hcju/proofwikips huggingface dataset:
Axiom:Axioms of Uncertainty/Axiom 7
Axiom:Axioms of Uncertainty/Axiom 7 100 81788 431710 2019-10-18T00:15:50Z Prime.mover 59 Created page with "== Axiom == {{:Axiom:Axioms of Uncertainty}}  ${H_n}$ fulfils the following axiom:  :$\map {H_{m n} } {\dfrac 1 {m n}, \dfrac 1 {m n}, \dotsc..." wikitext text/x-wiki == Axiom == {{:Axiom:Axioms of Uncertainty}} ${H_n}$ fulfils the following axiom:  :$\map {H_{m n} } {\dfrac 1 {m n}, \dfrac 1 {m n}, \dotsc, \dfrac 1 {m n} } = \map {H_m} {\dfrac 1 m, \dfrac 1 m, \dotsc, \dfrac 1 m} + \map {H_n} {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n}$  == Sources == * {{BookReference|Codes and Cryptography|1988|Dominic Welsh|prev = Axiom:Axioms of Uncertainty/Axiom 6|next = Axiom:Axioms of Uncertainty/Axiom 8}}: $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty
Axiom:Axioms of Uncertainty/Axiom 8
== Axiom == {{:Axiom:Axioms of Uncertainty}} ${H_n}$ fulfils the following axiom:  Let: :$p = p_1 + p_2 + \dotsb + p_m$ :$q = q_1 + q_2 + \dotsb + q_n$ such that: :each of $p_i$ and $q_j$ are non-negative :$p + q = 1$ Then: :$\map {H_{m + n} } {p_1, p_2, \dotsc, p_m, q_1, q_2, \dotsc q_n} = \map {H_2} {p, q} + p \map {H_m} {\dfrac {p_1} p, \dfrac {p_2} p, \dotsc, \dfrac {p_m} p} + q \map {H_n} {\dfrac {q_1} q, \dfrac {q_2} q, \dotsc, \dfrac {q_n} q}$  == Sources == * {{BookReference|Codes and Cryptography|1988|Dominic Welsh|prev = Axiom:Axioms of Uncertainty/Axiom 7|next = Function that Satisfies Axioms of Uncertainty}}: $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty
Complex Exponential Function is Entire
Complex Exponential Function is Entire 0 81953 432464 2019-10-27T15:09:41Z Caliburn 3218 Created page with "== Theorem ==  Let $\exp: \C \to \C$ be the complex exponential function.   Then $\exp$ is Definition:Entire Functio..." wikitext text/x-wiki == Theorem ==  Let $\exp: \C \to \C$ be the complex exponential function.  Then $\exp$ is entire.   == Proof == By the definition of the complex exponential function, $\exp$ admits a power series expansion about $0$:  :$\displaystyle \exp z = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\exp$ is entire it suffices to show that this series is everywhere convergent. Note that this power series is of the form:  :$\displaystyle \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ with $\xi = 0$.  Therefore, by Radius of Convergence of Power Series over Factorial: Complex Case, we have that the former power series is everywhere convergent.  Hence the result.
