hcju/proofwikiqa huggingface dataset:
Let $\struct {S, \preceq}$ be an ordered set. Let $\cl: S \to S$ be a mapping. === Definition 1 implies Definition 2 === Let $\cl$ be an inflationary, increasing and idempotent mapping. It is necessary to show that for all $x, y \in S$: :$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$ ==== Necessary Condition ==== {{begin-eqn}} {{eqn | l = x       | o = \preceq       | r = \map \cl y }} {{eqn | ll= \leadsto       | l = \map \cl x       | o = \preceq       | r = \map \cl {\map \cl y}       | c = $\cl$ is increasing }} {{eqn | ll= \leadsto       | l = \map \cl x       | o = \preceq       | r = \map \cl y       | c = $\cl$ is idempotent }} {{end-eqn}} {{qed|lemma}} ==== Sufficient Condition ==== Suppose that $\map \cl x \preceq \map \cl y$. {{begin-eqn}} {{eqn | l = x       | o = \preceq       | r = \map \cl x       | c = $\cl$ is inflationary }} {{eqn | ll= \leadsto       | l = x       | o = \preceq       | r = \map \cl y       | c = $\preceq$ is transitive by dint of being an ordering }} {{end-eqn}} {{qed|lemma}} === Definition 2 implies Definition 1 === Suppose that: :$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$ It is necessary to show that $\cl$ is inflationary, increasing and idempotent. ==== Inflationary ==== Let $x \in S$. {{begin-eqn}} {{eqn | l = \map \cl x       | o = \preceq       | r = \map \cl x       | c = $\preceq$ is reflexive by dint of being an ordering }} {{eqn | ll= \leadsto       | l = x       | o = \preceq       | r = \map \cl x       | c = by hypothesis: $\map \cl x \preceq \map \cl y \implies x \preceq \map \cl y$ }} {{end-eqn}} That is, $\cl$ is inflationary. {{qed|lemma}} ==== Increasing ==== It has been demonstrated that $\preceq$ is inflationary. Let $x, y \in S$ such that $x \preceq y$. Then: {{begin-eqn}} {{eqn | l = y       | o = \preceq       | r = \map \cl y       | c = $\cl$ is inflationary }} {{eqn | ll= \leadsto       | l = x       | o = \preceq       | r = \map \cl y       | c = $\preceq$ is transitive by dint of being an ordering }} {{eqn | ll= \leadsto       | l = \map \cl x       | o = \preceq       | r = \map \cl y       | c = by hypothesis: $x \preceq \map \cl y \implies \map \cl x \preceq \map \cl y$ }} {{end-eqn}} That is, $\cl$ is  increasing. {{qed|lemma}} ==== Idempotent ==== It has been demonstrated that $\preceq$ is inflationary. Let $x \in S$. Then: {{begin-eqn}} {{eqn | l = \map \cl x       | o = \preceq       | r = \map \cl x       | c = $\preceq$ is reflexive by dint of being an ordering }} {{eqn | ll= \leadsto       | l = \map \cl {\map \cl x}       | o = \preceq       | r = \map \cl x       | c = by hypothesis: $x \preceq \map \cl y \implies \map \cl x \preceq \map \cl y$ }} {{end-eqn}} Then as $\cl$ is inflationary: :$\map \cl x \preceq \map \cl {\map \cl x}$ As $\preceq$ is antisymmetric by dint of being an ordering: :$\map \cl {\map \cl x} = \map \cl x$ That is, $\cl$ is idempotent. {{qed|lemma}} Thus $\cl$ has been shown to be inflationary, increasing and idempotent as required. {{qed}} Category:Closure Operators ce90f9wdgohpt6lal5cs0negwu37hj7
