hcju/proofwikiqa huggingface dataset:
Let $\mathbf a = \left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert\mathbf u$. Then: {{begin-eqn}} {{eqn | l=\cos \angle \mathbf u, \mathbf a       | r=\frac {\mathbf u \cdot \mathbf a} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= from Cosine Formula for Dot Product }} {{eqn | r=\frac {\mathbf u \cdot \left({ \left\Vert{ \mathbf u }\right\Vert \mathbf v + \left\Vert{ \mathbf v }\right\Vert \mathbf u }\right)} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a} \right\Vert}       | c= }} {{eqn | r=\frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf u }\right) } {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= from Properties of Dot Product }} {{eqn | r=\frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf u }\right\Vert^2} {\left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= from Dot Product of Vector with Itself }} {{eqn | r=\frac {\mathbf u \cdot \mathbf v + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert} {\left\Vert{ \mathbf a }\right\Vert}       | c= }} {{end-eqn}} {{begin-eqn}} {{eqn | l=\cos \angle \mathbf a, \mathbf v       | r=\frac {\mathbf v \cdot \mathbf a} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= from Cosine Formula for Dot Product }} {{eqn | r=\frac {\mathbf v \cdot \left({ \left\Vert{ \mathbf u }\right\Vert \mathbf v + \left\Vert{ \mathbf v }\right\Vert \mathbf u }\right)} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= }} {{eqn | r=\frac {\left\Vert{ \mathbf u }\right\Vert \left({ \mathbf v \cdot \mathbf v }\right) + \left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right)} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= from Properties of Dot Product }} {{eqn | r=\frac {\left\Vert{ \mathbf v }\right\Vert \left({ \mathbf u \cdot \mathbf v }\right) + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert^2} {\left\Vert{ \mathbf v }\right\Vert \left\Vert{ \mathbf a }\right\Vert}       | c= Dot Product of Vector with Itself }} {{eqn | r=\frac {\mathbf u \cdot \mathbf v + \left\Vert{ \mathbf u }\right\Vert \left\Vert{ \mathbf v }\right\Vert} {\left\Vert{ \mathbf a }\right\Vert}       | c= }} {{end-eqn}} Comparing the two expressions gives us: :$\cos \angle \mathbf u, \mathbf a = \cos \angle \mathbf a, \mathbf v$ Since the angle used in the dot product is always taken to be between $0$ and $\pi$ and cosine is injective on this interval (from Shape of Cosine Function), we have: :$\angle \mathbf u, \mathbf a = \angle \mathbf a, \mathbf v$ The result follows. {{qed}}
From Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $\struct {G, \circ}$. Suppose $\struct {\set g, \circ}$ is a subgroup of $\struct {G, \circ}$. From Group is not Empty, $e \in \set g$. Thus it follows trivially that $\struct {\set g, \circ} = \struct {\set e, \circ}$. That is, $\struct {\set e, \circ}$ is the only subgroup of $\struct {G, \circ}$ of order $1$. {{qed}}
From Injection from Finite Set to Itself is Surjection, $f$ is a surjection. As $f$ is thus both an injection and a surjection, $f$ is a bijection by definition. Thus as $f$ is a bijection to itself, it is by definition a permutation. {{qed}}
From Surjection iff Right Inverse, $f$ has a right inverse $g: S \to S$. From Right Inverse Mapping is Injection, $g$ is an injection. From Injection from Finite Set to Itself is Permutation, $g$ is a permutation and so a bijection. From Inverse of Bijection is Bijection, $f$ is also a bijection. Thus as $f$ is a bijection to itself, it is by definition a permutation. {{qed}}