Elementary Symmetric Function/Examples/Recursion
== Example of Elementary Symmetric Function: Recursion ==  Let $\set {z_1, z_2, \ldots, z_{n + 1} }$ be a set of $n + 1$ values, duplicate values permitted. Then for $1 \le m \le n$: :$\map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } } = z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} }$  == Proof == Case $m = 1$ holds because $e_0$ is $1$ and $e_1$ is the sum of the elements. Assume $2 \le m \le n$. Define four sets: :$A = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n + 1}$ :$B = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_{m - 1} \le n, p_m = n + 1}$ :$C = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n}$ :$D = \set {\set {p_1, \ldots, p_{m - 1} } : 1 \le p_1 < \cdots < p_{m - 1} \le n}$ Then $A = B \cup C$ and $B \cap C = \O$ implies: :$\ds \sum_A z_{p_1} \cdots z_{p_m} = \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}$ Simplify: :$\ds \sum_B z_{p_1} \cdots z_{p_m} = z_{n + 1} \sum_D z_{p_1} \cdots z_{p_{m - 1} }$ Notation for Elementary Symmetric Functions: :$\ds \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } } = \sum_A z_{p_1} \cdots z_{p_m}$ :$\ds \sum_D z_{p_1} \cdots z_{p_{m - 1} } = \map {e_{m - 1} } {\set {z_1, \ldots, z_n} }$ :$\ds \sum_C z_{p_1} \cdots z_{p_m} = \map {e_m} {\set {z_1, \ldots, z_n} }$ Assemble the preceding equations: {{begin-eqn}} {{eqn | l = \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }        | r = \sum_A z_{p_1} \cdots z_{p_m} }} {{eqn | r = \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m} }} {{eqn | r = z_{n + 1} \sum_D z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m} }} {{eqn | r = z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} } }} {{end-eqn}}
Inverse of Vandermonde Matrix/Corollary
 == Corollary to Inverse of Vandermonde Matrix ==  Define for variables $\set {y_1,\ldots, y_k}$ elementary symmetric functions: {{begin-eqn}} {{eqn | l = \map {e_m} {\set {y_1, \ldots, y_k} }       | r = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le k } y_{j_1} y_{j_2} \cdots y_{j_m}       | c =  for $m = 0, 1, \ldots, k$ }} {{end-eqn}} Let $\set {x_1, \ldots, x_n}$ be a set of distinct values. Let $W_n$ and $V_n$ be Vandermonde matrices of order $n$: :$W_n = \begin{bmatrix}   1         &  x_1      & \cdots & x_1^{n-1} \\   1         & x_2       & \cdots & x_2^{n-1} \\ \vdots      & \vdots    & \ddots & \vdots    \\   1         & x_1^{n-1} & \cdots & x_n^{n-1} \\ \end{bmatrix}, \quad V_n = \begin{bmatrix}   x_1   & x_2    & \cdots & x_n    \\   x_1^2 & x_2^2  & \cdots & x_n^2  \\ \vdots  & \vdots & \ddots & \vdots \\   x_1^n & x_2^n  & \cdots & x_n^n  \\ \end{bmatrix}$ Let their matrix inverses be written as $W_n^{-1} = \begin{bmatrix} b_{ij} \end{bmatrix}$ $V_n^{-1} = \begin{bmatrix} c_{ij} \end{bmatrix}$. Then: {{begin-eqn}} {{eqn   | l = b_{ij}   | r = \dfrac {\paren {-1}^{n - i} \map {e_{n - i} } {\set {x_1, \ldots, x_n} \setminus \set {x_j} } } {\prod_{m \mathop = 1, m \mathop \ne j }^n \paren {x_j - x_m} }   | c = for $i, j = 1, \ldots, n$ }} {{eqn   | l = c_{ij}   | r = \dfrac 1 {x_i} \, b_{j i}   | c = for $i, j = 1, \ldots, n$ }} {{end-eqn}}  == Proof == The details appear in Inverse of Vandermonde Matrix/Proof 1, same notation.
