hcju/proofwikiqa huggingface dataset:
Thus by Way Below Closure is Directed in Bounded Below Join Semilattice: :$\forall x \in S:x^\ll$ is directed. Thus by definition pf complete lattice: :$L$ is up-complete. Let $x \in S$. Define $W := \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$ By definition of way below closure: :$x$ is upper bound for $x^\ll$ By definition of supremum: :$\sup \left({x^\ll}\right) \preceq x$ We will prove that :$W \subseteq x^\ll$ Let $d \in W$. By definition of $W$: :$\exists V: d = \inf V \land x \in V \land V \in \sigma\left({L}\right)$ By Scott Topology equals to Scott Sigma: :$\tau = \sigma\left({L}\right)$ By definition: :$V$ is open. By Infimum of Open Set is Way Below Element in Complete Scott Topological Lattice: :$d \ll x$ Thus by definition of way below closure: :$d \in x^\ll$ {{qed|lemma}} By Supremum of Subset: :$\sup W \preceq \sup\left({x^\ll}\right)$ By assumption: :$x \preceq \sup\left({x^\ll}\right)$ Hence by definition of antisymmetry: :$x = \sup \left({x^\ll}\right)$ {{qed}}
From the definition of a proper class, $\mathrm P$ is not a set. The rest then follows from the definition of a class. {{qed}} Category:Class Theory d2ap0spvwqtlgud2le5hshqan5ni1qq
By definition of greatest common divisor: :$R \divides S$ in $E \sqbrk X$ By Polynomial Forms over Field is Euclidean Domain, there exist $A, B \in F \sqbrk X$ such that: :$A P + B Q = R$ Because $S \divides P, Q$: :$S \divides R$ in $E \sqbrk X$ By $R \divides S$ and $S \divides R$: :$R = S$ {{qed}} Category:Field Theory Category:Polynomial Theory Category:Greatest Common Divisor 1wj176ibkir1rzcynfd9ji9gdsnz8qp
Let $x = z + 40$. Then: {{begin-eqn}} {{eqn | r = \left({z + 40}\right)^2 - 79 \left({z + 40}\right) + 1601       | o =        | c =  }} {{eqn | r = z^2 + 2 \times 40 z + 40^2 - 79 z - 79 \times 40 + 1601       | c =  }} {{eqn | r = z^2 + 80 z + 1600 - 79 z - 3160 + 1601       | c =  }} {{eqn | r = z^2 + z + 41       | c =  }} {{end-eqn}} Thus it can be seen that this is an application of Euler Lucky Number $41$. {{qed}}
Let $\phi:G\times X\to X$ denote the group action. Let $g\in G$. The map $\phi_g : X \to X : x\mapsto \phi(g,x)$ is continuous because $\phi$ is. Its inverse is given by $\phi_{g^{-1}}$, which is continuous as well. Thus $\phi_g$ is an homeomorphism of $X$. {{qed}}
If $G$ acts continuously, then by Continuous Group Action is by Homeomorphisms, $G$ acts by homeomorphisms Let $G$ act by homeomorphisms Let $\phi: G \times X \to X$ denote the group action. For $g \in G$, denote $\phi_g : X \to X : x \mapsto \map \phi {g, x}$ Let $U \subset X$ be open. By Inverse Image of Set Under Mapping from Product of Sets: :$\displaystyle \map {\phi^{-1} } U = \bigcup_{g \mathop \in G} \paren {\set g \times \map {\phi_g^{-1} } U}$ By definition of product topology, $\map {\phi^{-1} } U$ is open in $G \times X$. Thus $\phi$ is continuous. {{qed}}
From Euler Phi Function of Prime: :$\map \phi p = p - 1$ As $3$ is a prime number it follows that: :$\map \phi 3 = 3 - 1 = 2$ {{qed}}
{{begin-eqn}} {{eqn | l = 10110_2 + 01101_2       | r = 100011_2       | c =  }} {{eqn | ll= \leadsto       | l = 100011_2 + 110001_2       | r = 1010100_2       | c =  }} {{end-eqn}} It remains to be shown that a binary number of this form does not become a palindromic number. Let $d_n$ denote $n$ repetitions of a binary digit $d$ in a number. Thus: :$10111010000$ can be expressed as: :$101_3010_4$ '''Beware''' that the subscript, from here on in, does not denote the number base. It is to be shown that the reverse-and-add process applied to: :$101_n010_{n + 1}$ leads after $4$ iterations to: :$101_{n + 1}010_{n + 2}$ Thus: {{begin-eqn}} {{eqn | l = 101_n010_{n + 1} + 0_{n + 1}101_k01       | r = 110_n101_{n - 1}01       | c =  }} {{eqn | ll= \leadsto       | l = 110_n101_{n - 1}01 + 101_{n - 1}010_n11       | r = 101_{n + 1}010_{n + 1}       | c =  }} {{eqn | ll= \leadsto       | l = 101_{n + 1}010_{n + 1} + 0_{n + 1}101_{n + 1}01       | r = 110_{n - 1}10001_{n - 1}01       | c =  }} {{eqn | ll= \leadsto       | l = 110_{n - 1}10001_{n - 1}01 + 101_{n - 1}00010_{n - 1}11       | r = 101_{n + 1}010_{n + 2}       | c =  }} {{end-eqn}} As neither $101_n010_{n + 1}$ nor $101_{n + 1}010_{n + 2}$ are palindromic numbers, nor are any of the intermediate results, the result follows. {{qed}}
The edges are given as having lengths $44$, $117$ and $240$. The faces are therefore: :$44 \times 117$ :$44 \times 240$ :$117 \times 240$ The diagonals of these faces are given by Pythagoras's Theorem as follows: {{begin-eqn}} {{eqn | l = 44^2 + 117^2       | r = 15 \, 625       | c =  }} {{eqn | r = 125^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 44^2 + 240^2       | r = 59 \, 536       | c =  }} {{eqn | r = 244^2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = 117^2 + 240^2       | r = 71 \, 289       | c =  }} {{eqn | r = 267^2       | c =  }} {{end-eqn}} However, its space diagonal is calculated as: {{begin-eqn}} {{eqn | l = 44^2 + 117^2 + 240^2       | r = 73 \, 225       | c =  }} {{eqn | r = \paren {270 \cdotp 6012 \ldots}^2       | c =  }} {{end-eqn}} which, as can be seen, is not an integer. {{ProofWanted|Still to be proved that this is the smallest.}}
{{begin-eqn}} {{eqn | l = \sigma \left({44}\right)       | r = 84       | c = {{SigmaLink|44}} }} {{eqn | l = \sigma \left({45}\right)       | r = 78       | c = {{SigmaLink|45}} }} {{eqn | l = \sigma \left({46}\right)       | r = 72       | c = {{SigmaLink|46}} }} {{eqn | l = \sigma \left({47}\right)       | r = 48       | c = Sigma Function of Prime Number: $47$ is prime }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sigma \left({104}\right)       | r = 210       | c = {{SigmaLink|104}} }} {{eqn | l = \sigma \left({105}\right)       | r = 192       | c = {{SigmaLink|105}} }} {{eqn | l = \sigma \left({106}\right)       | r = 162       | c = {{SigmaLink|106}} }} {{eqn | l = \sigma \left({107}\right)       | r = 108       | c = Sigma Function of Prime Number: $107$ is prime }} {{end-eqn}} {{qed}}
From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. {{begin-eqn}} {{eqn | l = \tau \left({44}\right)       | r = 6       | c = $\tau$ of $44$ }} {{eqn | l = \tau \left({45}\right)       | r = 6       | c = $\tau$ of $45$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({75}\right)       | r = 6       | c = $\tau$ of $75$ }} {{eqn | l = \tau \left({76}\right)       | r = 6       | c = $\tau$ of $76$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({98}\right)       | r = 6       | c = $\tau$ of $98$ }} {{eqn | l = \tau \left({99}\right)       | r = 6       | c = $\tau$ of $99$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({116}\right)       | r = 6       | c = $\tau$ of $116$ }} {{eqn | l = \tau \left({117}\right)       | r = 6       | c = $\tau$ of $117$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({147}\right)       | r = 6       | c = $\tau$ of $147$ }} {{eqn | l = \tau \left({148}\right)       | r = 6       | c = $\tau$ of $148$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({171}\right)       | r = 6       | c = $\tau$ of $171$ }} {{eqn | l = \tau \left({172}\right)       | r = 6       | c = $\tau$ of $172$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({242}\right)       | r = 6       | c = $\tau$ of $242$ }} {{eqn | l = \tau \left({243}\right)       | r = 6       | c = $\tau$ of $243$ }} {{eqn | l = \tau \left({244}\right)       | r = 6       | c = $\tau$ of $244$ }} {{eqn | l = \tau \left({245}\right)       | r = 6       | c = $\tau$ of $245$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tau \left({332}\right)       | r = 6       | c = $\tau$ of $332$ }} {{eqn | l = \tau \left({333}\right)       | r = 6       | c = $\tau$ of $333$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sqbrk {\paren {\mathbf A^*}^* }_{i j}       | r = \overline {\sqbrk {\mathbf A^*}_{j i} }       | c = {{Defof|Hermitian Conjugate}} }} {{eqn | r = \overline {\paren {\overline {\sqbrk {\mathbf A}_{i j} } } }       | c = {{Defof|Hermitian Conjugate}} }} {{eqn | r = \sqbrk {\mathbf A}_{i j}       | c = Complex Conjugation is Involution }} {{end-eqn}} So: :$\paren {\mathbf A^*}^* = \mathbf A$ {{qed}} Category:Linear Algebra Category:Involutions 1vdgyl2fjuo3qnxxdisxrywovolmnma
Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes greater than $11$ ordered by size. Then $S = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, \dots}$. By Bertrand-Chebyshev Theorem: :$s_{n + 1} \le 2 s_n$ for all $n \in \N$. We observe that every integer $n$ where $45 < n \le 45 + s_{11} = 92$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_{10}} = \set {11, 13, 17, 19, 23, 29, 31, 37, 41, 43}$. Hence the result by Richert's Theorem. {{qed|lemma}} Here is a demonstration of our claim: {{begin-eqn}} {{eqn | l = 46       | r = 17 + 29 }} {{eqn | l = 47       | r = 11 + 17 + 19 }} {{eqn | l = 48       | r = 19 + 29 }} {{eqn | l = 49       | r = 13 + 17 + 19 }} {{eqn | l = 50       | r = 19 + 31 }} {{eqn | l = 51       | r = 11 + 17 + 23 }} {{eqn | l = 52       | r = 23 + 29 }} {{eqn | l = 53       | r = 13 + 17 + 23 }} {{eqn | l = 54       | r = 23 + 31 }} {{eqn | l = 55       | r = 13 + 19 + 23 }} {{eqn | l = 56       | r = 19 + 37 }} {{eqn | l = 57       | r = 11 + 17 + 29 }} {{eqn | l = 58       | r = 17 + 41 }} {{eqn | l = 59       | r = 13 + 17 + 29 }} {{eqn | l = 60       | r = 19 + 41 }} {{eqn | l = 61       | r = 13 + 19 + 29 }} {{eqn | l = 62       | r = 19 + 43 }} {{eqn | l = 63       | r = 13 + 19 + 31 }} {{eqn | l = 64       | r = 23 + 41 }} {{eqn | l = 65       | r = 11 + 17 + 37 }} {{eqn | l = 66       | r = 23 + 43 }} {{eqn | l = 67       | r = 13 + 17 + 37 }} {{eqn | l = 68       | r = 31 + 37 }} {{eqn | l = 69       | r = 13 + 19 + 37 }} {{eqn | l = 70       | r = 11 + 17 + 19 + 23 }} {{eqn | l = 71       | r = 17 + 23 + 31 }} {{eqn | l = 72       | r = 13 + 17 + 19 + 23 }} {{eqn | l = 73       | r = 19 + 23 + 31 }} {{eqn | l = 74       | r = 11 + 13 + 19 + 31 }} {{eqn | l = 75       | r = 13 + 19 + 43 }} {{eqn | l = 76       | r = 11 + 17 + 19 + 29 }} {{eqn | l = 77       | r = 17 + 19 + 41 }} {{eqn | l = 78       | r = 11 + 17 + 19 + 31 }} {{eqn | l = 79       | r = 17 + 19 + 43 }} {{eqn | l = 80       | r = 13 + 17 + 19 + 31 }} {{eqn | l = 81       | r = 17 + 23 + 41 }} {{eqn | l = 82       | r = 11 + 17 + 23 + 31 }} {{eqn | l = 83       | r = 19 + 23 + 41 }} {{eqn | l = 84       | r = 13 + 17 + 23 + 31 }} {{eqn | l = 85       | r = 19 + 23 + 43 }} {{eqn | l = 86       | r = 13 + 19 + 23 + 31 }} {{eqn | l = 87       | r = 17 + 29 + 41 }} {{eqn | l = 88       | r = 11 + 17 + 29 + 31 }} {{eqn | l = 89       | r = 19 + 29 + 41 }} {{eqn | l = 90       | r = 13 + 17 + 29 + 31 }} {{eqn | l = 91       | r = 19 + 31 + 41 }} {{eqn | l = 92       | r = 13 + 19 + 29 + 31 }} {{end-eqn}} {{qed}}
