avewright/proofwiki-math huggingface dataset:
Theorem: Are All Perfect Numbers Even?/Progress/Minimum Size

It had been established by $1986$ that an odd perfect number, if one were to exist, would have over $200$ digits.
By $1997$ that lower bound had been raised to $300$ digits.
By $2012$ that lower bound had been raised again to $1500$ digits.

Proof:
{{ProofWanted|Details}}
Theorem: Divisibility of Sum of 3 Fourth Powers

Let $n \in \Z_{\ge 0}$ be the sum of three $4$th powers.
Then:
:$n$ is divisible by $5$ {{iff}} all three addends are also divisible by $5$
:$n$ is divisible by $29$ {{iff}} all three addends are also divisible by $29$.

Proof:
Let $n = a^4 + b^4 + c^4$ for $a, b, c \in \Z$.
Theorem: Squares of form 2 n^2 - 1

The sequence of integers $\left\langle{n}\right\rangle$ such that $2 n^2 - 1$ is square begins:
:$1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, \ldots$
{{OEIS|A001653}}

Proof:
{{ProofWanted|Follows somehow from the fact that these numbers are the hypotenuses of Definition:Almost Isosceles Pythagorean Triangles.}}
Theorem: Prime-Generating Quadratics of form 2 a squared plus p

The quadratic form:
:$2 a^2 + p$
yields prime numbers for $a = 0, 1, \ldots, p - 1$ for values of $p$:
:$3, 5, 11, 29$

Proof:
=== 3 ===
{{:Prime-Generating Quadratics of form 2 a squared plus p/3}}
Theorem: Prime-Generating Quadratics of form 2 a squared plus p/3

The quadratic form:
:$2 a^2 + 3$
yields prime numbers for $a = 0, 1, 2$ but not for $a = 3$.

Proof:
{{begin-eqn}}
{{eqn | l = 2 \times 0^2 + 3
      | r = 0 + 3
}}
{{eqn | r = 3
      | c = which is prime
}}
{{eqn | l = 2 \times 1^2 + 3
      | r = 2 + 3
      | c = 
}}
{{eqn | r = 5
      | c = which is prime
}}
{{eqn | l = 2 \times 2^2 + 3
      | r = 2 \times 4 + 3
      | c = 
}}
{{eqn | r = 8 + 3
      | c = 
}}
{{eqn | r = 11
      | c = which is prime
}}
{{eqn | l = 2 \times 3^2 + 3
      | r = 2 \times 9 + 3
      | c = 
}}
{{eqn | r = 18 + 3
      | c = 
}}
{{eqn | r = 21
      | c = which is not prime
}}
{{end-eqn}}
Theorem: Prime-Generating Quadratics of form 2 a squared plus p/5

The quadratic form:
:$2 a^2 + 5$
yields prime numbers for $a = 0, 1, \ldots, 4$.

Proof:
{{begin-eqn}}
{{eqn | l = 2 \times 0^2 + 5
      | r = 0 + 5
}}
{{eqn | r = 5
      | c = which is prime
}}
{{eqn | l = 2 \times 1^2 + 5
      | r = 2 + 5
      | c = 
}}
{{eqn | r = 7
      | c = which is prime
}}
{{eqn | l = 2 \times 2^2 + 5
      | r = 2 \times 4 + 5
      | c = 
}}
{{eqn | r = 8 + 5
      | c = 
}}
{{eqn | r = 13
      | c = which is prime
}}
{{eqn | l = 2 \times 3^2 + 5
      | r = 2 \times 9 + 5
      | c = 
}}
{{eqn | r = 18 + 5
      | c = 
}}
{{eqn | r = 23
      | c = which is prime
}}
{{eqn | l = 2 \times 4^2 + 5
      | r = 2 \times 16 + 5
      | c = 
}}
{{eqn | r = 32 + 5
      | c = 
}}
{{eqn | r = 37
      | c = which is prime
}}
{{end-eqn}}
Theorem: Prime-Generating Quadratics of form 2 a squared plus p/11

The quadratic form:
:$2 a^2 + 11$
yields prime numbers for $a = 0, 1, \ldots, 10$.

Proof:
{{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}}
Theorem: Directed Suprema Preserving Mapping is Increasing

Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $f: S \to S$ be a mapping that preserves directed suprema.
Then $f$ is an increasing mapping.