Equation of Cardioid/Parametric
Equation of Cardioid/Parametric 0 82574 435490 2019-11-17T00:32:52Z Prime.mover 59 Created page with "== Theorem ==  Let $C$ be a cardioid embedded in a Cartesian coordinate plane such that: :its Definition:..." wikitext text/x-wiki == Theorem ==  Let $C$ be a cardioid embedded in a Cartesian coordinate plane such that: :its stator of radius $a$ is positioned with its center at $\tuple {a, 0}$ :there is a cusp at the origin. Then $C$ can be expressed by the parametric equation: :$\begin {cases} x = 2 a \cos t \paren {1 + \cos t} \\ y = 2 a \sin t \paren {1 + \cos t} \end {cases}$  == Proof == :525px Let $P = \polar {x, y}$ be an arbitrary point on $C$. From Polar Form of Equation of Cardioid, $C$ is expressed as a polar equation as: :$r = 2 a \paren {1 + \cos \theta}$ We have that: :$x = r \cos \theta$ :$y = r \sin \theta$ Replace $\theta$ with $t$ and the required parametric equation is the result.  == Sources == * {{MathWorld|Cardioid|Cardioid}}
Irrational Number divided by Rational Number is Irrational
Irrational Number divided by Rational Number is Irrational 0 82682 435876 2019-11-20T16:53:26Z Caliburn 3218 Created page with "== Theorem ==  Let $x$ be a irrational number.   Let $y$ be a non-zero Definition:Rational Number|rational..." wikitext text/x-wiki == Theorem ==  Let $x$ be a irrational number.  Let $y$ be a non-zero rational number. Then:  :$\dfrac x y$ is irrational.  == Proof ==  $\dfrac x y$ is rational number. Then there exists an integer $p_1$ and a natural number $q_1$ such that:  :$\dfrac x y = \dfrac {p_1} {q_1}$ That is:  :$x = \dfrac {p_1} {q_1} y$ From the fact that $y$ is rational, we similarly have that there exists an integer $p_2$ and a natural number $q_2$ such that: :$y = \dfrac {p_2} {q_2}$ Then:  :$x = \dfrac {p_1 p_2} {q_1 q_2}$ From Integer Multiplication is Closed, we have that $p_1 p_2$ is an integer.  From Natural Number Multiplication is Closed, we have that $q_1 q_2$ is a natural number. Let $p_3 = p_1 p_2$ and $q_3 = q_1 q_2$. Then $x$ is expressible in the form:  :$x = \dfrac {p_3} {q_3}$ where $p_3$ is an integer and $q_3$ is a natural number. This, however, implies that $x$ is rational, which is a contradiction. By Proof by Contradiction, we conclude that $\dfrac x y$ is irrational.
Equation of Confocal Ellipses/Formulation 1
== Definition ==  The equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$a$ and $b$ are (strictly) positive constants such that $a^2 > b^2$ :$\lambda$ is a (strictly) positive parameter such that $b^2 > -\lambda$ defines the set of all confocal ellipses whose foci are at $\tuple {\pm \sqrt {a^2 - b^2}, 0}$.  == Proof == Let $a$ and $b$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $E$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2  + \lambda} = 1$ As $b^2 > -\lambda$ it follows that: :$b^2 + \lambda > 0$ and as $a^2 > b^2$: :$a^2 + \lambda > 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {r^2} + \dfrac {y^2} {s^2} = 1$ where: :$r^2 = a^2 + \lambda$ :$s^2 = b^2 + \lambda$ From Equation of Ellipse in Reduced Form, this is the equation of an ellipse in reduced form. It follows that: :$\tuple {\pm \sqrt {a^2 + \lambda}, 0}$ are the positions of the vertices of $E$ :$\tuple {0, \pm \sqrt {b^2 + \lambda} }$ are the positions of the covertices of $E$ From Focus of Ellipse from Major and Minor Axis, the positions of the foci of $E$ are given by: :$\paren {a^2 + \lambda} - \paren {b^2 + \lambda} = c^2$ where $\tuple {\pm c, 0}$ are the positions of the foci of $E$. Thus we have: {{begin-eqn}} {{eqn | l = c^2       | r = \paren {a^2 + \lambda} - \paren {b^2 + \lambda}       | c = Focus of Ellipse from Major and Minor Axis }} {{eqn | r = a^2 - b^2       | c =  }} {{end-eqn}} Hence the result.  == Also see == * Equation of Confocal Conics * Equation of Confocal Hyperbolas
Equation of Confocal Hyperbolas/Formulation 1
