hcju/proofwikiqa huggingface dataset:
Let $F$ be a closed set in $X$. Suppose $S \subseteq F$. === $S^-$ is contained in $F$=== Let $L$ be a limit point of $S$. Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S \setminus \set L$ which converges to $L$. In other words: :$\forall n \in \N : x_n \in S \setminus \set L$. Furthermore: :$S \setminus \set L \subseteq S \subseteq F$ Since $F$ is closed, $L \in F$. So all limit points of $S$ belong to $F$. Hence, $S^- \subseteq F$. {{qed|lemma}} === $S^-$ is closed === Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S^-$. Let $\sequence {x_n}_{n \mathop \in \N}$ converge to $L$: :$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {x_n - L} < \epsilon$ We can have either $L \in S$ or $L \notin S$. Suppose $L \in S$. :By definition, $L \in S^-$. Suppose $L \notin S$. Define a new sequence $x_n'$ using $x_n$ as follows: :$(1): \quad$ if $x_n \in S$, then $x_n' := x_n$; :$(2): \quad$ if $x_n \notin S$, then $x_n$ is a limit point of $S$. ::Then there is open ball $\ds \map {B_{\frac 1 n}} {x_n}$ which has an element of $S$. ::Define $x_n'$ such that $x_n' \in S$ and $x_n' \in \map {B_{\frac 1 n}} {x_n}$ Suppose $x_n \in S$. Then: :$\norm {x_n' - L} = \norm {x_n - L}$ Suppose $x_n \notin S$. Then: {{begin-eqn}} {{eqn | l = \norm {x_n' - L}       | r = \norm {x_n' - x_n + x_n - L} }} {{eqn | o = \le       | r = \norm {x_n' - x_n} + \norm {x_n - L}       | c = Norm axiom: triangle inequality }} {{eqn | o = <       | r = \frac 1 n + \norm {x_n - L} }} {{eqn | o = <       | r = \frac 1 n + \epsilon }} {{end-eqn}} Thus, $\sequence {x_n'}_{n \mathop \in \N}$ is a sequence in $S \setminus \set L$ which converges to $L$. So $L$ is a limit point of $S$: :$L \in S^-$. By definition, $S^-$ is closed. {{qed|lemma}} === $S^-$ is the smallest closed set containing $S$ === {{AimForCont}} there exists a closed set $Q$ smaller than $S^-$ which contains $S$. $S^-$ differs from $S$ only by limit points of $S$. If $Q$ is smaller than $S^-$, it has to contain fewer limit points of $S$ than $S^-$. Hence, $Q$ would not contain all its limit points. By definition, $Q$ would not be closed. This is a contradiction. {{qed}}
Let $a = 0$, $b = 1$. Define $f : \closedint 0 1 \to \R$ by:  :$\map f x = \begin{cases}1 & x \in \Q \\ 0 & x \not \in \Q\end{cases}$ For each finite subdivision $P$ of $\closedint 0 1$, write:  :$P = \set {x_0, x_1, \ldots, x_n }$ with: :$0 = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = 1$ For each such subdivision also write: :$\displaystyle \map {V_f} P = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$ It suffices to find a sequence of finite subdivisions $\sequence {P_n}$ such that:  :$\displaystyle \lim_{n \mathop \to \infty} \map {V_f} {P_n} = \infty$ by the definition of bounded variation. For each $n \in \N$, let $P_n$ be a finite subdivision of size $n + 2$. Let $\sequence {x_n}_{0 \mathop \le i \mathop \le n + 2}$ be the sequence forming this subdivision. Note that from Between two Rational Numbers exists Irrational Number: :it is possible to select an irrational number strictly between two rational numbers. Similarly from Between two Real Numbers exists Rational Number: :it is possible to select a rational number strictly between two irrational numbers. We can therefore define $x_i$ for $1 \le i \le n + 1$ as follows: :if $i$ is odd, let $x_i$ be an irrational number between $x_{i - 1}$ and $b$ :if $i$ is even, let $x_i$ be a rational number between $x_{i - 1}$ and $b$. Let $x_0 = 0$ and $x_{n + 2} = 1$ to complete the definition. Then: {{begin-eqn}} {{eqn	| l = \map {V_f} {P_n} 	| r = \sum_{i \mathop = 1}^{n + 2} \size {\map f {x_i} - \map f {x_{i - 1} } } }} {{eqn	| o = \ge 	| r = \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } } }} {{end-eqn}} We omit the term $i = n + 2$ for convenience. Note that: :if $x_i$ is rational then $x_{i + 1}$ is irrational. That is: :if $\map f {x_i} = 1$ then $\map f {x_{i + 1} } = 0$. Similarly: :if $x_i$ is irrational then $x_{i + 1}$ is rational. That is: :if $\map f {x_i} = 0$ then $\map f {x_{i + 1} } = 1$. So for all $1 \le i \le n + 1$, we have: :$\size {\map f {x_i} - \map f {x_{i - 1} } } = 1$ Hence: :$\displaystyle \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } } = n + 1$ So: :$\displaystyle \map {V_f} {P_n} \ge n + 1$ Giving: :$\displaystyle \lim_{n \mathop \to \infty} \map {V_f} {P_n} = \infty$ as required. {{qed}}
From Element is Loop iff Rank is Zero: :$x$ is a loop {{iff}} $\map \rho {\set x} = 0$ where $\rho$ is the rank function of $M$. Now: {{begin-eqn}} {{eqn | r = x \in \map \sigma \O       | o = }} {{eqn | ll= \leadstoandfrom       | r = x \sim \O       | o =       | c = {{Defof|Closure Operator (Matroid)|Closure Operator}} }} {{eqn | ll= \leadstoandfrom       | r = \map \rho {\set x} = \map \rho \O       | o =       | c = {{Defof|Depends Relation (Matroid)|Depends Relation}} }} {{eqn | ll= \leadstoandfrom       | r = \map \rho {\set x} = 0       | o =       | c = Rank of Empty Set is Zero }} {{end-eqn}} The result follows. {{qed}}
By definition of a loop: :$x$ is a loop {{iff}} $\set x \notin \mathscr I$ From Singleton is Dependent implies Rank is Zero: :if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$ From Singleton is Independent implies Rank is One: :if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$ It follows that: :$\set x \notin \mathscr I$ {{iff}} $\map \rho {\set x} = 0$ {{qed}}
Let $x$ be a loop. By definition of a loop: :$\set x \notin \mathscr I$ By definition of a dependent subset: :$\set x$ is a dependent subset From Singleton of Element is Subset: :$\set x \subseteq A$ From Superset of Dependent Set is Dependent: :$A$ is a dependent subset {{qed}}
By definition of the closure of $A$: :$x \in \map \sigma A$ {{iff}} $x \sim A$ where $\sim$ is the depends relation on $M$. By definition of the depends relation: :$x \sim A$ {{iff}} $\map \rho {A \cup \set x} = \map \rho A$ where $\rho$ is the rank funtion on $M$. So it remains to show that: :$\map \rho {A \cup \set x} = \map \rho A$ By definition of the rank function: :$\map \rho {A \cup \set x} = \max \set {\size X : X \subseteq A \cup \set x \land X \in \mathscr I}$ From Max Equals an Operand: :$\exists X \in \mathscr I : X \subseteq A \cup \set x \land \size X = \map \rho {A \cup \set x}$ From the contrapositive statement of Set is Dependent if Contains Loop: :$x \notin X$ Now: {{begin-eqn}} {{eqn | l = X       | r = \paren{A \cup \set x} \cap X       | c = Intersection with Subset is Subset }} {{eqn | r = \paren{A \cap X} \cup \paren {\set x \cap X}       | c = Intersection Distributes over Union }} {{eqn | r = \paren{A \cap X} \cup \O       | c = Intersection With Singleton is Disjoint if Not Element }} {{eqn | r = \paren{A \cap X}       | c = Union with Empty Set }} {{eqn | ll = \leadsto       | l = X       | o = \subseteq       | r = A       | c = Intersection with Subset is Subset }} {{end-eqn}} From Max yields Supremum of Operands: :$\size X \le \max \set {\size Y : Y \subseteq A \land Y \in \mathscr I} = \map \rho A$ From Rank Function is Increasing: :$\map \rho A \le \map \rho {A \cup \set x} = \size X$ Thus: :$\map \rho A = \size X = \map \rho {A \cup \set x}$ {{qed}}
