nyuuzyou/wikis huggingface dataset:
Upper Bound of Order of Non-Abelian Finite Simple Group/Corollary

== Corollary to Upper Bound of Order of Non-Abelian Finite Simple Group ==
Let $H$ be a finite group of even order.
Let $u \in H$ be a self-inverse element of $H$.
Then there are finitely many types of finite simple group $G$ such that:
:$G$ has a self-inverse element $t \in G$
:$\map {C_G} t \cong H$
== Proof ==
First suppose that $G$ is abelian.
Then by Abelian Group is Simple iff Prime, $\order G = 2$.
So let $G$ be non-abelian.
From Upper Bound of Order of Non-Abelian Finite Simple Group:
:$\order G \le \paren {\dfrac {\order H \paren {\order H + 1} } 2}!$
which depends completely upon the given group $H$.
The result follows from Finite Number of Groups of Given Finite Order.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.15$
Category:Centralizers
Category:Simple Groups
Category:Self-Inverse Elements
Category:Finite Groups
Upper Bound of Ordinal Sum

== Theorem ==
Let $x$ and $y$ be ordinals.
Suppose $x > 1$.
Let $\sequence {a_n}$ be a finite sequence of ordinals such that:
:$a_n < x$ for all $n$
Let $\sequence {b_n}$ be a strictly decreasing finite sequence of ordinals such that:
:$b_n < y$ for all $n$
Then:
:$\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$
== Proof ==
The proof shall proceed by finite induction on $n$:
For all $n \in \N_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n x^{b_i} a_i < x^y$
=== Basis for the Induction ===
:$\map P 0$ says that:
:$\ds \sum_{i \mathop = 1}^0 x^{b_i} a_i < x^y$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{i \mathop = 1}^k x^{b_i} a_i < x^y$
Then we need to show:
:$\ds \sum_{i \mathop = 1}^{k \mathop + 1} x^{b_i} a_i < x^y$
=== Induction Step ===
This is our induction step:
By the inductive hypothesis:
:$\ds \sum_{i \mathop = 1}^k x^{b_{i + 1}} a_{i + 1} < x^{b_1}$ since $b_1$ is greater than all $b_i$ for $i > 1$.
Then:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \sum_{i \mathop = 1}^n a_i x^{b_i} < x^y$
== Sources ==
* : $\S 8.43$
Category:Ordinal Arithmetic
Upper Bound of Pell Number

== Theorem ==
For all $n \in \N$:
:$P_n \le \paren {1 + \sqrt 2}^{n - 1}$
where:
:$P_n$ is the $n$th Pell number
== Proof ==
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$P_n \le \paren {1 + \sqrt 2}^{n - 1}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$P_1 = 1 = \paren {1 + \sqrt 2}^0$
It is also necessary to demonstrate $\map P 2$ is true:
:$P_2 = 2 \le \paren {1 + \sqrt 2}^1$
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P j$ is true for all $1 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$P_j \le \paren {1 + \sqrt 2}^{j - 1}$ for $1 \le j \le k$
from which it is to be shown that:
:$P_{k + 1} \le \paren {1 + \sqrt 2}^k$
=== Induction Step ===
This is the induction step:
As we have already shown $\map P 1$ and $\map P 2$, we just need to prove the result for $k \ge 2$.
So $\map P 1 \land \map P 2 \land \dots \land \map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: P_n \le \paren {1 + \sqrt 2}^{n - 1}$
Category:Pell Numbers
Upper Bounds are Equivalent implies Suprema are equal

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $X, Y$ be subsets of $S$.
Assume that
:$X$ admits a supremum
and
:$\forall x \in S: x$ is upper bound for $X \iff x$ is upper bound for $Y$
Then $\sup X = \sup Y$
== Proof ==
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:
:$\sup X \preceq b$
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:
:$\sup X = \sup Y$
== Sources ==
* 
Category:Suprema
Upper Bounds for Prime Numbers

