hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = 2 \times 0^2 + 11       | r = 0 + 11 }} {{eqn | r = 11       | c = which is prime }} {{eqn | l = 2 \times 1^2 + 11       | r = 2 + 11       | c =  }} {{eqn | r = 13       | c = which is prime }} {{eqn | l = 2 \times 2^2 + 11       | r = 2 \times 4 + 11       | c =  }} {{eqn | r = 8 + 11       | c =  }} {{eqn | r = 19       | c = which is prime }} {{eqn | l = 2 \times 3^2 + 11       | r = 2 \times 9 + 11       | c =  }} {{eqn | r = 18 + 11       | c =  }} {{eqn | r = 29       | c = which is prime }} {{eqn | l = 2 \times 4^2 + 11       | r = 2 \times 16 + 11       | c =  }} {{eqn | r = 32 + 11       | c =  }} {{eqn | r = 43       | c = which is prime }} {{eqn | l = 2 \times 5^2 + 11       | r = 2 \times 25 + 11       | c =  }} {{eqn | r = 50 + 11       | c =  }} {{eqn | r = 61       | c = which is prime }} {{eqn | l = 2 \times 6^2 + 11       | r = 2 \times 36 + 11       | c =  }} {{eqn | r = 72 + 11       | c =  }} {{eqn | r = 83       | c = which is prime }} {{eqn | l = 2 \times 7^2 + 11       | r = 2 \times 49 + 11       | c =  }} {{eqn | r = 98 + 11       | c =  }} {{eqn | r = 109       | c = which is prime }} {{eqn | l = 2 \times 8^2 + 11       | r = 2 \times 64 + 11       | c =  }} {{eqn | r = 128 + 11       | c =  }} {{eqn | r = 139       | c = which is prime }} {{eqn | l = 2 \times 9^2 + 11       | r = 2 \times 81 + 11       | c =  }} {{eqn | r = 162 + 11       | c =  }} {{eqn | r = 173       | c = which is prime }} {{eqn | l = 2 \times 10^2 + 11       | r = 2 \times 100 + 11       | c =  }} {{eqn | r = 200 + 11       | c =  }} {{eqn | r = 211       | c = which is prime }} {{end-eqn}}
Let $x, y \in D$ such that :$x \preceq y$ Then by definition of reflexivity: :$\forall a,b \in \left\{ {x, y}\right\}: \exists z \in \left\{ {x, y}\right\}: a \preceq z \land b \preceq z$ By definition: :$\left\{ {x, y}\right\}$ is directed. By definition of mapping preserves directed suprema: :$f$ preserves the supremum of $\left\{ {x, y}\right\}$. By definition of join semilattice: :$\left\{ {x, y}\right\}$ admits a supremum. {{begin-eqn}} {{eqn | l = \sup \left({f\left[{\left\{ {x, y}\right\} }\right]}\right) | r = f\left({\sup \left\{ {x, y}\right\} }\right) | c = definition of mapping preserves the supremum }} {{eqn | r = f\left({x \vee y}\right) | c = definition of join }} {{eqn | r = f\left({y}\right) | c = Preceding iff Join equals Larger Operand }} {{end-eqn}} By Image of Pair under Mapping: :$f\left[{\left\{ {x, y}\right\} }\right] = \left\{ {f\left({x}\right), f\left({y}\right)}\right\}$ Thus by definitions of supremum and upper bound: :$f\left({x}\right) \preceq f\left({y}\right)$ {{qed}}
{{begin-eqn}} {{eqn | l = 0       | r = \int_a^b \paren 0 h \rd x }} {{eqn | r = \int_a^b \paren {-\map {\frac \d {\d x} } {P h'} + Q h} h \rd x }} {{eqn | r = \int_a^b Q h^2 \rd x - \int_a^b \map {\frac \d {\d x} } {P h'} h \rd x }} {{eqn | r = \int_a^b Q h^2 \rd x - \bigintlimits {P h' h} a b + \int_a^b P h' \rd h        | c = Integration by Parts }} {{eqn | r = \int_a^b Q h^2 \rd x + \int_a^b P h'^2 \rd x  }} {{eqn | r = \int_a^b \paren {P h'^2 + Q h^2} \rd x }} {{end-eqn}} {{qed}}
Exactly one of $n, n + 1, n + 2, n + 3$ is divisible by $4 = 2^2$. Thus, by definition, one of these is not square-free. {{qed}}
