nyuuzyou/wikis huggingface dataset:
Sums of Squares/Mistake/Second Edition

== Source Work ==
:
:$\mathbf C$: Additive Number Theory
::$\mathbf {C 20}$: Sums of Squares
== Mistake ==
:Apart from $256$ examples, the largest of which is $1167$, every number can be expressed as the sum of at most five composite numbers.
== Correction ==
It should say:
:Apart from $256$ examples, the largest of which is $1167$, every number can be expressed as the sum of at most five squares of composite numbers.
(emphasis added)
This has been corrected for .
== Also see ==
* Numbers Not Expressible as Sum of no more than $5$ Squares of Composite Numbers
== Sources ==
* : $\mathbf C$: Additive Number Theory: $\mathbf {C 20}$: Sums of Squares
Category:Mistakes/Unsolved Problems in Number Theory
Sums of Squares in Lines of Order 3 Magic Square

== Theorem ==
Consider the order 3 magic square:
: The sums of the squares of the top and bottom rows are equal, and differ by $18$ from the sums of the squares of the middle row
: The sums of the squares of the left and right columns are equal , and differ by $18$ from the sums of the squares of the middle column.
== Proof ==
For the rows:
For the colums:
== Sources ==
* : $9$
* : $9$
Category:Magic Squares
Sums of Squares of Diagonals of Order 3 Magic Square

== Theorem ==
Consider the order 3 magic square:
The sums of the squares of the diagonals, when expressed as $3$-digit decimal numbers, are equal to the sums of the squares of those same diagonals of that same order 3 magic square when reversed.
== Proof ==
For the top-left to bottom-right diagonals:
For the bottom-left to top-right diagonals:
== Also see ==
* Sums of Squares of Lines of Order 3 Magic Square
== Sources ==
* : $9$
* : $9$
Category:Magic Squares
Sums of Squares of Lines of Order 3 Magic Square

== Theorem ==
Consider the order 3 magic square:
: The sums of the squares of the rows, when expressed as $3$-digit decimal numbers, are equal to the sums of the squares of those same rows of that same order 3 magic square when reflected in a vertical axis:
$\quad\begin{array}{|c|c|c|}
\hline 6 & 7 & 2 \\
\hline 1 & 5 & 9 \\
\hline 8 & 3 & 4 \\
\hline
\end{array}$
Similarly:
: The sums of the squares of the columns, when expressed as $3$-digit decimal numbers, are equal to the sums of the squares of those same columns of that same order 3 magic square when reflected in a horizontal axis:
$\quad\begin{array}{|c|c|c|}
\hline 4 & 3 & 8 \\
\hline 9 & 5 & 1 \\
\hline 2 & 7 & 6 \\
\hline
\end{array}$
== Proof ==
For the rows:
For the columns:
== Also see ==
* Sums of Squares of Diagonals of Order 3 Magic Square
== Sources ==
* : $9$
* : $9$
Category:Magic Squares
Sums of both 2 and 3 Consecutive Squares

== Theorem ==
The following are the smallest positive integers that are the sum of both $2$ and $3$ consecutive non-zero square numbers:
:$365, 35 \, 645, 3 \, 492 \, 725, 342 \, 251 \, 285, 33 \, 537 \, 133 \, 085, 3 \, 286 \, 296 \, 790 \, 925, \ldots$
== Proof ==
== Sources ==
* : $2$: Hexes and Stars
* : $365$
Category:Sums of Squares
Sun Tzu Suan Ching/Examples/Example 1

== Example of Problem from ==
:A woman was washing dishes in a river, when an official whose business was overseeing the waters demanded of her:
:"Why are there so many dishes here?"
:"Because a feasting was entertained in the house," the woman replied.
:Thereuon the official enquired the number of guests.
:"I don't know," the woman said, "how many guests there had been;
:but every $2$ used a dish for rice between them;
:every $3$ a dish for broth;
:every $4$ a dish for meat;
:and there were $65$ dishes in all."
== Solution ==
:$60$ guests.
== Proof ==
Let $x$ be the number of guests.
Then:
== Sources ==
* 
* 
* : Sun Tsu Suan Ching: $69$
Category:Sun Tzu Suan Ching
Sun Tzu Suan Ching/Examples/Example 2

