hcju/proofwikiqa huggingface dataset:
From Factorisation of $z^n - a$: :$z^{2 n} - y^{2 n} = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha^k y}$ where $\alpha$ is a primitive complex $2 n$th roots of unity, for example: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / \paren {2 n} }       | c =  }} {{eqn | r = \cos \dfrac {2 \pi} {2 n} + i \sin \dfrac {2 \pi} {2 n}       | c =  }} {{eqn | r = \cos \dfrac \pi n + i \sin \dfrac \pi n       | c =  }} {{end-eqn}} From Complex Roots of Unity occur in Conjugate Pairs: :$U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$ where $U_{2 n}$ denotes the complex $2 n$th roots of unity: :$U_{2 n} = \set {z \in \C: z^{2 n} = 1}$ The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$. The case $k = n$ is taken care of by setting $\alpha^k = -1$, from whence we have the factor $x + y$. Taking the product of each of the remaining factors of $x^{2 n} - y^{2 n}$ in pairs: {{begin-eqn}} {{eqn | l = \paren {x - \alpha^k y} \paren {x - \alpha^{2 n - k} y}       | r = \paren {x - \alpha^k y} \paren {x - \overline {\alpha^k} y}       | c = Complex Roots of Unity occur in Conjugate Pairs }} {{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k y \overline {\alpha^k} y       | c =  }} {{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2       | c = Modulus in Terms of Conjugate }} {{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + y^2       | c = Modulus of Complex Root of Unity equals 1 }} {{eqn | r = x^2 - x y \paren {\cos \dfrac {k \pi} n + i \sin \dfrac {k \pi} n + \cos \dfrac {k \pi} n - i \sin \dfrac {k \pi} n} + y^2       | c = Definition of $\alpha$ }} {{eqn | r = x^2 - 2 x y \cos \dfrac {k \pi} n + x y       | c = simplification }} {{end-eqn}} Hence the result. {{qed}}
From Factorisation of $z^n + a$: :$z^{2 n} + y^{2 n} = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha_k y}$ where $\alpha_k$ are the complex $2n$th roots of negative unity: {{begin-eqn}} {{eqn | l = \alpha_k       | r = e^{i \paren {2 k + 1} \pi / {2 n} }       | c = from Roots of Complex Number }} {{eqn | r = \map \cos {\dfrac {\paren {2 k + 1} \pi} {2 n} } + i \, \map \sin {\dfrac {\paren {2 k + 1} \pi} {2 n} }       | c = $k \in \set {0, 1, 2, \ldots, 2 n - 1}$ }} {{end-eqn}} Then we have that: :$U_{2 n} = \set {\tuple {\alpha_0, \alpha_{2 n - 1} }, \tuple {\alpha_1, \alpha_{2 n - 2} }, \ldots, \tuple {\alpha_k, \alpha_{2 n - k - 1} }, \ldots, \tuple {\alpha_{n - 1}, \alpha_n } }$ where $U_{2 n}$ denotes the complex $2n$th roots of negative unity: :$U_{2 n} = \set {z \in \C: z^{2 n} = -1}$ Taking the product, $p_k$, of the factors of $x^{2 n} + y^{2 n}$ in pairs: {{begin-eqn}} {{eqn | l = p_k       | r = \paren {x - \alpha_k y} \paren {x - \alpha_{2 n - k - 1} y}       | c =  }} {{eqn | r = \paren {x - \alpha_k y} \paren {x - \overline {\alpha_k} y}       | c = Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs }} {{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \alpha_k y \overline {\alpha_k} y       | c =  }} {{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \cmod {\alpha_k}^2 y^2       | c = Modulus in Terms of Conjugate }} {{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + y^2       | c = Modulus of Complex Root of Negative Unity equals 1 }} {{eqn | r = x^2 - x y \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + y^2       | c = Definition of $\alpha_k$ }} {{eqn | r = x^2 - 2 x y \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + y^2       | c = simplification }} {{end-eqn}} However {{begin-eqn}} {{eqn | l = p_{n - k}       | r = x^2 - 2 x y \cos \dfrac {\paren {2 n - 2 k + 1} \pi} {2 n} + y^2       | c = }} {{eqn | r = x^2 - 2 x y \cos \dfrac {\paren {-2 k + 1} \pi + 2 n \pi} {2 n} + y^2       | c = }} {{eqn | r = x^2 + 2 x y \cos \dfrac {\paren {-2 k + 1} \pi} {2 