hcju/proofwikiqa huggingface dataset:
By Bottom is Way Below Any Element: :$\bot \ll \bot$ where $\ll$ is the way below relation. By definition: :$\bot$ is compact. Thus by definition of compact subset: :$\bot \in K\left({L}\right)$ {{qed}}
Consider $p \ne m$. Then $\mathbf {AB}$ is defined, but $\mathbf {BA}$ is not. So: :$p \ne m \implies \mathbf {AB} \ne \mathbf {BA}$ Now consider $p = m$, and $m \ne n$. Then $\mathbf {A B}$ is an $m \times m$ matrix, while $\mathbf {BA}$ is an $n \times n$ matrix. $p \ne m \implies \mathbf {AB} \ne \mathbf {BA}$ regardless of whether $m = n$ is true or not. Hence: :$p \ne m \lor m \ne n \implies \mathbf {AB} \ne \mathbf {BA}$ Now consider $p = m$, $m = n$, and $n \ne 1$. It remains to be proved that: :$\mathbf {AB} \ne \mathbf {BA}$ This is demonstrated in Matrix Multiplication is not Commutative. Hence we have that $p \ne m \lor m \ne n \implies \mathbf {AB} \ne \mathbf {BA}$ regardless of whether $n = 1$ is true or not, so it is easily seen that: :$p \ne m \lor m \ne n \lor n \ne 1 \implies \exists \mathbf A \in \MM_{m \times n}, \mathbf B \in \MM_{n \times p}: \mathbf {AB} \ne \mathbf {BA} \dashv \vdash \\ \left(\forall \mathbf A \in \MM_{m \times n}, \mathbf B \in \MM_{n \times p}: \mathbf {AB} = \mathbf {BA}\right) \implies \lnot \left({p \ne m \lor m \ne n \lor n \ne 1}\right)$ by Rule of Transposition. The existential quantifier is needed here because there exist square matrices that commute under matrix multiplication, for example the unit matrix. :$\lnot \left({p \ne m \lor m \ne n \lor n \ne 1}\right) \dashv \vdash p = m \land m = n \land n = 1$ by a generalization of De Morgan's Law. Finally, by the properties of an equivalence relation: :$(1): \quad \paren {\forall \mathbf A \in \MM_{m \times n}, \mathbf B \in \MM_{n \times p}: \mathbf {AB} = \mathbf {BA} } \implies p = m = n = 1$ Now to prove the converse: Let $\mathbf A \in \MM_{m \times n}$, $\mathbf B \in \MM_{n \times p}$, and suppose that $p = m = n = 1$, then because $R$ is a commutative ring: :$\mathbf {AB} = A_{11} B_{11} = B_{11} A_{11} = \mathbf {BA}$ and so: :$(2): \quad p = m = n = 1 \implies \paren {\forall \mathbf A \in \MM_{m \times n}, \mathbf B \in \MM_{n \times p}: \mathbf {AB} = \mathbf {BA} }$ Combining $(1)$ and $(2)$ yields the result. {{qed}} Category:Conventional Matrix Multiplication Category:Commutativity Category:Proofs by Induction 55e8727xfy0op1ydlh1pgqwl9tzoc1f
Proof by Counterexample: Let $\mathcal R_1 := \left({S, S, R_1}\right)$ and $\mathcal R_2 := \left({S, S, R_2}\right)$ be relations defined as: Let: :$S = \left\{ {0, 1, 2}\right\}$ :$R_1 = \left\{ {\left({0, 1}\right)}\right\}$ :$R_2 = \left\{ {\left({1, 2}\right)}\right\}$ We have that: :$\mathcal R_1 \circ \mathcal R_2 = \left({S, S, \left\{ {\left({0, 2}\right)}\right\} }\right\}$ while: :$\mathcal R_2 \circ \mathcal R_1 = \left({S, S, \varnothing}\right\}$ {{qed}}
:$y \in x^{\mathrm{compact} }$ {{iff}} :$y \preceq x$ and $y$ is compact by definition of compact closure {{iff}} :$y \in x^\preceq$ and $y$ is compact by definition of lower closure of element {{iff}} :$y \in x^\preceq$ and $y \in K\left({L}\right)$ by definition of compact subset {{iff}} :$y \in x^\preceq \cap K\left({L}\right)$ by definition of intersection Thus by definition of set equality: :the result holds. {{qed}}