=== Necessary Condition === Let $x$ be a loop. By definition of a loop: :$\set x$ is a dependent subset of $S$ Let $A \subseteq \set x$ be a dependent subset. From Power Set of Singleton: :$\powerset {\set x} = \set{\O, \set x}$ By matroid axiom $(\text I 1)$: :$\O$ is an independent subset Then: :$A = \set x$ It follows that: :$\set x$ is a minimal dependent subset of $S$. Then $\set x$ be a circuit by definition.  {{qed|lemma}} === Sufficient Condition === Let $\set x$ be a circuit. By definition of a circuit: :$\set x$ is a minimal dependent subset of $S$. In particular, $\set x$ is a dependent subset of $S$. Then $\set x$ is a loop by definition. {{qed}}
=== Necessary Condition === Let $B \in \mathscr B$ such that $x \in B$. From Singleton of Element is Subset: :$\set x \subseteq B$ By definition of a base: :$B \in \mathscr I$ From matroid axiom $(\text I 2)$: :$\set x \in \mathscr I$ Then $\set x$ is not a dependent subset by definition. It follows that $x$ is not a loop by definition. {{qed|lemma}} === Sufficient Condition === Let $x$ not be a loop. By definition of a loop: :$x$ is not a dependent subset By definition of a dependent subset: :$x \in \mathscr I$ From Independent Subset is Contained in Base: :$\exists B \in \mathscr B: \set x \subseteq B$ By definition of a subset: :$x \in B$ {{qed}}
=== Necessary Condition === Let $x$ and $y$ be parallel. By definition of parallel: :$\set x$ is independent  :$\set y$ is independent  :$\set {x, y}$ is dependent Let $A \subseteq \set {x, y}$ be dependent. Thus: :$A \ne \set x, \set y$ By matroid axiom $(\text I 1)$: :$\O$ is independent Thus: :$A \ne \O$ From Power Set of Doubleton: :$\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$ Thus: :$A = \set {x, y}$ By definition of a minimal set: :$\set {x, y}$ is a minimal dependent subset It follows that $\set {x, y}$ is a circuit by definition. {{qed|lemma}} === Sufficient Condition === Let $\set {x, y}$ by a circuit. By definition of a circuit: :$\set {x, y}$ is a minimal dependent subset By definition of a subset: :$\set x, \set y \subseteq \set {x, y}$ By definition of a minimal dependent subset:  :$\set x, \set y$ are independent subsets It follows that $x$ and $y$ are parallel by definition. {{qed}}
Let $x$ be parallel to $y$ and $y$ be parallel to $z$. By definition of parallel: :$\set x$, $\set y$, $\set z$ are independent subsets :$\set {x, y}$, $\set {y, z}$ are dependent subsets To show that $x$ is parallel to $z$ it remains to show that: :$\set {x, z}$ is dependent {{AimForCont}} $\set {x, z}$ is independent. By matroid axiom $(\text I 3)$: :$\exists w \in \set{x, z} \setminus \set y : \set{w, y} \in \mathscr I$ By definition of the doubleton: :$\set{x, y} \in \mathscr I \lor \set{z, y} \in \mathscr I$ This contradicts the assumption that $\set {x, y}$, $\set {y, z}$ are dependent subsets. It follows that: :$\set {x, z}$ is dependent {{qed}}