Note that $32, 36, 40$ and $104, 108, 112$ are all divisible by $4 = 2^2$, so are by definition not square-free. Then inspecting each number in turn: {{begin-eqn}} {{eqn | l = 29       | o =        | c = is prime }} {{eqn | l = 30       | r = 2 \times 3 \times 5       | c = and so is square-free }} {{eqn | l = 31       | o =        | c = is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 33       | r = 3 \times 11       | c = and so is square-free }} {{eqn | l = 34       | r = 2 \times 17       | c = and so is square-free }} {{eqn | l = 35       | r = 5 \times 7       | c = and so is square-free }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 37       | o =        | c = is prime }} {{eqn | l = 38       | r = 2 \times 19       | c = and so is square-free }} {{eqn | l = 39       | r = 3 \times 13       | c = and so is square-free }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 41       | o =        | c = is prime }} {{eqn | l = 42       | r = 2 \times 3 \times 7       | c = and so is square-free }} {{eqn | l = 43       | o =        | c = is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 101       | o =        | c = is prime }} {{eqn | l = 102       | r = 2 \times 3 \times 17       | c = and so is square-free }} {{eqn | l = 103       | o =        | c = is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 105       | r = 3 \times 5 \times 7       | c = and so is square-free }} {{eqn | l = 106       | r = 2 \times 53       | c = and so is square-free }} {{eqn | l = 107       | o =        | c = is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 109       | o =        | c = is prime }} {{eqn | l = 110       | r = 2 \times 5 \times 11       | c = and so is square-free }} {{eqn | l = 111       | r = 3 \times 37       | c = and so is square-free }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 113       | o =        | c = is prime }} {{eqn | l = 114       | r = 2 \times 3 \times 19       | c = and so is square-free }} {{eqn | l = 115       | r = 5 \times 27       | c = and so is square-free }} {{end-eqn}} {{qed}}
Let $x, y \in S$. Thus {{begin-eqn}} {{eqn | l = f \sqbrk {\set {x, y} }       | r = f \sqbrk {\set x \cup \set y}       | c = Union of Unordered Tuples }} {{eqn | r = f \sqbrk {\set x} \cup f \sqbrk {\set y}       | c = Image of Union under Mapping }} {{eqn | r = \set {\map f x} \cup \set {\map f y}       | c = Image of Singleton under Mapping }} {{eqn | r = \set {\map f x, \map f y}       | c = Union of Unordered Tuples }} {{end-eqn}} {{qed}}
We will prove that :$\forall X \subseteq M: \forall Y \subseteq S: Y = \left\{ {x \in S: x}\right.$ is upper bound for $\left.{X \land \map f x \preceq x}\right\} \implies \inf_L Y \in M$ Let $X \subseteq M$, $Y \subseteq S$ such that :$Y = \left\{ {x \in S: x}\right.$ is upper bound for $\left.{X \land \map f x \preceq x}\right\}$ We will prove that :$\map f {\inf Y}$ is lower bound for $Y$. Let $y \in Y$. By definition of $Y$: :$\map f y \preceq y$ By definitions of infimum and lower bound: :$\inf Y \preceq y$ By definition of increasing mapping: :$\map f {\inf Y} \preceq \map f y$ Thus by definition of transitivity: :$\map f {\inf Y} \preceq y$ {{qed|lemma}} We will prove that :$\map f {\map f {\inf Y} }$ is upper bound for $X$. Let $m \in X$. We will prove that :$m$ is lower bound for $Y$. Let $y \in Y$. By definition of $Y$: :$\map f y$ is upper bound for $X$ and $\map f y \preceq y$ By definition of upper bound: :$m \preceq \map f y$ Thus by definition of transitivity: :$m \preceq y$ {{qed|lemma}} By definition of infimum: :$m \preceq \inf Y$ By definition of increasing mapping: :$\map f m \preceq \map f {\inf Y}$ By definition of subset: :$m \in M$ By definition of $M$: :$m = \map f m$ By definition of increasing mapping: :$m \preceq \map f {\map f {\inf Y} }$ {{qed|lemma}} By definition of infimum: :$\map f {\inf Y} \preceq \inf Y$ By definition of increasing mapping: :$\map f {\map f {\inf Y} } \preceq \map f {\inf Y}$ By definition of $Y$: :$\map f {\inf Y} \in Y$ By definitions of infimum and lower bound: :$\inf Y \preceq \map f {\inf Y}$ By definition of antisymmetry: :$\inf Y = \map f {\inf Y}$ Thus by definition of $M$: :$\inf Y \in M$ {{qed|lemma}} Define: :$Y_0 = \left\{ {y \in S: \map f y}\right.$ is upper bound for $\left.{M \land \map f y \preceq y}\right\}$ By lemma: :$\inf Y_0 \in M$ We will prove that :$\forall X \subseteq M: X$ admits a supremum in $P$. Let $X \subseteq M$. Define: :$Y = \left\{ {y \in S: \map f y}\right.$ is upper bound for $\left.{M \land \map f y \preceq y}\right\}$ Define $z = \inf Y$. By lemma: :$z \in M$ We will prove that :$z$ is upper bound for $X$ in $P$. Let $m \in X$. By analogy: :$m$ is lower bound for $Y$ in $L$. By definition of infimum: :$m \preceq \inf Y$ Thus by definition of ordered subset: :$m \precsim z$ {{qed|lemma}} We will prove that :$\forall x \in M: x$ is upper bound for $X$ in $P \implies z \precsim x$ Let $x \in M$ such that :$x$ is upper bound for $X$. By definition of $M$: :$x = \map f x$ By definition of ordered subset: :$\map f x$ is upper bound for $X$ in $L$. By definition of $Y$: :$x \in Y$ By definitions of infimum and lower bound: :$\inf Y \preceq x$ Thus by definition of ordered subset: :$z \precsim x$ {{qed|lemma}} Hence $X$ admits a supremum in $P$. {{qed|lemma}} Thus by {{Complete iff Admits All Suprema]]: :$P$ is a complete lattice. {{qed}}
We use induction on $l$. The base case $l = 0$ is trivial. Let $l \ge 0$ be such that a solution $x_{k + l}$ exists and is unique up to a multiple of $p^{k + l}$. Choose a solution $x_{k + l}$ satisfying: :$f \left({x_{k + l} }\right) \equiv 0 \pmod {p^{k + l} }$ :$x_{k + l} \equiv x_k \pmod {p^k}$ By Congruence by Divisor of Modulus, each solution $x_{k + l + 1}$ is also a solution of the previous congruence. By uniqueness, it has to satisfy $x_{k + l + 1}\equiv x_{k + l} \pmod {p^{k + l} }$, hence is of the form $x_{k+  l} + t p^{k + l}$ with $t \in \Z$. Let $d = \deg f$. We have, for all $t \in \Z$: {{begin-eqn}} {{eqn | o = =       | l = f \left({x_{k + l} + t p^{k + l} }\right)       | r = f \left({x_{k + l} }\right) + t p^{k + l} f' \left({x_{k + l} }\right) + \left({t p^{k + l} }\right)^2 m       | c = for some $m \in \Z$, by Taylor Expansion for Polynomials: Order 1 }} {{eqn | o = \equiv       | r = f \left({x_{k+l} }\right) + t p^{k + l} f' \left({x_{k + l} }\right) \pmod {p^{k + l + 1} }       | c =  }} {{end-eqn}} Because $f' \left({x_{k + l} }\right) \equiv f' \left({x_k}\right) \not \equiv 0 \pmod p$, $f' \left({x_{k + l} }\right)$ is invertible modulo $p$. {{MissingLinks|invertible modulo $p$}} Thus $x_{k + l} + t p^{k + l}$ is a solution modulo $p^{k + l + 1}$ {{iff}}: :$t \equiv - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod p$ Thus, necessarily: :$x_{k + l + 1} \equiv x_{k + l} - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod {p^{k + l + 1} }$ which proves the existence and uniqueness. By induction, we have shown uniqueness and existence for all $l \ge 0$, as well as the relations: :$x_{k + l + 1} \equiv x_{k + l} - \dfrac {f \left({x_{k + l} }\right)} {f' \left({x_{k + l} }\right)} \pmod {p^{k + l + 1} }$ :$x_{k + l + 1} \equiv x_{k + l} \pmod {p^{k + l} }$ {{qed}}