== Theorem ==
Let $p: \N \to \N$ be the prime enumeration function.
Then $\forall n \in \N$, the value of $\map p n$ is bounded above.
In particular:
=== Result 1 ===
=== Result 2 ===
=== Result 3 ===
== Motivataion ==
In the first two cases it can be seen that the limit found is wildly extravagantly large.
However, they are results that are easily established, and they have their uses.
Category:Number Theory
Category:Prime Numbers
Category:Upper Bounds for Prime Numbers
Upper Bounds for Prime Numbers/Result 1

== Theorem ==
Let $p: \N \to \N$ be the prime enumeration function.
Then $\forall n \in \N$, the value of $\map p n$ is bounded above.
In particular:
:$\forall n \in \N: \map p n \le 2^{2^{n - 1} }$
== Proof ==
Proof by strong induction:
Let us write $p_n = \map p n$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map p n \le 2^{2^{n - 1} }$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $2 \le 2^{2^0} = 2$.
This is our basis for the induction.
=== Induction Hypothesis ===
Suppose that, for some $k \in \N$, each of $\map P 1, \map P 2, \ldots, \map P k$ is true.
It remains to show that it logically follows that $\map P {k + 1}$ is true.
That is, that:
:$\map p {k + 1} \le 2^{2^k}$
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
The result follows by the Second Principle of Mathematical Induction.
== Motivation ==
It can be seen that the limit found is wildly extravagantly large.
However, it is an easily established result, and it has its uses.
== Sources ==
* : $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $7$
Category:Upper Bounds for Prime Numbers
Upper Bounds for Prime Numbers/Result 2

== Theorem ==
Let $p: \N \to \N$ be the prime enumeration function.
Then $\forall n \in \N$, the value of $\map p n$ is bounded above.
In particular:
:$\forall n \in \N: \map p n \le \paren {p \paren {n - 1} }^{n - 1} + 1$
== Proof ==
Let us write $p_n = \map p n$.
Let us take $N = p_1 p_2 \cdots p_n + 1$.
By the same argument as in Euclid's Theorem, we have that either $N$ is prime, or it is not.
If $N$ is prime, then either $N = p_{n + 1}$ or not, in which case $N > p_{n + 1}$.
In the second case, $N$ has a prime factor not in $\set {p_1, p_2, \ldots, p_n}$
Therefore it must have a prime factor greater than any of $\set {p_1, p_2, \ldots, p_n}$.
In any case, the next prime after $p_n$ can be no greater than $p_1 p_2 \cdots p_n + 1$.
Hence the result.
== Note ==
It can be seen that the limit found is wildly extravagantly large.
However, it is an easily established result, and it has its uses.
Category:Upper Bounds for Prime Numbers
Upper Bounds for Prime Numbers/Result 3

== Theorem ==
Let $p: \N \to \N$ be the prime enumeration function.
Then $\forall n \in \N$, the value of $p \left({n}\right)$ is bounded above.
In particular:
: $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$
== Proof ==
Let us write $p_n = p \left({n}\right)$.
From Bertrand's Conjecture, for each $n \ge 2$ there exists a prime $p$ such that $n < p < 2 n$.
For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
: $p_n < 2^n$
$P(1)$ is the statement:
:$p_1 = 2 = 2^1$
As this does not fulfil the criterion:
:$p \left({n}\right) < 2^n$
it is not included in the result.
=== Basis for the Induction ===
$P(2)$ is true, as this just says:
:$p_2 = 3 < 2^2 = 4$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
:$p_k < 2^k$
Then we need to show:
:$p_{k+1} < 2^{k+1}$
=== Induction Step ===
This is our induction step:
Suppose that $p_k < 2^k$ for some $k \ge 2$.
By Bertrand's Conjecture, there exists a prime $q$ such that $2^k < q < 2^{k+1}$.
So:
: $p < 2^k < q < 2^{k+1}$
But as $q$ is a prime exceeding $p_k$, it means:
: $p_{k+1} \le q < 2^{k+1}$
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
: $\forall n \in \N_{>1}: p \left({n}\right) < 2^n$
Category:Upper Bounds for Prime Numbers
Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure

== Theorem ==
Let $L = \left({S, \preceq}\right)$ be an ordered set.
Let $\left({T, \precsim}\right)$ be an ordered subset of $L$.
Let $t \in T$.
Then $t^\succsim = T \cap t^\succeq$
== Proof ==
By definition of ordered subset:
:$T \subseteq S$
We will prove that
:$t^\succsim \subseteq T \cap t^\succeq$
Let $x \in t^\succsim$
By definition of upper closure of element:
:$x \in T$ and $t \precsim x$
By definition of ordered subset:
:$t \preceq x$
By definition of upper closure of element:
:$x \in t^\succeq$
Thus by definition of intersection:
:$x \in T \cap t^\succeq$
We will prove that
:$T \cap t^\succeq \subseteq t^\succsim$
Let $x \in T \cap t^\succeq$
By definition of intersection:
:$x \in T$ and $x \in t^\succeq$
By definition of upper closure of element:
:$t \preceq x$
By definition of ordered subset:
:$t \precsim x$
Thus by definition of upper closure of element:
:$x \in t^\succsim$
By definition of set equality:
:$t^\succsim = T \cap t^\succeq$
Category:Upper Closures
Upper Closure is Closure Operator

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T^\succeq$ be the upper closure of $T$ for each $T \subseteq S$.
Then $\cdot^\succeq$ is a closure operator.
== Proof ==
=== Inflationary ===
Let $T \subseteq S$.
Let $t \in T$.
Then since $T \subseteq S$, $t \in S$ by the definition of subset.
Since $\preceq$ is reflexive, $t \preceq t$.
Thus by the definition of upper closure, $t \in T^\succeq$.
Since this holds for all $t \in T$, $T \subseteq T^\succeq$.
Since this holds for all $T \subseteq S$:
:$\cdot^\succeq$ is inflationary.
=== Order-Preserving===
Let $T \subseteq U \subseteq S$.
Let $x \in T^\succeq$.
Then by the definition of upper closure: for some $t \in T$, $t \preceq x$.
By the definition of subset:
:$t \in U$
Thus by the definition of upper closure:
:$x \in U^\succeq$
Since this holds for all $x \in T^\succeq$:
:$T^\succeq \subseteq U^\succeq$
Since this holds for all $T$ and $U$:
:$\cdot^\succeq$ is order-preserving.
=== Idempotent ===
Let $T \subseteq S$.
By Upper Closure is Upper Section, $T^\succeq$ is an upper section.
Thus by Upper Section: Definition 3:
:$\paren {T^\succeq}^\succeq = T^\succeq$
Since this holds for all $T$:
:$\cdot^\succeq$ is idempotent.
== Also see ==
* Lower Closure is Closure Operator
Category:Upper Closures
Category:Closure Operators
Upper Closure is Compact in Topological Lattice

== Theorem ==
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:
:$x \in \bigcup \GG$
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 $\FF$.
== Sources ==
* 
Category:Topological Order Theory
Category:Compact Spaces
Upper Closure is Decreasing

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $x, y$ be elements of $S$ such that
:$x \preceq y$
then $y^\succeq \subseteq x^\succeq$
where $y^\succeq$ denotes the upper closure of $y$.
== Proof ==
Let $z \in y^\succeq$.
By definition of upper closure of element:
:$y \preceq z$
By definition of ordering, $\preceq$ is transitive.
From $x \preceq y$ and $y \preceq z$:
:$x \preceq z$
Thus again by definition of upper closure of element:
:$z \in x^\succeq$
== Sources ==
* 
* 
Category:Upper Closures
Upper Closure is Smallest Containing Upper Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $U = T^\succeq$ be the upper closure of $T$.
Then $U$ is the smallest upper section containing $T$ as a subset.
== Proof ==
Follows from Upper Closure is Closure Operator and Set Closure is Smallest Closed Set/Closure Operator.
== Also see ==
* Lower Closure is Smallest Containing Lower Section
Category:Upper Closures
Category:Upper Sections
Upper Closure is Strict Upper Closure of Immediate Predecessor