Proof:
Let $x, y \in D$ such that
:$x \preceq y$
Then by definition of reflexivity:
:$\forall a, b \in \set {x, y}: \exists z \in \set {x, y}: a \preceq z \land b \preceq z$
By definition:
:$\set {x, y}$ is directed.
By definition of mapping preserves directed suprema:
:$f$ preserves the supremum of $\set {x, y}$.
By definition of join semilattice:
:$\set {x, y}$ admits a supremum.
{{begin-eqn}}
{{eqn | l = \map \sup {f \sqbrk {\set {x, y} } }
      | r = \map f {\sup \set {x, y} }
      | c = {{Defof|Mapping Preserves Supremum|subdef = Subset}}
}}
{{eqn | r = \map f {x \vee y}
      | c = {{Defof|Join (Order Theory)}}
}}
{{eqn | r = \map f y
      | c = Preceding iff Join equals Larger Operand
}}
{{end-eqn}}
By Image of Doubleton under Mapping:
:$f \sqbrk {\set {x, y} } = \set {\map f x, \map f y}$
Thus by definitions of supremum and upper bound:
:$\map f x \preceq \map f y$
{{qed}}
Theorem: Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 1

Let the function $\map h x$ satisfy the equation:
:$-\map {\dfrac \d {\d x} } {P h'} + Q h = 0$
Let $\map h x$ have the boundary conditions:
:$\map h a = \map h b = 0$
Then:
:$\ds \int_a^b \paren {P h'^2 + Q h^2} \rd x = 0$

Proof:
{{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}}
Theorem: 4 Consecutive Integers cannot be Square-Free

Let $n, n + 1, n + 2, n + 3$ be four consecutive positive integers.
At least one of these is not square-free.

Proof:
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}}
Theorem: Sequence of 4 Consecutive Square-Free Triplets

The following sets of $4$ consecutive triplets of integers, with one integer between each triplet, are square-free:
:$29, 30, 31; 33, 34, 35; 37, 38, 39; 41, 42, 43$
:$101, 102, 103; 105, 106, 107; 109, 110, 111; 113, 114, 115$

Proof:
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}}
Theorem: Image of Doubleton under Mapping

Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall x, y \in S: f \sqbrk {\set {x, y} } = \set {\map f x, \map f y}$

Proof:
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}}
Theorem: Hensel's Lemma/First Form

Let $p$ be a prime number.
Let $k > 0$ be a positive integer.
Let $\map f X \in \Z \sqbrk X$ be a polynomial.
Let $x_k \in \Z$ such that:
{{begin-eqn}}
{{eqn | l = \map f {x_k}
      | o = \equiv
      | r = 0
      | rr= \pmod {p^k}
}}
{{eqn | l = \map {f'} {x_k}
      | o = \not \equiv
      | r = 0
      | rr= \pmod p
}}
{{end-eqn}}
{{explain|$f'$}}
Then for every integer $l \ge 0$, there exists an integer $x_{k + l}$ such that:
{{begin-eqn}}
{{eqn | l = \map f {x_{k + l} }
      | o = \equiv
      | r = 0
      | rr= \pmod {p^{k + l} }
}}
{{eqn | l = x_{k + l}
      | o = \equiv
      | r = x_k
      | rr= \pmod {p^k}
}}
{{end-eqn}}
and any two integers satisfying these congruences are congruent modulo $p^{k + l}$.
Moreover, for all $l \ge 0$ and any solutions $x_{k + l}$ and $x_{k + l + 1}$:
{{begin-eqn}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l} - \frac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } }
      | rr= \pmod {p^{k + l + 1} }
}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l}
      | rr= \pmod {p^{k + l} }
}}
{{end-eqn}}

