hcju/proofwikiqa huggingface dataset:
We have: {{begin-eqn}} {{eqn|l = \vec{p r} + \vec{r q}      |r = \left({r - p}\right) + \left({q - r}\right)      |c = Definition Vector in Affine Space }} {{eqn|r = \left({r + \left({q - r}\right)}\right) - p      |c = Definition of Affine Space: axiom $(A3)$ }} {{eqn|r = q - p      |c = Definition of Affine Space: axiom $(A1)$ }} {{eqn|r = \vec{p q}      |c = Definition Vector in Affine Space }} {{end-eqn}} {{Qed}} {{namedfor|Michel Chasles|cat=Chasles}} Category:Affine Geometry 8ncowekvl3l3sifurb0613kogovvmkk
=== Proof of Surjection === Let $p \in \mathcal E$. Let $v = p - p_0 \in V$. Let $\left({\lambda_1, \ldots, \lambda_n}\right)$ be coordinates of $v$ in the basis $\left({e_1, \ldots, e_n}\right)$. Then: {{begin-eqn}} {{eqn|l = p_0 + \sum_{i \mathop = 1}^n \lambda_ie_i      |r = p_0 + v }} {{eqn|r = p_0 + \left({p - p_0}\right) }} {{eqn|r = p }} {{end-eqn}} thus demonstrating that $\Theta_\mathcal R$ is a surjection. === Proof of Injection === Let: :$\Theta_\mathcal R \left({\lambda_1, \ldots, \lambda_n}\right) = \Theta_\mathcal R \left({\mu_1, \ldots, \mu_n}\right)$ That is: :$\displaystyle p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i = p_0 + \sum_{i \mathop = 1}^n \mu_i e_i$ Then by $(3)$ of Properties of Affine Spaces: :$\displaystyle \sum_{i \mathop = 1}^n \lambda_i e_i = \sum_{i \mathop = 1}^n \mu_i e_i$ By Expression of Vector as Linear Combination from Basis is Unique: : $\lambda_i = \mu_i$ for $i = 1, \ldots, n$. Hence the result. {{Qed}} Category:Affine Geometry o23me65r2z1l0vvbrkmcidnuk3vnftm
Let $r$ be any point in $\mathcal E$. Set: :$\displaystyle q = r + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i}$ We are required to prove that for any other point $m \in \mathcal E$: :$\displaystyle q = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}$ So: {{begin-eqn}} {{eqn | l = m + \sum_{i \mathop = 1}^n \lambda_i \vec{m p_i}       | r = m + \sum_{i \mathop = 1}^n \lambda_i \left({\vec{m r} + \vec{r p_i} }\right)       | c = Chasles' Relation }} {{eqn | r = m + \left(\sum_{i \mathop = 1}^n \lambda_i\right) \vec{m r} + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i} }} {{eqn | r = m + \vec{m r} + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i}       | c = by the assumption $\displaystyle \sum_{i \mathop = 1}^n \lambda_i = 1$ }} {{eqn | r = r + \sum_{i \mathop = 1}^n \lambda_i \vec{r p_i}       | c = Axiom $(1)$ for an affine space }} {{eqn | r = q       | c = Definition of $q$ }} {{end-eqn}} Hence the result. {{Qed}} Category:Affine Geometry 0pqxk9fc9gk6u8augbzooca1p170whd
The proof proceeds by induction on $n$, the number of terms in the sequence. We first define a propositional function, $P$, as follows: For each $n \in \N$ such that $n \ge 2$, let $P(n)$ be the proposition that if both of the following hold: : $\langle x_k \rangle_{k \in \left\{ {1, 2, \dots, n}\right\} }$ is a sequence of $n$ terms : $\forall k \in \left\{ {1, 2, \dots, n-1}\right\}: x_k \mathrel {\mathcal R} x_{k+1}$ then $x_1 \mathrel {\mathcal R} x_n$. === Basis for the Induction === The case $n = 2$ is verified as follows: In this case, the sequence has only two elements, $x_1$ and $x_2$. Thus by the premise, $x_1 \mathrel {\mathcal R} x_2$, so $P(2)$ holds. This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$ with $n \ge 2$. Suppose that $P(n)$ holds. This is our induction hypothesis. === Induction Step === This is our induction step: Let $\langle x_k \rangle_{k \in \left\{ {1, 2, \dots, n, n+1}\right\} }$ be a sequence of $n+1$ terms. For each $k \in \left\{ {1, 2, \dots, n-1, n}\right\}$, let $x_k \mathrel{\mathcal R} x_{k+1}$. In particular, $x_n \mathrel{\mathcal R} x_{n+1}$. By the induction hypothesis: :$x_1 \mathrel{\mathcal R} x_n$. Thus since $\mathcal R$ is transitive: :$x_1 \mathrel {\mathcal R} x_{n+1}$ We conclude that $P(n+1)$ holds. The result follows by the Principle of Mathematical Induction. {{qed}} Category:Transitive Relations a0r2ufbrr8fyp3k1s5ddqbgb6lgb7ar
The result follows from: :Transitive Closure of Reflexive Relation is Reflexive :Reflexive Closure of Transitive Relation is Transitive :Composition of Compatible Closure Operators {{qed}} Category:Reflexive Closures Category:Transitive Closures bw3oi7eog31sgwy7klra98nk1gd0sr8
Let $\left({x, y}\right) \in \mathcal R \cap \mathcal R^{-1}$ By definition of intersection: :$\left({x, y}\right) \in \mathcal R$ :$\left({x, y}\right) \in \mathcal R^{-1}$ By definition of inverse relation: :$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R^{-1}$ :$\displaystyle \left({x, y}\right) \in \mathcal R^{-1} \implies \left({y, x}\right) \in \left ({\mathcal R^{-1}} \right )^{-1}$ By Inverse of Inverse Relation the second statement may be rewritten: :$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R^{-1}$ :$\left({x, y}\right) \in \mathcal R^{-1} \implies \left({y, x}\right) \in \mathcal R$ Then by definition of intersection: :$\left({y, x}\right) \in \mathcal R \cap \mathcal R^{-1}$ Hence $\mathcal R \cap \mathcal R^{-1}$ is symmetric. {{qed}}
Let $Q = \bigcap \AA$. By the definition of closure operator, $f$ is inflationary, order-preserving, and idempotent. Let $A \in \AA$. By Intersection is Largest Subset, $Q \subseteq A$. Since $f$ is order-preserving, $\map f Q \subseteq \map f A$. By the definition of closed set, $\map f A = A$ Thus $\map f Q \subseteq A$. This holds for all $A \in \AA$. Thus by Intersection is Largest Subset: :$\map f Q \subseteq \bigcap \AA$ Since $\bigcap \AA = Q$: :$\map f Q \subseteq Q$ Since $f$ is inflationary: :$Q \subseteq \map f Q$ Thus by definition of set equality: :$Q = \map f Q$ Therefore $Q$ is closed with respect to $f$. {{qed}} Category:Closure Operators Category:Set Intersection 0wx60nxfhxnwbqfzj1mo6nccgfu2jok
By the definition of closure operator, $\cl$ is idempotent. Therefore $\map \cl {\map \cl T} = \map \cl T$, so $\map \cl t$ is closed. {{qed}} Category:Closure Operators eclqthg2av1s8vwraog48pwy1xzukub
By Intersection of Relation with Inverse is Symmetric Relation: :$\RR \cap \RR^{-1}$ is a symmetric relation. Suppose for some $\SS \in \powerset \RR$ that $S$ is symmetric and not equal to $\RR \cap \RR^{-1}$. We will show that it is a proper subset of $\RR \cap \RR^{-1}$. Suppose $\tuple {x, y} \in \SS$. Then as $\SS \subseteq \RR$: :$\tuple {x, y} \in \RR$ As $\SS$ is symmetric: :$\tuple {x, y} \in \SS$ So as $\SS \subseteq \RR$: :$\tuple {x, y} \in \RR$ By definition of inverse relation: :$\tuple {x, y} \in \RR^{-1}$ By definition of intersection: :$\tuple {x, y} \in \RR \cap \RR^{-1}$ Thus: :$\tuple {x, y} \in \SS \implies \tuple {x, y} \in \RR \cap \RR^{-1}$ By definition of subset: :$\SS \subseteq \RR \cap \RR^{-1}$ Finally, as we assumed $\SS \ne \RR \cap \RR^{-1}$: :$\SS \subset \RR \cap \RR^{-1}$ Hence the result. {{qed}} Category:Relation Theory Category:Symmetric Relations 6k95y20lcui80oc5hcca1henqlhj01p