== Definition ==  The equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$ where: :$\tuple {x, y}$ denotes an arbitrary point in the cartesian plane :$a$ and $b$ are (strictly) positive constants such that $a^2 > b^2$ :$\lambda$ is a (strictly) positive parameter such that $b^2 < -\lambda < a^2$ defines the set of all confocal hyperbolas whose foci are at $\tuple {\pm \sqrt {a^2 + b^2}, 0}$.  == Proof == Let $a$ and $b$ be arbitrary (strictly) positive real numbers fulfilling the constraints as defined. Let $E$ be the locus of the equation: :$(1): \quad \dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} = 1$ As $b^2 < -\lambda$ it follows that: :$b^2 + \lambda < 0$ and as $-\lambda < a^2$: :$a^2 + \lambda > 0$ Thus $(1)$ is in the form: :$\dfrac {x^2} {r^2} - \dfrac {y^2} {s^2} = 1$ where: :$r^2 = a^2 + \lambda$ :$s^2 = -\lambda + b^2$ From Equation of Hyperbola in Reduced Form, this is the equation of an hyperbola in reduced form. It follows that: :$\tuple {\pm \sqrt {a^2 + \lambda}, 0}$ are the positions of the vertices of $E$ :$\tuple {0, \pm \sqrt {b^2 - \lambda} }$ are the positions of the covertices of $E$ From Focus of Hyperbola from Transverse and Conjugate Axis, the positions of the foci of $E$ are given by: :$\paren {a^2 + \lambda} + \paren {b^2 - \lambda} = c^2$ where $\tuple {\pm c, 0}$ are the positions of the foci of $E$. Thus we have: {{begin-eqn}} {{eqn | l = c^2       | r = \paren {a^2 + \lambda} + \paren {b^2 - \lambda}       | c = Focus of Hyperbola from Transverse and Conjugate Axis }} {{eqn | r = a^2 + b^2       | c =  }} {{end-eqn}} Hence the result.  == Also see == * Equation of Confocal Conics * Equation of Confocal Ellipses
Method of Undetermined Coefficients/Exponential of Sine and Cosine
== Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form: :$\map R x = e^{a x} \paren {\alpha \sin b x + \beta \cos b x}$ The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$(2): \quad y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$.  Substitute a trial solution of similar form, either: :$e^{a x} \paren {A \sin b x + B \cos b x}$ or replace the  of $(1)$ by: :$\paren {\alpha - i \beta} e^{i \paren {a + i b} x}$ find a solution, and take the real part. If $e^{a x} \sin b x$ and $e^{a x} \cos b x$ appear in the general solution to $(2)$, then insert a factor of $x$: :$x e^{a x} \paren {A \sin b x + B \cos b x}$ or: :$x \paren {\alpha - i \beta} e^{i \paren {a + i b} x}$  == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Linear Second Order ODE/y'' + 4 y = 3 sin 2 x|next = Method of Undetermined Coefficients/Product of Polynomial and Exponential}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(i)}$
Method of Undetermined Coefficients/Product of Polynomial and Exponential
== Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form: :$\map R x = e^{a x} \paren {\map f x}$ where $\map f x$ is a real polynomial function. The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$(2): \quad y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$.  Substitute a trial solution of similar form: :$e^{a x} \paren {\map g x}$ where $\map g x$ is a real polynomial function with undetermined coefficients of as high a degree as $f$. Then: :differentiate twice {{WRT|Differentiation}} $x$ :establish a set of simultaneous equations by equating powers :solve these equations for the coefficients. If $e^{a x} \paren {\map g x}$ appears in the general solution to $(2)$, then add a further degree to $g$. The last step may need to be repeated if that last polynomial also appears as a general solution to $(2)$.  == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Method of Undetermined Coefficients/Exponential of Sine and Cosine|next = Method of Undetermined Coefficients/Product of Polynomial and Function of Sine and Cosine}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(ii)}$
Method of Undetermined Coefficients/Product of Polynomial and Function of Sine and Cosine
== Proof Technique == Consider the nonhomogeneous linear second order ODE with constant coefficients: :$(1): \quad y'' + p y' + q y = \map R x$ Let $\map R x$ be of the form: :$\map R x = \paren {\alpha \cos b x + \beta \sin b x} \paren {\map f x}$ where $\map f x$ is a real polynomial function. The '''Method of Undetermined Coefficients''' can be used to solve $(1)$ in the following manner. == Method and Proof == Let $\map {y_g} x$ be the general solution to: :$(2): \quad y'' + p y' + q y = 0$ From Solution of Constant Coefficient Homogeneous LSOODE, $\map {y_g} x$ can be found systematically. Let $\map {y_p} x$ be a particular solution to $(1)$. Then from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$\map {y_g} x + \map {y_p} x$ is the general solution to $(1)$. It remains to find $\map {y_p} x$.  