Let $y \in x^{\mathrm {compact} }$. By definition of compact closure: :$y \preceq x$ and $y$ is compact. By definition of compact: :$y \ll y$ where $\ll$ denotes the way below relation. By Preceding and Way Below implies Way Below and definition of reflexivity: :$y \ll x$ Thus by definition of way below closure: :$y \in x^\ll$ {{qed}}
Thus by assumption: :$x^\ll$ is a non-empty set. Let $y, z \in x^\ll$ By definition of way below closure: :$y \ll x$ and $z \ll x$ By Join is Way Below if Operands are Way Below :$y \vee z \ll x$ By definition of way below closure: :$y \vee z \in x^\ll$ By Join Succeeds Operands: :$y \preceq y \vee z$ and $z \preceq y \vee z$ Thus by definition :$x^\ll$ is directed. {{qed}}
=== Definition 1 implies Definition 2 === Let $\mathcal R$ be a relation which fulfils the condition: :$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \notin \mathcal R$ Then: {{begin-eqn}} {{eqn | o =        | r = \left({x, y}\right) \in \mathcal R       | c =  }} {{eqn | o = \implies       | r = \left({y, x}\right) \notin \mathcal R       | c = by hypothesis }} {{eqn | o = \implies       | r = \left({x, y}\right) \notin \mathcal R^{-1}       | c = {{Defof|Inverse Relation}} }} {{eqn | o = \implies       | r = \left({x, y}\right) \in \complement \left({\mathcal R^{-1} }\right)       | c = {{Defof|Set Complement}} }} {{eqn | o = \implies       | r = \mathcal R \subseteq \complement \left({\mathcal R^{-1} }\right)       | c = {{Defof|Subset}} }} {{eqn | o = \implies       | r = \mathcal R \cap \mathcal R^{-1} = \varnothing       | c = Empty Intersection iff Subset of Complement }} {{end-eqn}} Hence $\mathcal R$ is asymmetric by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\mathcal R$ be a relation which fulfils the condition: : $\mathcal R \cap \mathcal R^{-1} = \varnothing$ Then: {{begin-eqn}} {{eqn | o =       | r = \left({x, y}\right) \in \mathcal R }} {{eqn | o = \implies       | r = \left({x, y}\right) \notin \mathcal R^{-1}       | c = as $\mathcal R \cap \mathcal R^{-1} = \varnothing$ }} {{end-eqn}} Hence $\mathcal R$ is asymmetric by definition 1. {{Qed}} Category:Symmetric Relations 1mhxk5edgp3hw40o98b5cq68wprwz8q
We will prove that :$\forall b \in S: b$ is upper bound for $Y \implies \sup X \preceq b$ Let $b \in S$ such that :$b$ is upper bound for $Y$. By assumption: :$b$ is upper bound for $X$. Thus by definition of supremum: :$\sup X \preceq b$ {{qed|lemma}} By definition of supremum: :$\sup X$ is upper bound for $X$. By assumption: :$\sup X$ is upper bound for $Y$. Thus by definition of supremum: :$\sup X = \sup Y$ {{qed}}
{{begin-eqn}} {{eqn | l = \tuple {t, s}       | o = \in       | r = \RR_2^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = \tuple {s, t}       | o = \in       | r = \RR_2       | c = {{Defof|Inverse Relation}} }} {{eqn | ll= \leadsto       | l = \tuple {s, t}       | o = \in       | r = \RR_1       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | l = \tuple {t, s}       | o = \in       | r = \RR_1^{-1}       | c = {{Defof|Inverse Relation}} }} {{end-eqn}} The result follows by definition of subset. {{qed}} Category:Subsets Category:Inverse Relations go17ymoftd92q7gvsjnuspu7i5xiec5
=== First Implication === Let $x$ be upper bound for $X$, Let $z \in X$. By definition of upper bound: :$z \preceq x$ Thus by definition of transitivity: :$z \preceq y$ {{qed|lemma}} === Second Implication === This follows by mutatis mutandis. {{qed}}