We use induction on $l$. The base case $l = 0$ is trivial. Let $l \ge 0$ be such that a solution $x_{k + l}$ exists and is unique up to a multiple of $b^{k + l}$. Choose a solution $x_{k + l}$. Each solution $x_{k + l + 1}$ is also a solution of the previous congruence. By uniqueness, it has to satisfy $x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$, hence is of the form $x_{k + l} + t b^{k + l}$ with $t \in \Z$. Let $d = \deg f$. We have: {{begin-eqn}} {{eqn | l = \map f {x_{k + l} + t b^{k + l} }       | r = \map f {x_{k + l} } + t b^{k + l} \map {f'} {x_{k + l} } + \paren {t b^{k + l} }^2 m       | c = for some $m \in \Z$, by Taylor Expansion for Polynomials: Order 1 }} {{eqn | o = \equiv       | r = \map f {x_{k + l} } + t b^{k + l} \map {f'} {x_{k + l} } \pmod {b^{k + l + 1} }       | c =  }} {{end-eqn}} Because $\map {f'} {x_{k + l} } \equiv \map {f'} {x_k} \pmod b$, $\map {f'} {x_{k + l} }$ is invertible modulo $b$. Thus $x_{k + l} + t b^{k + l}$ is a solution modulo $b^{k + l + 1}$ {{iff}}: :$t \equiv - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod b$ Thus, necessarily: :$x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$ which proves the existence and uniqueness. By induction, we have shown uniqueness and existence for all $l \ge 0$, as well as the relations: :$x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {b^{k + l + 1} }$ :$x_{k + l + 1} \equiv x_{k + l} \pmod {b^{k + l} }$ {{qed}}
Let $n = 2^k$. Then: {{begin-eqn}} {{eqn | l = \map \sigma n       | r = \dfrac {2^{k + 1} - 1} {2 - 1}       | c = Sigma Function of Power of Prime }} {{eqn | r = 2 \times 2^k - 1       | c =  }} {{eqn | r = 2 n - 1       | c =  }} {{end-eqn}} {{qed}} Category:Sigma Function Category:2 rlenuaadjh0jsyktdilz01eai5wuvm1
=== Definitions 1 and 2 are equivalent === {{begin-eqn}} {{eqn | l = \bigcup_{t \mathop \in T} \eqclass t \sim       | r = \set {s \in S: \exists t \in T: s \in \eqclass t \sim}       | c = {{Defof|Union of Family of Subsets}} }} {{eqn | r = \set {s \in S: \exists t \in T: s \sim t}       | c = {{Defof|Equivalence Class}} }} {{end-eqn}} {{qed|lemma}} === Definitions 1 and 3 are equivalent === {{begin-eqn}} {{eqn | l = q^{-1} \sqbrk {q \sqbrk T}       | r = q^{-1} \sqbrk {\set {\eqclass t \sim: t \in T} }       | c = {{Defof|Image of Subset under Mapping}} }} {{eqn | r = \set {s \in S: \map q s \in \set {\eqclass t \sim: t \in T} }       | c = {{Defof|Preimage of Subset under Mapping}} }} {{eqn | r = \set {s \in S: \eqclass s \sim \in \set {\eqclass t \sim: t \in T} }       | c = {{Defof|Quotient Mapping}} }} {{eqn | r = \set {s \in S: \exists t \in T: \eqclass s \sim = \eqclass t \sim}       | c =  }} {{eqn | r = \set {s \in S: \exists t \in T: s \sim t}       | c = Equivalence Class holds Equivalent Elements }} {{end-eqn}} {{qed}} Category:Equivalence Relations fqismd5jnqigd2l2wlz8vmkpf6mgl02