Let $\gcd \set {m, n} = d$. From Integers Divided by GCD are Coprime: there exists $m', n' \in \Z$ such that $m = d m'$, $n = d n'$. Then: {{begin-eqn}} {{eqn | l = \paren {g^m}^{n'}       | r = \paren {g^{d m'} }^{n'}       | c = Definition of $m'$ }} {{eqn | r = \paren {g^{d n'} }^{m'}       | c =  }} {{eqn | r = \paren {g^n}^{m'}       | c = Definition of $n'$ }} {{eqn | r = e^{m'}       | c = $n$ is the order of $g$ }} {{eqn | r = e       | c =  }} {{end-eqn}} By Element to Power of Multiple of Order is Identity: :$\order {g^m} \divides n'$. {{AimForCont}} $\order {g^m} = n'' < n'$. By Bézout's Lemma: :$\exists x, y \in \Z: m x + n y = d$ {{begin-eqn}} {{eqn | l = g^{d n''}       | r = g^{\paren {m x + n y} n''}       | c =  }} {{eqn | r = \paren {g^{m n''} }^x \paren {g^n}^{y n''}       | c =  }} {{eqn | r = e^x \paren {g^n}^{y n''}       | c = $n''$ is the order of $g^m$ }} {{eqn | r = e^x e^{y n''}       | c = $n$ is the order of $g$ }} {{eqn | r = e       | c =  }} {{end-eqn}} But $d n'' < d n' = n$, contradicting the fact that $n$ is the order of $g$. Therefore: :$\order {g^m} = n'$ Recalling the definition of $n'$: :$\order {g^m} = \dfrac n {\gcd \set {m, n} }$ as required. {{qed}}
From Existence of Cyclic Group of Order n, there exists a cyclic group whose order is $n$. In particular, a concrete example of such a group is demonstrated in Roots of Unity under Multiplication form Cyclic Group. {{qed}}
Let $\map \mu X$ denote the Lebesgue measure of a set $X$ of real numbers. We have that: :$(1): \quad$ $\map \mu X$ is a countably additive function :$(2): \quad$ $\map \mu X$ is translation-invariant :$(3): \quad$ From Measure of Interval is Length, $\map \mu {\closedint a b} = b - a$ for every closed interval $\closedint a b$. For all real numbers in the closed unit interval $\mathbb I = \closedint 0 1$, define the relation $\sim$ such that: :$\forall x, y \in \mathbb I: x \sim y \iff x - y \in \Q$ where $\Q$ is the set of rational numbers. That is, $x \sim y$ {{iff}} their difference is rational. By Non-Measurable Set of Real Numbers: Lemma 1, $\sim$ is an equivalence relation. For each $x \in \mathbb I$, let $\eqclass x \sim$ denote the equivalence class of $x$ under $\sim$. By the Axiom of Choice, it is possible to choose $1$ element from each equivalence class. Thus is created a set $M \subset \mathbb I$ such that: :for each $x \in R$ there exists a unique $y \in M$ and $r \in \Q$ such that $x = y + r$. Let: :$M_r = \set {y + r: y \in M}$ for each $r \in \Q$. Thus $\R$ is partitioned into countably many disjoint sets: :$(1): \quad \displaystyle \R = \bigcup \set {M_r: r \in \Q}$ Suppose $M$ were Lebesgue measurable. First, $\map \mu M = 0$ is impossible, because from $(1)$ this would mean  $\map \mu \R = 0$. Suppose $\map \mu M > 0$. Then: {{begin-eqn}} {{eqn | l = \map \mu {\closedint 0 1}       | o = \ge       | r = \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }       | c =  }} {{eqn | r = \sum_{r \mathop \in \Q \mathop \land 0 \mathop \le r \mathop \le 1} \map \mu {M_r}       | c =  }} {{eqn | r = \infty       | c = as each $M_r$ would have to have the same measure as $M$ }} {{end-eqn}} So $\map \mu M > 0$ is also impossible. Hence $M$ is not Lebesgue measurable. {{qed}} {{refactor|Put this statement into a "example" page or something}} The set $M$ described here is an example of a Vitali set: a subset of the real numbers which has no Lebesgue measure. {{AoC}} {{Namedfor|Giuseppe Vitali|cat = Vitali}}
For any group $\struct {G, \circ}$ of order $n$ and any set of $n$ elements, $X$ can be the underlying set of a group which is isomorphic to $\struct {G, \circ}$, as follows: Choose a bijection $\phi: G \to X$. Define the group operation $*$ on $X$ by the rule: :$\map \phi {g_1} * \map \phi {g_2} = \map \phi {g_1 \circ g_2}$ for all $g_1, g_2 \in G$. By definition, $\phi$ is an isomorphism. From Isomorphism Preserves Groups it follows that $\struct {X, *}$ is a group. Thus all groups of order $n$ of all possible types appear among all possible arrangements of binary operation. But from Count of Binary Operations on Set, this count is $n^{\paren {n^2} }$. This is the upper bound to the number of types of group of order $n$. {{qed}}
