avewright/proofwiki-math huggingface dataset:
Theorem: 37 is Second Number whose Period of Reciprocal is 3

$37$ is the $2$nd positive integer (after $27$) the decimal expansion of whose reciprocal has a period of $3$:
:$\dfrac 1 {37} = 0 \cdotp \dot 02 \dot 7$

Proof:
From Reciprocal of $37$:
{{:Reciprocal of 37}}
Counting the digits, it is seen that this has a period of recurrence of $3$.
It can be determined by inspection of all smaller integers that this is indeed the $2$nd to have a period of $3$.
{{qed}}
Theorem: Centered Hexagonal Number as Sum of Triangular Numbers

Let $C_n$ be the $n$th centered hexagonal number.
Then:
:$C_n = 6 T_{n - 1} + 1$
where $T_{n - 1}$ denotes the $n - 1$th triangular number.

Proof:
{{begin-eqn}}
{{eqn | l = C_n
      | r = 3 n \paren {n - 1} + 1
      | c = Closed Form for Centered Hexagonal Numbers
}}
{{eqn | r = 6 \paren {\dfrac {\paren {n - 1} n} 2} + 1
      | c = 
}}
{{eqn | r = 6 T_{n - 1} + 1
      | c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
{{qed}}
Theorem: Closed Form for Centered Hexagonal Numbers

Let $C_n$ be the $n$th centered hexagonal number.
Then:
:$C_n = 3 n \paren {n - 1} + 1$

Proof:
By the definition of centered hexagonal number:
{{begin-eqn}}
{{eqn | l = C_n
      | r = 1 + \sum_{k \mathop = 1}^{n - 1} 6 k
      | c = {{Defof|Centered Hexagonal Number}}
}}
{{eqn | r = 1 + 6 \sum_{k \mathop = 1}^{n - 1} k
      | c = 
}}
{{eqn | r = 6 \paren {\dfrac {n \paren {n - 1} } 2} + 1
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 3 n \paren {n - 1} + 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Positive Even Integers not Expressible as Sum of 2 Composite Odd Numbers

The positive even integers which cannot be expressed as the sum of $2$ composite odd numbers are:
:$2, 4, 6, 8, 10, 12, 14, 16, 20, 22, 26, 28, 32, 38$
{{OEIS|A118081}}

Proof:
The smallest composite odd numbers are $9$ and $15$, so trivially $2$ to $16$ and $20$ to $22$ cannot be expressed as the sum of $2$ composite odd numbers.
We have:
{{begin-eqn}}
{{eqn | l = 18
      | r = 9 + 9
}}
{{eqn | l = 24
      | r = 9 + 15
}}
{{eqn | l = 30
      | r = 21 + 9
}}
{{eqn | r = 15 + 15
}}
{{eqn | l = 34
      | r = 9 + 25
}}
{{eqn | l = 36
      | r = 9 + 27
}}
{{end-eqn}}
It remains to investigate $26, 28$ and $32$.
This will be done by progressively subtracting smaller composite odd numbers from them, and noting that the difference is not composite.
{{begin-eqn}}
{{eqn | l = 26 - 9
      | r = 17
      | c = which is prime
}}
{{eqn | l = 26 - 15
      | r = 11
      | c = which is prime
}}
{{eqn | l = 26 - 21
      | r = 5
      | c = which is prime
}}
{{eqn | l = 26 - 25
      | r = 1
      | c = which is not composite
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 28 - 9
      | r = 19
      | c = which is prime
}}
{{eqn | l = 28 - 15
      | r = 13
      | c = which is prime
}}
{{eqn | l = 28 - 21
      | r = 7
      | c = which is prime
}}
{{eqn | l = 28 - 25
      | r = 3
      | c = which is prime
}}
{{eqn | l = 28 - 27
      | r = 1
      | c = which is not composite
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 32 - 9
      | r = 32
      | c = which is prime
}}
{{eqn | l = 32 - 15
      | r = 17
      | c = which is prime
}}
{{eqn | l = 32 - 21
      | r = 11
      | c = which is prime
}}
{{eqn | l = 32 - 25
      | r = 7
      | c = which is prime
}}
{{eqn | l = 32 - 27
      | r = 5
      | c = which is prime
}}
{{end-eqn}}
It remains to be demonstrated that all even integers greater than $38$ can be expressed as the sum of $2$ composite odd numbers.
We note that $9 + 6 k$ is odd and a multiple of $3$.
Numbers $18$ and greater of the form $6 n$ can be expressed as:
:$\left({9 + 6 k}\right) + 9$
Numbers $34$ and greater of the form $6 n + 4$ can be expressed as:
:$\left({9 + 6 k}\right) + 25$
Numbers $44$ and greater of the form $6 n + 2$ can be expressed as:
:$\left({9 + 6 k}\right) + 35$
We have that $40$ and $42$ are of the form $6 n + 4$ and $6 n$ respectively.
Hence all even integers greater than $38$ are accounted for.
{{qed}}
Theorem: Upper Closure is Compact in Topological Lattice