{{NotZFC}} By the axiom of class comprehension, let $B$ be the class defined as: {{begin-eqn}} {{eqn | l = B       | r = \set {x: x \in A \land \map \phi x} }} {{eqn | r = \set {x \in A: \map \phi x}       | c = Set-Builder Notation }} {{end-eqn}} {{AimForCont}} that $B$ is not a set. Then $B$ must be a proper class. It is easily seen that $B \subseteq A$. So by the Axiom of Powers: :$B \in \powerset A$ where $\powerset A$ is denotes the power set of $A$. But by Proper Class is not Element of Class, this is a contradiction. Therefore by contradiction it follows that $B$ is a set. {{qed}} Category:Gödel-Bernays Class Theory k64flenmqw83ze9rur8epx3x2emm1v9
{{begin-eqn}} {{eqn | l = H_n       | r = n \paren {2 n - 1}       | c = Closed Form for Hexagonal Numbers }} {{eqn | r = \frac {2 n \paren {2 n - 1} } 2       | c =  }} {{eqn | r = \frac {2 n \paren {2 n - 2 + 1} } 2       | c =  }} {{eqn | r = 4 \frac {n \paren {n - 1} } 2 + \frac {2 n} 2       | c =  }} {{eqn | r = 4 T_{n - 1} + n       | c = Closed Form for Triangular Numbers }} {{end-eqn}} Hence the result. {{qed}}
=== Necessary Condition === Let $\mathbf W$ be an arbitrary differentiable symmetric matrix. Then {{begin-eqn}} {{eqn | l = 0       | r = \int_a^b \map {\frac \d {\d x} } {\mathbf h \mathbf W \mathbf h} \rd x }} {{eqn | r = \int_a^b \mathbf h \mathbf W' \mathbf h \rd x + 2 \int_a^b \mathbf h' \mathbf W \mathbf h \rd x }} {{end-eqn}} Suppose, $\mathbf W$ is such that: :$\mathbf Q + \mathbf W' = \mathbf W \mathbf P^{-1} \mathbf W$ Then: {{begin-eqn}} {{eqn | l = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + \mathbf h \mathbf Q \mathbf h} \rd x       | r = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + 2 \mathbf h' \mathbf W \mathbf h + \mathbf h \mathbf Q \mathbf h + \mathbf h \mathbf W' \mathbf h} \rd x  }} {{eqn | r = \int_a^b \paren {\mathbf h' \mathbf P \mathbf h' + 2 \mathbf h' \mathbf W \mathbf h + \mathbf h \mathbf W \mathbf P^{-1} \mathbf W \mathbf h} \rd x }} {{eqn | r = \int_a^b \paren {\mathbf P^{1/2} \mathbf h' + \mathbf P^{1/2} \mathbf h' \mathbf P^{-1/2} \mathbf W \mathbf h + \mathbf P^{-1/2} \mathbf W \mathbf h}^2 \rd x }} {{end-eqn}} {{explain|Clarify the below -- what is being used to show what?}} Note that: :$\mathbf P^{1/2} \mathbf h' + \mathbf P^{-1/2} \mathbf W \mathbf h \ne 0$ unless: :$\forall x\in\closedint a b: \map {\mathbf h} x = 0$ However, this contradicts the absence of conjugate points. Hence: :$K > 0$ {{qed|lemma}} === Sufficient Condition === Consider the following functional: :$\ds \int_a^b \paren {K t + \mathbf h' \paren {1 - t} \mathbf h'} \rd x$ The corresponding Euler's equations are: :$-\map {\dfrac \d {\d x} } {t \mathbf P \mathbf h' + \paren {1 - t} \mathbf h'} + t \mathbf Q \mathbf h = 0$ Suppose the interval $\closedint a b$ contains a point $\tilde a$ conjugate to $a$. Then the determinant $\size {h_{ij} }$ vanishes. Therefore there exists a linear combination of $h_i$ not identically equal to zero such that $\map {\mathbf h} {\tilde a} = 0$. Furthermore, since the Euler's equations are continuous {{WRT}} $t$, so is the solution of this equation. {{AimForCont}} $\tilde a = b$. By lemma, $K$ vanishes. This contradicts the positive definiteness of $K$. Therefore, $\tilde a \ne b$. Thus, for $t = 1$ the conjugate point may only reside in $\openint a b$. {{finish|to be continued}}