== Example of Problem from ==
:There are certain things whose number is unknown.
:Repeatedly divided by $3$, the remainder is $2$;
:by $5$ the remainder is $3$,
:and by $7$ the remainder is $2$.
:What will be the number?
== Solution ==
The number of objects could be any one of the numbers:
:$23 + 105 n$
where $n \in \N$ is an arbitrary natural number.
== Proof ==
From Chinese Remainder Theorem: $x \equiv 2 \pmod 3, 3 \pmod 5, 2 \pmod 7$ we have that:
:$x \equiv 23 \pmod {105}$
The result follows.
== Sources ==
* 
* : Sun Tsu Suan Ching: $70$
Category:Sun Tzu Suan Ching
Sun Tzu Suan Ching/Examples/Example 3

== Example of Problem from ==
:There are $3$ sisters, of whom the eldest comes home once every $5$ days,
:the middle every $4$ days,
:and the youngest in every $3$ days.
:In how many days will all the $3$ meet together?
== Solution ==
:$60$ days.
== Proof ==
The answer is found by:
:$\map \lcm {4, 5, 3}$
== Sources ==
* 
* 
* : Sun Tsu Suan Ching: $71$
Category:Sun Tzu Suan Ching
Sundry Coset Results

== Theorems ==
Let $G$ be a group and let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let:
: $x H$ denote the left coset of $H$ by $x$;
: $H y$ denote the right coset of $H$ by $y$.
Then the following results apply:
=== Element in Coset iff Product with Inverse in Subgroup ===
=== Cosets are Equal iff Product with Inverse in Subgroup ===
=== Cosets are Equal iff Element in Other Coset ===
=== Coset Equals Subgroup iff Element in Subgroup ===
=== Elements in Same Coset iff Product with Inverse in Subgroup ===
=== Regular Representation on Subgroup is Bijection to Coset ===
Category:Cosets
Superabundant Number/Examples

== Examples of Superabundant Numbers ==
=== 4 ===
=== 6 ===
=== 12 ===
=== 24 ===
=== 36 ===
=== 48 ===
=== 60 ===
=== 120 ===
=== 180 ===
=== 240 ===
Category:Superabundant Numbers
Superabundant Numbers are Infinite in Number

== Theorem ==
There are infinitely many superabundant numbers.
== Proof ==
 the set $S$ of superabundant numbers is finite.
Let $m$ be the greatest element of $S$.
By definition of superabundant, $m$ has the largest abundancy index of all the elements of $S$.
Consider the integer $2 m$.
From Abundancy Index of Product is greater than Abundancy Index of Proper Factors, $2 m$ has a higher abundancy index than $m$.
There are two possibilities:
:$(1): \quad 2 m$ is the smallest integer greater that $n$ which has a higher abundancy index than $m$.
By definition, that would make $m$ superabundant.
:$(2) \quad$ There exists a finite set $T := \set {n \in \Z: m < n < 2 m: \map A n > \map A m}$, where $\map A n$ denotes the abundancy index of $n$.
The smallest element $t$ of $T$ therefore has an abundancy index greater than all smaller positive integers.
Thus by definition $t$ is superabundant.
In either case, there exists a superabundant number not in $S$.
Thus $S$ cannot contain all superabundant numbers.
But this contradicts our initial assumption that the set $S$, containing all superabundant numbers is finite.
It follows by Proof by Contradiction that $S$ is infinite.
Hence the result.
== Sources ==
* : $12$
* : $12$
Category:Superabundant Numbers
Superfactorial/Examples/3

== Example of Superfactorial ==
The superfactorial of $3$ is given by:
:$3 \$ = 3!^{3!^{3!} } = 6^{6^6} = 6^{46 \, 656}$
== Sources ==
* : $3!^{3!^{3!} }$
Category:Superfactorials
Superinductive Class under Strictly Progressing Mapping is Proper Class