=== Necessary Condition === Let $x$ be a fixed point of $f$. Then: : $f \left({x}\right) = x$ and so by the definition of image of mapping: : $x \in f \left[{S}\right]$ {{qed|lemma}} === Sufficient Condition === Let $x \in f \left[{S}\right]$. Then by the definition of image: :$\exists y \in S: f \left({y}\right) = x$ Then: {{begin-eqn}} {{eqn | l = x       | r = f \left({y}\right)       | c =  }} {{eqn | ll= \implies       | l = f \left({x}\right)       | r = f \left({f \left({y}\right)}\right)       | c = Definition of Mapping }} {{eqn | r = f \left({y}\right)       | c = Definition of Idempotent Mapping }} {{eqn | r = x       | c =  }} {{end-eqn}} Thus by definition $x$ is a fixed point of $f$. {{qed}} Category:Idempotent Mappings qpaykkfuia110sfxp5wukrwvj13eviq
Proof by Counterexample: Let $S = \left\{{a, b, c}\right\}$ where $a$, $b$, and $c$ are distinct. Let: : ${\preceq} = \left\{{\left({a, a}\right), \left({b, b}\right), \left({c, c}\right), \left({a, c}\right), \left({b, c}\right)}\right\}$: Then $\preceq$ is an ordering, but $\preceq^\leftrightarrow$ is not transitive, as follows: $\preceq$ is reflexive because it contains the diagonal relation on $S$. That $\preceq$ is transitive and antisymmetric can be verified by inspecting all ordered pairs of its elements. Thus $\preceq$ is an ordering. Now consider $\preceq^\leftrightarrow$, the symmetric closure of $\preceq$: : ${\preceq^\leftrightarrow} = {\preceq} \cup {\preceq}^{-1} = \left\{{\left({a, a}\right), \left({b, b}\right), \left({c, c}\right), \left({a, c}\right), \left({c, a}\right), \left({b, c}\right), \left({c, b}\right)}\right\}$ by inspection. Now $\left({a, c}\right) \in {\preceq^\leftrightarrow}$ and $\left({c, b}\right) \in {\preceq^\leftrightarrow}$, but $\left({a, b}\right) \notin {\preceq^\leftrightarrow}$. Thus $\preceq^\leftrightarrow$ is not transitive. {{qed}} Category:Closure Operators e6yrejzm7xnk4pm1f0q9c2yo0ztm679
Let $x \in S$. By the premise: :$\map f {\map g x} \in g \sqbrk {f \sqbrk S}$ Since $f \sqbrk S \subseteq S$: :$\map f {\map g x} \in g \sqbrk S$ Thus for some $y \in S$: :$\map f {\map g x} = \map g y$ Since $g$ is idempotent: :$\map g {\map g y} = \map g y$ By the choice of $y$: :$\map g {\map f {\map g x} } = \map g {\map g y} = \map g y = \map f {\map g x}$ Thus: :$\map f {\map g {\map f {\map g x} } } = \map f {\map f {\map g x} } = \map f {\map g x}$ That is, $f \circ g$ is idempotent. {{qed}} Category:Idempotent Mappings Category:Composite Mappings 188o1vw8v5s7jzlw3xg1s414okta2j3
=== $(2)$ implies $(1)$ === Follows from Composition of Commuting Idempotent Mappings is Idempotent. {{qed|lemma}} === $(1)$ implies $(2)$ === Suppose that $f \circ g$ and $g \circ f$ are idempotent. Then $\paren {f \circ g} \circ \paren {f \circ g} = f \circ g$. By Composition of Mappings is Associative and the definition of composition, we have for each $x \in S$: :$\map f {\map g {\map f {\map g x} } } = \map f {\map g x}$ Because $\preceq$ is an ordering and hence reflexive: :$\map f {\map g {\map f {\map g x} } } \preceq \map f {\map g x}$ Since $f$ is inflationary: :$\map g {\map f {\map g x} } \preceq \map f {\map g {\map f {\map g x} } }$ Thus since $\preceq$ is an ordering and hence transitive: :$\map g {\map f {\map g x} } \preceq \map f {\map g x}$ Since $g$ is inflationary: :$\map f {\map g x} \preceq \map g {\map f {\map g x} }$ Thus since $\preceq$ is an ordering and hence antisymmetric: :$\map g {\map f {\map g x} } = \map f {\map g x}$ Since this holds for all $x \in S$, Equality of Mappings shows that: :$g \circ f \circ g = f \circ g$   The same argument, with the roles of $f$ and $g$ reversed, shows that: :$f \circ g \circ f = g \circ f$ Combining everything, we obtain: :$f \circ g = f \circ \paren {g \circ f \circ g} = f \circ \paren {g \circ f} = g \circ f$ Thus $f \circ g = g \circ f$, so $f$ and $g$ commute. {{qed|lemma}} === $(2)$ implies $(3)$ === Equals substitute for equals. {{qed|lemma}} === $(3)$ implies $(1)$ === Follows from Composition of Idempotent Mappings. {{qed}} Category:Mapping Theory jsjv4nlgl8koimfkb9zq9tqoq98tor4