Thus by Compact Closure is Directed: :$\forall x \in S:x^{\mathrm{compact} }$ is directed where $x^{\mathrm{compact} }$ denotes the compact closure of $x$. Thus by definition of complete lattice: :$L$ is up-complete. Let $x \in S$. By definition of lower closure of element: :$x$ is upper closure for $x^\preceq$ By definition of supremum: :$\sup \left({x^\preceq}\right) \preceq x$ By Compact Closure is Intersection of Lower Closure and Compact Subset: :$x^{\mathrm{compact} } = x^\preceq \cap K\left({L}\right)$ By Intersection is Subset: :$x^{\mathrm{compact} } \subseteq x^\preceq$ By Supremum of Subset: :$\sup \left({x^{\mathrm{compact} } }\right) \preceq \sup \left({x^\preceq}\right)$ By definition of transitivity: :$\sup \left({x^{\mathrm{compact} } }\right) \preceq x$ Aiming for a contradiction suppose that :$x \ne \sup \left({x^{\mathrm{compact} } }\right)$ We will prove that :$x \notin \left({\left({x^{\mathrm{compact} } }\right) }\right)^\preceq$ Aiming for a contradiction suppose that :$x \in \left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq$ By definition of lower closure of element: :$x \preceq \sup \left({x^{\mathrm{compact} } }\right)$ Thus by definition of antisymmetry: :this contrasicts $x \ne \sup \left({x^{\mathrm{compact} } }\right)$ {{qed|lemma}} By definition of relative complement: :$x \in \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$ By Complement of Lower Closure of Element is Open in Scott Topological Ordered Set: :$\complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$ is open. By definition of basis: :$\complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right) = \bigcup\left\{ {G \in \mathcal B: G \subseteq \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)}\right\}$ By definition of union :$\exists X \in \left\{ {G \in \mathcal B: G \subseteq \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)}\right\}: x \in X$ Then :$X \in \mathcal B \land X \subseteq \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$ By definition of $\mathcal B$: :$\exists k \in K\left({L}\right): X = k^\succeq$ By definition of compact subset: :$k$ is compact. By definition of upper closure of element: :$k \preceq x$ By definition of compact closure: :$k \in x^{\mathrm{compact} }$ By definitions of supremum and upper bound: :$k \preceq \sup \left({x^{\mathrm{compact} } }\right)$ By definition of upper closure of element: :$\sup \left({x^{\mathrm{compact} } }\right) \in X$ By gefinition of subset: :$\sup \left({x^{\mathrm{compact} } }\right) \in \complement_S\left({\left({\sup \left({x^{\mathrm{compact} } }\right) }\right)^\preceq}\right)$ By definition of difference: :$\sup \left({x^{\mathrm{compact} } }\right) \notin \left({\sup \left({x^{\mathrm{compact} } }\right)}\right)^\preceq$ Thus by definition of lower closure of element: :this contradicts $\sup \left({x^{\mathrm{compact} } }\right) \in \left({\sup \left({x^{\mathrm{compact} } }\right)}\right)^\preceq$ {{qed}}
{{NotZFC}} === Necessary Condition === Suppose that $A$ is a proper class. Then by the Axiom of Limitation of Size there exists a bijection $f: A \to \mathrm U$, where $\mathrm U$ is the universal class. There also exists a bijection $g: \mathrm P \to \mathrm U$. By Inverse of Bijection is Bijection, there exists a bijection $g^{-1}: \mathrm U \to \mathrm P$ From Composite of Bijections is Bijection, it follows that $f \circ g^{-1}: A \to \mathrm P$ is a bijection. {{qed|lemma}} === Sufficient Condition === Suppose that there exists a bijection $f: A \to \mathrm P$. Because $\mathrm P$ is a proper class, it follows from the Axiom of Limitation of Size that there exists a bijection $g: \mathrm P \to \mathrm U$. Then by Composite of Bijections is Bijection, $f \circ g: A \to \mathrm U$ is a bijection from $A$ to the universe. And so by the Axiom of Limitation of Size, $A$ is proper. Hence the result. {{qed}} Category:Gödel-Bernays Class Theory Category:Class Mappings trp116vjw0pctf1ati55c925yy9bax1