== Theorem ==
Let $A$ be a class.
Let $g: A \to A$ be a strictly progressing mapping on $A$.
Let $A$ be superinductive under $g$.
Then $A$ cannot be a set, and thus is a proper class.
== Proof ==
 $A$ is a set.
Then from Set which is Superinductive under Progressing Mapping has Fixed Point, $A$ has a fixed point.
However, we have that $g$ is a strictly progressing mapping on $A$.
Hence $g$ has no fixed point in $A$.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.9$
Category:Superinductive Classes
Category:Strictly Progressing Mappings
Superset Relation is Compatible with Subset Product

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $\circ_\PP$ be the subset product on $\powerset S$, the power set of $S$.
Then the superset relation $\supseteq$ is compatible with $\circ_\PP$.
== Proof ==
By Subset Relation is Compatible with Subset Product, the subset relation $\subseteq$ is compatible with $\circ_\PP$.
From Inverse of Subset Relation is Superset, the inverse of $\subseteq$ is $\supseteq$.
The result follows from Inverse of Relation Compatible with Operation is Compatible.
Category:Compatible Relations
Superset of Absorbing Set is Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $A$ be an absorbing set.
Let $B \supseteq A$. 
Then $B$ is absorbing.
== Proof ==
Let $x \in X$.
Since $A$ is absorbing, there exists $t \in \R_{> 0}$ such that:
:$x \in t A$ for $\cmod \alpha \ge t$.
Since $A \subseteq B$, we obtain:
:$x \in t B$ for $\cmod \alpha \ge t$.
Category:Absorbing Sets
Superset of Co-Countable Set

== Theorem ==
Every superset of a co-countable set is co-countable.
== Proof ==
Let $S$ be a set.
Let $A$ be co-countable in $S$, and let $B$ be such that $A \subseteq B \subseteq S$.
From Relative Complement inverts Subsets, it follows that:
:$\complement_S \left({B}\right) \subseteq \complement_S \left({A}\right)$
As $A$ is co-countable, $\complement_S \left({A}\right)$ is countable.
By Subset of Countably Infinite Set is Countable, it follows that $\complement_S \left({B}\right)$ is also countable.
Therefore, $B$ is also co-countable, and the result follows.
== Also see ==
* Subset of Countably Infinite Set is Countable
Category:Set Theory
Superset of Dependent Set is Dependent

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A, B \subseteq S$ such that $A \subseteq B$
If $A$ is a dependent subset then $B$ is a dependent subset.
=== Corollary ===
== Proof ==
From the contrapositive statement of matroid axiom $(\text I 2)$:
:$A \notin \mathscr I \implies B \notin \mathscr I$
By the definition of a dependent subset:
:If $A$ is not an dependent subset then $B$ is not an dependent subset.
Category:Matroid Dependent Subsets
Superset of Dependent Set is Dependent/Corollary

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $A \subseteq S$.
Let $x \in A$.
If $x$ is a loop then $A$ is dependent.
== Proof ==
Let $x$ be a loop.
By definition of a loop:
:$\set x \notin \mathscr I$
By definition of a dependent subset:
:$\set x$ is a dependent subset
From Singleton of Element is Subset:
:$\set x \subseteq A$
From Superset of Dependent Set is Dependent:
:$A$ is a dependent subset
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Dependent Subsets
Category:Matroid Loops
Superset of Infinite Set is Infinite

== Theorem ==
Let $S$ be an infinite set.
Let $T \supseteq S$ be a superset of $S$.
Then $T$ is also infinite.
== Proof ==
Suppose $T$ were finite.
Then by Set Finite iff Injection to Initial Segment of Natural Numbers, there is an injection $f: T \to \N_{<n}$ for some $n \in \N$.
But then by Restriction of Injection is Injection, also the restriction of $f$ to $S$:
:$f {\restriction_S}: S \to \N_{<n}$
is an injection.
Again by Set Finite iff Injection to Initial Segment of Natural Numbers, this contradicts the assumption that $S$ is infinite.
Hence $T$ is infinite.
== Sources ==
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Exercise $2$
Category:Set Theory
Superset of Linearly Dependent Set is Linearly Dependent

