \forall x \in S :        | l = \map {\paren {c_\lambda * f} } x       | r = \map {c_\lambda} x * \map f x       | c = Definition of $c_\lambda * f$  }} {{eqn | r = \lambda * \map f x       | c = Definition of $c_\lambda$  }} {{eqn | r = \map {\paren {\lambda * f} } x       | c = Definition of $\lambda * f$  }} {{end-eqn}} From Equality of Mappings: :$c_\lambda * f = \lambda * f$ Similarly: :$f * c_\lambda = f * \lambda$ The result follows. {{qed}} Category:Combination Theorem for Continuous Mappings to Topological Semigroup svvbv7qo3ii4ai6autzd3mnufqqmfa2
The weight of $B$ is the magnitude of the force exerted on it by the influence of the gravitational field it is in. By Newton's Second Law of Motion, that force is given by: :$\mathbf W = -m g \mathbf k$ where: :$g$ is the value of the acceleration due to gravity at the surface of Earth :$\mathbf k$ is a unit vector directed vertically upwards. Hence the magnitude of $\mathbf W$ is given by: :$W = \size {-m g \mathbf k} = m g$ {{qed}}
By definition of a topological ring, $\struct {R, +, \tau_{_R} }$ is a topological group. From Product Rule for Continuous Mappings to Topological Group: :$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
By definition of a topological ring: :$\struct {R, +, \tau_{_R} }$ is a topological group. From Inverse Rule for Continuous Mappings to Topological Group: :$-g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
By definition of a topological ring: :$\struct {R, *, \tau_{_R} }$ is a topological semigroup. From Multiple Rule for Continuous Mappings to Topological Semigroup: :$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings. {{qed}}
By definition of a topological ring: :$\struct {R, *, \tau_{_R} }$ is a topological semigroup. From Product Rule for Continuous Mappings to Topological Semigroup: :$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
By definition of a topological division ring, $\struct {R, +, *, \tau_{_R} }$ is a topological ring. From Sum Rule for Continuous Mappings into Topological Ring: :$f + g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
By definition of a topological division ring: :$\struct {R, +, *, \tau_{_R} }$ is a topological ring. From Multiple Rule for Continuous Mappings to Topological Ring: :$\lambda * f, f * \lambda: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ are continuous mappings. {{qed}}
By definition of a topological division ring: :$\struct {R, +, *, \tau_{_R} }$ is a topological ring. From Product Rule for Continuous Mappings to Topological Ring: :$f * g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
Let $R^* = R \setminus \set 0$. Let $\tau^*$ be the subspace topology on $R^*$. By definition of a topological division ring: :$\phi: \struct {R^*, \tau^*} \to \struct {R, \tau_{_R} }$ such that $\forall x \in R^*: \map \phi x = x^{-1}$ is a continuous mapping Let $g^*: \struct {U, \tau_{_U} } \to \struct {R^*, \tau^*}$ be the restriction of $g$ to $U \subseteq R$. From Restriction of Continuous Mapping is Continuous, $g^*$ is a continuous mapping.   From Composite of Continuous Mappings is Continuous, the composition $\phi \circ g^* : \struct {U, \tau_{_U} } \to \struct {R, \tau_{_R} }$ is continuous. Now: {{begin-eqn}} {{eqn | lo= \forall x \in U :       | l = \map {\paren {g^{-1} } } x       | r = \map g x^{-1}       | c = Definition of $g^{-1}$ }} {{eqn | r = \map \phi {\map g x}       | c = Definition of $\phi$ }} {{eqn | r = \map \phi {\map {g^*} x}       | c = Since $\map g x \neq 0$  }} {{eqn | r = \map {\paren {\phi \circ g^*} } x       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} From Equality of Mappings: :$g^{-1} = \phi \circ g^*$ The result follows. {{qed}} Category:Combination Theorem for Continuous Mappings to Topological Division Ring 0ausrn4zvo920c3rfoac6m9adr6f0sg