Substitute a trial solution of similar form, either: :$\paren {\alpha \cos b x + \beta \sin b x} \paren {\map g x}$ or replace the  of $(1)$ by: :$\paren {\alpha - i \beta} e^{i \paren {a + i b} x} \paren {\map g x}$ find a solution, and take the real part. In the above, $\map g x$ is a real polynomial function with undetermined coefficients of as high a degree as $f$. Then: :differentiate twice {{WRT|Differentiation}} $x$ :establish a set of simultaneous equations by equating powers :solve these equations for the coefficients. If $\paren {\alpha \cos b x + \beta \sin b x} \paren {\map g x}$ appears in the general solution to $(2)$, then add a further degree to $g$. The last step may need to be repeated if that last polynomial also appears as a general solution to $(2)$.  == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Method of Undetermined Coefficients/Product of Polynomial and Exponential|next = Method of Undetermined Coefficients/Sum of Several Terms}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.6$ Particular solution: some further cases $\text{(iii)}$
Solution of Constant Coefficient Linear nth Order ODE
{{refactor|Extract the homogeneous case into its own page and improve its rigour and its structure. Then address the particular case by means of Method of Undetermined Coefficients as per usual.}} == Proof Technique == Consider the linear second order ODE with constant coefficients: :$(1): \quad \displaystyle \sum_{k \mathop = 0}^n a_k \dfrac {\d^k y} {d x^k} = \map R x$ where $a_k$ is a constant for $0 \le k \le n$ and $\map R x$ is a function of $x$. The general solution to $(1)$ can be found as follows. ;Find the roots $m_1, m_2, \ldots, m_n$ of the auxiliary equation: :$(2): \quad \displaystyle \sum_{k \mathop = 0}^n a_k m^k = 0$. If $(2)$ has distinct roots, then the general solution $\map {y_g} x$ is of the form: :$y_g = \sum_{k \mathop = 0}^n A_k e^{m_k x}$ If there are repeated roots of $(2)$, further needs to be done. Let $m_j$ be a repeated root of $(2)$ with multiplicity $r$. Then the  of $(2)$ can be written: :$\map P m \paren {m - m_j}^r$ where $\map P m$ is a polynomial which does not contain the factor $m - m_j$. Then the $r$ instances of $m_j$ give rise to the solutions: :$(4): \quad y = A_{j_0} e^{m_j x} + A_{j_1} x e^{m_j x} + \dotsb + A_{j_{r - 1} } x^{r - 1} e^{m_j x}$ == Proof == Let the reduced equation of $(1)$ be written: :$(3): \quad \paren {a_n D^n + a_{n - 1} D^{n - 1} + \dotsb + a_1 D + a_0} y = 0$ By factoring the  we get: :$\map P D \paren {D - m_j}^r y = 0$ It remains to be shown that $(4)$ is a solution to $(3)$. Each of the terms is of the form $p e^{m_j x} u$, where $u$ is a power of $x$ and $p$ is a constant. So: {{begin-eqn}} {{eqn | l = \map {\paren {D - m_j} } {p e^{m_j x} u}       | r = \map D {p e^{m_j x} u} - m_j p e^{m_j x} u       | c =  }} {{eqn | r = p e^{m_j x} D u       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map {\paren {D - m_j}^2} {p e^{m_j x} u}       | r = \map {\paren {D - m_j} } {p e^{m_j x} D u}       | c =  }} {{eqn | r = p e^{m_j x} \map D {D u}       | c =  }} {{eqn | r = p e^{m_j x} \map {D^2} u       | c =  }} {{end-eqn}} and so on until: {{begin-eqn}} {{eqn | l = \map {\paren {D - m_j}^r} {p e^{m_j x} u}       | r = p e^{m_j x} D^r u       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} because $D^r u = 0$ when $u = 1, x, x^2, \ldots, x^{r - 1}$.  == Sources == * {{BookReference|Elementary Differential Equations & Operators|1958|G.E.H. Reuter|prev = Definition:Reduced Equation of Linear ODE with Constant Coefficients|next = ODE/(D^4 - 1) y = sin x}}: Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 3$. Equations of higher order and systems of first order equations: $\S 3.1$ The $n$th order equation
Lindelöf's Lemma/Lemma/Lemma/Lemma
{{rename|This is such a basic result it really should not be hidden as obscurely as a lemma of a lemma of a lemma of a lemma.}} == Lemma ==  Let $S$ be countable set. Let $T$ be a set. Let $T$ be in one-to-one correspondence with $S$. Then $T$ is countable.  == Proof == $S$ is countable. Therefore, $S$ is in one-to-one correspondence with a subset of the natural numbers by a definition of countable set. $T$ is in one-to-one correspondence with $S$. Therefore, $T$ is in one-to-one correspondence with a subset of the natural numbers by Composite of Bijections is Bijection. Accordingly, $T$ is countable by a definition of countable.