Let $L = \struct {S, \preceq, \tau}$ be a topological lattice.
Suppose that:
:for every subset $X$ of $S$ if $X$ is open, then $X$ is upper.
Let $x \in S$.
Then $x^\succeq$ is compact
where $x^\succeq$ denotes the upper closure of $x$.

Proof:
Let $\FF$ be a set of subsets of $S$ such that:
:$\FF$ is open cover of $x^\succeq$
By definition of cover:
:$x^\succeq \subseteq \bigcup \FF$
By definitions of upper closure of element and reflexivity:
:$x \in x^\succeq$
By definition of subset:
:$x \in \bigcup \FF$
By definition of union:
:$\exists Y \in \FF: x \in Y$
Define $\GG = \set Y$.
By definition of open cover:
:$Y$ is open.
We will prove that:
:$x^\succeq \subseteq \bigcup \GG$
Let $y \in x^\succeq$.
By definition of upper closure of element:
:$x \preceq y$
By Union of Singleton:
:$\bigcup \GG = Y$
By assumption:
:$Y$ is upper.
Thus by definition of upper section:
:$x \in \bigcup \GG$
{{qed|lemma}}
Then by definition:
:$\GG$ is cover of $x^\succeq$
By definitions of singleton and subset:
:$\GG \subseteq \FF$
By definition:
:$\GG$ is subcover of $\FF$.
By Singleton is Finite:
:$\GG$ is finite.
Thus by definition:
:$\GG$ is finite subcover of $\FF$.
{{qed}}
Theorem: Strictly Positive Real Numbers are Closed under Multiplication

The set $\R_{>0}$ of strictly positive real numbers is closed under multiplication:
:$\forall a, b \in \R_{> 0}: a \times b \in \R_{> 0}$

Proof:
Let $b > 0$.
From {{Real-number-axiom|O2}}:
:$a > c \implies a \times b > c \times b$
Thus setting $c = 0$:
:$a > 0 \implies a \times b > 0 \times b$
But from Real Zero is Zero Element:
:$0 \times b = 0$
Hence the result:
:$a, b > 0 \implies a \times b > 0$
{{qed}}
Theorem: Positive Power Function on Non-negative Reals is Strictly Increasing

Let $a \in \Q_{> 0}$ be a strictly positive rational number.
Let $f_a: \R_{\ge 0} \to \R$ be the real function defined as:
:$\map {f_a} x = x^a$
Then $f_a$ is strictly increasing.