By definition of a topological division ring: :$\struct {R, +, *, \tau_{_R} }$ is a topological ring. From Negation Rule for Continuous Mappings to Topological Ring: :$-g: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
By definition of a topological ring: :$\struct {R, +, \tau_{_R} }$ is a topological group. From Multiple Rule for Continuous Mappings to Topological Group: :$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
By definition of a topological division ring: :$\struct {R, +, *, \tau_{_R} }$ is a topological ring. From Translation Rule for Continuous Mappings to Topological Ring: :$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
{{begin-eqn}} {{eqn | lo= \forall x \in s:       | l = \map {\paren {* \circ \paren {f \times g} } } x       | r = \map * {\map {\paren {f \times g} } x }       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map * {\tuple {\map f  x, \map g x } }       | c = Definition of $f \times g$ }} {{eqn | r = \map f  x * \map g x       | c = Definition of $*$ }} {{eqn | r = \map {\paren {f \oplus g} } x       | c = Definition of $f \oplus g$ }} {{end-eqn}} From Equality of Mappings, $* \circ \paren {f \times g} = f \oplus g$. {{qed}} Category:Abstract Algebra Category:Pointwise Operations 235ggisywe8kwl84xmlun4doguq13g6
{{begin-eqn}} {{eqn | l = \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} }       | r = \sum_{m \mathop = - \infty}^\infty \map {J_m} x t^m       | c = Generating Function for Bessel Function of the First Kind of Order n of x }} {{eqn | l = \dfrac 1 {2 \pi i} \int_C t^{- n - 1} \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} } \d t       | r = \dfrac 1 {2 \pi i} \int_C t^{- n - 1} \sum_{n \mathop = - \infty}^\infty \map {J_m} x t^m \d t       | c = where $C$ denotes a contour encircling the origin once  }} {{eqn | r = \sum_{n \mathop = - \infty}^\infty \dfrac {\map {J_m} x} {2 \pi i} \int_C t^{m - n - 1} \d t       | c =  }} {{eqn | r = \dfrac {\map {J_n} x} {2 \pi i} 2 \pi i       | c = Residue Theorem }} {{eqn | r = \map {J_n} x       | c =  }} {{end-eqn}} Take $C$ to be the unit circle. Write $t = e^{- i \theta}$ and let $\theta$ be decreasing from $\pi$ to $- \pi$, so that we integrate along $C$ counterclockwise. Then: {{begin-eqn}} {{eqn | ll = \leadsto       | l = \map {J_n} x       | r = \dfrac 1 {2 \pi i} \int_\pi^{- \pi} e^{- i \theta \paren {- n - 1} } \map \exp {\dfrac x 2 \paren {e^{- i \theta} - e^{i \theta} } } \d e^{- i \theta}       | c =  }} {{eqn | r = \dfrac 1 {2 \pi} \int_{- \pi}^\pi \map \exp {- x i \sin \theta + i n \theta} \d \theta       | c = Sine in terms of Hyperbolic Sine }} {{eqn | r = \dfrac 1 {2 \pi} \paren {\int_0^\pi \map \exp {i n \theta - i x \sin \theta} \d \theta + \int_{- \pi}^0 \map \exp {i n \theta - i x \sin \theta} \d \theta}       | c =  }} {{eqn | r = \dfrac 1 {2 \pi} \paren {\int_0^\pi \map \exp {i n \theta - i x \sin \theta} \d \theta - \int_\pi^0 \map \exp {i n \paren {- \theta} - i x \map \sin {- \theta} } \d \theta}       | c = substitution of $- \theta$ }} {{eqn | r = \dfrac 1 {2 \pi} \paren {\int_0^\pi \map \exp {i n \theta - i x \sin \theta} \d \theta + \int_0^\pi \map \exp {- \paren {i n \theta - i x \sin \theta} } \d \theta}       | c = }} {{eqn | r = \dfrac 1 \pi \int_0^\pi \map \cos {n \theta - x \sin \theta} \d \theta       | c = Cosine in terms of Hyperbolic Cosine }} {{end-eqn}} {{qed}}