First we show that $\mathcal C_h$ induces a closure operator on $S$. Let $\mathcal A \subseteq \mathcal C_h$. By Intersection is Largest Subset: : $\mathcal A \subseteq \mathcal C_f$ and: : $\mathcal A \subseteq \mathcal C_g$ Thus by Intersection of Closed Sets is Closed/Closure Operator: :$\bigcap \mathcal A \in \mathcal C_f$ and :$\bigcap \mathcal A \in \mathcal C_g$ Thus by the definition of set intersection: : $\bigcap \mathcal A \in \mathcal C_h$ Thus by Closure Operator from Closed Sets, $C_h$ induces a closure operator $h$ on $S$. Now we will show that $f \circ g = h$. Having established that, it can be seen that $g \circ f = h$ will hold by reversing the variable names. Let $T \subseteq S$. By definition of closed set: :$f \left({g \left({T}\right)}\right) \in \mathcal C_f$ :$g \left({T}\right) \in \mathcal C_g$ By the premise: : $f \left({g \left({T}\right)}\right) \in \mathcal C_g$ Thus by definition of set intersection: : $f \left({g \left({T}\right)}\right) \in \mathcal C_f \cap \mathcal C_g = C_h$ So $f \left({g \left({T}\right)}\right)$ is closed with respect to $h$. By Set Closure is Smallest Closed Set/Closure Operator: : $h \left({T}\right) \subseteq f \left({g \left({T}\right)}\right)$ By definition of closed set: : $h \left({T}\right) \in C_h$ Thus by the definition of set intersection: : $h \left({T}\right) \in C_f$ and : $h \left({T}\right) \in C_g$ By Set Closure is Smallest Closed Set/Closure Operator: : $g \left({T}\right) \subseteq h \left({T}\right)$ By Closure Operator: Axiom $(2)$ $f$ is order-preserving: : $f \left({g \left({T}\right)}\right) \subseteq f \left({h \left({T}\right)}\right)$ Recall that $h \left({T}\right) \in C_f$. By definition of closed set: : $f \left({h \left({T}\right)}\right) = h \left({T}\right)$ Thus: : $f \left({g \left({T}\right)}\right) \subseteq h \left({T}\right)$ We have that: :$h \left({T}\right) \subseteq f \left({g \left({T}\right)}\right)$ So by definition of set equality: :$h \left({T}\right) = f \left({g \left({T}\right)}\right)$ {{qed}} Category:Closure Operators fhkw1vv19pbm9uivq9sbv61y6amq2cu
By the definition of closure operator, $\cl$ is idempotent. Therefore: :$\map \cl {\map \cl x} = \map \cl x$ It follows by definition that $\map \cl x$ is a closed element. {{qed}}
=== Inflationary === $x$ is a lower bound of $x^\succeq$. Hence by Lower Bound for Subset, $x$ is also a lower bound of $C \cap x^\succeq$. By the definition of smallest element, $x \preceq \map \cl x$. {{qed|lemma}} === Order-Preserving === Suppose that $x \preceq y$. Then: :$C \cap y^\succeq \subseteq C \cap x^\succeq$ By Smallest Element of Subset: :$\map \cl x \preceq \map \cl y$ === Idempotent === Let $x \in S$. For each $x \in S$: :$\map \cl x = \map \min {C \cap x^\succeq}$ Thus: :$\map \cl x \in \paren {C \cap x^\succeq} \subseteq C$ That is to say, $\map \cl x$ is its own smallest successor in $C$. Thus: :$\map \cl x = \map \cl {\map \cl x}$ {{qed|lemma}} When $x \in C$, $x$ is the minimum of $C \cap x^\succeq$ Hence, elements of $C$ are closed elements with respect to $\cl$. Suppose that $x$ is closed with respect to $\cl$. Then: :$x = \map \min {C \cap x^\succeq}$ so in particular: :$x \in C$ {{qed}} Category:Closure Operators Category:Order Theory hzff01klrhedb7dqje56pkzdwsi8ljk