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $b$ be the immediate successor element of $a$:
Then:
:$a^\succ = b^\succcurlyeq$
where:
:$a^\succ$ is the strict upper closure of $a$
:$b^\succcurlyeq$ is the upper closure of $b$.
== Proof ==
Let:
:$x \in a^\succ$
By the definition of strict upper closure:
:$a \prec x$
By the definition of total ordering:
:$x \prec b$ or $x \succcurlyeq b$
If $x \prec b$ then $a \prec x \prec b$, contradicting the premise.
Thus:
:$x \succcurlyeq b$
and so:
:$x \in b^\succcurlyeq$
By definition of subset:
:$a^\succ \subseteq b^\succcurlyeq$
Let:
:$x \in b^\succcurlyeq$
By the definition of upper closure:
:$b \preccurlyeq x$
Since $a \prec b$, Extended Transitivity shows that $a \prec x$.
Thus:
:$x \in a^\succ$
By definition of subset:
:$b^\succcurlyeq \subseteq a^\succ$
Therefore by definition of set equality:
:$a^\succ = b^\succcurlyeq$
Category:Total Orderings
Category:Upper Closures
Upper Closure is Upper Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a subset of $S$.
Let $U$ be the upper closure of $T$.
Then $U$ is an upper section.
== Proof ==
Let $a \in U$.
Let $b \in S$ with $a \preceq b$.
By the definition of upper closure, there is a $t \in T$ such that $t \preceq a$.
By transitivity, $t \preceq b$.
Thus, again by the definition of upper closure:
:$b \in U$
Since this holds for all such $a$ and $b$, $U$ is an upper section.
== Also see ==
* Lower Closure is Lower Section
Category:Upper Sections
Category:Upper Closures
Upper Closure of Coarser Subset is Subset of Upper Closure

== Theorem ==
Let $L = \left({S, \preceq}\right)$ be a preordered set.
Let $A, B$ be subsets of $S$ such that
:$A$ is coarser than $B$.
Then $A^\succeq \subseteq B^\succeq$
== Proof ==
Let $x \in A^\succeq$
By definition of upper closure of subset:
:$\exists y \in A: y \preceq x$
By definition of coarser subset:
:$\exists z \in B: z \preceq y$
By definition of transitivity:
:$z \preceq x$
Thus by definition of upper closure of subset:
:$x \in B^\succeq$
== Sources ==
* 
* 
Category:Preorder Theory
Category:Upper Closures
Upper Closure of Element is Filter

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $s$ be an element of $S$.
Then:
:$s^\succeq$ is a filter in $\struct {S, \preceq}$
where $s^\succeq$ denotes the upper closure of $s$.
== Proof ==
By Singleton is Directed and Filtered Subset
:$\set s$ is a filtered subset of $S$
By Filtered iff Upper Closure Filtered:
:$\set s^\succeq$ is a filtered subset of $S$
By Upper Closure is Upper Section:
:$\set s^\succeq$ is a upper section in $S$
By Upper Closure of Singleton
:$\set s^\succeq = s^\succeq$
By definition of reflexivity:
:$s \preceq s$
By definition of upper closure of element:
:$s \in s^\succeq$
Thus by definition:
:$s^\succeq$ is non-empty, filtered and upper.
Thus by definition:
:$s^\succeq$ is a filter in $\struct {S, \preceq}$
== Sources ==
* 
* 
* 
Category:Upper Closures
Upper Closure of Element is Way Below Open Filter iff Element is Compact