From Corollary to Normed Division Ring Operations are Continuous: :$\struct{R, +, *, \tau_{_R} }$ is a topological division ring. From Translation Rule for Continuous Mappings to Topological Division Ring: :$\lambda + f: \struct {S, \tau_{_S} } \to \struct {R, \tau_{_R} }$ is a continuous mapping. {{qed}}
From Corollary to Normed Division Ring Operations are Continuous, $\struct{R, +, *, \tau_{_R}}$ is a topological division ring. From Negation Rule for Continuous Mappings to Topological Division Ring, $- g : \struct{S, \tau_{_S}} \to \struct{R, \tau_{_R}}$ is a continuous mapping. {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\d f^*} {\d p}       | r = -\frac {\d \map f {\map x p} } {\d p} + \frac {\map \d {p \map x p} } {\d p}       | c = {{Defof|Legendre Transform}} }} {{eqn | r = -f' \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}       | c = Product Rule for Derivatives }} {{eqn | r = -p \frac {\d x} {\d p} + x + p \frac {\d x} {\d p}       | c = Definition of $p$ }} {{eqn | r = x }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \frac {\d^2 f^*} {\d p^2}       | r = \map {x'} p }} {{eqn | r = \frac 1 {\map {p'} x}       | c = Derivative of Inverse Function }} {{eqn | r = \frac 1 {\map {f''} x}       | c = Definition of $p$ }} {{end-eqn}} We have that $\map f x$ is real and strictly convex. Hence, by Real Function is Strictly Convex iff Derivative is Strictly Increasing, $\map {f'} x$ is strictly increasing. Then: :$\map {f''} x > 0$ Therefore, the first derivative of $f^*$ is strictly increasing. By Real Function is Strictly Convex iff Derivative is Strictly Increasing, $f^*$ is strictly convex. {{qed}}
By definition of a topological ring, $ \struct {R, + , \tau}$ is a topological group. From Right and Left Regular Representations in Topological Group are Homeomorphisms: :$\lambda_x, \,\rho_x: \struct {R, \tau} \to \struct {R, \tau}$ are homeomorphisms with inverses $\,\lambda_{-x}, \,\rho_{-x}: \struct {R, \tau} \to \struct {R, \tau}$ respectively. {{qed}} Category:Topological Rings 0imlmciqm2d3kw2gprai012lniwq176
Let $I_{_R} : R \to R$ be the identity mapping on $R$. For all $y \in R$, let $y * I_{_R} : R \to R$ be the mapping defined by: :$\forall z \in R: \map {\paren {y * I_{_R} } } z = y * \map {I_{_R}} z$ For all $y \in R$, let $I_{_R} * y : R \to R$ be the mapping defined by: :$\forall z \in R: \map {\paren {I_{_R} * y}} z = \map {I_{_R}} z * y$ === Lemma 1 === {{:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms/Lemma 1}}{{qed|lemma}} From Identity Mapping is Continuous, $I_{_R} : \struct{R, \tau} \to \struct{R, \tau}$ is continuous. From Multiple Rule for Continuous Mappings into Topological Ring: :$x * I_{_R}, \, I_{_R} * x, x^{-1} * I_{_R}, \,I_{_R} * x^{-1}$ are continuous. === Lemma 2 === {{:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms/Lemma 2}}{{qed|lemma}} By definition of homeomorphism, $\lambda_x, \,\rho_x  : \struct{R, \tau} \to \struct{R, \tau}$ are homeomorphisms with inverse mappings $\lambda_{x^{-1} }, \,\rho_{x^{-1} }  : \struct{R, \tau} \to \struct{R, \tau}$ respectively. {{qed}} Category:Topological Rings Category:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms noehn8x3mq7q7i7opzpaacklwzlns8f