By the definition of closure operator, $f$ is inflationary. Thus $x \preceq \map f x$. By definition, $\map f x$ is closed. So $\map f x$ is a closed element that succeeds $x$. We will now show that it is the smallest such. Let $k$ be a closed element of $S$ such that $x \preceq k$. Since $f$ is a closure operator, it is increasing. Therefore $\map f x \preceq \map f k$. Since $k$ is closed, $\map f k = k$. Thus $\map f x \preceq k$. {{qed}} Category:Closure Operators 1955mm4t98ditsvzrbvqb6rtpq7y9zu
Let $x \in S$. Let $C$ be the set of closed elements of $S$ (with respect to either $f$ or $g$) that succeed $x$. By Closure is Smallest Closed Successor, $f \left({x}\right)$ and $g \left({x}\right)$ are smallest closed successors of $x$. That is, $f \left({x}\right)$ and $g \left({x}\right)$ are smallest elements of $C \cap \bar\uparrow x$, where $\bar\uparrow x$ is the upper closure of $x$. By Smallest Element is Unique, $f \left({x}\right) = g \left({x}\right)$. Since this holds for all $x \in S$, $f = g$ by Equality of Mappings. {{qed}} Category:Closure Operators 2kotny5ma94x5wbj6yj7udjqagueqdr
Let $\struct {S, \preceq}$ be an ordered set. Let $\cl: S \to S$ be a closure operator on $S$. Let $x \in S$. By the definition of closure operator, $\cl$ is idempotent. Thus by Fixed Point of Idempotent Mapping: :An element of $S$ is a fixed point of $\cl$ {{iff}} it is in the image of $\cl$. Thus the above definitions are equivalent. {{qed}} Category:Closure Operators 51loub76vimcrpl17rqj45v5kghjuk3
Whatever flavour of number under discussion, the algebraic structure $\struct {\mathbb K, +, \times}$ in which this number sits is at least a semiring. The binary operation that is multiplication is therefore closed on that algebraic structure. Therefore: : $\forall x \in \mathbb K: x \times x \in \mathbb K$ {{qed}} Category:Numbers h9o91tir35n4ut6k92frbx01j9dqge4
By the definition of relation composition and subset we have that statement $(1)$ may be written as: :$(1')\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, y) \in B  \implies  (x, z) \in C }\right)$ {{explain|Actually, that only gets us to $\forall x, z \in S: ((\exists y: (y, z) \in A  \land  (x, y) \in B) \implies (x,z) \in C)$.}} Using a different arrangement of variable names, statement $(2)$ can be written: :$(2')\quad \forall x, y, z \in S: \left({ (z, y) \in A^{-1} \land (x, z) \in \overline C \implies (x, y) \in \overline B }\right)$ By the definition of the inverse and the complement of a relation we can rewrite this as: :$(2'')\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, z) \notin C)  \implies  (x, y) \notin B }\right)$ Similarly, statement $(3)$ can be written: :$(3')\quad \forall x, y, z \in S: \left({ (x, z) \in \overline C \land (y, x) \in B^{-1} \implies (y, z) \in \overline A }\right)$ By the definition of the inverse and the complement of a relation we can rewrite this as: :$(3'')\quad \forall x, y, z \in S: \left({ (x, z) \notin C \land (x, y) \in B \implies (y, z) \notin A }\right)$ So in all we have: :$(1')\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, y) \in B  \implies  (x, z) \in C }\right)$ :$(2'')\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, z) \notin C)  \implies  (x, y) \notin B }\right)$ :$(3'')\quad \forall x, y, z \in S: \left({ (x, z) \notin C \land (x, y) \in B \implies (y, z) \notin A }\right)$ {{finish}}