== Theorem ==
Let $S$ be a set of elements of a unitary module.
Let $S$ contain a subset $T$ which is linearly dependent.
Then $S$ is also linearly dependent.
== Proof ==
Let $S$ be a linearly independent set.
Let $T$ be a subset of $S$.
By Subset of Linearly Independent Set is Linearly Independent, $T$ is also linearly independent.
Thus by Proof by Contraposition, if $T$ is linearly dependent, then so must $S$ be.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
Category:Unitary Modules
Category:Linear Dependence
Category:Proofs by Contraposition
Superset of Neighborhood in Metric Space is Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$ be a point in $M$.
Let $N$ be a neighborhood of $a$ in $M$.
Let $N \subseteq N' \subseteq A$.
Then $N'$ is a neighborhood of $a$ in $M$.
== Proof ==
By definition of neighborhood:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$
where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$ in $M$.
By Subset Relation is Transitive:
:$\map {B_\epsilon} a \subseteq N'$
The result follows by definition of neighborhood of $a$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 3$
Category:Neighborhoods
Superset of Neighborhood in Topological Space is Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Let $N \subseteq N' \subseteq S$.
Then $N'$ is a neighborhood of $x$ in $T$.
That is:
:$\forall x \in S: \forall N \in \NN_x: N' \supseteq N \implies N' \in \NN_x$
where $\NN_x$ is the neighborhood filter of $x$.
== Proof ==
By definition of neighborhood:
:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
where $U$ is an open set of $T$.
By Subset Relation is Transitive:
:$U \subseteq N'$
The result follows by definition of neighborhood of $x$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 3$
Category:Neighborhoods
Superset of Order Generating is Order Generating

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X, Y$ be subsets of $S$ such that
:$X$ is order generating
and
:$X \subseteq Y$
Then $Y$ is order generating.
== Proof ==
Let $x \in S$.
Thus by definition of complete lattice:
:$x^\succeq \cap Y$ admits an infimum.
By definition of complete lattice:
:$x^\succeq$ admits an infimum
and
:$x^\succeq \cap X$ admits an infimum.
By Intersection is Subset:
:$x^\succeq \cap Y \subseteq x^\succeq$
By Infimum of Subset:
:$\map \inf {x^\succeq} \preceq \map \inf {x^\succeq \cap Y}$
By Infimum of Upper Closure of Element:
:$x \preceq \map \inf {x^\succeq \cap Y}$
By Set Intersection Preserves Subsets/Corollary:
:$x^\succeq \cap X \subseteq x^\succeq \cap Y$
By Infimum of Subset:
:$\map \inf {x^\succeq \cap Y} \preceq \map \inf {x^\succeq \cap X}$
By definition of order generating:
:$\map \inf {x^\succeq \cap Y} \preceq x$
Thus by definition of antisymmetry:
:$x = \map \inf {x^\succeq \cap Y}$
== Sources ==
* 
* 
Category:Complete Lattices
Category:Order Generating Subsets
Superset of Unsatisfiable Set is Unsatisfiable

== Theorem ==
Let $\LL$ be a logical language.
Let $\mathscr M$ be a formal semantics for $\LL$.
Let $\FF$ be an $\mathscr M$-unsatisfiable set of formulas from $\LL$.
Let $\FF'$ be a superset of $\FF$.
Then $\FF'$ is also $\mathscr M$-unsatisfiable.
== Proof ==
By assumption, $\FF$ is unsatisfiable.
Suppose now $\FF'$ were satisfiable.
Then it would follow from Subset of Satisfiable Set is Satisfiable that $\FF$ were also satisfiable.
We conclude that $\FF'$ must be unsatisfiable.
Category:Formal Semantics
Superspace of Homeomorphic Subspaces may not have Homeomorphism to Itself containing Subspace Homeomorphism