Let $G_1$ and $G_2$ be groups of order $p$. From Prime Group is Cyclic, $G_1$ and $G_2$ are both cyclic groups. From Cyclic Groups of Same Order are Isomorphic, $G_1$ and $G_2$ are isomorphic. Thus by definition, $G_1$ and $G_2$ are of the same type. Hence the result. {{qed}}
By definition: :$\RR \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}$ :$\map \RR x = \set {y \in T:\tuple {x, y} \in \RR}$ First: {{begin-eqn}} {{eqn | l = y        | o = \in       | r = \RR \sqbrk X }} {{eqn | ll= \leadsto       | l = \exists x \in X: \tuple {x, y}       | o = \in       | r = \RR       | c = Definition of $\RR \sqbrk X$ }} {{eqn | ll= \leadsto       | l = \tuple {x, y}       | o = \in       | r = \bigcup_{x \mathop \in X} \set {\tuple {x, y} \in \RR}       | c = {{Defof|Set Union}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \bigcup_{x \mathop \in X} \set {y \in T: \tuple {x, y} \in \RR}       | c = {{Defof|Relation}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \bigcup_{x \mathop \in X} \map \RR x       | c = Definition of $\map \RR x$ }} {{eqn | ll= \leadsto       | l = \RR \sqbrk X       | o = \subseteq       | r = \bigcup_{x \mathop \in X} \map \RR x       | c = {{Defof|Subset}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = \bigcup_{x \mathop \in X} \map \RR x       | c =  }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \bigcup_{x \mathop \in X} \set {y \in T: \tuple {x, y} \in \RR}       | c = Definition of $\map \RR x$ }} {{eqn | ll= \leadsto       | l = \exists x \in X: \tuple {x, y}       | o = \in       | r = \RR       | c = {{Defof|Set Union}} }} {{eqn | ll= \leadsto       | l = y        | o = \in       | r = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}       | c = {{Defof|Relation}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \RR \sqbrk X       | c = Definition of $\RR \sqbrk X$ }} {{eqn | ll= \leadsto       | l = \bigcup_{x \mathop \in X} \map \RR x       | o = \subseteq       | r = \RR \sqbrk X       | c = {{Defof|Subset}} }} {{end-eqn}} So: :$\displaystyle \bigcup_{x \mathop \in X} \map \RR x \subseteq \RR \sqbrk X$ and: :$\displaystyle \RR \sqbrk X \subseteq \bigcup_{x \mathop \in X} \map \RR x$ The result follows by definition of set equality. {{qed}} Category:Relation Theory Category:Subsets Category:Set Union r4xlio6zh4r18w9i8hn8mtcy1wot3bh
By definition, $\RR^{-1} \subseteq T \times S$ is a relation on $T \times S$. Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly. {{qed}} Category:Relation Theory qe7ehp1ro09ic0fjgyeilan3myg8mvl
By definition, $f^{-1} \subseteq T \times S$ is a relation on $T \times S$. Thus Image of Subset under Relation equals Union of Images of Elements can be applied directly. {{qed}} Category:Preimages under Mappings Category:Subsets Category:Set Union a3ik6tclkfa5owlyn526l44dveapcw9
By definition, a mapping is a relation. Thus Image of Domain of Relation is Image Set applies. {{qed}}
By definition, a mapping is a relation. Thus Image of Singleton under Relation applies. {{Qed}}
By definition, a mapping is a relation. Thus Image of Subset under Relation equals Union of Images of Elements applies. {{Qed}}
From Image of Intersection under Mapping: :$f \sqbrk {S \cap T} \subseteq f \sqbrk S \cap f \sqbrk T$ From Empty Set is Subset of All Sets: :$f \sqbrk {S \cap T} = \O$ From Image of Subset under Relation is Subset of Image: Corollary 2: :$S \cap T = \O$ {{qed}}