As $f$ is a mapping, it is by definition also a many-to-one relation. It follows from Inverse of Many-to-One Relation is One-to-Many that its inverse $f^{-1}$ is a one-to-many relation. Thus Image of Intersection under One-to-Many Relation: Family of Sets can be applied for $\mathcal R = f^{-1}$: :$\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} T_i} = \bigcap_{i \mathop \in I} \mathcal R \sqbrk {T_i}$ where $\mathcal R \sqbrk {T_i}$ denotes the image of $T_i$ under $\mathcal R$. {{Qed}}
=== Sufficient Condition === Let $L$ be arithmetic. Let $a, x, y \in S$ such that :$a \ll x$ and $a \ll y$ By Algebraic iff Continuous and For Every Way Below Exists Compact Between: :$\exists c \in K\left({L}\right): a \preceq c \preceq x$ and :$\exists k \in K\left({L}\right): a \preceq k \preceq y$ where $K\left({L}\right)$ denotes the compact subset of $L$. By Meet Semilattice is Ordered Structure: :$c \wedge k \preceq x \wedge y$ By definition of arithmetic: :$K\left({L}\right)$ is meet closed. By definition of meet closed: :$c \wedge k \in K\left({L}\right)$ By definition of compact subset: :$c \wedge k$ is compact. By definition of compact: :$c \wedge k \ll c \wedge k$ By Meet is Idempotent: :$a \wedge a = a$ By Meet Semilattice is Ordered Structure: :$a \preceq c \wedge k$ Thus by Preceding and Way Below implies Way Below :$a \ll x \wedge y$ {{qed|lemma}} === Necessary Condition === Let $\ll$ be a multiplicative relation. Thus $L$ is algebraic. It remains to prove that :$K\left({L}\right)$ is meet closed. Let $x, y \in K\left({L}\right)$. By definition of compact subset: :$x$ and $y$ are compact. By definition of compact: :$x \ll x$ and $y \ll y$ By Way Below Relation is Auxiliary Relation: :$\ll$ is auxiliary relation. By Multiplicative Auxiliary Relation iff Congruent: :$x \wedge y \ll x \wedge y$ By definition: :$x \wedge y$ is compact. Thus by definition of compact subset: :$x \wedge y \in K\left({L}\right)$ {{qed}}
From Finite Subsets form Ideal, the set of finite subsets of $I$ form an ideal of $I$. From Elements with Support in Ideal form Submagma of Direct Product, $T$ is a submagma of $S$. Category:Direct Products a8j7iq8367fpt48etb26zeo5nb77b67
By Product of Cyclotomic Polynomials: :$\displaystyle \prod_{d \mathop \divides n} \map {\Phi_d} x = x^n - 1$ for all $n \in \N$. The nonzero rational forms form an abelian group under multiplication. By the Möbius inversion formula for abelian groups, this implies: :$\displaystyle \map {\Phi_n} x = \prod_{d \mathop \divides n} \paren {x^d - 1}^{\map \mu {n / d} }$ for all $n \in \N$. {{qed}} Category:Cyclotomic Polynomials ic3t80hx736i6vqfc3xi1j06ekyufsm
By Inverse of Mapping is One-to-Many Relation, $f^{-1}$ is a one-to-many relation: :$f^{-1} \subseteq T \times S$ whose domain is the image set of $f$. By definition of composition of relations: :$f \circ f^{-1} := \set {\tuple {x, z} \in T \times T: \exists y \in S: \tuple {x, y} \in \mathcal f^{-1} \land \tuple {y, z} \in f}$ Thus $f \circ f^{-1}$ is a relation on $T \times T$. Let $x \in \Img f$. Then: :$\exists y \in S: \tuple {x, y} \in f^{-1}$ As $f$ is a mapping, and so by definition a many-to-one relation; :$\map f y = x$ for all $y$ such that $\tuple {x, y} \in f^{-1}$. That is: :$\forall x \in \Img f: f \circ \map {f^{-1} } x = x$ Hence the result by definition of identity mapping. {{qed}}