Proof:
By the power rule for derivatives:
:$\map {D_x} {x^a} = a x^{a - 1}$
By power of positive real number is positive, it is seen that:
:$x > 0 \implies x^{a - 1} > 0$
By Strictly Positive Real Numbers are Closed under Multiplication, it follows that $\map {D_x} {x^a} > 0$ for all $x \in \openint 0 {+\infty}$.
Hence by Derivative of Monotone Function, $f_a$ is strictly increasing
{{qed}}
Category:Examples of Strictly Increasing Real Functions
Category:Powers
mbm4cinna10as6bbtc2r9qdqdrbedwp
Theorem: Smallest Consecutive Even Numbers such that Added to Divisor Count are Equal

$30$ is the smallest positive even integer $n$ with the property:
{{begin-eqn}}
{{eqn | l = n + \map {\sigma_0} n
      | r = m
      | c = 
}}
{{eqn | l = \paren {n + 2} + \map {\sigma_0} {n + 2}
      | r = m
      | c = 
}}
{{eqn | l = \paren {n + 4} + \map {\sigma_0} {n + 4}
      | r = m
      | c = 
}}
{{end-eqn}}
where:
:$m \in \Z_{>0}$ is some positive integer
:$\map {\sigma_0} n$ is the divisor count function: the number of divisors of $n$.
In this case, where $n = 30$, we have that $m = 38$.

Proof:
From Divisor Count Function from Prime Decomposition, we have:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where the prime decomposition of $n$ is:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
{{begin-eqn}}
{{eqn | l = 2 + \map {\sigma_0} 2
      | r = 2 + 2
      | rr= = 4
      | c = as $2 = 2^1$
}}
{{eqn | l = 4 + \map {\sigma_0} 4
      | r = 4 + 7
      | rr= = 7
      | c = as $4 = 2^2$
}}
{{eqn | l = 6 + \map {\sigma_0} 6
      | r = 6 + 4
      | rr= = 10
      | c = as $6 = 2^1 3^1$
}}
{{eqn | l = 8 + \map {\sigma_0} 8
      | r = 8 + 4
      | rr= = 12
      | c = as $8 = 2^3$
}}
{{eqn | l = 10 + \map {\sigma_0} {10}
      | r = 10 + 4
      | rr= = 14
      | c = as $10 = 2^1 5^1$
}}
{{eqn | l = 12 + \map {\sigma_0} {12}
      | r = 12 + 6
      | rr= = 18
      | c = as $12 = 2^2 3^1$
}}
{{eqn | l = 14 + \map {\sigma_0} {14}
      | r = 14 + 4
      | rr= = 18
      | c = as $14 = 2^1 7^1$
}}
{{eqn | l = 16 + \map {\sigma_0} {16}
      | r = 16 + 5
      | rr= = 21
      | c = as $16 = 2^4$
}}
{{eqn | l = 18 + \map {\sigma_0} {18}
      | r = 18 + 6
      | rr= = 24
      | c = as $18 = 2^1 3^2$
}}
{{eqn | l = 20 + \map {\sigma_0} {20}
      | r = 20 + 6
      | rr= = 26
      | c = as $20 = 2^2 5^1$
}}
{{eqn | l = 22 + \map {\sigma_0} {22}
      | r = 22 + 4
      | rr= = 26
      | c = as $22 = 2^1 11^1$
}}
{{eqn | l = 24 + \map {\sigma_0} {24}
      | r = 24 + 8
      | rr= = 32
      | c = as $24 = 2^3 3^1$
}}
{{eqn | l = 26 + \map {\sigma_0} {26}
      | r = 26 + 4
      | rr= = 30
      | c = as $26 = 2^1 13^1$
}}
{{eqn | l = 28 + \map {\sigma_0} {28}
      | r = 28 + 6
      | rr= = 34
      | c = as $28 = 2^2 7^1$
}}
{{eqn | l = 30 + \map {\sigma_0} {30}
      | r = 30 + 8
      | rr= = 38
      | c = as $30 = 2^1 3^1 5^1$
}}
{{eqn | l = 32 + \map {\sigma_0} {32}
      | r = 32 + 6
      | rr= = 38
      | c = as $32 = 2^5$
}}
{{eqn | l = 34 + \map {\sigma_0} {34}
      | r = 34 + 4
      | rr= = 38
      | c = as $34 = 2^1 17^1$
}}
{{end-eqn}}
{{qed}}
Theorem: 2-Digit Positive Integer equals Product plus Sum of Digits iff ends in 9

Let $n$ be a $2$-digit positive integer.
Then:
:$n$ equals the sum added to the product of its digits
{{iff}}:
:the last digit of $n$ is $9$.