A mapping is by definition a left-total relation. Therefore Preimage of Image under Left-Total Relation is Superset applies: :$A \subseteq S \implies A \subseteq f^{-1} \sqbrk {f \sqbrk A}$ Thus: :$\set x \subseteq S \implies \set x \subseteq f^{-1} \sqbrk {f \sqbrk A}$ Hence the result. {{qed}}
A mapping is by definition a relation. Therefore Image of Preimage under Mapping: Corollary applies: :$B \subseteq \Img S \implies \paren {f \circ f^{-1} } \sqbrk B = B$ Thus: :$\set y \subseteq T \implies f^{-1} \sqbrk {f \sqbrk {\set y} } = \set y$ Hence the result. {{qed}}
An injection is a type of one-to-one relation, and therefore also a one-to-many relation. Therefore One-to-Many Image of Set Difference applies: :$\mathcal R \left[{S_1}\right] \setminus \mathcal R \left[{S_2}\right] = \mathcal R \left[{S_1 \setminus S_2}\right]$ {{iff}} $\mathcal R$ is one-to-many. We have that $f$ is a mapping and therefore a many-to-one relation. So $f$ is a one-to-many relation {{iff}} $f$ is also an injection. It follows that: :$\forall S_1, S_2 \subseteq S: f \left[{S_1}\right] \setminus f \left[{S_2}\right] = f \left[{S_1 \setminus S_2}\right]$ {{iff}} $f$ is an injection. {{qed}}
{{begin-eqn}} {{eqn | l = \map \complement {\bigcup_{i \mathop \in I} S_i}       | r = \mathbb U \setminus \paren {\bigcup_{i \mathop \in I} S_i}       | c = {{Defof|Set Complement}} }} {{eqn | r = \bigcap_{i \mathop \in I} \paren {\mathbb U \setminus S_i}       | c = De Morgan's Laws for Set Difference: Difference with Union }} {{eqn | r = \bigcap_{i \mathop \in I} \map \complement {S_i}       | c = {{Defof|Set Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | l = \map \complement {\bigcap_{i \mathop \in I} S_i}       | r = \mathbb U \setminus \paren {\bigcap_{i \mathop \in I} S_i}       | c = {{Defof|Set Complement}} }} {{eqn | r = \bigcup_{i \mathop \in I} \paren {\mathbb U \setminus S_i}       | c = De Morgan's Laws for Set Difference: Difference with Intersection }} {{eqn | r = \bigcup_{i \mathop \in I} \map \complement {S_i}       | c = {{Defof|Set Complement}} }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Let $\Bbb S$ converge to a limit. Then by definition: :$\ds \limsup_{n \mathop \to \infty} E_n = \liminf_{n \mathop \to \infty} E_n$ and so by definition of set equality: :$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$ {{qed|lemma}} === Necessary Condition === Suppose that: :$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$ From Limit Superior includes Limit Inferior: :$\ds \liminf_{n \mathop \to \infty} E_n \subseteq \limsup_{n \mathop \to \infty} E_n$ whether or not $\Bbb S$ converges to a limit. By definition of set equality: :$\ds \limsup_{n \mathop \to \infty} E_n = \liminf_{n \mathop \to \infty} E_n$ and so by definition $\Bbb S$ converges to a limit. {{qed}} Category:Limits Superior of Set Sequences Category:Limits Inferior of Set Sequences is5qz1szkina5763lfyl7zdoq1ymgtz
Let $A, B \in \mathcal R$. From Set Difference as Symmetric Difference with Intersection: : $A * \left({A \cap B}\right) = A \setminus B$ By hypothesis: :$A \cap B \in \mathcal R$ and: :$A * \left({A \cap B}\right) \in \mathcal R$ and so: :$A \setminus B \in \mathcal R$ {{qed}} Category:Set Intersection Category:Set Difference Category:Symmetric Difference Category:Set Systems cgvyaoug8dsj4nteepq16t572lqwan0
Let $A, B \in \mathcal R$. From Union as Symmetric Difference with Intersection: : $\left({A * B}\right) * \left({A \cap B}\right) = A \cup B$ By hypothesis: :$A \cap B \in \mathcal R$ and: :$\left({A * B}\right) * \left({A \cap B}\right)$ and so: :$A \cup B \in \mathcal R$ {{qed}} Category:Set Intersection Category:Set Union Category:Symmetric Difference Category:Set Systems lztpo7atayaw90asuztd0i0luzejh0o