=== $(1)$ iff $(2)$ === By the definition of relation composition and subset we have that statement $(1)$ may be written as: :$(1'):\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, y) \in B  \implies  (x, z) \in C }\right)$ Using a different arrangement of variable names, statement $(2)$ can be written: :$(2'):\quad \forall x, y, z \in S: \left({ (z, y) \in A^{-1} \land (x, z) \in \overline C \implies (x, y) \in \overline B }\right)$ By the definition of inverse relation and the complement of a relation we can rewrite this as: :$(2''):\quad \forall x, y, z \in S: \left({ (y, z) \in A  \land  (x, z) \notin C)  \implies  (x, y) \notin B }\right)$ We shall use the method of truth tables. The two statements will be equivalent iff the columns under the main connectives, which is $\implies$ in each case, are identical. '''Statement 1:''' $\begin{array}{ccccc} ((y, z) \in A & \land & (x, y) \in B) & \implies & (x, z) \in C \\ \hline T & T & T & T & T \\ T & T & T & F & F \\ T & F & F & T & T \\ T & F & F & T & F \\ F & F & T & T & T \\ F & F & T & T & F \\ F & F & F & T & T \\ F & F & F & T & F \\ \end{array}$ '''Statement 2:''' $\begin{array}{ccccc} ((y, z) \in A & \land & (x, z) \notin C) & \implies & (x, y) \notin B \\ \hline T & F & F & T & F \\ T & T & T & F & F \\ T & F & F & T & T \\ T & T & T & T & T \\ F & F & F & T & F \\ F & F & T & T & F \\ F & F & F & T & T \\ F & F & T & T & T \\ \end{array}$ {{qed|lemma}} === $(2)$ iff $(3)$ === {{finish}}
We are required to show that: :$\forall x \in R_m, y \in R_n: x \circ y \in R_{m \cdot n}$ First suppose that $m = n = e$ are both the identity. In this case, $R_m = R_n = R$. Since by definition, $R$ is closed under $\circ$, it follows that :$\forall x \in R, y \in R: x \circ y \in R$ as required. Now suppose that either $m \neq e$ or $n \neq e$. After possibly exchanging $m$ and $n$, we may as well assume that $n \neq e$. In particular, $R_n = \mathbf 0$ is the zero ring. So if $y \in R_n$, then $y = 0$. Therefore, for every $x \in R_m$, by Ring Product with Zero, we must have  :$x \circ y = x \circ 0 = 0$ Since $R_{m \cdot n}$ is an abelian group it must by definition contain $0$. Therefore $x \circ y \in R_{m \cdot n}$ as required. {{Qed}} Category:Ring Theory l0jo8rrsxtsep3cb7ip1u31nhyifjba
Since $x$ is a fixed point of $g$: : $g \left({x}\right) = x$ Thus: : $f \left({g \left({x}\right)}\right) = f \left({x}\right)$ Since $x$ is a fixed point of $f$: : $f \left({x}\right) = x$ It follows that: : $\left({f \circ g}\right) \left({x}\right) = f \left({g \left({x}\right)}\right) = x$ Thus $x$ is a fixed point of $f \circ g$. {{qed}} Category:Mapping Theory 5xe9ki10ex12x7axmxtisa1dllpha31
The proof proceeds by mathematical induction on $n$, the number of mappings. === Base Case === {{questionable|Flawed. The base case needs to be the two-element case for obvious reasons.}} If $n = 1$, then $g = f_0$. Since $x$ is a fixed point of $f_0$, it is also a fixed point of $g$. {{qed|lemma}} === Inductive Step === Suppose that the theorem holds for $n$.  We will show that it holds for $n+1$. Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_{n+1}$. Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1} \circ f_n$ be the composition of all the $f_i$s. Since the theorem holds for $n$, $x$ is a fixed point of $f_0 \circ f_1 \circ \dots \circ f_{n-1}$. By Composition of Mappings is Associative: : $g = \left({ f_0 \circ f_1 \circ \dots \circ f_{n-1} }\right) \circ f_n$ Thus by Fixed Point of Mappings is Fixed Point of Composition (for two mappings), $x$ is a fixed point of $g$. {{qed}} Category:Mapping Theory h1jv5hvio53fngdgbwydje3btthk4dg