Let $G = \vec{\mathcal G}$ be the difference space of $\mathcal G$. We are required to show that the following axioms are satisfied: {{begin-axiom}} {{axiom | n = 1         | q = \forall p, q \in \mathcal G         | m = p + \left({q - p}\right) = q }} {{axiom | n = 2         | q = \forall p \in \mathcal G: \forall u, v \in G         | m = \left({p + u}\right) + v = p + \left({u + v}\right) }} {{axiom | n = 3         | q = \forall p, q \in \mathcal G: \forall u \in G         | m = \left({p - q}\right) + u = \left({p + u}\right) - q }} {{end-axiom}} Proof of $(1)$: Let $p = \left({p', p''}\right), q = \left({q', q''}\right) \in \mathcal G$. We have: {{begin-eqn}} {{eqn | l = p + \left({q - p}\right)       | r = \left({p', p''}\right) + \left({\left({q', q''}\right) - \left({p', p''}\right)}\right) }} {{eqn | r = \left({p', p''}\right) + \left({q' - p', q'' - p''}\right)       | c = Definition of $-$ in Product Space }} {{eqn | r = \left({p' + \left({q' - p'}\right), p'' + \left({q'' - p''}\right)}\right)        | c = Definition of $+$ in Product Space }} {{eqn | r = \left({q' , q''}\right)        | c = Axiom $(1)$ in Affine Spaces $\mathcal E$, $\mathcal F$ }} {{end-eqn}} {{qed|lemma}} Proof of $(2)$: Let $p = \left({p', p''}\right) \in \mathcal G$. Let $u = \left({u', u''}\right), v = \left({v', v''}\right) \in G$. We have: {{begin-eqn}} {{eqn | l = \left({p + u}\right) + v       | r = \left({\left({p', p''}\right) + \left({u', u''}\right)}\right) + \left({v', v''}\right) }} {{eqn | r = \left({p' + u', p'' + u''}\right) + \left({v', v''}\right)       | c = Definition of $+$ in Product Space }} {{eqn | r = \left({\left({p' + u'}\right) + v', \left({p'' + u''}\right) + v''}\right)       | c = Definition of $+$ in Product Space }} {{eqn | r = \left({p' + \left({u' + v'}\right), p'' + \left({u'' + v''}\right)}\right)       | c = Axiom $(2)$ in Affine Spaces $\mathcal E$, $\mathcal F$ }} {{eqn | r = \left({p', p''}\right) + \left({\left({u', u''}\right) + \left({v', v''}\right)}\right)       | c = Definition of $+$ in Product Space }} {{eqn | r = p + \left({u + v}\right) }} {{end-eqn}} {{qed|lemma}} Proof of $(3)$: Let $p = \left({p', p''}\right), q = \left({q', q''}\right) \in \mathcal G$. Let $u = \left({u', u''}\right) \in G$. We have: {{begin-eqn}} {{eqn | l = \left({p - q}\right) + u       | r = \left({\left({p', p''}\right) - \left({q', q''}\right)}\right) + \left({u', u''}\right) }} {{eqn | r = \left({\left({p' - q'}\right) + u', \left({p'' - q''}\right) + u''}\right)       | c = Definition of $+,-$ in Product Space }} {{eqn | r = \left({\left({p' + u'}\right) - q', \left({p'' + u''}\right) - q''}\right)       | c = Axiom $(3)$ in Affine Spaces $\mathcal E$, $\mathcal F$ }} {{eqn | r = \left({\left({p', p''}\right) + \left({u', u''}\right)}\right) - \left({q', q''}\right)        | c = Definition of $+,-$ in Product Space }} {{eqn | r = \left({p - q}\right) + u }} {{end-eqn}} {{Qed}} Category:Affine Geometry flidbfjyz98ggjlutx77hmw1swmiuyy