n} + y^2       | c = Cosine of Angle plus Straight Angle: $\map \cos {x + \pi} = -\cos x$ }} {{eqn | r = x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2       | c = Cosine Function is Even: $\map \cos {- x} = \cos x$ }} {{end-eqn}} Consider the permutation: :$\sigma = \begin{pmatrix}   1 & 2     & \cdots & k     & \cdots & n - 1 & n \\   n & n - 1 & \cdots & n - k & \cdots & 2     & 1 \end{pmatrix}$ From Permutation of Indices of Product: :$\displaystyle \prod_{\map R k} p_k = \prod_{\map R {\map \sigma k} } p_{\map \sigma k}$ Hence: {{begin-eqn}} {{eqn | l = x^{2 n} + y^{2 n}       | r = \displaystyle \prod_{k \mathop = 1}^n p_k       | c = }} {{eqn | r = \displaystyle \prod_{k \mathop = 1}^n p_{n - k}       | c = }} {{eqn | r = \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2}       | c = }} {{end-eqn}} {{Proofread|I believe this to be correct, but would welcome a sanity check.}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \map {S_r} x       | r = \set {y \in R : \norm {y - x} = r}       | c = {{Defof|Sphere in Normed Division Ring}} }} {{eqn | r = \set {y \in R : \norm {y - x} \le r} \cap \set {y \in R : \norm{y - x} \ge r} }} {{eqn | r = \map { {B_r}^-} x \cap \paren {R \setminus \map {B_r} x}       | c = {{Defof|Open Ball of Normed Division Ring}} and {{Defof|Closed Ball of Normed Division Ring}} }} {{end-eqn}} Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$. By Open Balls Are Clopen then $\map {B_r} x$ is both open and closed in $d$. By Metric Induces Topology then $R \setminus \map {B_r} x$ is is both open and closed in $d$. By Closed Balls Are Clopen then $\map { {B_r}^-} x$ is both open and closed in $d$. By Metric Induces Topology then the intersection of a finite number of open sets is open.    Hence $\map {S_r} x$ is open in the metric space $\struct {R, d}$. By Intersection of Closed Sets is Closed in Topological Space then $\map {S_r} x$ is closed in $\struct {R, d}$. The result follows. {{qed}}
Proof by Counterexample Take the quadratic equation: :$(1): \quad x^2 + 1 = 0$ From the Quadratic Formula, the solution to $(1)$ is: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {-0 \pm \sqrt {0^2 - 4 \times 1 \times 1} } {2 \times 1}       | c =  }} {{eqn | r = \pm \sqrt {-1}       | c =  }} {{end-eqn}} But there is no real number $x$ such that: :$x^2 = -1$ Hence the result. {{qed}}
{{ProofWanted|Needs a proper explanation of the nature of Definition:Projection (Analytic Geometry)}}
=== $(1)$ implies $(2)$ === Let $D$ be a connected set by definition 1. Then by definition: :Every pair of points in $D$ can be joined by a staircase contour. But a staircase contour is a polygonal path all of whose points are in $D$. Thus $D$ is a connected set by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $D$ be a connected set by definition 2. Then by definition: :Every pair of points in $D$ can be joined by a polygonal path $P$ all points of which are in $D$. {{ProofWanted|It remains to be shown that $P$ can be replaced by a staircase contour.}} Thus $D$ is a connected set by definition 1. {{qed}} Category:Connected Sets (Complex Analysis) p3jkeb9p3mqn3pz1unf30ve2ae42xi3
Let $S$ be a subset of $R$. Let $x, y \in S: x \ne y$ Let $r \in \R_{>0} : r = \norm {x - y}$ Consider the open ball $\map {B_r} x$ such that: :$x \in \map {B_r} x$ :$y \notin \map {B_r} x$ By Open Balls are Clopen then $\map {B_r} x$ is both open and closed. By Complement of Clopen Set is Clopen then $R \setminus \map {B_r} x$ is open. Hence $S$ is not connected. The result follows. {{qed}}
Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$. {{begin-eqn}} {{eqn | l = \cmod {z_1 + z_2}       | o = \le       | r = \cmod {z_1} + \cmod {z_2}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2}       | o = \le       | r = \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2       | o = \le       | r = {a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2 + 2 \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}       | c = squaring both sides }} {{eqn | ll= \leadstoandfrom       | l = {a_1}^2 + 2 a_1 b_1 + {b_1}^2 + {a_2}^2 + 2 a_2 b_2 + {b_2}^2        | o = \le       | r = {a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2 + 2 \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}       | c = multiplying out }} {{eqn | ll= \leadstoandfrom       | l = a_1 b_1 + a_2 b_2        | o = \le       | r = \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}       | c = simplifying }} {{eqn | ll= \leadstoandfrom       | l = \paren {a_1 b_1 + a_2 b_2}^2       | o = \le       | r = \paren { {a_1}^2 + {a_2}^2} \paren { {b_1}^2 + {b_2}^2}       | c = Cauchy's Inequality }} {{eqn | ll= \leadstoandfrom       | l = {a_1}^2 {b_1}^2 + 2 a_1 b_1 a_2 b_2 + {a_2}^2 {b_2}^2       | o = \le       | r = {a_1}^2 {b_1}^2 + {a_2}^2 {b_2}^2 + {a_1}^2 {b_2}^2 + {a_2}^2 {b_1}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 2 a_1 b_1 a_2 b_2       | o = \le       | r = {a_1}^2 {b_2}^2 + {a_2}^2 {b_1}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 0       | o = \le       | r = \paren {a_1 b_2}^2 - 2 \paren {a_1 b_2} \paren {a_2 b_1} + \paren {a_2 b_1}^2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = 0       | o = \le       | r = \paren {a_1 b_2 - a_2 b_1}^2       | c =  }} {{end-eqn}} The final statement is a tautology, and all implications are reversible. Hence the result. {{qed}}
Let $z_1 = a_1 + i a_2, z_2 = b_1 + i b_2$. Then from the definition of the modulus, the above equation translates into: :$\paren {\paren {a_1 + b_1}^2 + \paren {a_2 + b_2}^2}^{\frac 1 2} \le \paren { {a_1}^2 + {a_2}^2}^{\frac 1 2} + \paren { {b_1}^2 + {b_2}^2}^{\frac 1 2}$ This is a special case of Minkowski's Inequality, with $n = 2$. {{qed}}
Let $z_1$ and $z_2$ be represented by the points $A$ and $B$ respectively in the complex plane. From Geometrical Interpretation of Complex Addition, we can construct the parallelogram $OACB$ where: :$OA$ and $OB$ represent $z_1$ and $z_2$ respectively :$OC$ represents $z_1 + z_2$. :400px As $OACB$ is a parallelogram, we have that $OB = AC$. The lengths of $OA$, $AC$ and $OC$ are: {{begin-eqn}} {{eqn | l = OA       | r = \cmod {z_1} }} {{eqn | l = AC       | r = \cmod {z_2} }} {{eqn | l = OC       | r = \cmod {z_1 + z_2} }} {{end-eqn}} But $OA$, $OB$ and $OC$ form the sides of a triangle. The result then follows directly from Sum of Two Sides of Triangle Greater than Third Side. {{qed}}
We have: {{begin-eqn}} {{eqn | l = AB       | r = z_2 - z_1       | c = Geometrical Interpretation of Complex Subtraction }} {{eqn | r = \paren {x_2 + i y_2} - \paren {x_1 + i y_1}       | c =  }} {{eqn | r = \paren {x_2 - x_1} + \paren {y_2 - y_1} i       | c = {{Defof|Complex Subtraction}} }} {{eqn | ll= \leadsto       | l = \size {AB}       | r = \cmod {\paren {x_2 - x_1} + \paren {y_2 - y_1} i}       | c =  }} {{eqn | r = \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}       | c = {{Defof|Complex Modulus}} }} {{end-eqn}} {{qed}}
Suppose it is not the case that $a = b = 0$. Then: {{begin-eqn}} {{eqn | l = a z_1 + b z_2       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a \paren {x_1 + i y_1} + b \paren {x_2 + i y_2}       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a x_1 + b x_2       | r = 0       | c =  }} {{eqn | l = a y_1 + b y_2       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a x_1       | r = -b x_2       | c =  }} {{eqn | l = a y_1       | r = -b y_2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \dfrac {y_1} {x_1}       | r = \dfrac {y_2} {x_2}       | c =  }} {{end-eqn}} and $z_1$ and $z_2$ are parallel. {{qed}}