If $f \in \Z \sqbrk X$ then $\cont f \in \Z$ by definition of content. Conversely, suppose that: :$f = a_d X^d + \cdots + a_1 X + a_0 \notin \Z \sqbrk X$ Let $m = \min \set {n \in \N : n f \in \Z \sqbrk X}$. Then, by definition of content: :$\cont f = \dfrac 1 m \gcd \set {m a_d, \ldots, m a_0}$ So $\cont f \in \Z$ would mean that this GCD is a multiple of $m$. This, however, means that for each $i$, $\dfrac {m a_i} m = a_i$ is an integer, which contradicts our assumption that $f \notin \Z \sqbrk X$. {{Qed}} Category:Content of Polynomial tle7se59a7q909dvaxtit66n56tgio1
Since $f$ is monic, it can be written as: :$f = X^r + \cdots + a_1 X + a_0$ Let $n = \inf \set {n \in \N : n f \in \Z \sqbrk X}$. Let $d = \cont {n f}$. Then by definition of content: :$d = \gcd \set {n, n a_{r - 1}, \ldots, n a_1, n a_0}$ Therefore, by definition of GCD, $d$ divides $n$. So say $n = k d$ with $k \in \Z$. Then:  :$\cont f = \dfrac d {k d} = \dfrac 1 k$ as required. {{Qed}} Category:Content of Polynomial Category:Monic Polynomials kblu9qebsdbv86z0a5zd2jhnb86ubfh
Let $q = \dfrac a b$ with $a, b \in \Z$. Let $\Z \sqbrk X$ denote the ring of polynomials over $\Z$. Let $n \in \Z$ such that $n f \in \Z \sqbrk X$.  Then we have: :$b n \paren {q f} = a n f \in \Z \sqbrk X$ By the definition of content, and using that $a \in \Z$: :$\cont {b n q f} = a \cont {n f}$ {{handwaving|why is $b n$ the infimum mentioned in the definition?}} By definition of content: :$\cont {q f} = \dfrac 1 {b n} \cont {b n q f}$ Combining the above with the definition of $\cont f$: :$\cont {q f} = \dfrac a b \dfrac 1 n \cont {n f} = q \cont f$ {{Qed}} Category:Polynomial Theory Category:Content of Polynomial jydm79bz6zwzzp09mabap9kjb8p52c8
=== Necessary Condition === Follows from Fixed Point of Mappings is Fixed Point of Composition. {{qed|lemma}} === Sufficient Condition === Let $h = f \circ g$. Let $x$ be a fixed point of $h$. Then by the definition of composition: : $f \left({g \left({x}\right)}\right) = x$ Since $f$ is inflationary: : $x \preceq g \left({x}\right)$ Suppose for the sake of contradiction that $x \ne g \left({x}\right)$. Then $x \prec g \left({x}\right)$. Since $f$ is also inflationary: : $g \left({x}\right) \preceq f \left({g \left({x}\right)}\right)$ Thus by Extended Transitivity: : $x \prec f \left({g \left({x}\right)}\right)$ But this contradicts the assumption that $x$ is a fixed point of $f \circ g$. Therefore, $x = g \left({x}\right)$, and $x$ is a fixed point of $g$. Suppose for the sake of contradiction that $f \left({x}\right) \ne x$. Then $x \prec f \left({x}\right)$. As we have shown that $x = g \left({x}\right)$, it follows that: : $x \prec f \left({g \left({x}\right)}\right)$ But this contradicts assumption that $x$ is a fixed point of $f \circ g$. Hence, $x$ is also a fixed point of $f$. {{qed}} Category:Order Theory fgaz4yr9gvacrktsbz58kviss2ovnsl
Follows directly from Invertible Elements of Monoid form Subgroup of Cancellable Elements. {{qed}}