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x \in S$.
Then:
:$x^\succeq$ is a way below open filter on $L$
:$x$ is compact
== Proof ==
=== Sufficient Condition ===
Suppose:
:$x^\succeq$ is a way below open filter on $L$
By definitions of upper closure of element and reflexivity:
:$x \in x^\succeq$
By definition of way below open:
:$\exists y \in x^\succeq: y \ll x$
By definition of upper closure of element:
:$x \preceq y$
By Preceding and Way Below implies Way Below and definition of reflexivity:
:$x \ll x$
By definition:
:$x$ is compact.
=== Necessary Condition ===
Let $x$ be compact.
By Upper Closure of Element is Filter:
:$x^\succeq$ is filter on $L$.
It remains to show that:
:$x^\succeq$ is way below open.
Let $u \in x^\succeq$.
By definition of upper closure of element:
:$x \preceq u$
and by definition of reflexivity:
:$x \in x^\succeq$
By definition of compact:
:$x \ll x$
By Preceding and Way Below implies Way Below:
:$x \ll u$
Thus
:$\exists z \in x^\succeq: z \ll u$
== Sources ==
* 
* 
Category:Way Below Relation
Upper Closure of Element without Element is Filter implies Element is Meet Irreducible

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $x \in S$.
Let
:$x^\succeq \setminus \set x$ be a filter in $L$.
Then $x$ is meet irreducible.
== Proof ==
Let $a, b \in S$.
:
:$x = a \wedge b$ and $x \ne a$ and $x \ne b$
By Meet Precedes Operands:
:$x \preceq b$ and $x \preceq a$
By definition of upper closure of element:
:$b, a \in x^\succeq$
By definitions of singleton and difference:
:$b, a \in x^\succeq \setminus \set x$
By definition of filtered:
:$\exists z \in x^\succeq \setminus \set x: z \preceq a \land z \preceq b$
By definition of infimum:
:$z \preceq x$
By definition of upper section:
:$x \in x^\succeq \setminus \set x$
Thus this contradicts $x \in \set x$ by definition of singleton.
== Sources ==
* 
* 
Category:Order Theory
Category:Meet Irreducible Elements
Upper Closure of Singleton

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $s$ be an element of $S$.
Then:
:$\set s^\succeq = s^\succeq$
where:
:$\set s^\succeq$ denotes the upper closure of $\set s$
:$s^\succeq$ denotes the upper closure of $s$
== Proof ==
== Sources ==
* 
Category:Upper Closures
Upper Closure of Subset is Subset of Upper Closure

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $X, Y$ be subsets of $S$.
Then
:$X \subseteq Y \implies X^\succeq \subseteq Y^\succeq$
where $X^\succeq$ denotes the upper closure of $X$.
== Proof ==
Let $X \subseteq Y$.
Let $x \in X^\succeq$.
By definition of upper closure of subset:
:$\exists y \in X: y \preceq x$
By definition of subset:
:$y \in Y$
Thus by definition of upper closure of subset:
:$x \in Y^\succeq$
== Sources ==
* 
Category:Order Theory
Category:Upper Closures
Upper Darboux Integral Never Smaller than Lower Darboux Integral

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a bounded real function.
The lower Darboux integral of $f$ over $\closedint a b$ is less than or equal to the upper Darboux integral of $f$ over the same bounds.
That is:
:$\ds \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x$
== Proof ==
Let the value of the lower Darboux integral be $L$, and the value of the upper Darboux integral be $U$.
, suppose $L > U$.
Then $\epsilon = \dfrac {L - U} 2$ is positive.
By the definitions of lower Darboux integral and upper Darboux integral:
:$\ds \sup_P \map L P > \inf_P \map U P$
where $P$ ranges over all subdivisions of $\closedint a b$, and $\map L P$ and $\map U P$ are the lower Darboux sum and upper Darboux sum, respectively.
By the Characterizing Property of Supremum of Subset of Real Numbers, there exists are partition $P_L$ such that:
:$\map L {P_L} > L - \epsilon = \dfrac {L + U} 2$
Likewise, by the Characterizing Property of Infimum of Subset of Real Numbers, there is a partition $P_U$ such that:
:$\map U {P_U} < U + \epsilon = \dfrac {U + L} 2 $
Thus:
:$\map U {P_U} < \map L {P_L}$
But, by Upper Sum Never Smaller than Lower Sum for any Pair of Subdivisions, this is a contradiction.
Therefore, $L \le U$.
Category:Lower Darboux Integral
Category:Upper Darboux Integral
Category:Integral Calculus
Upper Darboux Sum Never Smaller than Lower Darboux Sum