=== 1 implies 2 === Let $T = \overline T$. By definition of saturation: : $T = \displaystyle \bigcup_{t \mathop \in T} \left[\!\left[{t}\right]\!\right]$ so we can take $U = T$. {{qed}} === 1 implies 3 === Let $T = \overline T$. By definition of saturation: :$T = q^{-1} \left[{q \left[{T}\right]}\right]$ so we can take $V = q \left[{T}\right]$. {{qed}} === 2 implies 1 === Let $T = \displaystyle\bigcup_{u \mathop \in U} \left[\!\left[{u}\right]\!\right]$ with $U \subset S$. Let $s \in S$ and $t \in T$ such that $s \sim t$. By definition of union: :$\exists u \in U : t \in \left[\!\left[{u}\right]\!\right]$ By definition of equivalence class: :$t \sim u$ Because $\sim$ is transitive: :$s \sim u$ By definition of equivalence class: :$s \in \left[\!\left[{u}\right]\!\right]$ Thus: :$s \in T$ Because $s$ was arbitrary: :$\overline T \subset T$ By Set is Contained in Saturation Under Equivalence Relation: :$T \subset \overline T$ Thus: :$T = \overline T$ {{qed}} === 3 implies 1 === Let $V$ be a subset of the quotient mapping of $S$ by $\sim$: :$V \subset S / \sim$ Let $T$ be the preimage of $V$ under $q$: :$T = q^{-1} \left[{V}\right]$ By Quotient Mapping is Surjection and Image of Preimage of Subset under Surjection equals Subset: :$q \left[{q^{-1} \left[{V}\right]}\right] = V$ Thus: {{begin-eqn}} {{eqn | l = q^{-1} \left[{q \left[{T}\right]}\right]       | r = q^{-1} \left[{q \left[{q^{-1} \left[{V}\right]}\right]}\right]       | c =  }} {{eqn | r = q^{-1} \left[{V}\right]       | c =  }} {{eqn | r = T       | c =  }} {{end-eqn}} Thus $T$ equals its saturation. {{qed}} Category:Equivalence Relations 4ja7mfpgss0uloxlhwfj062o9xv383f
By Bottom is Compact: :$\bot_L$ is a compact element, where $\bot_L$ denotes the smallest element in $L$. By definition of compact subset: :$\bot_L \in K \left({L} \right)$ By definition of the smallest element: :$\forall x \in K\left({L}\right): \bot_L \preceq x$ By definition of ordered subset: :$\forall x \in K\left({L}\right): \bot_L \precsim x$ Thus by definition: :$P$ is bounded below. Thus by Compact Subset is Join Subsemilattice: :$P$ is join semilattice. {{qed}}
{{begin-eqn}} {{eqn | l = 0       | r = \int_a^b \paren 0 h \rd x }} {{eqn | r = \int_a^b \paren {-\map {\frac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h } h \rd x }} {{eqn | r = \int_a^b t Q h^2 \rd x - \int_a^b h \map \d {\paren {t P + \paren {1 - t} } h'} }} {{eqn | r = \int_a^b t Q h^2 \rd x - h \bigintlimits {\paren {t P + \paren {1 - t} } h'} a b + \int_a^b \paren {\paren {t P + \paren {1 - t} } h'} \rd h       | c = Integration by Parts }} {{eqn | r = \int_a^b t Q h^2 \rd x + \int_a^b \paren {\paren {t P + \paren {1 - t} } h'} h' \rd x }} {{eqn | r = \int_a^b \paren {\paren {P h'^2 + Q h^2} t + \paren {1 - t} h'^2} \rd x }} {{end-eqn}} {{qed}}
By Necessary Condition for Twice Differentiable Functional to have Minimum, $J$ is minimised by $y = \map {\hat y} x$ if: :$\delta^2 J \sqbrk {\hat y; h} \ge 0$  for all admissable real functions $h$. By lemma 1 of Legendre's Condition, :$\ds \delta^2 J \sqbrk {y; h} = \int_a^b \paren {P h'^2 + Q h^2} \rd x$ where: :$P = F_{y' y'}$ {{explain|and what is $Q$?}} By Nonnegative Quadratic Functional implies no Interior Conjugate Points, $\openint a b$ does not contain any conjugate points {{WRT}} $J$. {{qed}} {{Namedfor|Carl Gustav Jacob Jacobi|cat = Jacobi}}