Let $p$ be irreducible in $D$. Let $U_D$ be the group of units of $D$. By definition, an irreducible element is not a unit. So from Principal Ideals in Integral Domain: :$\ideal p \subset D$ Suppose the principal ideal $\ideal p$ is not maximal. Then there exists an ideal $K$ of $D$ such that: :$\ideal p \subset K \subset R$ Because $D$ is a principal ideal domain: :$\exists x \in R: K = \ideal x$ Thus: :$\ideal p \subset \ideal x \subset D$ Because $\ideal p \subset \ideal x$: :$x \divides p$ by Principal Ideals in Integral Domain. That is: :$\exists t \in D: p = t \circ x$ But $p$ is irreducible in $D$, so $x \in U_D$ or $t \in U_D$. That is, either $x$ is a unit or $x$ is an associate of $p$. But since $K \subset D$: :$\ideal x \ne D$ so $x \notin U_D$ by Principal Ideals in Integral Domain. Also, since $\ideal p \subset \ideal x$: :$\ideal p \ne \ideal x$ so $x$ is not an associate of $p$, by Principal Ideals in Integral Domain. This contradiction shows that $\ideal p$ is a maximal ideal of $D$. {{qed}}
Let $\left({p}\right)$ be a maximal ideal of $D$. Let $p = f g$ be any factorization of $p$. We must show that one of $f, g$ is a unit. Suppose that neither of $f, g$ is a unit. First it will be shown that: :$\left({p}\right) \subsetneqq \left({f}\right)$ Let $x \in \left({p}\right)$. That is: :$\exists q \in D: x = p q$ Then: :$x = f g q \in \left({f}\right)$ so: :$\left({p}\right) \subseteq \left({f}\right)$ Now suppose $f \in \left({p}\right)$. Then: :$\exists r \in D: f = r p$ and so from $p = f g$ above: :$f = r g f$ Therefore: :$r g = 1$ and so $g$ is a unit. This is a contradiction. Thus: :$f \notin \left({p}\right)$ and clearly: :$f \in \left({f}\right)$ so: :$\left({p}\right) \subsetneqq \left({f}\right)$ as claimed. {{Handwaving|"Clearly" needs to be replaced by a link to the definition which specifies this fact.}} Therefore, since $\left({p}\right)$ is maximal, we must have: :$\left({f}\right) = D$ But we assumed that $f$ is not a unit. So there is no $h \in D$ such that $f h = 1$. Therefore: :$1 \notin \left({f}\right) = \left\{{f h: h \in D}\right\}$ and: :$\left({f}\right) \subsetneqq D$ This is a contradiction. Therefore at least one of $f, g$ must be a unit. This completes the proof. {{qed}}
We have that $\displaystyle \sum_{k \mathop = 0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_{\ge 0}$. Set $m = 0$: : $\displaystyle \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$ Thus: :$\displaystyle \forall a_k \in D: \sum_{k \mathop = 0}^0 a_k \circ x^k \in D$ It follows directly that $D$ is a subring of $D \left[{x}\right]$ by applying the Subring Test on elements of $D$. {{qed}}
Let $\phi: \R \to \R/{\sim}$ be the quotient mapping. Then: :$\forall x \in \Q: \map \phi x = \Q$ :$\forall x \in \mathbb I: \map \phi x = \mathbb I$ {{AimForCont}} $\set {\mathbb I} \in \tau_\sim$. Then by the definition of the quotient topology: :$\O \subsetneqq \mathbb I = \phi^{-1} \sqbrk {\set {\mathbb I} } \in \tau$ Thus by Rationals are Everywhere Dense in Topological Space of Reals, $\mathbb I$ contains a rational number. This is a contradiction. {{AimForCont}} $\set \Q \in \tau_\sim$. Then: :$\O \subsetneqq \Q = \phi^{-1} \sqbrk {\set \Q} \in \tau$ Thus by Irrationals are Everywhere Dense in Topological Space of Reals, $\Q$ contains a irrational number. This is a contradiction. As $\R / {\sim}$ has exactly two elements, its only non-empty proper subsets are $\set \Q$ and $\set {\mathbb I}$. As neither of these sets is $\tau_\sim$-open, $\struct {\R / {\sim}, \tau_\sim}$ is indiscrete. {{qed}} Category:Quotient Spaces Category:Real Number Line with Euclidean Topology 7xylwfsreu0d6btn0gh71xdhpx4qwml