Proof:
Let $n = 10 x + y$ where $0 < x \le 9, 0 \le y \le 9$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x + y} + \paren {x y}
      | r = 10 x + y
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x y - 9 x
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x \paren {y - 9}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = y
      | r = 9
      | c = as $x \ne 0$
}}
{{end-eqn}}
{{qed}}
Theorem: Infimum of Open Set is Way Below Element in Complete Scott Topological Lattice

Let $T = \struct {S, \preceq, \tau}$ be a complete topological lattice with Scott topology.
Let $X$ be an open subset of $S$,
Let $x \in X$.
Then $\inf X \ll x$
where $\ll$ denotes the way below relation.

Proof:
By Open iff Upper and with Property (S) in Scott Topological Lattice:
:$X$ is upper and has property (S).
Let $D$ be a directed subset of $S$ such that
:$x \preceq \sup D$
By definition of upper section:
:$\sup D \in X$
By definition of property (S):
:$\exists y \in D: \forall d \in D: y \preceq d \implies d \in X$
By definitions of infimum and complete lattice:
:$\inf X$ is a lower bound for $X$.
By definition pf reflexivity:
:$y \in X$
Thus by definition of lower bound:
:$\exists d \in D: \inf X \preceq d$
Thus by definition pf way below relation:
:$\inf X \ll x$
{{qed}}
Theorem: Cyclic Permutations of 5-Digit Multiples of 41

Let $n$ be a multiple of $41$ with $5$ digits.
Let $m$ be an integer formed by cyclically permuting the digits of $n$.
Then $m$ is a multiple of $41$.

Proof:
First we note that $10^5 - 1 = 41 \times 271 \times 9$
$10$ generates exactly $5$ elements in $\Z_{41}$  Subgroup Generated by One Element is Cyclic
{{explain|what follows from what? Grammar confusing}}
{{begin-eqn}}
{{eqn | q = \forall k \in \N
      | l = 10^{0 + 5 k}
      | o = \equiv
      | r = 1
      | rr= \pmod {41}
}}
{{eqn | l = 10^{1 + 5 k}
      | o = \equiv
      | r = 10
      | rr= \pmod {41}
}}
{{eqn | l = 10^{2 + 5 k}
      | o = \equiv
      | r = 18
      | rr= \pmod {41}
}}
{{eqn | l = 10^{3 + 5 k}
      | o = \equiv
      | r = 16
      | rr= \pmod {41}
}}
{{eqn | l = 10^{4 + 5 k}
      | o = \equiv
      | r = 37
      | rr= \pmod {41}
}}
{{end-eqn}}
In $\Z_{41}$: $ \set {10^1, 10^2, 10^3, 10^4, 10^5} \leadsto \set {10, 18, 16, 37, 1}$
Let $n$ be a multiple of $41$ with $5$ digits.
Then we have:
{{begin-eqn}}
{{eqn | l = 41 \times c
      | r = a_4 \times 10^4 + a_3 \times 10^3 + a_2 \times 10^2 + a_1 \times 10^1 + a_0
}}
{{end-eqn}}
Let $m$ be an integer formed by cyclically permuting the digits of $n$.
Then we have:
{{begin-eqn}}
{{eqn | l = 10 \times 41 \times c 
      | r = 10 \times \paren {a_4 \times 10^4 + a_3 \times 10^3 + a_2 \times 10^2 + a_1 \times 10^1 + a_0}
      | c = multiply original number by $10$
}}
{{eqn | r =  a_4 \times 10^5 + a_3 \times 10^4 + a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1
      | c = 
}}
{{eqn | r =  a_3 \times 10^4 + a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1 + a_4 \times 10^0
      | c = $10^5$ and $10^0 \equiv 1 \pmod {41}$
}}
{{end-eqn}}
Therefore $m$ is a multiple of $41$.
{{qed}}
Theorem: Smallest Prime Number not Difference between Power of 2 and Power of 3

$41$ is the smallest prime number which is not the difference between a power of $2$ and a power of $3$.