=== Definition 1 implies Definition 2 === Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$: : $(RS \, 1_1): \quad \mathcal R \ne \varnothing$ : $(RS \, 2_1): \quad A \cap B \in \mathcal R$ : $(RS \, 3_1): \quad A * B \in \mathcal R$ As $\mathcal R$ is non-empty, there exists some $A \in \mathcal R$. From Symmetric Difference with Self is Empty Set: :$A * A = \varnothing$ By hypothesis $A * A \in \mathcal R$ and so $\varnothing \in \mathcal R$. Thus criterion $(RS \, 1_2)$ is fulfilled. From Closure of Intersection and Symmetric Difference imply Closure of Set Difference it follows that criterion $(RS \, 2_2)$ is fulfilled. From Closure of Intersection and Symmetric Difference imply Closure of Union it follows that criterion $(RS \, 3_2)$ is fulfilled. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$: : $(RS \, 1_2): \quad \varnothing \in \mathcal R$ : $(RS \, 2_2): \quad A \setminus B \in \mathcal R$ : $(RS \, 3_2): \quad A \cup B \in \mathcal R$ We have that $\varnothing \in \mathcal R$ and so $\mathcal R$ is non-empty. Thus criterion $(RS \, 1_1)$ is fulfilled. By hypothesis, $\mathcal R$ is closed under $\setminus$ and $\cup$. Thus: :$\forall A, B \in \mathcal R: \left({A \setminus B}\right) \cup \left({B \setminus A}\right) \in \mathcal R$ But by the definition of symmetric difference: :$A * B := \left({A \setminus B}\right) \cup \left({B \setminus A}\right)$ Thus: :$\forall A, B \in \mathcal R: A * B \in \mathcal R$ and so $\mathcal R$ is closed under symmetric difference. Thus criterion $(RS \, 3_1)$ is fulfilled. From Union minus Symmetric Difference equals Intersection: :$\forall A, B \in \mathcal R: \left({A \cup B}\right) \setminus \left({A * B}\right) = A \cap B$ Thus $\mathcal R$ is closed under set intersection. Thus criterion $(RS \, 2_1)$ is fulfilled. {{qed|lemma}} === Definition 2 iff Definition 3 === Let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$: : $(RS \, 1_2): \quad \varnothing \in \mathcal R$ : $(RS \, 2_2): \quad A \setminus B \in \mathcal R$ : $(RS \, 3_2): \quad A \cup B \in \mathcal R$ Criteria $(RS \, 1_3)$ and $(RS \, 2_3)$ are fulfilled immediately. Consider $A, B \in \mathcal R: A \cap B = \varnothing$. Then as $A, B \in \mathcal R$ it follows by $(RS \, 3_2)$ that $A \cup B \in \mathcal R$ and so $(RS \, 3_3)$ is fulfilled. Now let $\mathcal R$ be a system of sets such that for all $A, B \in \mathcal R$: : $(RS \, 1_3): \quad \varnothing \in \mathcal R$ : $(RS \, 2_3): \quad A \setminus B \in \mathcal R$ : $(RS \, 3_3): \quad A \cap B = \varnothing \implies A \cup B \in \mathcal R$ Again, criteria $(RS \, 1_2)$ and $(RS \, 2_2)$ are fulfilled immediately. Let $A, B \in \mathcal R$. Then from Set Difference Union Second Set is Union: :$A \cup B = \left({A \setminus B}\right) \cup B$ From Set Difference Intersection with Second Set is Empty Set: :$\left({A \setminus B}\right) \cap B = \varnothing$ Thus from $(RS \, 3_3)$: ::$A \cup B = \left({A \setminus B}\right) \cup B \in \mathcal R$ {{qed}}
Let $\sequence {E_n : n \in \N}$ be a sequence of sets. Let: :$\ds B_n := \bigcap_{j \mathop = n}^\infty E_j$ Let: :$\ds \liminf_{n \mathop \to \infty} E_n := \bigcup_{n \mathop = 0}^\infty B_n$ that is, $\ds \liminf_{n \mathop \to \infty} \ E_n$ according to definition 1. Let: :$E := \set {x: x \in E_i \text{ for all but finitely many } i}$ that is, $\ds \liminf_{n \mathop \to \infty} E_n$ according to definition 2. By definition of set equality, it is enough to prove: :$\ds \liminf_{n \mathop \to \infty} E_n \subseteq E$ and: :$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$ From Set Complement inverts Subsets, we have: :$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n} \iff \liminf_{n \mathop \to \infty} E_n \subseteq E$ The strategy will therefore be to prove: :$\ds E \subseteq \liminf_{n \mathop \to \infty} E_n$ and: :$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n}$ === Definition $2$ is contained in Definition $1$ === Let $x \in E$. By definition $x$ is in all but a finite number of $E_i$. Let $m \in \Z_{\ge 0}$ be the largest integer such that $x \notin E_m$. $m$ has to exist, because the set $\set {i \in \N: x \notin E_i}$ is a finite set. Let $M \in \Z$ such that $m < M$. Then as $x \in E_j$ for all $j > m$ it follows by definition of set intersection that: :$x \in B_M = \ds \bigcap_{j \mathop = M}^\infty E_j$ From the definition of set union, it follows that: :$x \in \ds \bigcup_{n \mathop = 0}^\infty B_n = \liminf_{n \mathop \to \infty} E_n$ Hence: :$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$ {{qed|lemma}} === Complement of Definition 2 is contained in Complement of Definition 1 === Let $x \in \map \complement E$ That is, $x \notin E$. The definition of $E$ grants the existence of a strictly increasing sequence $\sequence {i_n}$ of natural numbers, such that: :$\forall n \in \N: x \notin E_{i_n}$ It follows that for every $k \in \N$, there exists an $n \in \Z$ with $i_n > k$. Subsequently, from the definition of set intersection: :$B_k = \ds \bigcap_{j \mathop = k}^\infty E_j \subseteq E_{i_n}$ and hence $x \notin B_k$. As $k$ was arbitrary, we have: :$x \notin \ds \bigcup_{k \mathop = 0}^\infty B_k = \liminf_{n \mathop \to \infty} E_n$ It follows that: :$\ds \map \complement E \subseteq \map \complement {\liminf_{n \mathop \to \infty} E_n}$ {{qed|lemma}} Therefore, we have established that :$x \in \ds \liminf_{n \mathop \to \infty} E_n \iff x \in E$ From the definition of set equality, it follows that: :$\ds \liminf_{n \mathop \to \infty} E_n = E$ {{qed}}
Begin by defining: :$\ds B_n := \bigcup_{j \mathop = n}^\infty E_j$ Then by definition 1: :$\ds \limsup_{n \to \infty} E_n = \bigcap_{n \mathop = 0}^\infty B_n$ === First Direction === Let $x$ belong to $E_i$ for infinitely many $i \in \N$. Let $\map \phi n$ be the sequence consisting of these numbers in increasing order. Then for any number $k$, there exists a number $a$ such that $\map \phi a \ge k$. Hence: :$\ds x \in E_{\map \phi a} \subseteq \bigcup_{j \mathop = k}^\infty E_j = B_k$ Since $k$ was arbitrary, it follows that $x \in B_n$ for each $n$. So: :$\ds x \in \limsup_{n \mathop \to \infty} E_n$ === Second Direction === Let: :$\ds x \in \bigcap_{n \mathop = 0}^\infty B_n$ {{AimForCont}} $x$ occurs in only finitely many $E_i$. Then there is a largest value of $i$ (call it $i_0$) for which the membership holds. Hence: :$x \notin \paren {E_{i_0 + 1} \cup E_{i_0 + 2} \cup \ldots} = B_{i_0 + 1}$ Therefore: :$\ds x \notin \bigcap_{n \mathop = 0}^\infty B_n$ This contradicts our assumption about $x$. Hence $x$ belongs to infinitely many terms of the sequence. {{qed}}
From Union is Idempotent: :$S \cup S = S$ and :$\varnothing \cup \varnothing = \varnothing$ Then from Union with Empty Set: :$S \cup \varnothing = S$ So $\left\{{S, \varnothing}\right\}$ is closed under union. From Relative Complement of Empty Set: :$\complement_S \left({\varnothing}\right) = S$ and from Relative Complement with Self is Empty Set: :$\complement_S \left({S}\right) = \varnothing$ so $\left\{{S, \varnothing}\right\}$ is closed under complement. Hence the result, by definition of algebra of sets. {{qed}} Category:Algebras of Sets 13vhg5iqcdka84nj8odh2tinqsasb7e
Let $A, B \in \RR$. {{begin-eqn}} {{eqn | l = A \setminus B       | r = A \cap \map \complement B       | c = Set Difference as Intersection with Complement }} {{eqn | r = \map \complement {\map \complement A \cup B}       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}} As both set union and complement are closed in $\RR$ the result follows. {{qed}}