By Quotient Space of Real Line may be Indiscrete, there is an equivalence relation $\sim$ on $\R$ such that the quotient space $\struct {\R / {\sim}, \tau_\sim}$ has two points and is indiscrete. It follows directly from the definition of Kolmogorov space that $\struct {\R / {\sim}, \tau_\sim}$ is not a Kolmogorov space. {{qed}} Category:Quotient Spaces Category:Real Number Line with Euclidean Topology Category:T0 Spaces 1k2eb023ogto02shmzysst4j3e45btq
Let $Y = \R / {\sim}$. Let $\phi: \R \to Y$ be the quotient mapping. Note that: :$\map \phi x = \set x$ if $x$ is irrational. :$\map \phi x = \Q$ if $x$ is rational. === Kolmogorov === If $x$ is irrational, then $\phi^{-1} \sqbrk {Y \setminus \set x} = \R \setminus \set x$. Thus $Y \setminus \set x$ is open in $Y$. Let $p, q \in Y$ such that $p \ne q$. Then $\set p$ or $\set q$ must be a singleton containing an irrational number. {{WLOG}}, suppose that $\set p$ is a singleton containing an irrational number. Then as shown above, $Y \setminus P$ is open in $Y$. Thus so $p$ and $q$ are distinguishable. Since this holds for any two points in $Y$, the space is Kolmogorov. {{qed|lemma}} === Not Fréchet === {{AimForCont}} $\set \Q$ is closed in $Y$. By Identification Mapping is Continuous, $\phi$ is continuous. Thus $\phi^{-1} \sqbrk {\set \Q} = \Q$ is closed in $\R$. But this contradicts the fact that $\Q \subsetneqq \R$ and Rationals are Everywhere Dense in Topological Space of Reals. Thus the singleton $\set \Q$ is not closed in $Y$. Hence $\struct {Y, \tau_\sim}$ is not a Fréchet space. {{qed}} Category:Quotient Spaces Category:Real Number Line with Euclidean Topology Category:T0 Spaces Category:T1 Spaces axy3wekfsjuhd6sbkmjrs6p3rxqg1bj
An element is closed with respect to a closure operator iff it is a fixed point of that operator. Since $f$ and $g$ are closure operators, they are inflationary. Thus the result follows from Fixed Point of Composition of Inflationary Mappings. {{qed}} Category:Closure Operators hwt0nny0li0kzk0oud7lzfroivzhwel
{{BeginTableau| \vdash ((p \land q) \lor (\lnot p \land q)) \lor ( (p \land \lnot q) \lor (\lnot p \land \lnot q)) }} {{ExcludedMiddle|1|p \lor \lnot p}} {{ExcludedMiddle|2|q \lor \lnot q}} {{Conjunction|3||(p \lor \lnot p) \land (q \lor \lnot q)|1|2}} {{SequentIntro|4||((p \lor \lnot p) \land q) \lor ((p \lor \lnot p) \land \lnot q)|3|Conjunction Distributes over Disjunction}} {{Assumption|5|(p \lor \lnot p) \land q}} {{SequentIntro|6|5|(p \land q) \lor (\lnot p \land q)|5|Conjunction Distributes over Disjunction}} {{Implication|7||(p \lor \lnot p) \land q \implies (p \land q) \lor (\lnot p \land q)|5|6 }} {{Assumption|8|(p \lor \lnot p) \land \lnot q}} {{SequentIntro|9|8| (p \land \lnot q) \lor (\lnot p \land \lnot q)|8| Conjunction Distributes over Disjunction }} {{Implication|10|| (p \lor \lnot p) \land \lnot q \implies (p \land \lnot q) \lor (\lnot p \land \lnot q)|8|9}} {{SequentIntro|11|| ((p \lor \lnot p) \land q) \lor ((p \lor \lnot p) \land \lnot q) \implies ((p \land q) \lor (\lnot p \land q)) \lor ( (p \land \lnot q) \lor (\lnot p \land \lnot q))|7,10|Constructive Dilemma}} {{ModusPonens|12|| ((p \land q) \lor (\lnot p \land q)) \lor ( (p \land \lnot q) \lor (\lnot p \land \lnot q))|11|4 }} {{EndTableau}} {{qed}} {{LEM}} Category:Propositional Logic 48jjx3jhedwa77osqylwqo68vmo427a