=== Lemma === {{:Distinct Matroid Elements are Parallel iff Each is in Closure of Other/Lemma}}{{qed|lemma}} === Necessary Condition === Let $x$ and $y$ be parallel. By definition of parallel: :$\set x$ and $\set y$ are independent  :$\set {x, y}$ is dependent By definition of a loop: :$x$ and $y$ are not loops  From Lemma: :$x \in \map \sigma {\set y}$ :$y \in \map \sigma {\set x}$ It has been shown that conditions $(1), (2)$ and $(3)$ above hold. {{qed|lemma}} === Sufficient Condition === Let conditions $(1), (2)$ and $(3)$ above hold. By definition of a loop: :$\set x$ and $\set y$ are independent  From Lemma: :$\set {y, x} \notin \mathscr I$ It follows that $x$ is parallel to $y$ by definition.  {{qed}}
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$. Let $x \in \map \sigma A$ Let $y$ be parallel to $x$. By the definitions of the closure operator and depends depends: :$\map \rho A = \map \rho {A \cup \set x}$ and: :$y \in \map \sigma A$ {{iff}} $\map \rho A = \map \rho {A \cup \set y}$ From Rank Function is Increasing: :$\map \rho {A \cup \set x} = \map \rho A \le \map \rho {A \cup \set y} \le \map \rho {A \cup \set {x,y}}$ To show $\map \rho A = \map \rho {A \cup \set y}$ it is sufficient to show: :$\map \rho {A \cup \set {x,y}} = \map \rho {A \cup \set x}$ From Rank Function is Increasing: :$\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x,y}}$ By the definition of parallel: :$\set x$ is independent From Independent Subset is Contained in Maximal Independent Subset: :$\exists X \in \mathscr I : \set x \subseteq X \subseteq A \cup \set x : \size X = \map \rho {A \cup \set x}$ From Subset Relation is Transitive: :$X \subseteq A \cup \set {x, y}$ From Independent Subset is Contained in Maximal Independent Subset: :$\exists Y \in \mathscr I : X \subseteq Y \subseteq A \cup \set {x,y} : \size Y = \map \rho {A \cup \set {x,y}}$ By definition of a subset: :$x \in Y$ By definition of parallel: :$\set{x, y}$ is dependent From Superset of Dependent Set is Dependent: :$\set{x, y} \not \subseteq Y$ Then: :$y \notin Y$ So: :$Y \subseteq A \cup \set x$ By definition of the rank function: :$\size Y \le \map \rho {A \cup \set x}$ As $\size Y = \map \rho {A \cup \set {x,y}}$ then: :$\map \rho {A \cup \set {x,y}} \le \map \rho {A \cup \set x}$ As $\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x,y}}$ then: :$\map \rho {A \cup \set {x,y}} = \map \rho {A \cup \set x}$ The result follows. {{qed}}
Let $x, y \in A$.   From Doubleton of Elements is Subset: :$\set{x, y} \subseteq A$ By the definition of parallel elements: :$\set {x, y}$ is dependent From Superset of Dependent Set is Dependent: :$A$ is dependent {{qed}}
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$. We proceed by Proof by Contraposition. That is, it is shown that: :if $x \notin A$ then either $x$ is not a loop or $A$ is not a flat subset Let $x \notin A$. Let $x$ be a loop. By definition of a loop: :$\set x$ is a dependent subset. From Rank Function is Increasing: :$\map \rho A \le \map \rho {A \cup \set x}$ Let $X \in \mathscr I$ such that $X \subseteq A \cup \set x$. From Superset of Dependent Set is Dependent: :$\set x \not \subseteq X$ From Singleton of Element is Subset: :$x \notin X$ So: :$X \subseteq A$ By definition of the rank function: :$\size X \le \map \rho A$ From Max yields Supremum of Parameters: :$\map \rho {A \cup \set x} = \max \set{\size X : X \in \mathscr I \land X \subseteq A \cup \set x} \le \map \rho A$ Then: :$\map \rho A = \map \rho {A \cup \set x}$ If follows that $A$ is not a flat subset by definition. It has been shown that: :if $x \notin A$ then either $x$ is not a loop or $A$ is not a flat subset The theorem holds by the Rule of Transposition. {{qed}}