Proof:
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:
{{begin-eqn}}
{{eqn | l = \map f {x_{k + l} }
      | o = \equiv
      | r = 0
      | rr= \pmod {p^{k + l} }
}}
{{eqn | l = x_{k + l}
      | o = \equiv
      | r = x_k
      | rr= \pmod {p^k}
}}
{{end-eqn}}
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 | l = \map f {x_{k + l} + t p^{k + l} }
      | r = \map f {x_{k + l} } + t p^{k + l} \map {f'} {x_{k + l} } + \paren {t p^{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 p^{k + l} \map {f'} {x_{k + l} }
      | rr= \pmod {p^{k + l + 1} }
}}
{{end-eqn}}
Because $\map {f'} {x_{k + l} } \equiv \map {f'} {x_k} \not \equiv 0 \pmod p$, $\map {f'} {x_{k + l} }$ 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}}:
:$t \equiv - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod p$
Thus, necessarily:
:$x_{k + l + 1} \equiv x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } } \pmod {p^{k + l + 1} }$
which proves the existence and uniqueness.
By induction, we have shown uniqueness and existence for all $l \ge 0$, as well as the relations:
{{begin-eqn}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l} - \dfrac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } }
      | rr= \pmod {p^{k + l + 1} }
}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l}
      | rr= \pmod {p^{k + l} }
}}
{{end-eqn}}
{{qed}}
Theorem: Hensel's Lemma for Composite Numbers

Let $b \in \Z \setminus \set {-1, 0, 1}$ be an integer.
Let $k > 0$ be a positive integer.
Let $\map f X \in \Z \sqbrk X$ be a polynomial.
Let $x_k \in \Z$ such that:
:$\map f {x_k} \equiv 0 \pmod {b^k}$
:$\gcd \set {\map {f'} {x_k}, b} = 1$
Then for every integer $l \ge 0$, there exists an integer $x_{k + l}$ such that:
:$\map  f {x_{k + l} } \equiv 0 \pmod {b^{k + l} }$
:$x_{k + l} \equiv x_k \pmod {b^k}$
and any two integers satisfying these congruences are congruent modulo $b^{k + l}$.
Moreover, for all $l \ge 0$ and any solutions $x_{k + l}$ and $x_{k + l + 1}$:
{{begin-eqn}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l} - \frac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } }
      | rr= \pmod {b^{k + l + 1} }
}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l}
      | rr= \pmod {b^{k + l} }
}}
{{end-eqn}}

Proof:
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} }
      | rr= \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:
{{begin-eqn}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l} - \frac {\map f {x_{k + l} } } {\map {f'} {x_{k + l} } }
      | rr= \pmod {b^{k + l + 1} }
}}
{{eqn | l = x_{k + l + 1}
      | o = \equiv
      | r = x_{k + l}
      | rr= \pmod {b^{k + l} }
}}
{{end-eqn}}
{{qed}}
Theorem: Divisor Sum of Power of 2

Let $n \in \Z_{>0}$ be a power of $2$.
Then:
:$\map {\sigma_1} n = 2 n - 1$

Proof:
Let $n = 2^k$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} n
      | r = \dfrac {2^{k + 1} - 1} {2 - 1}
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = 2 \times 2^k - 1
      | c = 
}}
{{eqn | r = 2 n - 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Divisor Sum Function
Category:2
5adh8ba8ttxo6k4732otbkyj39696is
Theorem: Equivalence of Definitions of Saturation Under Equivalence Relation

Let $\sim$ be an equivalence relation on a set $S$.
Let $T \subset S$ be a subset.
{{TFAE|def = Saturation (Equivalence Relation)|view = saturation}}

Proof:
=== 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}}
Theorem: Equivalence of Definitions of Saturated Set Under Equivalence Relation

Let $\sim$ be an equivalence relation on a set $S$.
Let $T \subset S$ be a subset.
{{TFAE|def = Saturated Set (Equivalence Relation)|view = saturated set|context = Equivalence Relation}}

Proof:
=== 1 implies 2 ===
Let $T = \overline T$.
By definition of saturation:
:$T = \ds \bigcup_{t \mathop \in T} \eqclass t {}$
so we can take $U = T$.
{{qed}}
Theorem: Compact Subset is Bounded Below Join Semilattice

Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $P = \left({K\left({L}\right), \precsim}\right)$ be an ordered subset of $L$,
where $K\left({L}\right)$ denotes the compact subset of $L$.
Then $P$ is a bounded below join semilattice.