We have that: :$\displaystyle \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} + 2 = 4 \times 10^n + \sum_{k \mathop = 0}^{n - 1} 9 \times 10^k$ using the Basis Representation Theorem. It remains to be demonstrated that: :$\displaystyle 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} = \sum_{k \mathop = 1}^n 9 \times 10^k + 4$ again using the Basis Representation Theorem. Thus: {{begin-eqn}} {{eqn | l = 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7}       | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 2 \times 9 \times 10^k + 2 \times 7 }} {{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 18 \times 10^k + 14 }} {{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 1}^{n - 1} 10 \times 10^k + 10 }} {{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 0}^{n - 1} 10 \times 10^k }} {{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 1}^n 10^k }} {{eqn | r = 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + 10^n + \sum_{k \mathop = 1}^{n - 1} 10^k }} {{eqn | r = 9 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 4 }} {{eqn | r = \sum_{k \mathop = 1}^n 9 \times 10^k + 4 }} {{end-eqn}} {{qed}}
Because $\displaystyle \sum_{n \mathop = 1}^\infty \left({f_n - 1}\right)$ converges locally uniformly absolutely on $D$, the series converges locally uniformly absolutely. Thus $f = \displaystyle \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely. By Uniform Limit of Analytic Functions is Analytic, $f$ is analytic The last claim follows from Zeroes of Infinite Product of Analytic Functions {{qed}}
Let $\map P n$ be the property: :there exists a prime number $p$ such that $n \le p \le \dfrac {9 n} 8$. First note that $\map P n$ does not hold for the following $n < 48$: :$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14, 15, 19, 21, 23, 24, 25, 31, 32, 47$ Taking $47$ as an example: {{begin-eqn}} {{eqn | l = 48       | r = 2^4 \times 3       | c = is not prime }} {{eqn | l = 49       | r = 7^2       | c = is not prime }} {{eqn | l = 50       | r = 2 \times 5^2       | c = is not prime }} {{eqn | l = 51       | r = 3 \times 17       | c = is not prime }} {{eqn | l = 52       | r = 2^2 \times 13       | c = is not prime }} {{end-eqn}} We have that: : $\dfrac {9 \times 47} 8 = 52 \cdotp 875$ and so it is seen that $\map P {47}$ does not hold. {{ProofWanted|It remains to be shown that $\map P n$ holds for all $n \ge 48$ as well.}}
Let $\epsilon>0$. Because $f$ is uniformly continuous, there exist $\delta>0$ such that $d(f(x),f(y))<\epsilon$ for $d(x,y)<\delta$. Because $(g_n)$ converges uniformly , there exist $N>0$ such that $d(g_n(x),g(x))<\delta$ for $n>N$ and all $x\in X$. Thus $d(f(g_n(x)),f(g(x)))<\epsilon$ for $n>N$ and all $x\in X$. Thus $(f\circ g_n)$ converges uniformly to $f\circ g$. {{qed}} Category:Uniform Convergence Category:Uniform Continuity 7gh6m2n2qwtuakwms8rmeir9cx5x6aa
Let $\epsilon > 0$. For $x, y \in \C$ with $\map \Re x, \map \Re y \le a$: {{begin-eqn}} {{eqn | l = \cmod {e^x - e^y}       | r = \cmod {e^y} \cdot \cmod {e^{x - y} - 1} }} {{eqn | r = e^{\map \Re y} \cdot \cmod {e^{x - y} - 1}       | c = Absolute Value of Complex Exponential }} {{eqn | o = \le       | r = e^a \cdot \cmod {e^{x - y} - 1}       | c = Exponential is Strictly Increasing }} {{end-eqn}} Because Exponential Function is Continuous, there exists $\delta > 0$ such that $\cmod {e^z - 1} < \epsilon$ for $\cmod z < \delta$. Thus if $\cmod {x - y} < \delta$, $\cmod {e^x - e^y} < e^a \epsilon$. Thus $\exp$ is uniformly continuous. {{qed}} Category:Exponential Function Category:Uniform Continuity h0ifso4s6wse84h58wc3segqt32krs2
Because $T$ is weakly locally compact, each point of $S$ has a compact neighborhood in $T$. Because $f_n \to f$ compactly, each point of $S$ has a (compact) neighborhood on which $f_n \to f$ uniformly. Thus $f_n$ converges locally uniformly to $f$. {{qed}}