Proof:
First we have:
{{begin-eqn}}
{{eqn | l = 2
      | r = 3^1 - 2^0
      | c = 
}}
{{eqn | l = 3
      | r = 2^2 - 3^0
      | c = 
}}
{{eqn | l = 5
      | r = 3^2 - 2^2
      | c = 
}}
{{eqn | l = 7
      | r = 3^2 - 2^1
      | c = 
}}
{{eqn | l = 11
      | r = 3^3 - 2^4
      | c = 
}}
{{eqn | l = 13
      | r = 2^4 - 3^1
      | c = 
}}
{{eqn | l = 17
      | r = 3^4 - 2^6
      | c = 
}}
{{eqn | l = 19
      | r = 3^3 - 2^3
      | c = 
}}
{{eqn | l = 23
      | r = 3^3 - 2^2
      | c = 
}}
{{eqn | l = 29
      | r = 2^5 - 3^1
      | c = 
}}
{{eqn | l = 31
      | r = 2^5 - 3^0
      | c = 
}}
{{eqn | l = 37
      | r = 2^6 - 3^3
      | c = 
}}
{{end-eqn}}
{{AimForCont}} $41 = 2^n - 3^m$.
We have that $n > 3$.
Thus:
:$2^n \equiv 0 \pmod 8$
and as:
:$41 \equiv 1 \pmod 8$
we have:
:$1 \equiv -3^m \pmod 8$
which is not possible:
For any integer $k$, 
{{begin-eqn}}
{{eqn | l = -3^{2 k} 
      | r = -9^k
}}
{{eqn | o = \equiv 
      | r = -1^k
      | rr = \pmod 8
      | c = Congruence of Powers
}}
{{eqn | o = \equiv 
      | r = -1
      | rr = \pmod 8
}}
{{eqn | l = -3^{2 k + 1} 
      | r = 3 \paren {-3^{2 k} }
}}
{{eqn | o = \equiv 
      | r = 3 \paren {-1}
      | rr = \pmod 8
      | c = Modulo Multiplication is Well-Defined
}}
{{eqn | o = \equiv 
      | r = -3
      | rr = \pmod 8
}}
{{end-eqn}}
So for any integer $m$, $1 \not \equiv -3^m \pmod 8$.
Now suppose $41 = 3^m - 2^n$.
We have that $m > 1$ and $n > 2$.
Taking $\mod 3$ on both sides:
{{begin-eqn}}
{{eqn | l = 41 
      | r = 3^m - 2^n
}}
{{eqn | ll = \leadsto
      | l = -1
      | o = \equiv 
      | r = - 2^n
      | rr = \pmod 3
      | c = 
}}
{{eqn | ll = \leadsto
      | l = 1
      | o = \equiv 
      | r = 2^n
      | rr = \pmod 3
      | c = 
}}
{{eqn | o = \equiv 
      | r = \paren {-1}^n
      | rr = \pmod 3
      | c = Congruence of Powers
}}
{{end-eqn}}
which shows that $n$ is even.
Taking $\mod 4$ on both sides:
{{begin-eqn}}
{{eqn | l = 41 
      | r = 3^m - 2^n
}}
{{eqn | ll = \leadsto
      | l = 1
      | o = \equiv 
      | r = 3^m
      | rr = \pmod 4
      | c = 
}}
{{eqn | o = \equiv 
      | r = \paren {-1}^m
      | rr = \pmod 4
      | c = Congruence of Powers
}}
{{end-eqn}}
which shows that $m$ is even as well.
Now we take $\mod 5$ on both sides:
{{begin-eqn}}
{{eqn | l = 41 
      | r = 3^m - 2^n
}}
{{eqn | ll = \leadsto
      | l = 1
      | o = \equiv 
      | r = 3^m - 2^n
      | rr = \pmod 5
      | c = 
}}
{{eqn | o = \equiv 
      | r = 9^{m / 2} - 4^{n / 2}
      | rr = \pmod 5
      | c = 
}}
{{eqn | o = \equiv 
      | r = \paren {-1}^{m / 2} - \paren {-1}^{n / 2}
      | rr = \pmod 5
      | c = Congruence of Powers
}}
{{end-eqn}}
But $\paren {-1}^{m / 2} - \paren {-1}^{n / 2}$ can only be $0$ or $\pm 2$, not $1$, which is a contradiction.
Therefore $41$ is not the difference between a power of $2$ and a power of $3$.
{{qed}}
Theorem: Prime-Generating Quadratic of form 2 x squared minus 1000 x minus 2609