Define the following sums: :$\displaystyle A_n = \sum_{i \mathop = 1}^n a_i$ :$\displaystyle B_n = \sum_{i \mathop = 1}^n b_i$ Let $\epsilon > 0$ and $\mu = \dfrac {\epsilon} 2$.  By the definition of convergent sequences, there exists $k \in \N$ such that: :$\forall n > k: \paren {L - \mu} b_n < a_n < \paren {L + \mu} b_n$ Rewrite the sum as: {{begin-eqn}} {{eqn | l = A_n = a_1 + a_2 + \cdots + a_k + a_{k + 1} + \cdots + a_n       | o = }} {{eqn | ll= \leadsto       | l = a_1 + a_2 + \cdots + a_k + \paren {L - \mu} \paren {b_{k + 1} + \cdots + b_n}       | o = <       | m = a_1 + a_2 + \cdots + a_n       | mo= <       | r = a_1 + a_2 + \cdots + a_k + \paren {L + \mu} \paren {b_{k + 1} + \cdots + b_n} }} {{eqn | ll= \leadsto       | l = A_k + \paren {L - \mu} \paren {B_n - B_k}       | o = <       | m = A_n       | mo= <       | r = A_k + \paren {L + \mu} \paren {B_n - B_k} }} {{end-eqn}} Divide above by $B_n$: :$\dfrac {A_k + \paren {L - \mu} B_k} {B_n} + \paren {L - \mu} < \dfrac {A_n} {B_n} < \paren {L + \mu} + \dfrac {A_k + \paren {L + \mu} B_k} {B_n}$ Let $k$ be fixed.  From Reciprocal of Null Sequence and Combination Theorem for Sequences, the sequence $\sequence {\dfrac {A_k + \paren {L \pm \epsilon} B_k} {B_n} }$ converges to zero. By the definition of convergent sequences, there exists $N > k > 0$ such that: :$\size {\dfrac {A_k + \paren {L \pm \mu} B_k} {B_n} } < \mu$ for all $n > N$ Substitute the above into the inequality and obtain: {{begin-eqn}} {{eqn | o =         | r = L - 2 \mu < \frac {A_n} {B_n} < L + 2 \mu }} {{eqn | l = \leadstoandfrom       | o =        | r = \size {\frac {A_n} {B_n} - L} < \epsilon       | c = for all $n > N$ }} {{end-eqn}} Hence by the definition of convergent sequences the result follows. {{qed}}
By Bottom is Way Below Any Element: :$\bot \ll \bot$ where $\ll$ denotes the way below relation. Hence $\bot$ is a compact element. {{qed}}
By definition of quotient topology, $\map p U$ is open in $X / \sim$ {{iff}} $\map {p^{-1} } {\map p U}$ is open in $X$. {{qed}}
=== Sufficient Condition === Assume that :$x^{\mathrm{compact} }$ is principal ideal in $P$. By definitions of compact subset and compact closure: :$x^{\mathrm{compact} } \subseteq K \left({L} \right)$ By definition of principal ideal: :$\exists y \in x^{\mathrm{compact} }: y$ is upper bound for $x^{\mathrm{compact} }$ in $P$. By definition of ordered subset: :$y$ is upper bound for $x^{\mathrm{compact} }$ in $L$. By definition of supremum: :$\sup_L \left({x^{\mathrm{compact} } }\right) \preceq y$ By definition of algebraic: :$L$ satisfies axiom of K-approximation. By axiom of K-approximation: :$x \preceq y$ By definition of compact closure: :$y \preceq x$ and $y$ is a compact element. Thus by definition of antisymmetry: :$x$ is a compact element. {{qed|lemma}} === Necessary Condition === Let $x$ be a compact element. By definition of compact subset: :$x \in K\left({L}\right)$ We will prove that :$x^{\mathrm{compact} } = x^\precsim$ Let $y \in K\left({L}\right)$. :$y \in x^{\mathrm{compact} }$ {{iff}} :$y \preceq x$ by definitions of compact closure and compact subset {{iff}} :$y \precsim x$ by definition of ordered subset {{iff}} :$y \in x^\precsim$ {{qed|lemma}} Hence $x^{\mathrm{compact} }$ is principal ideal in $P$. {{qed}}