=== Definition 1 implies Definition 2 === Let $\struct {S, \preceq}$ be an order complete set by definition 1. Let $H \subseteq S$ have a lower bound. Let $K$ be the set of all lower bounds of $H$. Then $K \ne \O$. By definition of lower bound: :$\forall x \in K: \forall y \in H: x \le y$ and so all elements of $H$ are upper bounds of $K$. Thus by hypothesis $K$ admits a supremum. Let $k = \map \sup K$. By definition of supremum, $k$ precedes every upper bound of $K$. In particular, $k$ precedes every element of $H$. Thus $k$ is a lower bound of $H$. But we have that $k$ is an upper bound of $K$. That is, $k$ succeeds every lower bound of $H$. That is, $k$ is an infimum of $H$. Thus $\struct {S, \preceq}$ is an order complete set by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\struct {S, \preceq}$ be an order complete set by definition 2. Let $H \subseteq S$ have an upper bound. Let $K$ be the set of all upper bounds of $H$. Then $K \ne \O$. By definition of upper bound: :$\forall x \in K: \forall y \in H: y \le x$ and so all elements of $H$ are lower bounds of $K$. Thus by hypothesis $K$ admits an infimum. Let $k = \map \inf K$. By definition of infimum, $k$ succeeds every lower bound of $K$. In particular, $k$ succeeds every element of $H$. Thus $k$ is an upper bound of $H$. But we have that $k$ is a lower bound of $K$. That is, $k$ precedes every upper bound of $H$. That is, $k$ is a supremum of $H$. Thus $\struct {S, \preceq}$ is an order complete set by definition 1. {{qed}}
By hypothesis, let the intersection of $K$ with every set of the form: :$\set {y \in T: u \prec y \prec v: u, v \in T, u \prec v}$ be non-empty. {{AimForCont}} $f \ne g$. Then: :$\exists x \in S: \map f x \ne \map g x$ {{WLOG}}, suppose that $\map f x < \map g x$. Let: :$a \in H$ such that $a \preceq x$ :$b \in H$ such that $b \succeq x$ We have that $f$ and $g$ are increasing. Thus: :$\map f a \preccurlyeq \map f x$ :$\map g b \succcurlyeq \map g x$ Thus no element of $H$ maps to the set: :$\set {y \in T: \map f x \prec y \prec \map g x}$ That is, the intersection of $K$ with this set, which is of the form: :$\set {y \in T: u \prec y \prec v: u, v \in T, u \prec v}$ is empty. This contradicts our hypothesis. Thus, by Proof by Contradiction: :$f = g$ {{qed}}
=== Necessary Condition === Suppose $f$ is such that it is not the case that: :for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$. Proof by Counterexample: Let $S = \R_{>0}$ be the set of all (strictly) positive real numbers. Let $H \subseteq S$ be the open real interval $H = \openint 0 1$. Let $T = H$. By the axiomatic definition of real numbers, $S$ and $T$ are totally ordered. From the Continuum Property, $T$ is order complete. Let $f: H \to T$ be the identity mapping. Note that while $H$ has an upper bound in $S$, for example $1$ Let $y \in T$ be an upper bound of $f \sqbrk H$. By construction: :$y < 1$ and so: :$\exists \epsilon \in \R_{>0}: y + \epsilon = 1$ Thus: :$y < y + \dfrac \epsilon 2 < 1$ and so there exists $y' = y + \dfrac \epsilon 2 \in T$ such that: :$y < y'$ and such that: :$y' = \map f {y'} \in f \sqbrk H$ So $y$ is not an upper bound of $f \sqbrk H$. Thus $f \sqbrk H$ has no upper bound in $T$. Let $g: S \to T$ be an extension of $f$ to $S$. Consider $\map g 1$. We have that $1$ is an upper bound of $H$. Let $\map g 1 = y$. As $y \in \openint 0 1$ it follows that: :$y > 1$ and so: :$\exists \epsilon \in \R_{>0}: y + \epsilon = 1$ But then: :$y < y + \dfrac \epsilon 2 < 1$ and so: :$\map g 1 < \map g {y + \dfrac \epsilon 2}$ and so $g$ is not increasing. Thus it has been demonstrated that if it is not the case that: :for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$ where: :$\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets and: :$T$ is order complete and: :$f: H \to T$ is an increasing mapping from $H$ to $T$. and: :$H \subseteq S$ is a subset of $S$ then it is not necessarily the case that $f$ always has an extension to $S$ which is increasing. {{qed|lemma}} === Sufficient Condition === Let $f$ be such that: :for all $A \subseteq H$: if $A$ is bounded in $S$, then $f \sqbrk A$ is bounded in $T$. An increasing mapping $g: S \to T$ is to be constructed such that $g$ is an extension of $f$. So, let $A \subseteq H$ such that $A$ is bounded in $S$. Let $x \in A$. Consider the set: :$A' = \set {y \in A: y \preceq x}$ $A'$ is bounded in $S$: :bounded below by a lower bound of $A$ :bounded above by $y$. Thus by hypothesis $f \sqbrk {A'}$ is bounded in $T$. A lower bound of $f \sqbrk {A'}$ is also a lower bound of $f \sqbrk A$. Similarly for upper bounds. {{explain|{{BookReference|General Topology|1955|John L. Kelley}}, from which this has been taken, is unclear here exactly what is being demonstrated by the above.}} For each $x \in S$ let $L_x$ be defined as: :$L_x = \set {y \in H: y \le x}$ Note that $L_x$ is bounded above by $x$ by definition of bounded above. Suppose $L_x = \O$. Then $x$ is a lower bound for $H$. Thus $f \sqbrk H$ has an infimum $v$, and $\map g x$ can be defined as: :$\map g x = v$ Suppose $L_x \ne O$. We have that $x$ is an upper bound of $L_x$. Hence by hypothesis $f \sqbrk {L_x}$ has an upper bound. As $T$ is order complete, $f \sqbrk {L_x}$ admits a supremum. Thus $\map g x$ can be defined as: :$\map g x = \sup \set {f \sqbrk {L_x} }$ It remains to be proved that $g$ is an increasing mapping. {{finish}}
Let $\map f x = x^n$ for $x \in \R, n \in \N$. Let $a \in \R$. By definition of the derivative: :$\displaystyle \map {f'} a = \lim_{x \mathop \to a} \frac {\map f x - \map f a} {x - a} = \lim_{x \mathop \to a} \frac {x^n - a^n} {x - a}$ === Case $\text I$ === For $n = 0$ it is possible to do: {{begin-eqn}} {{eqn | l = \map {f'} a       | r = \lim_{x \mathop \to a} \frac {x^0 - a^0} {x - a}       | c =  }} {{eqn | r = \lim_{x \mathop \to a} \frac 0 {x - a} }} {{eqn | r = 0       | c =  }} {{end-eqn}} We have that: :$0 \cdot x^{0 - 1} = 0$ for all $x \ne 0$. So in this case: :$\map {f'} x = n x^{n - 1}$ {{qed|lemma}} === Case $\text {II}$ === For $n = 1$ we have: :$\map f x = x$ From Derivative of Identity Function: :$\map {f'} x = 1$ Then we note that: :$1 \cdot x^{1 - 1} = 1$ So for the case $n = 1$: :$\map {f'} x = n x^{n - 1}$ {{qed|lemma}} === Case $\text {III}$ === $\R$ is a commutative ring, so for $n \ge 2$ it is possible to do: {{begin-eqn}} {{eqn | l = \map {f'} a       | r = \lim_{x \mathop \to a} \frac {x^n - a^n} {x - a}       | c =  }} {{eqn | r = \lim_{x \mathop \to a} \frac {\paren {x - a} \sum_{j \mathop = 0}^{n - 1} x^{n - j - 1} a^j} {x - a}       | c = Difference of Two Powers }} {{eqn | r = \lim_{x \mathop \to a} \sum_{j \mathop = 0}^{n - 1} x^{n - j - 1} a^j }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} a^j       | c = Real Polynomial Function is Continuous }} {{eqn | r = \sum_{j \mathop = 0}^{n - 1} a^{n - 1} }} {{eqn | r = \sum_{j \mathop = 1}^n a^{n - 1}       | c = shifting the index from $0$ to $1$ }} {{eqn | r = n a^{n - 1} }} {{end-eqn}} This holds for all $a \in \R$, so: :$\map {f'} x = n x^{n - 1}$ {{qed}}