== Theorem ==
Let $\closedint a b$ be a closed interval of the set $\R$ of real numbers.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a finite subdivision of $\closedint a b$.
Let $f: \R \to \R$ be a real function.
Let $f$ be bounded on $\closedint a b$.
Let $\map L P$ be the lower Darboux sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$.
Let $\map U P$ be the upper Darboux sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$.
Then $\map L P \le \map U P$.
== Proof ==
For all $\nu \in 1, 2, \ldots, n$, let $\closedint {x_{\nu - 1} } {x_\nu}$ be a closed subinterval of $\closedint a b$.
As $f$ is bounded on $\closedint a b$, it is bounded on $\closedint {x_{\nu - 1} } {x_\nu}$.
So, let $m_\nu$ be the infimum and $M_\nu$ be the supremum of $\map f x$ on the interval $\closedint {x_{\nu - 1} } {x_\nu}$.
By definition, $m_\nu \le M_\nu$.
So $m_\nu \paren {x_\nu - x_{\nu - 1} } \le M_{\nu} \paren {x_\nu - x_{\nu - 1} }$.
It follows directly that $\ds \sum_{\nu \mathop = 1}^n m_\nu \paren {x_\nu - x_{\nu - 1} } \le \sum_{\nu \mathop = 1}^n M_\nu \paren {x_\nu - x_{\nu - 1} }$.
== Sources ==
* : $2.5$: The Riemann Integral
Category:Real Analysis
Upper Limit of Number of Unit Fractions to express Proper Fraction from Greedy Algorithm

== Theorem ==
Let $\dfrac p q$ denote a proper fraction expressed in canonical form.
Let $\dfrac p q$ be expressed as the sum of a finite number of distinct unit fractions using Fibonacci's Greedy Algorithm.
Then $\dfrac p q$ is expressed using no more than $p$ unit fractions.
== Proof ==
Let $\dfrac {x_k} {y_k}$ and $\dfrac {x_{k + 1} } {y_{k + 1} }$ be consecutive stages of the calculation of the unit fractions accordingly:
:$\dfrac {x_k} {y_k} - \dfrac 1 {\ceiling {y_n / x_n} } = \dfrac {x_{k + 1} } {y_{k + 1} }$
By definition of Fibonacci's Greedy Algorithm:
:$\dfrac {x_{k + 1} } {y_{k + 1} } = \dfrac {\paren {-y_k} \bmod {x_k} } {y_k \ceiling {y_k / x_k} }$
It is established during the processing of Fibonacci's Greedy Algorithm that:
:$\paren {-y_k} \bmod {x_k} < x_k$
Hence successive numerators decrease by at least $1$.
Hence there can be no more unit fractions than there are natural numbers between $1$ and $p$.
Hence the result.
== Historical Note ==
== Sources ==
* : Egyptian Fractions
Category:Fibonacci's Greedy Algorithm
Upper Section is Convex

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be an upper section.
Then $T$ is convex in $S$.
== Proof ==
Let $a, c \in T$.
Let $b \in S$.
Let $a \preceq b \preceq c$.
Since:
:$a \in T$
:$a \preceq b$
:$T$ is an upper section
it follows that:
:$b \in T$
This holds for all such $a$, $b$, and $c$.
Therefore, by definition, $T$ is convex in $S$.
== Also see ==
* Lower Section is Convex
Category:Upper Sections
Category:Convex Sets (Order Theory)