Let $y \in R$. {{begin-eqn}} {{eqn | lo = \forall z \in R :       | l = \map {\paren {y * I_{_R} } } z       | r = y * \map {I_{_R} } z       | c = Definition of $y * I_{_R}$ }} {{eqn | r = y * z       | c = {{Defof|Identity Mapping}} $I_{_R}$ }} {{eqn | r = \map {\lambda_y} z       | c = {{Defof|Left Regular Representation}} $\lambda_y$ }} {{end-eqn}} From Equality of Mappings: :$\lambda_y = y * I_{_R}$ {{begin-eqn}} {{eqn | lo = \forall z \in R :       | l = \map {\paren {I_{_R} * y} } z       | r = \map {I_{_R} } z * y       | c = Definition of $I_{_R} * y$ }} {{eqn | r = z * y       | c = {{Defof|Identity Mapping}} $I_{_R}$ }} {{eqn | r = \map {\rho_y} z       | c = {{Defof|Right Regular Representation}} $\rho_y$ }} {{end-eqn}} From Equality of Mappings: :$\rho_y = I_{_R} * y$ {{qed}} Category:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms sms45d5uflbwydhlv0na0ib9xetzfpt
Consider the composite of $x * I_{_R}$ with $x^{-1} * I_{_R}$. {{begin-eqn}} {{eqn | lo= \forall y \in R :       | l = \map {\paren {\paren{x * I_{_R} } \circ \paren{x^{-1} * I_{_R} } } } y       | r = \map {\paren{x * I_{_R} } } {\map {\paren{x^{-1} * I_{_R} } } y }       | c =  }} {{eqn | r = \map {\paren{x * I_{_R} } } {x^{-1} * y }       | c =  }} {{eqn | r = x * \paren {x^{-1} * y }       | c =  }} {{eqn | r = \paren {x * x^{-1} } * y       | c =  }} {{eqn | r = y       | c =  }} {{eqn | r = \map {I_{_R} } y       | c =  }} {{end-eqn}} From Equality of Mappings, $\paren{x * I_{_R} } \circ \paren{x^{-1} * I_{_R} } = I_{_R}$. Consider the composite of $x^{-1} * I_{_R}$ with $x * I_{_R}$. {{begin-eqn}} {{eqn | lo= \forall y \in R :       | l = \map {\paren {\paren{x^{-1} * I_{_R} } \circ \paren{x * I_{_R} } } } y       | r = \map {\paren{x^{-1} * I_{_R} } } {\map {\paren{x * I_{_R} } } y }       | c =  }} {{eqn | r = \map {\paren{x^{-1} * I_{_R} } } {x * y }       | c =  }} {{eqn | r = x^{-1} * \paren {x * y }       | c =  }} {{eqn | r = \paren {x^{-1} * x } * y       | c =  }} {{eqn | r = y       | c =  }} {{eqn | r = \map {I_{_R} } y       | c =  }} {{end-eqn}} From Equality of Mappings, $\paren{x^{-1} * I_{_R} } \circ \paren{x * I_{_R} } = I_{_R}$. Hence $x * I_{_R}$ has a left inverse and a right inverse. From definition 2 of bijection, $x * I_{_R}$ is a bijection and the inverse is $x^{-1} * I_{_R}$. Similarly, $I_{_R} * x$ is a bijection and the inverse is $I_{_R} * x^{-1}$. {{qed}} Category:Multiplicative Regular Representations of Units of Topological Ring are Homeomorphisms oop1dt37z3mbpykkpuihm2nhxq9gxlz
Let $n \in \Z$. From Open Ball in P-adic Numbers is Closed Ball: :$\map {B_{p^{-n} } } a = \map {B^{\,-}_{p^{-\paren {n + 1} } } } a$ From Closed Balls of P-adic Number: :$\map { B^{\,-}_{p^{-\paren {n + 1} } } } a = a + p^{n + 1} \Z_p$ The result follows. {{qed}}
Let $\BB_a$ be the set of all open balls of $Q_p$ centered on $a$. That is: :$\BB_a = \set{\map {B_\epsilon} a : \epsilon \in \R_{>0}}$ From Open Balls Centered on P-adic Number is Countable: :$\BB_a = \set {\map {B_{p^{-n} } } a : n \in Z}$ From Open Balls form Local Basis for Point of Metric Space, $\BB_a$ is a local basis of $a$. Recall that the $p$-adic norm on the $p$-adic numbers is a non-Archimedean norm by definition. From Open Balls are Clopen In Non-Archimedean Division Ring the set $\BB_a = \set {\map {B_{p^{-n} } } a : n \in \Z}$ is a local basis of clopen sets. {{qed}} Category:Topology of P-adic Numbers 9dx10xa54h2l6450crtl10fwnvqu9nq