The situation is modelled by considering the ratios of the areas of: :a square to the circle in which it is inscribed :a square to the circle around which it has been circumscribed. Let a square $S$ be inscribed in a circle $C$ of radius $r$. Let $A_c$ and $A_s$ be the areas of $C$ and $S$ respectively. From Area of Circle: :$A_c = \pi r^2$ The diameter of $S$ is $2 r$. Thus from Pythagoras's Theorem its side is of length $r \sqrt 2$. From Area of Square: :$A_s = 2 r^2$ Thus: :$\dfrac {A_s} {A_c} = \dfrac {2 r^2} {\pi r^2} = \dfrac 2 \pi \approx 0.6366 \ldots$ {{qed|lemma}} Let a square $S$ be circumscribed around a circle $C$ of radius $r$. Let $A_c$ and $A_s$ be the areas of $C$ and $S$ respectively. From Area of Circle: :$A_c = \pi r^2$ The side of $S$ is of length $2 r$. From Area of Square: :$A_s = 4 r^2$ Thus: :$\dfrac {A_c} {A_s} = \dfrac {\pi r^2} {4 r^2} = \dfrac \pi 4 \approx 0.7853 \ldots$ {{qed|lemma}} Thus a round peg takes up more space ($0.7853 \ldots$) of a square hole than a square peg takes up ($0.6366 \ldots$) of a round hole. {{qed}}
By Arithmetic iff Way Below Relation is Multiplicative in Algebraic Lattice: :$\ll$ is a multiplicative relation. Thus by Way Below Relation is Multiplicative implies Pseudoprime Element is Prime: :the result holds. {{qed}}
By If Every Element Pseudoprime is Prime then Way Below Relation is Multiplicative: :$\ll$ is a multiplicative relation. where $\ll$ denotes the way below relation. Thus by Arithmetic iff Way Below Relation is Multiplicative in Algebraic Lattice: :the result holds. {{qed}}
=== Necessary Condition === Let $f: \R \to \R$ be a real function with a periodic element $L$. Then: {{begin-eqn}} {{eqn | l = \map f x       | r = \map f {n L + r} \quad 0 < r < \size L       | c = {{Defof|Quotient (Algebra)/Real|Quotient}} }} {{eqn | r = \map f r       | c = General Periodicity Property }} {{eqn | r = \map f {x \bmod L} }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $f: \R \to \R$ be a real function such that for all $x \in \R$: :$\map f {x \bmod L} = \map f x$ Let $n = \floor {\dfrac x L}$. Then: {{begin-eqn}} {{eqn | l = \map f {x + L}       | r = \map f {\paren {x + L} \bmod L} }} {{eqn | r = \map f {x + L - \floor {\frac {x + L} L} L} }} {{eqn | r = \map f {x + L - \paren {n + 1} L} }} {{eqn | r = \map f {x - n L} }} {{eqn | r = \map f {x \bmod L} }} {{eqn | r = \map f x }} {{end-eqn}} {{qed}} Category:Periodic Functions rvzzuzuf3lgl96kwzfirjj3a10rp7y9
From the definition of $x$ to the power of $r$: :$x^r = \map \exp {r \ln x}$ The result follows from Exponential of Real Number is Strictly Positive. {{qed}} Category:Power of Positive Real Number is Positive h0m8t9tu4wm2hmo6bal3kzgkd208761
The case when $x = \varnothing$ By Empty is Bottom of Lattice of Power Set: :$x = \bot$ where $\bot$ denotes the bottom of $L$. By Bottom is Way Below Any Element: :$x \ll x$ where $\ll$ denotes the way below relation. Thus by definition :$x$ is a finite set {{iff}} $x$ is a compact element. {{qed|lemma}} The case when $x \ne \varnothing$: === Sufficient Condition when Non-empty === Let $x$ be a finite set. We will prove that :for every a set $Y$ of subsets of $X$ such that $x \subseteq \bigcup Y$ ::then