Let $z$ be a point on the $L$. Then: :$z - z_1 = b \paren {z - z_2}$ where $b$ is some real number. Then: {{begin-eqn}} {{eqn | l = b       | r = \frac {z - z_1} {z - z_2}       | c =  }} {{eqn | ll= \leadsto       | l = \map \arg {\frac {z - z_1} {z_2 - z_1} }       | r = \arg b       | c =  }} {{eqn | r = 0       | c = as $b$ is real }} {{end-eqn}} {{qed}}
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane. Let $z$ be an arbitrary point on $L$ represented by the point $P$. :500px From Geometrical Interpretation of Complex Addition: {{begin-eqn}} {{eqn | l = OA + AP       | r = OP       | c =  }} {{eqn | ll= \leadsto       | l = z_1 + AP       | r = z       | c =  }} {{eqn | ll= \leadsto       | l = AP       | r = z - z_1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = OA + AB       | r = OB       | c =  }} {{eqn | ll= \leadsto       | l = z_1 + AB       | r = z_2       | c =  }} {{eqn | ll= \leadsto       | l = AB       | r = z_2 - z_1       | c =  }} {{end-eqn}} As $AP$ and $AB$ are collinear: :$AP = t AB$ and so: :$z - z_1 = t \paren {z_2 - z_1}$ The given expressions follow after algebra. {{qed}}
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane. Let $z$ be an arbitrary point on $L$ represented by the point $P$. :500px As $AP$ and $AB$ are collinear: :$m AP = n PB$ and so: :$m \paren {z - z_1} = n \paren {z_2 - z_1}$ The result follows. {{qed}}
To show that $\OO$ is a subring the Subring Test is used by showing: :$(1): \quad \OO \ne \O$ :$(2): \quad \forall x, y \in \OO: x + \paren {-y} \in \OO$ :$(3): \quad \forall x, y \in \OO: x y \in \OO$ '''(1)''' By Norm of Unity, :$\norm{1_R} = 1$ Hence: :$1_R \in \OO \ne \O$ {{qed|lemma}} '''(2)''' Let $x, y \in \OO$. Then: {{begin-eqn}} {{eqn | l = \norm {x + \paren{-y} }        | o = \le       | r = \max \set {\norm x, \norm{-y} }       | c = {{NormAxiom|4}} }} {{eqn | r = \max \set {\norm x, \norm y}       | c = Norm of Negative }} {{eqn | o = \le       | r = 1       | c = Since $x, y \in \OO$ }} {{end-eqn}} Hence: :$x + \paren {-y} \in \OO$ {{qed|lemma}} '''(3)''' Let $x, y \in \OO$. Then: {{begin-eqn}} {{eqn | l = \norm{x y}        | o = \le       | r = \norm x \norm y       | c = {{NormAxiom|2}} }} {{eqn | o = \le       | r = 1       | c = Since $x, y \in \OO$ }} {{end-eqn}} Hence: :$x y \in \OO$ {{qed|lemma}} By Subring Test it follows that $\OO$ is a subring of $R$. Since $1_R \in S$ and $1_R$ is the unity of $R$ then $1_R$ is the unity of $\OO$. By Division Ring has No Proper Zero Divisors then $R$ has no proper zero divisors. Hence $\OO$ has no proper zero divisors. {{qed}}
First it is shown that $\PP$ is an ideal of $\OO$ by applying Test for Ideal. That is, it is shown that: :$(1): \quad \PP \ne \O$ :$(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$ :$(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$ '''(1)''' By {{NormAxiom|1}}: :$\norm {0_R} = 0$ Hence: :$0_R \in \PP \ne \O$ {{qed|lemma}} '''(2)''' Let $x, y \in \PP$. Then: {{begin-eqn}} {{eqn | l = \norm {x + \paren{-y} }        | o = \le       | r = \max \set {\norm x, \norm{-y} }       | c = {{NormAxiom|4}} }} {{eqn | r = \max \set {\norm x, \norm y}       | c = Norm of Negative }} {{eqn | o = <       | r = 1       | c = Since $x, y \in \PP$ }} {{end-eqn}} Hence: :$x + \paren {-y} \in \PP$ {{qed|lemma}} '''(3)''' Let $x \in \PP, y \in \OO$. Then: {{begin-eqn}} {{eqn | l = \norm{x y}        | o = \le       | r = \norm x \norm y       | c = {{NormAxiom|2}} }} {{eqn | o = <       | r = 1       | c = Since $x \in \PP, y \in \OO$ }} {{end-eqn}} Hence: :$x y \in \PP$ {{qed|lemma}} By Test for Ideal it follows that $\PP$ is an ideal of $\OO$. By Maximal Left and Right Ideal iff Quotient Ring is Division Ring the statements '''(a)''', '''(b)''' and '''(c)''' above are equivalent. It is now shown that statement '''(a)''' holds. Let $J$ be a left ideal of $\OO$: :$\PP \subsetneq J \subset \OO$ Let $x \in J \setminus \PP$, then: :$\norm x = 1$ By Norm of Inverse then: :$\norm {x^{-1} } = 1 / \norm x = 1 / 1 = 1$  Hence: :$x^{-1} \in \OO$ Since $J$ is a left ideal then: :$x^{-1} x = 1_R \in J$ Thus: :$\forall y \in \OO: y \cdot 1_R = y \in J$ That is, $J = \OO$ Hence $\PP$ is a maximal left ideal. The result follows. {{qed}}