=== Lemma === Let $\struct {S, \preccurlyeq}$ be an ordered set. Let $C_i$ be a '''complete meet subsemilattice''' of $S$. Then $C = \displaystyle \bigcap_{i \mathop \in I} C_i$ is also a '''complete meet subsemilattice'''. === Proof === Let $D \subseteq C$. By Intersection is Largest Subset, $D \subseteq C_i$ for each $i \in I$. Thus $D$ has an infimum in $S$ and $\inf D \in C_i$ for each $i \in I$. By the definition of intersection, $\inf D \in C$. {{qed|lemma}} By the lemma, $C$ is a '''complete meet semilattice'''. Let $x \in S$. Then $C \cap x^\succcurlyeq$ has an infimum in $S$ which lies in $C$, where $x^\succcurlyeq$ is the upper closure of $x$. By the definition of infimum: :$x \preceq \inf \struct {C \cap x^\succcurlyeq}$ so this infimum is in fact the smallest element of $C \cap x^\succcurlyeq$. Thus $C$ induces a closure operator on $S$ by Closure Operator from Closed Elements. {{qed}} Category:Closure Operators 5pdwkakzph1uw2ae1711bkuw0g0c9gu
We are required to show that: {{begin-axiom}} {{axiom|n = 1        |q = \forall p, q \in \mathcal E        |m = p + \left({q - p}\right) = q }} {{axiom|n = 2        |q = \forall p \in \mathcal E: \forall u, v \in E        |m = \left({p + u}\right) + v = p + \left({u \mathop{+} v}\right) }} {{axiom|n = 3        |q = \forall p, q \in \mathcal E: \forall u \in E        |m = \left({p - q}\right) \mathop{+} u = \left({p + u}\right) - q }} {{end-axiom}} By the definition of the standard affine structure, the addition and subtraction operations are simply those in the vector space $E$. That is, we want to show that: {{begin-axiom}} {{axiom|n = 1        |q = \forall u,v \in E        |m = u + \left({v - u}\right) = v }} {{axiom|n = 2        |q = \forall u,v,w \in E        |m = \left({u + v}\right) + w = u + \left({v \mathop{+} w}\right) }} {{axiom|n = 3        |q = \forall u,v,w \in E        |m = \left({v - u}\right) \mathop{+} w = \left({v + w}\right) - u }} {{end-axiom}} By definition the addition operation on a vector space is commutative and associative. But all three axioms are immediate consequences of commutativity and associativity. This concludes the proof. {{Qed}} {{MissingLinks|particularly to addition, subtraction etc.}} Category:Affine Geometry 1mwu9z9v1i4kpuxgvjfopf2qxpv9z8i
=== Antireflexivity === Follows from Reflexive Reduction is Antireflexive. {{qed|lemma}} === Transitivity === Suppose $\tuple {x, y}, \tuple {y, z} \in \RR^\ne$. By antireflexivity $x \ne y$ and $y \ne z$. We consider the two remaining cases. ==== Case 1: $x = z$ ==== If $x = z$ then: :$\tuple {x, y}, \tuple {y, x} \in \RR^\ne$ and so: :$\tuple {x, y}, \tuple {y, x} \in \RR$ Then by the antisymmetry of $\RR$: :$x = y$ and: :$\tuple {x, x} \in \RR^\ne$ which contradicts that $\RR^\ne$ is antireflexive. ==== Case 2: $x \ne z$ ==== By the transitivity of $\RR$: :$\tuple {x, z} \in \RR$ and by $x$ and $z$ being distinct: :$\tuple {x, z} \notin \Delta_S$ It follows by the definition of reflexive reduction: :$\tuple {x, z} \in \RR^\ne$ Hence $\RR^\ne$ is transitive. {{qed}}
By definition, an ordering is both reflexive and transitive. The result then follows from Reflexive Reduction of Transitive Antisymmetric Relation is Strict Ordering. {{qed}}