Let $n \in \Z$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {B^{\,-}_{p^{-n} } } a       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \norm {x - a}_p       | o = \le       | r = p^{-n}       | c = {{Defof|Closed Ball in P-adic Numbers}} }} {{eqn | ll= \leadstoandfrom       | l = p^n \norm {x - a}_p       | o = \le       | r = 1       | c = Multiply both sides of equation by $p^n$ }} {{eqn | ll= \leadstoandfrom       | l = \norm {p^{-n} }_p \norm {x - a}_p       | o = \le       | r = 1       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | ll= \leadstoandfrom       | l = \norm {p^{-n} \paren {x - a} }_p       | o = \le       | r = 1       | c = {{NormAxiom|2}} }} {{eqn | ll= \leadstoandfrom       | l = p^{-n} \paren {x - a}       | o = \in       | r = \Z_p       | c = {{Defof|P-adic Integers|$p$-adic integers}} }} {{eqn | ll= \leadstoandfrom       | l = x - a       | o = \in       | r = p^n \Z_p       | c = {{Defof|Principal Ideal}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = a + p^n \Z_p       | c = {{Defof|Left Coset}} }} {{end-eqn}} From set equality: :$\map {B^-_{p^{-n} } } a = a + p^n \Z_p$ The result follows. {{qed}}
Let $n \in \Z$. From Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers: :$\map {B^-_{p^{-n} } } a = \displaystyle \bigcup_{i \mathop = 0}^{p - 1} \map {B^-_{p^{-\paren {n + 1} } } } {a + i p}$ where $\set {\map {B^-_{p^{-\paren {n + 1 } } } } {a + i p^n} : i = 0, \dots, p - 1}$ is a set of pairwise disjoint closed balls. From Open Ball in P-adic Numbers is Closed Ball: :$\map {B^-_{p^{-n} } } a = \displaystyle \bigcup_{i \mathop = 0}^{p - 1} \map {B_{p^{-n} } } {a + i p}$ where $\set {\map {B_{p^{-n} } } {a + i p^n} : i = 0, \dots, p - 1}$ is a set of pairwise disjoint Open balls. The result follows. {{qed}}
For all $\epsilon \in \R_{>0}$: :let $\map {B^-_\epsilon} a$ denote the closed ball of $a$ of radius $\epsilon$.  Let $n \in \Z$. Then: {{begin-eqn}} {{eqn | l = \map {S_{p^{-n} } } a       | r = \map {B^-_{p^{-n} } } a \setminus \map {B_{p^{-n} } } a       | c = Sphere is Set Difference of Closed and Open Ball in P-adic Numbers }} {{eqn | r = \paren {\bigcup_{i \mathop = 0}^{p - 1} \map {B_{p^{-n} } } {a + i p^n} } \setminus \map {B_{p^{-n} } } a       | c = Closed Ball is Disjoint Union of Open Balls in P-adic Numbers }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n} \cup \map {B_{p^{-n} } } {a + 0 \cdot p^n} } \setminus \map {B_{p^{-n} } } a       | c = Union is Associative and commutative }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n} \cup \map {B_{p^{-n} } } a } \setminus \map {B_{p^{-n} } } a       | c = $a + 0 \cdot p^n = a$ }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^{p-1} \map {B_{p^{-n} } } {a + i p^n} } \setminus \map {B_{p^{-n} } } a       | c = Set Difference with Union is Set Difference }} {{eqn | r = \bigcup_{i \mathop = 1}^{p - 1} \paren {\map {B_{p^{-n} } } {a + i p^n} \setminus \map {B_{p^{-n} } } a }       | c = Set Difference is Right Distributive over Union }} {{end-eqn}} From Closed Ball is Disjoint Union of Open Balls in P-adic Numbers: :$\set {\map {B_{p^{-n} } } {a + i p^n}: i = 0, \dots, p - 1}$ is a set of pairwise disjoint open balls. Continuing from above: {{begin-eqn}} {{eqn | l = \map {S_{p^{-n} } } a       | r = \bigcup_{i \mathop = 1}^{p - 1} \paren {\map {B_{p^{-n} } } {a + i p^n} \setminus \map {B_{p^{-n} } } a}       | c =  }} {{eqn | r = \bigcup_{i \mathop = 1}^{p - 1} \map {B_{p^{-n} } } {a + i p^n}        | c = Set Difference with Disjoint Set }} {{end-eqn}} {{qed}}