Let $z \in S$. Then, for example: :$\cmod z \le 2$ That is, $S$ is contained entirely within a circle of radius $2$ whose center is at the origin. {{qed}}
We have that $S$ is countably infinite and also bounded. Hence by the Bolzano-Weierstrass Theorem $S$ has at least one limit point. Let $\epsilon \in \R_{>0}$. Let $\map {N_\epsilon} z$ be the $\epsilon$-neighborhood of $z$. Let $n \in \N$ such that $n > \dfrac 1 \epsilon$. Then $\cmod {\dfrac i n} < \epsilon$ and so: : $\dfrac i n \in \map {N_\epsilon} z$ where $\map {N_\epsilon} {z_0}$ denotes the $\epsilon$-neighborhood of $z$. Hence, by definition, $z$ is a limit point of $S$. Now let $z_1 \in \C$ such that $z_1 \ne 0$. Let $z \in S$ such that $z \ne z_1$. Then: :$\cmod {z_1 - z} > 0$ Let $\epsilon \in \R_{>0}$ such that $\epsilon < \cmod {z_1 - z}$. Then: :$z \notin \map {N_\epsilon} {z_1}$ By making $\epsilon$ smaller than the minimum $\cmod {z_1 - z}$ for $z \in S$, it is seen that: :$\forall z \in S: z \ne z_1 \implies z \notin \map {N_\epsilon} {z_1}$ and so $z_1$ is not a limit point of $S$. {{qed}}
From Sequence of Imaginary Reciprocals: Limit Points, $S$ has one limit point $z = 0$. But: :$\nexists n \in \N: \dfrac i n = 0$ so $0 \notin S$. As $S$ does not contain (all) its limit point(s), it follows by definition that $S$ is not closed. {{qed}}
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$. Then: {{begin-eqn}} {{eqn | l = \OO       | r = \set {\dfrac a b \in \Q : \norm {\dfrac a b}_p \le 1}       | c = {{Defof|Valuation Ring Induced by Non-Archimedean Norm}} }} {{eqn | o = }} {{eqn | r = \set{\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} \ge 0}       | c = {{Defof|P-adic Norm}} }} {{eqn | o = }} {{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b \ge 0}       | c = {{Defof|P-adic Valuation|$p$-adic Valuation on Rationals}} }} {{eqn | o = }} {{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a \ge \map {\nu_p} b} }} {{end-eqn}} Let $\dfrac a b \in \Q$ be in canonical form.  Then $a \perp b$ Suppose $p \divides b$.  Then $p \nmid a$. Hence: :$\map {\nu_p} b \gt 0 = \map {\nu_p} a$ Suppose $p \nmid b$.  Then: :$\map {\nu_p} a \ge 0 = \map {\nu_p} b$ So:  :$\map {\nu_p} a \ge \map {\nu_p} b$ {{iff}} $p \nmid b$   Hence: :$\OO = \set {\dfrac a b \in \Q : p \nmid b }$ {{qed}}
Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the $p$-adic valuation on $\Q$. Then: {{begin-eqn}} {{eqn | l = \PP       | r = \set {\dfrac a b \in \Q : \norm{\dfrac a b}_p < 1}       | c = {{Defof|Valuation Ideal Induced by Non-Archimedean Norm}} }} {{eqn | o = }} {{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} > 0}       | c = {{Defof|P-adic Norm|$p$-adic Norm}} }} {{eqn | o = }} {{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b > 0}       | c = {{Defof|P-adic Valuation|$p$-adic Valuation on Rationals}} }} {{eqn | o = }} {{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a > \map {\nu_p} b} }} {{end-eqn}} Let $\dfrac a b \in \Q$ be in canonical form.  Then $a \perp b$ Suppose $p \divides a$.  Then $p \nmid b$. Hence: :$\map {\nu_p} a > 0 = \map {\nu_p} b$ Suppose $p \nmid a$.  Then: :$\map {\nu_p} b \ge 0 = \map {\nu_p} a$ So:  :$\map {\nu_p} a > \map {\nu_p} b$ {{iff}} $p \nmid b$ and $p \divides a$    Hence: :$\PP = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$ So: {{begin-eqn}} {{eqn | l = \dfrac a b \in \PP       | o = \leadstoandfrom       | r = p \nmid b, p \divides a }} {{eqn | o = \leadstoandfrom       | r = p \nmid b, \exists a' \in \Z: a = p a'  }} {{eqn | o = \leadstoandfrom       | r = \exists a' \in \Z: a = p a',  \dfrac {a'} b \in \Z_{\ideal p}       | c = Valuation Ring of P-adic Norm on Rationals }} {{eqn | o = \leadstoandfrom       | r = \dfrac a b \in p \Z_{\ideal p} }} {{end-eqn}} Hence: :$\PP = p \Z_{\ideal p}$ {{qed}}