The quadratic function:
:$2 x^2 - 1000 x - 2609$
has $602$ prime values among its first $1000$ values.
Some of those prime numbers are negative.

Proof:
{{ProofWanted|Lots of tedious calculations needed to prove this.}}
Theorem: Scott Topology equals to Scott Sigma

Let $\struct {T, \preceq, \tau}$ be a up-complete topological lattice with Scott topology.
Then $\tau = \map \sigma {T, \preceq}$
where $\map \sigma L$ denotes the Scott sigma of $L$.

Proof:
This follows by Open iff Upper and with Property (S) in Scott Topological Lattice and definition Scott sigma.
{{qed}}
Theorem: Complement of Lower Closure is Prime Element in Inclusion Ordered Set of Scott Sigma

Let $L = \struct {S, \preceq, \tau}$ be a complete Scott topological lattice.
Let $D = \struct {\map \sigma L, \precsim}$ be an inclusion ordered set of the Scott sigma of $L$.
Let $x \in S$.
Then:
:$\relcomp S {x^\preceq}$ is a prime element in $D$
and:
:$\relcomp S {x^\preceq} \ne S$

Proof:
By Scott Topology equals to Scott Sigma:
:$\tau = \map \sigma L$
By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
:$x^\preceq = \set x^-$
where $\set x^-$ denotes the topological closure of $\set x$.
By Topological Closure of Singleton is Irreducible:
:$x^\preceq$ is topologically irreducible
Thus by Complement of Irreducible Topological Subset is Prime Element:
:$\relcomp S {x^\preceq}$ is a prime element in $D$.
By definitions of reflexivity and lower closure of element:
:$x \in x^\preceq$
By definitions of relative complement and difference:
:$x \notin \relcomp S {x^\preceq}$
Hence by definition set equality:
:$\relcomp S {x^\preceq} \ne S$
{{qed}}
Theorem: Topological Closure of Singleton is Irreducible

Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be a point of $T$.
Then:
:$\set x^-$ is irreducible
where $\set x^-$ denotes the topological closure of $\set x$.

Proof:
Follows from:
:Trivial Topological Space is Irreducible
:Closure of Irreducible Subspace is Irreducible
{{qed}}
Theorem: Topological Closure of Singleton is Irreducible

Let $T = \struct {S, \tau}$ be a topological space.
Let $x$ be a point of $T$.
Then:
:$\set x^-$ is irreducible
where $\set x^-$ denotes the topological closure of $\set x$.

Proof:
{{AimForCont}} that
:$\set x^-$ is not irreducible.
By Set is Subset of its Topological Closure:
:$\set x \subseteq \set x^-$
By definitions of singleton and Subset:
:$x \in \set x^-$
By definition of irreducible:
:$\exists X_1, X_2 \subseteq S: X_1, X_2$ are closed
and:
{{begin-eqn}}
{{eqn | l = \set x^-
      | r = X_1 \cup X_2
      | c = 
}}
{{eqn | lo= \text {and}
      | l = \set x^-
      | o = \ne
      | r = X_1
      | c = 
}}
{{eqn | lo= \text {and}
      | l = \set x^-
      | o = \ne
      | r = X_2
      | c = 
}}
{{end-eqn}}
By definition of union:
:$x \in X_1$ or $x \in X_2$
By definitions of singleton and subset:
:$\set x \subseteq X_1$ or $\set x \subseteq X_2$
By definition of closure :
:$\set x^- \subseteq X_1$ or $\set x^- \subseteq X_2$
By Set is Subset of Union:
:$X_1 \subseteq \set x$ and $X_2 \subseteq \set x$
Thus by definition of set equality:
:this contradicts $\set x^- \ne X_1$ and $\set x^- \ne X_2$
{{qed}}
Theorem: Element equals to Supremum of Infima of Open Sets that Element Belongs implies Topological Lattice is Continuous