Proof:
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}}
Theorem: Necessary and Sufficient Condition for Quadratic Functional to be Positive Definite/Lemma 2

Let $\map h x : \closedint a b \to \R$ be continuously differentiable $\forall x \in \closedint a b$.
Suppose the function $\map h x$ satisfies the equation:
:$-\map {\dfrac \d {\d x} } {\paren {t P + \paren {1 - t} } h'} + t Q h = 0$
subject to the boundary conditions:
:$\map h {a, t} = \map h {b, t} = 0$
Then:
:$\ds \int_a^b \paren {\paren {P h'^2 + Q h^2} t + \paren {1 - t} h'^2} \rd x = 0$

Proof:
{{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}}
Theorem: Jacobi's Necessary Condition

Let $J$ be a functional, such that:
:$J \sqbrk y = \ds \int_a^b \map F {x, y, y'} \rd x$
Let $\map y x$ correspond to the minimum of $J$.
Let:
:$F_{y'y'}>0$
along $\map y x$.
Then the open interval $\openint a b$ contains no points conjugate to $a$.

Proof:
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}}
Theorem: Stolz-Cesàro Theorem

Let $\sequence {a_n}$ be a sequence.
{{explain|Domain of $\sequence {a_n}$ -- $\R$ presumably but could it be $\C$?}}
Let $\sequence {b_n}$ be a sequence of (strictly) positive real numbers such that:
:$\ds \sum_{i \mathop = 0}^\infty b_n = \infty$
If:
:$\ds \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = L \in \R$
then also:
:$\ds \lim_{n \mathop \to \infty} \dfrac {a_1 + a_2 + \cdots + a_n} {b_1 + b_2 + \cdots + b_n} = L$

Proof:
Define the following sums:
:$\ds A_n = \sum_{i \mathop = 1}^n a_i$
:$\ds B_n = \sum_{i \mathop = 1}^n b_i$
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 $\sequence {\dfrac {A_k - \paren {L \pm \mu} B_k} {B_n} }$ 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 | o = \leadstoandfrom 
      | 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}}
Theorem: Bottom is Compact

Let $L$ be a bounded below ordered set.
Then $\bot$ is a compact element
where $\bot$ is the smallest element in $L$.

Proof:
By Bottom is Way Below Any Element:
:$\bot \ll \bot$
where $\ll$ denotes the way below relation.
Hence $\bot$ is a compact element.
{{qed}}
Theorem: Saturation Under Equivalence Relation in Terms of Graph

Let $\RR \subset S \times S$ be an equivalence relation on a set $S$.
Let $\pr_1, \pr_2 : S \times S \to S$ denote the projections.
Let $T\subset S$ be a subset.
Let $\overline T$ denote its saturation.
Then the following hold:
:$\overline T = \map {\pr_1} {\RR \cap \map {\pr_2^{-1} } T}$
:$\overline T = \map {\pr_2} {\RR \cap \map {\pr_1^{-1} } T}$

Proof:
Let $s \in S$.
We have:
{{begin-eqn}}
{{eqn | o = 
      | r = s \in \map {\pr_1} {\RR \cap \map {\pr_2^{-1} } T}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | o = 
      | r = \exists t \in S: \tuple {s, t} \in \RR \cap \map {\pr_2^{-1} } T
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | o = 
      | r = \exists t \in S : \tuple {s, t} \in \RR \cap (S \times T)
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | o = 
      | r = \exists t \in T : \tuple {s, t} \in \RR
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | o = 
      | r = s \in \overline T
      | c = {{Defof|Saturation Under Equivalence Relation}}
}}
{{end-eqn}}
A similar reasoning proves the second identity.
{{qed}}
Category:Equivalence Relations
cwsib8d0idi9a9r4g0b2nutugs7g2z7
Theorem: Projection of Subset is Open iff Saturation is Open

Let $\sim$ be an equivalence relation on a topological space $\struct {X, \tau}$.
Let $p$ denote the quotient mapping induced by $\sim$.
Let $\tau_\sim$ be the quotient topology on $X / \sim$ by $p$.
Let $\struct {X / \sim, \tau_\sim}$ be the quotient space of $X$ by $\sim$.
Let $U \subset X$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\map p U$ is open in $\struct {X / \sim, \tau_\sim}$}}
{{item|(2):|The saturation of $U$ is open in $\struct {X, \tau}$}}
{{end-itemize}}

Proof:
By definition of quotient topology, $\map p U$ is open in $\struct {X / \sim, \tau_\sim}$ {{iff}} $\map {p^{-1} } {\map p U}$ is open in $\struct {X, \tau}$.
{{qed}}