By Valuation Ring of P-adic Norm on Rationals: :$\Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b}$ is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$. By Valuation Ideal of P-adic Norm on Rationals: :$p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$ is the induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$. By definition, the induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the quotient ring $\Z_{\ideal p} / p \Z_{\ideal p}$. By Quotient Ring of Integers with Principal Ideal, $\F_p$ is isomorphic to $\Z / p \Z$, where $p \Z$ is the principal ideal of $\Z$ generated by $p$. To complete the proof it is sufficient to show that $\Z / p \Z$ is isomorphic to $\Z_{\ideal p} / p \Z_{\ideal p}$. By Integers form Subring of Valuation Ring of P-adic Norm on Rationals then $\Z$ is a subring of $\Z_{\ideal p}$. Let $\phi : \Z \to \Z_{\ideal p} / p \Z_{\ideal p}$ be the mapping defined by: :$\forall a \in \Z: \map \phi a = a + p \Z_{\ideal p}$ === Lemma 1 === {{:Residue Field of P-adic Norm on Rationals/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Residue Field of P-adic Norm on Rationals/Lemma 2}}{{qed|lemma}} === Lemma 3 === {{:Residue Field of P-adic Norm on Rationals/Lemma 3}}{{qed|lemma}} Hence $\phi$ is a ring epimorphism with: :$p \Z = \map \ker \phi$ By Quotient Ring of Kernel of Ring Epimorphism then $\Z / p \Z$ is isomorphic to $\Z_{\ideal p} / p \Z_{\ideal p}$ The result follows. {{qed}}
Consider the point $z = \dfrac i n \in S$. Let $\delta \in \R_{>0}$. Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$. Then $\map {N_\delta} z$ contains at least one point of $S$ ($i / n$ itself) as well as points which are not in $S$. Hence, by definition, $z$ is a boundary point of $S$. Let $z = 0$. Similarly, let $\map {N_\delta} z$,  Let $\delta \in \R_{>0}$. Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$. Let $n \in \N$ such that $n > \dfrac 1 \delta$. Then $\cmod {\dfrac i n} < \delta$ and so: : $\dfrac i n \in \map {N_\delta} z$ Thus every $\delta$-neighborhood of $z = 0$ also contains points of $S$ and points not in $S$. Hence, by definition, $z = 0$ is also a boundary point of $S$. {{qed}}
From Sequence of Imaginary Reciprocals: Boundary Points, every $z \in S$ is a boundary point of $S$. Thus no $z \in S$ is an interior point. {{qed}}
From Sequence of Imaginary Reciprocals: Interior, no $z \in S$ is an interior point. Hence $S$ cannot be open. {{qed}}
Let $z_1 \in S$ and $z_2 \in S$ be joined by a polygonal path $P$. Then there are points of $P$ which are not in $S$. Hence, by definition, $S$ is not connected. {{qed}}
From Sequence of Imaginary Reciprocals: Openness, $S$ is not an open set. From Sequence of Imaginary Reciprocals: Connectedness, $S$ is not connected. Hence, by definition, $S$ is not an open region. {{qed}}
Let $\phi: \N \to S$ be the mapping defined as: :$\forall n \in \N: \map \phi n = \dfrac i n$ $\phi$ is a bijection. Hence the result by definition of countably infinite. {{qed}}
From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$. But from Sequence of Imaginary Reciprocals: Closedness, $S$ is not closed. Hence the result by definition of compact. {{qed}}
From Topological Closure is Closed, $S^-$ is closed. From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$. It follows trivially that $S^-$ is also bounded in $\C$. Hence the result by definition of compact. {{qed}}