Let $L = \left({S, \preceq, \tau}\right)$ be a complete Scott topological lattice.
Let
:$\forall x \in S: x = \sup \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$
Then $L$ is continuous.

Proof:
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}}
Theorem: Proper Class is not Element of Class

Let $\mathrm P$ be a proper class.
Then $\mathrm P$ is not an element of any class.
That is:
:$\neg \exists A : \mathrm P \in A$

Proof:
From the definition of a proper class, $\mathrm P$ is not a set.
The result follows by definition of a class.
{{qed}}
Category:Class Theory
6wra5khkj4nktbsc7sdsz5nm70wmsjt
Theorem: GCD of Polynomials does not depend on Base Field

Let $E / F$ be a field extension.
Let $P, Q \in F \sqbrk X$ be polynomials.
Let:
:$\gcd \set {P, Q} = R$ in $F \sqbrk X$
:$\gcd \set {P, Q} = S$ in $E \sqbrk X$.
Then $R = S$.
In particular, $S \in F \sqbrk X$.

Proof:
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
Theorem: Prime-Generating Quadratic of form x squared - 79 x + 1601

The quadratic function:
:$x^2 - 79 x + 1601$
gives prime values for integer $x$ such that $0 \le x \le 79$.
The primes generated are repeated once each.

Proof:
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}}
Theorem: Continuous Group Action is by Homeomorphisms

Let $G$ be a topological group acting continuously on a topological space $X$.
Then $G$ acts by homeomorphisms.

Proof:
Let $\phi: G \times X \to X$ denote the group action.
Let $g \in G$.
We are given that $\phi$ is a bijection.
The map $\phi_g: X \to X : x \mapsto \map \phi {g, x}$ is continuous because $\phi$ is.
{{explain|Why?}}
By Inverse of Bijection is Bijection, the inverse of $\phi$ is given by $\phi_{g^{-1} }$ is also a bijection.
{{mistake|grammar in the above, inconsistency in linguistic presentation in the following}}
So the inverse of $\phi$ is continuous.
So inverse of $\phi_g$ is continuous.
Thus $\phi_g$ is an homeomorphism of $X$.
{{qed}}
Theorem: Discrete Group Acts Continuously iff Acts by Homeomorphisms

Let $G$ be a discrete group acting on a topological space $X$.
Then the following are equivalent:
:$G$ acts continuously
:$G$ acts by homeomorphisms

Proof:
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.
Then:
{{begin-eqn}}
{{eqn | l = \phi^{-1} \sqbrk U
      | r = \set { \tuple {g, x} \in G \times X : \map \phi {g, x} \in U }
      | c = {{Defof|Preimage of Mapping}} with respect to $\phi$
}}
{{eqn | r = \set { \tuple {g, x} \in G \times X : \map {\phi _g} x \in U }
}}
{{eqn | r = \set { \tuple {g, x} \in G \times X : x \in \phi _g^{-1} \sqbrk U }
      | c = {{Defof|Preimage of Mapping}} with respect to $\phi _g$
}}
{{eqn | r = \bigcup _{g \in G} \paren { \set g \times {\phi _g^{-1} } \sqbrk U }
}}
{{end-eqn}}
Since each $\set g \times {\phi _g^{-1} \sqbrk U }$ is open by {{Defof|Product Topology}}, $\phi^{-1} \sqbrk U$ is open.
Thus $\phi$ is continuous.
{{qed}}
Theorem: Euler Phi Function of 3

:$\map \phi 3 = 2$

Proof:
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}}