there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$ Let $Y$ be a set of subsets of $X$ such that :$x \subseteq \bigcup Y$ By definitions of union and subset: :$\forall e \in x: \exists u \in Y: e \in u$ By Axiom of Choice: :$\exists f:x \to Y: \forall e \in x: e \in f\left({e}\right)$ Define $Z = f\left[{x}\right]$ By Image of Mapping from Finite Set is Finite: :$Z$ is a finite set. By definition of image of mapping: :$Z \subseteq Y$ It remains to prove that :$x \subseteq \bigcup Z$ Let $e \in x$. By definitions of $f$ and image of mapping: :$e \in f\left({e}\right) \in Z$ Thus by definition of union: :$e \in \bigcup Z$ {{qed|lemma}} By Way Below in Lattice of Power Set: :$x \ll x$ Thus by definition of compact: :$x$ is compact. {{qed|lemma}} === Necessary Condition when Non-empty === Let $x$ be a compact element. By definition of subset: :$Y := \left\{ {\left\{ {y}\right\}: y \in x}\right\}$ is set of subsets of $X$. We will prove that :$x \subseteq \bigcup Y$ Let $y \in x$. By definition of $Y$: :$\left\{ {y}\right\} \in Y$ By definition of singleton: :$y \in \left\{ {y}\right\}$ Thus by definition of union: :$y \in \bigcup Y$ {{qed|lemma}} By definition of compact: :$x \ll x$ By Way Below in Lattice of Power Set: :there exists finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$ By Singleton is Finite: :$\forall A \in Z: A$ is a finite set. By Finite Union of Finite Sets is Finite: :$\bigcup Z$ is a finite set. Thus by Subset of Finite Set is Finite: :$x$ is a finite set. {{qed}}
Denote $\displaystyle L = \lim_{t \mathop \to +\infty} \map f t$. Define the constant mapping: :$\map C t = - L$ Further define: :$\map g t = \map f t + \map C t$ From: :Constant Function is of Exponential Order Zero, :Sum of Functions of Exponential Order, it is sufficient to prove that $g$ is of exponential order $0$. Fix $\epsilon > 0$ arbitrarily small. By definition of limit at infinity, there exists $c \in \R$ such that: :$\forall t > c: \size {\map f t - L} < \epsilon$ Therefore: {{begin-eqn}} {{eqn | lo= \forall t \ge c + 1 :       | l = \size {\map g t}       | r = \size {\map f t + \map C t} }} {{eqn | r = \size {\map f t - L} }} {{eqn | o = <       | r = \epsilon }} {{eqn | r = \epsilon \cdot e^0       | c = Exponential of Zero }} {{end-eqn}} The result follows from the definition of exponential order, with $M = c + 1$, $K = \epsilon$, and $a = 0$. {{qed}} Category:Exponential Order lvk8spvz050lbd2gahjror7ecar3duz
Let $U$ be an upper bound of $f$. Let $L$ be a lower bound of $f$. Let $K > \max \set {\size U, \size L}$. Then:  {{begin-eqn}} {{eqn | ll= \forall t \ge 1:       | l = \size {\map f t}       | o = <       | r = K       | c = {{Defof|Bounded Mapping}} }} {{eqn | r = K e^{0 t}       | c = Exponential of Zero }} {{end-eqn}} The result follows from the definition of exponential order with $M = 1$, $K = K$, and $a = 0$. {{qed}}
Follows from Function with Limit at Infinity of Exponential Order Zero. {{qed}} {{MissingLinks|limit of $\arctan$}} Category:Exponential Order bieu3d2ml5x5jwrc0bdhsm79yqd94si
Follows from Function with Limit at Infinity of Exponential Order Zero. {{qed}} {{MissingLinks|limit of $\arccot$}} Category:Exponential Order ri29ldq9urkxmn6a26kq9sxapo9m553
The limit: :$\displaystyle \lim_{x \mathop \to +\infty} \map \Si x = \lim_{x \mathop \to +\infty} \int_{t \mathop \to 0}^{t \mathop = x} \frac {\sin t} t \rd t$ is the Dirichlet Integral. {{qed}}