Let $n = 6 k$. Then: :$n = 2 \times 3 k$ and so $3 k$ is a factor of $n$. :$n = 3 \times 2 k$ and so $2 k$ is a factor of $n$. :$n = 6 \times k$ and so $k$ is a factor of $n$. But: :$n = k + 2 k + 3 k$ and so is the sum of a subset of its factors Hence the result by definition of semiperfect. {{qed}} Category:Semiperfect Numbers Category:6 3askxluwp6e4d1z6jxg3a37go0jam49
{{ProofWanted|use Group Acts by Homeomorphisms Implies Projection on Quotient Space is Open and Open Projection and Closed Graph Implies Quotient is Hausdorff}} Category:Topological Groups Category:Hausdorff Spaces 82ai0gmozp1x1kqo22rnjy79g5q6rxi
Let $P$ be a subset of the divisors of $n$ such that the sum of the elements of $P$ equals $n$. Let $\sigma = \displaystyle \sum_{p \mathop \in P} p$ be the sum of the elements of $P$. Let: :$Q = \set {k p: p \in P}$ be the set of elements of $P$ multiplied by $k$. We have by definition that: :$\forall p \in P: p \divides P$ where $p \divides P$ denotes divisibility. Then we have: :$\forall p \in P: k p \divides k P$ Thus $Q$ constitutes a subset of the divisors of $k n$. But: {{begin-eqn}} {{eqn | l = n k       | r = k \sigma       | c =  }} {{eqn | r = k \sum_{p \mathop \in P} p       | c =  }} {{eqn | r = \sum_{p \mathop \in P} k p       | c =  }} {{eqn | r = \sum_{q \mathop \in Q} q       | c =  }} {{end-eqn}} demonstrating that $Q$ constitutes a subset of the divisors of $k n$ equal to $k n$. Hence the result, by definition of semiperfect number. {{qed}} Category:Semiperfect Numbers bwwiv7sseq7wi2gkqx661fmxv4wrz02
Let $x \in c\left[{K\left({L}\right)}\right]$. By definition of image of set: :$\exists y \in K\left({L}\right): x = c\left({y}\right)$ and :$x \in c\left[{S}\right]$ By definition of compact subset: :$y$ is compact in $L$. By definition of compact element: :$y \ll y$ where $\ll$ denotes the way below relation. Define $P = \left({c \left[{S} \right], \precsim}\right)$ as an ordered subset of $L$. We will prove that :for every directed subset $D$ of $c\left[{S}\right]$: $x \precsim \sup_P D \implies \exists d \in D: d \precsim x$ Let $D$ be a directed subset of $c\left[{S}\right]$. By definition of ordered subset: :$D$ is directed in $L$. By definition of algebraic ordered set: :$L$ is up-complete. By definition of up-complete: :$D$ admits a supremum in $L$. {{begin-eqn}} {{eqn | l = c\left({ {\sup}_L D}\right) | r = {\sup}_L\left({c\left[{D}\right]}\right) | c = definition of mapping preserves directed suprema }} {{eqn | r = {\sup}_L D | c = definition of closure operator/idempotent }} {{end-eqn}} By definition of image of set: :$\sup_L D \in c\left[{S}\right]$ By Supremum in Ordered Subset: :$\sup_L D = \sup_P D$ By definition of ordered subset: :$x \preceq \sup_L D$ By definition of closure operator/inflationary: :$y \preceq x$ By definition of transitivity: :$y \preceq \sup_L D$ By definition of way below relation: :$\exists d \in D: y \preceq d$ By definition of subset: :$d \in c\left[{S}\right]$ By definition of closure operator/idempotent: :$d = c\left({d}\right)$ By definition of closure operator/increasing: :$x = c\left({y}\right) \preceq c\left({d}\right) = d$ Thus by definition of ordered subset: :$\exists d \in D: x \precsim d$ {{qed|lemma}} By definition of way below relation: :$x \ll_P x$ By definition: :$x$ is a compact element in $P$. Thus by definition of compact subset: :$x \in K\left({P}\right)$ {{qed}}