By the definition of the vectorialization of an affine space, the mapping $\Theta_\mathcal R : k^n \to \mathcal E$ defined by: :$\displaystyle \map {\Theta_\mathcal R} {\lambda_1, \ldots, \lambda_n} = p_0 + \sum_{i \mathop = 1}^n \lambda_i e_i$ is a bijection from $k^n$ to $\mathcal E$. Therefore, by Homomorphic Image of Vector Space, it suffices to prove that $\Theta_\mathcal R$ is a linear transformation. By General Linear Group is Group: :$\Theta_\mathcal R$ is a linear transformation {{iff}} its inverse ${\Theta_\mathcal R}^{-1}$ is a linear transformation. Therefore, it suffices to show that: :$\forall p, q \in \mathcal E, \mu \in k: \map { {\Theta_\mathcal R}^{-1} } {\mu \cdot p + q} = \mu \cdot \map { {\Theta_\mathcal R}^{-1} } p + \map { {\Theta_\mathcal R}^{-1} } g$ Thus: {{begin-eqn}} {{eqn | l = \map { {\Theta_\mathcal R}^{-1} } {\mu \cdot p + q}       | r = \map { {\Theta_\mathcal R}^{-1} } {\map {\Theta_\mathcal R} {\mu \cdot \map { {\Theta_\mathcal R}^{-1} } p} + q}       | c = Definition of $\mu \cdot p$ }} {{eqn | r = {\Theta_\mathcal R}^{-1} \map {\Theta_\mathcal R} {\mu \cdot \map { {\Theta_\mathcal R}^{-1} } p + \map { {\Theta_\mathcal R}^{-1} } q}       | c = Definition of $+$ in $\mathcal E$ }} {{eqn | r = \mu \cdot \map { {\Theta_\mathcal R}^{-1} } p + \map { {\Theta_\mathcal R}^{-1} } q       | c = because $\map { {\Theta_\mathcal R}^{-1} } {\Theta_\mathcal R}$ is the identity mapping }} {{end-eqn}} This is the required identity. {{Qed}} Category:Affine Geometry c1s8miwk3qonr3mfv14spg8gq6tz8co
By the definition of reflexive reduction: :$\mathcal R^\ne = \mathcal R \setminus \Delta_S$ where $\Delta_S$ denotes the diagonal relation on $S$. By Set Difference Intersection with Second Set is Empty Set: :$\left({\mathcal R \setminus \Delta_S}\right) \cap \Delta_S = \varnothing$ Hence by Relation is Antireflexive iff Disjoint from Diagonal Relation, $\mathcal R^\ne$ is antireflexive. {{qed}} Category:Reflexive Reductions 07pom45czxxv2l3lexaqcs1zqac5e8e
Let $\RR$ be transitive. Then by Transitive Relation is Antireflexive iff Asymmetric it follows directly that: :$(1): \quad$ If $\RR$ is antireflexive then it is asymmetric :$(2): \quad$ If $\RR$ is asymmetric then it is antireflexive. {{qed}} Category:Order Theory rp8tzzdaz1n1ad5581rq06y42eiwj2g
=== Subbandhood implies Idempotency === ==== Proving $\left({X \circ_{\mathcal P} X}\right) \subseteq X$ ==== Let $c \in X \circ_{\mathcal P} X$. By the definition of subset product for some $a, b \in X$ we have: :$a \circ b = c$ Suppose $c \notin X$. Then: :$a \circ b \notin X$ Which contradicts that $\left({X, \circ}\right)$ is a subband. ==== Proving $X \subseteq \left({X \circ_{\mathcal P} X}\right)$ ==== Let $a \in X$. By the definition of subset product: :$X \circ_{\mathcal P} X = \{ a \circ b: a, b \in X \}$ As $\circ$ is idempotent: :$a \circ a = a$. Thus: :$a \in \left({X \circ_{\mathcal P} X}\right)$ Hence by the definition of subset: :$X \subseteq \left({X \circ_{\mathcal P} X}\right)$ {{qed|lemma}} === Idempotency implies Subbandhood === Let $X \in \mathcal P \left({S}\right)$. Suppose $X$ is idempotent: That is suppose: :$X \circ_{\mathcal P} X = X$ Let $a, b \in X$. By the definition of subset product: :$X \circ_{\mathcal P} X = \{ a \circ b: a, b \in X \}$ Then $a \circ b \in X$. Hence $\left({X, \circ}\right)$ is a magma. By Restriction of Associative Operation is Associative it is a semigroup. Finally by Restriction of Idempotent Operation is Idempotent it is a band. {{qed}} Category:Abstract Algebra Category:Idempotence 71ppa6bwguziz1w5pc0ba576pb8l322