Lindelöf's Lemma/Lemma/Lemma
== Lemma ==  Let $R$ be a set of real intervals with rational numbers as endpoints. Let every interval in $R$ be of the same type of which there are four: $\openint \ldots \ldots$, $\closedint \ldots  \ldots$, $\hointr \ldots \ldots$, and $\hointl \ldots \ldots$. Then $R$ is countable.  == Proof == === Lemma 2 === {{:Lindelöf's Lemma/Lemma/Lemma/Lemma|Lemma 2}} {{qed|lemma}} By Rational Numbers are Countably Infinite, the rationals are countable. By Subset of Countably Infinite Set is Countable, a subset of the rationals is countable. The endpoint of an interval in $R$ is characterized by a rational number as every interval in $R$ is of the same type. Therefore, the set consisting of the left hand endpoints of every interval in $R$ is countable. Also, the set consisting of the right hand endpoints of every interval in $R$ is countable. The cartesian product of countable sets is countable. Therefore, the cartesian product of the sets consisting of the respectively left hand and  right hand endpoints of every interval in $R$ is countable. A subset of this cartesian product is in one-to-one correspondence with $R$. This subset is countable by Subset of Countably Infinite Set is Countable. $R$ is countable by Lemma 2 as $R$ is in one-to-one correspondence with a countable set.
Absolute Value of Cut is Greater Than or Equal To Zero Cut
== Definition == Let $\alpha$ be a cut. Let $\size \alpha$ denote the '''absolute value of $\alpha$'''. Then: :$\size \alpha \ge 0^*$ where: :$0^*$ denotes the rational cut associated with the (rational) number $0$ :$\ge$ denotes the ordering on cuts. == Proof == Let $\alpha \ge 0^*$. Then by definition $\size \alpha = \alpha \ge 0^*$. Let $\alpha < 0^*$. Then: :$\exists \beta: \beta + \alpha = 0^*$ Thus: :$\alpha = -\beta$ and it follows that $\beta > 0^*$. The result follows.  == Sources == * {{BookReference|Principles of Mathematical Analysis|1964|ed = 2nd|edpage = Second Edition|Walter Rudin|prev = Definition:Absolute Value of Cut|next = Absolute Value of Cut is Zero iff Cut is Zero}}: Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.24$. Definition
Absolute Value of Cut is Zero iff Cut is Zero
== Definition == Let $\alpha$ be a cut. Let $\size \alpha$ denote the '''absolute value of $\alpha$'''. Then: :$\size \alpha = 0^*$  $\alpha = 0^*$ where $0^*$ denotes the rational cut associated with the (rational) number $0$. == Proof == Let $\alpha 0^*$. Then by definition $\size \alpha = \alpha = 0^*$. Let $\alpha \ne 0^*$. Then either: :$\alpha > 0^*$ in which case $\size \alpha = \alpha > 0^*$ or: :$\alpha < 0^*$ in which case $\size \alpha = -\alpha > 0^*$ In either case $\size \alpha \ne 0^*$. The result follows.  == Sources == * {{BookReference|Principles of Mathematical Analysis|1964|ed = 2nd|edpage = Second Edition|Walter Rudin|prev = Absolute Value of Cut is Greater Than or Equal To Zero Cut|next = Definition:Multiplication of Cuts}}: Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.24$. Definition