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$. === Necessary Condition === Let $M$ contain no loops. By definition of a loop: :$\forall x \in S : \set x \in \mathscr I$ Let $x \in S \setminus \O$. Then: {{begin-eqn}} {{eqn | l = \map \rho {\O \cup \set x}       | r = \map \rho {\set x}       | c = Union with Empty Set }} {{eqn | r = \size {\set x}       | c = Rank of Independent Subset Equals Cardinality }} {{eqn | r = 1       | c = Cardinality of Singleton }} {{eqn | r = 0 + 1 }} {{eqn | r = \map \rho \O + 1       | c = Rank of Empty Set is Zero }} {{end-eqn}} It follows that $\O$ is a flat subset by definition. {{qed|lemma}} === Sufficient Condition === Let $\O$ be a flat subset. Let $x \in S$. From Set Difference with Empty Set is Self :$x \in S \setminus \O$ {{begin-eqn}} {{eqn | l = \map \rho {\set x}       | r = \map \rho {\O \cup \set x}       | c = Union with Empty Set }} {{eqn | r = \map \rho \O + 1       | c = {{Defof|Flat (Matroid)|Flat Subset}} }} {{eqn | r = 0 + 1       | c = Rank of Empty Set is Zero }} {{eqn | r = 1 }} {{end-eqn}} From Element is Loop iff Rank is Zero: :$x$ is not a loop It follows that $M$ contains no loops. {{qed}}
Let $\epsilon$ be a positive real number. Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$ Consider specifically the case $n = 1$. From the absolute continuity of $f$, we have that whenever $a \le x \le y \le b$ and: :$y - x < \delta$ we have: :$\size {\map f x - \map f y} < \epsilon$ Notice however that:  :$\size {\map f x - \map f y} = \size {\map f y - \map f x}$ We therefore have: :$\size {\map f x - \map f y} < \epsilon$ when: :$x - y < \delta$ as well. So, in fact, for all $x, y \in \closedint a b$ with: :$\size {x - y} < \delta$ we have: :$\size {\map f x - \map f y} < \epsilon$ Since $\epsilon$ was arbitrary: :$f$ is uniformly continuous. {{qed}} Category:Uniformly Continuous Real Functions Category:Absolutely Continuous Functions g3gmkhld8qrtscjl505md6oi47b9sqy
Suppose $y \in Y^-$. Then there is a sequence $\displaystyle \sequence {y_n}_{n \mathop \in \N} \in Y$ which converges to $y$. Suppose $y \in Y$ and $y$ is a limit point. Then one can define a constant sequence $\sequence {y_n}_{n \mathop \in \N} = y$. === Closed under restriction of vector addition === Let $x, y \in Y^-$. Let $\sequence {x_n}_{n \mathop \in \N}, \sequence {y_n}_{n \mathop \in \N}$ be sequences in $Y$. Suppose: :$\displaystyle \lim_{n \mathop \to \infty} x_n = x$ :$\displaystyle \lim_{n \mathop \to \infty} y_n = y$ Since $Y$ is a subspace: :$\forall n \in \N : x_n + y_n \in Y \subseteq Y^-$ Furthermore: :$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = x + y$ $Y^-$ is closed and contains its limit points. Hence, $x + y \in Y^-$. {{qed|lemma}} === Closed under restriction of scalar multiplication === Let $\alpha \in \Bbb K$, where $\Bbb K$ is a field. Let $y \in Y^-$. Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $Y$. Suppose: :$\displaystyle \lim_{n \mathop \to \infty} y_n = y$ Since $Y$ is a subspace: :$\forall n \in \N : \alpha \cdot y_n \in Y \subseteq Y^-$ Furthermore: :$\displaystyle \lim_{n \mathop \to \infty} \alpha \cdot y_n = \alpha \cdot y$. $Y^-$ is closed and contains its limit points. Hence: :$\alpha \cdot y \in Y^-$ {{qed|lemma}} === Nonemptiness === $Y$ is a subspace. Hence: :$0 \in Y \subseteq Y^-$ Thus $Y^-$ contains at least one element and is non-empty. {{qed|lemma}} Hence, $Y^-$ is closed subspace {{qed}}