{{improve|See talk page}} === Case 1: $T$ is the Empty Set === By: : Empty Set is Submagma of Magma : Restriction of Associative Operation is Associative : Restriction of Idempotent Operation is Idempotent it follows that $\left({\varnothing, \circ_\mathcal P}\right)$ is a subband of $\left({T, \circ_\mathcal P}\right)$. Let $X \in \varnothing$. Then by the definition of the empty set it follows that $\left({X, \circ}\right)$ is a subband of $\left({S, \circ}\right)$ vacuously. {{qed|lemma}} === Case 2: $T$ is Non-Empty === Let $X \in T$. Then by definition of a subband $X$ is idempotent under $\circ_\mathcal P$. That is: :$X \circ_\mathcal P X = X$ By Subband iff Idempotent under Induced Operation we have that $\left({X, \circ}\right)$ is a subband of $\left({S, \circ}\right)$. Hence the result. {{qed}} Category:Abstract Algebra rzucm5udvldrjrsxyvx0v6pqqp1dcu0
{{begin-eqn}} {{eqn | l = \paren {f \circ g} \circ \paren {f \circ g}       | r =  f \circ \paren {g \circ f} \circ g       | c = Composition of Mappings is Associative }} {{eqn | r = f \circ \paren {f \circ g} \circ g       | c = by hypothesis }} {{eqn | r = \paren {f \circ f} \circ \paren {g \circ g}       | c = Composition of Mappings is Associative }} {{eqn | r = f \circ g       | c = $f$ and $g$ are idempotent by hypothesis }} {{end-eqn}} {{qed}}
By Set of All Self-Maps is Semigroup, the set of all self-maps on $S$ forms a semigroup under composition. The result follows from Product of Commuting Idempotent Elements is Idempotent. {{qed}}
By Composition of Inflationary Mappings is Inflationary: : $f \circ g$ and $g \circ f$ are inflationary. By Composition of Increasing Mappings is Increasing: : $f \circ g$ and $g \circ f$ are increasing. Thus each of the two composite mappings will be a closure operator {{iff}} it is idempotent. Therefore the equivalences follow from Composition of Inflationary and Idempotent Mappings. {{qed}} Category:Closure Operators oucxe64yg144ymp7baexi0vhnaqojit
Let $x, y \in S$ with $x \preceq_S y$. Since $g$ is increasing: :$\map g x \preceq_T \map g y$ Since $f$ is increasing: :$\map f {\map g x} \preceq_U \map f {\map g y}$ By the definition of composition: :$\map {\paren {f \circ g} } x \preceq_U \map {\paren {f \circ g} } y$ Since this holds for all such $x$ and $y$, $f \circ g$ is increasing. {{qed}} Category:Increasing Mappings 9140tydzpn8oxb1j16avgpnzfzhls01
Let $x \in S$. {{begin-eqn}} {{eqn |n = 1       |l = x       |o = \preceq       |r = g \left({x}\right)       |c = $g$ is inflationary }} {{eqn |n = 2       |l = g \left({x}\right)       |o = \preceq       |r = f \left({g \left({x}\right)}\right)       |c = $f$ is inflationary }} {{eqn |l = x       |o = \preceq       |r = f \left({g \left({x}\right)}\right)       |c = $(1)$ and $(2)$ and $\preceq$ is an ordering and hence transitive }} {{eqn |ll = \implies        |l = x       |o = \preceq       |r = \left({f \circ g}\right) \left({x}\right)       |c = Definition of composition }} {{end-eqn}} Since this holds for all $x \in S$, $f \circ g$ is inflationary. {{qed}} Category:Order Theory 7kdh1yq2liq0v5xy1cfcyjy0dhfbzec