Let $x, y \in C^-$. Suppose $x, y$ are limit points. Then there are sequences $\sequence {x_n}_{n \mathop \in \N}, \sequence {x_n}_{n \mathop \in \N}$ in $C$, such that: :$\displaystyle \lim_{n \mathop \to \infty} x_n = x$ :$\displaystyle \lim_{n \mathop \to \infty} x_y = y$ Let $\alpha \in \closedint 0 1$. Then: {{begin-eqn}} {{eqn | l = \paren {1 - \alpha} x + \alpha y       | r = \paren {1 - \alpha} \lim_{n \mathop \to \infty} x_n + \alpha \lim_{n \mathop \to \infty} y_n }} {{eqn | r = \lim_{n \mathop \to \infty} \paren {\paren {1 - \alpha} x_n + \alpha y_n} }} {{end-eqn}} Since $C$ is convex: :$\displaystyle \forall n \in \N : \paren {1 - \alpha} x_n + \alpha y_n \in C \subseteq C^-$ $C^-$ is closed and contains its limit points. Hence: :$\paren {1 - \alpha} x + \alpha y \in C^-$ {{qed}}
It suffices to show the result for prime values of $n$. The case $n = 2$ is covered in 1 plus Square is not Perfect Power. So we consider the cases where $n$ is an odd prime. {{begin-eqn}} {{eqn | l = x^m       | r = y^n + 1 }} {{eqn | r = y^n - \paren {-1}^n       | c = as $n$ is odd }} {{eqn | r = \paren {y - \paren {-1} } \sum_{j \mathop = 0}^{n - 1} y^{n - j - 1} \paren {-1}^j       | c = Difference of Two Powers }} {{eqn | r = \paren {y + 1} \paren {\map Q y \paren {y + 1} + R}       | c = Division Theorem for Polynomial Forms over Field }} {{end-eqn}} where $Q$ is a polynomial in one unknown and $R$ is a degree zero polynomial, so $R$ is a constant. We have: {{begin-eqn}} {{eqn | l = R       | r = \map Q {-1} \paren {-1 + 1} + R }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - j - 1} \paren {-1}^j }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - 1} }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} 1       | c = as $n$ is odd }} {{eqn | r = n }} {{end-eqn}} Hence we have $x^m = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}$. Since $x$ is a prime, we have: :$x \divides y + 1$ :$x \divides \map Q y \paren {y + 1} + n$ Hence $x \divides n$. Since $x > 1$ and $n$ is a prime, we must have $x = n$. Now we write $y + 1 = x^\alpha$. Then we have: {{begin-eqn}} {{eqn | l = x^m       | r = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n} }} {{eqn | r = x^\alpha \paren {\map Q y x^\alpha + x} }} {{eqn | r = x^{\alpha + 1} \paren {\map Q y x^{\alpha - 1} + 1} }} {{end-eqn}} For $\alpha > 1$, $x \nmid \map Q y x^{\alpha - 1} + 1$. Hence $\alpha = 1$. This gives $y + 1 = x = n$. The equation now simplifies to: :$\paren {y + 1}^m = y^n + 1$ Expanding: {{begin-eqn}} {{eqn | l = \paren {y + 1}^m       | r = \sum_{j \mathop = 0}^m \binom m j y^j 1^{m - j}       | c = Binomial Theorem }} {{eqn | r = 1 + \sum_{j \mathop = 1}^m \binom m j y^j }} {{eqn | r = y^n + 1 }} {{eqn | ll = \leadsto       | l = \sum_{j \mathop = 1}^m \binom m j y^{j - 1}       | r = y^{n - 1} }} {{eqn | ll = \leadsto       | l = \binom m 1 y^0       | r = 0       | rr = \pmod y       | c = as $y > 1$ }} {{eqn | ll = \leadsto       | l = m       | r = 0       | rr = \pmod y       | c = Binomial Coefficient with One }} {{end-eqn}} hence we must have $y \divides m$. By Absolute Value of Integer is not less than Divisors, $y \le m$. Moreover, from $\displaystyle \sum_{j \mathop = 1}^m \binom m j y^{j - 1} = y^{n - 1}$ we also have: :$y^{n - 1} > \dbinom m m y^{m - 1} = y^{m - 1}$ Therefore we also have $n > m$. This gives $y = n - 1 \ge m$. The two inequalities forces $y = m$. Now our original equation is further simplified to: {{begin-eqn}} {{eqn | l = \paren {y + 1}^y       | r = y^{y + 1} + 1 }} {{eqn | ll = \leadsto       | l = \paren {1 + \frac 1 y}^y       | r = y + \frac 1 {y^y}       | c = Dividing both sides by $y^y$ }} {{end-eqn}} From Real Sequence (1 + x over n)^n is Convergent: :$\paren {1 + \dfrac 1 y}^y$ is increasing and has limit $e$. Then we have for all $y \in \N$: :$y + \dfrac 1 {y^y} < e < 3$ Since $\dfrac 1 {y^y} > 0$ and $y > 1$, we can only have $y = 2$. This gives the solution $3^2 = 2^3 + 1$, and there are no others. {{qed}}
Let $F$ be a field. Let $I \subset F$ be a non-null ideal of $F$. Let $a \in I$ be non-zero. Since $F$ is a field, $a^{-1}$ exists. We have that $1 = a^{-1} \cdot a \in I$. Since $1 \in I$, for every element $b \in F$: :$b = b \cdot 1 \in I$ we have that $I = F = \ideal 1$ if $I \ne \set 0$. Thus the only ideals of $F$ are $\ideal 0 = \set 0$ and $\ideal 1 = F$, which are both principal ideals. Hence $F$ is a principal ideal domain. {{qed}} Category:Principal Ideal Domains Category:Field Theory 82o7uz5ng4f9faonjp1mx046d66y0sm
=== Necessary Condition === {{:Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition}}{{qed}}
Let $x \in X \setminus D$. Suppose $D$ is dense in $X$. Then: :$\forall n \in N : \exists d_n \in D : d_n \in \map {B_{\frac 1 n}} x$ where $\displaystyle \map {B_{\frac 1 n}} x$ is an open ball. Let $\sequence {d_n}_{n \mathop \in \N}$ be a sequence in $D$. Then: :$\forall n \in \N : \norm {x - d_n} < \frac 1 n$ Hence, $x$ is a limit point of $D$. In other words, $x \in D^-$. We have just shown that: :$x \in X \setminus D \implies x \in D^-$ Hence: :$X \setminus D \subseteq D^-$. By definition of closure: :$D \subseteq D^-$ Therefore: {{begin-eqn}} {{eqn | l = X       | r = D \cup \paren {X \setminus D} }} {{eqn | o = \subseteq       | r = D^- }} {{eqn | o = \subseteq       | r = X }} {{end-eqn}} Thus: :$X = D^-$.
Let $X = D^-$. We have to show, that for every $x \in X$ there is an open ball with an element from $D^-$. We have that $X = D \cup \paren {X \setminus D}$. Suppose $x \in X \setminus D$. Then $x \in D^- \setminus D$. Hence, $x$ is a limit point of $D$. Therefore, there is a sequence $\sequence {d_n}_{n \mathop \in \N}$ in $D$ which converges to $x$. Thus: :$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \norm {x - d_N} < \epsilon$ In other words: :$\displaystyle d_N \in D \implies d_N \in \map {B_\epsilon} x$ Therefore: :$d_N \in D \cap \map {B_\epsilon} x$. Suppose $x \in D$. Let $\epsilon > 0$. Then $x \in \map {B_\epsilon} x \cap D$. From both parts and definition we conclude that $D^-$ is dense in $X$.
By the Reverse Triangle Inequality, we have: :$\cmod {\norm {x_n} - \norm x} \le \norm {x_n - x}$ Hence by the Squeeze Theorem and Convergent Sequence Minus Limit, $\norm {x_n} \to \norm x$ as $n \to \infty$. {{Qed}}