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.
Let $H_1$ and $H_2$ be a homeomorphic.
Then it may be the case that there does not exist a homeomorphism $g: T_1 \to T_2$ such that:
:$g \restriction_{H_1} = f$
where:
:$g \restriction_{H_1}$ is the restriction of $g$ to $H_1$
:$f: H_1 \to H_2$ is a homeomorphism.
== Proof ==
Proof by Counterexample:
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $H_1 := \set 0 \cup \closedint 1 2 \cup \set 3$, where $\closedint 1 2$ denotes the closed interval from $1$ to $2$.
Let $H_2 := \closedint 0 1 \cup \set 2 \cup \set 3$.
$H_1$ and $H_2$ are homeomorphic, as can be demonstrated by the mapping $\phi: H_1 \to \H_2$ defined as:
:$\forall x \in H_1: \map \phi x = \begin {cases} 2 & : x = 0 \\ 3 & : x = 3 \\ x - 1 & : x \in \closedint 1 2 \end {cases}$
which is trivially a homeomorphism.
So, let $g$ be a homeomorphism from $H_1$ and $H_2$.
Each of the singletons in $H_1$ has to map to one of the singletons in $H_2$.
As a result, $g$ is not monotone.
Let $f: \R \to \R$ be a homeomorphism such that $g$ is a restriction of $f$.
From Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone, $f$ is monotone.
Hence $f$ cannot be a bijection.
Hence $f$ cannot be a homeomorphism.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $8$
Category:Homeomorphisms (Topological Spaces)
Supplemental Chords are Perpendicular

== Theorem ==
Let $\CC$ be a circle.
Let $K_1$ and $K_2$ be a pair of supplemental chords of $\CC$.
Then $K_1$ and $K_2$ are perpendicular.
== Proof ==
This is merely a restatement of Thales' Theorem stated in terms of supplemental chords.
== Sources ==
* 
* 
Category:Supplemental Chords
Supplementary Interior Angles implies Parallel Lines

== Theorem ==
Given two infinite straight lines which are cut by a transversal, if the interior angles on the same side of the transversal are supplementary, then the lines are parallel.
== Proof ==
:
Let $AB$ and $CD$ be infinite straight lines.
Let $EF$ be a transversal that cuts them.
Let at least one pair of interior angles on the same side of the transversal be supplementary.
, let those interior angles be $\angle BGH$ and $\angle DHG$.
So, by definition, $\angle DHG + \angle BGH$ equals two right angles.
Also, from Two Angles on Straight Line make Two Right Angles, $\angle AGH + \angle BGH$ equals two right angles.
Then from Euclid's first and third common notion and Euclid's fourth postulate:
:$\angle AGH = \angle DHG$
Finally, by Equal Alternate Angles implies Parallel Lines:
:$AB \parallel CD$
== Sources ==
* : Book $\text{I}$. Propositions
Category:Transversals (Geometry)
Category:Parallel Lines
Category:Supplementary Angles
Suprema Preserving Mapping on Ideals Preserves Directed Suprema

== Theorem ==
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.
Let $f: S \to T$ be a mapping.
Let every filter $F$ in $\left({S, \preceq}\right)$, $f$ preserve the infimum on $F$.
Then $f$ preserves directed suprema.
== Proof ==
This follows by mutatis mutandis of the proof of Infima Preserving Mapping on Filters Preserves Filtered Infima.
== Sources ==
* 
* 
Category:Order Theory
Suprema Preserving Mapping on Ideals is Increasing

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered sets.
Let $f: S \to T$ be a mapping.
For every ideal $I$ in $\struct {S, \preceq}$, let $f$ preserve the supremum on $I$.
Then $f$ is increasing.
== Proof ==
Let $x, y \in S$ such that:
:$x \preceq y$
By Supremum of Singleton:
:$\set x$ and $\set y$ admit suprema in $\struct {S, \preceq}$
By Supremum of Lower Closure of Set:
:$\set x^\preceq$ and $\set y^\preceq$ admit suprema in $\struct {S, \preceq}$
where $\set x^\preceq$ denotes the lower closure of $\set x$
By Lower Closure of Singleton:
:$x^\preceq$ and $y^\preceq$ admit suprema in $\struct {S, \preceq}$
By Lower Closure of Element is Ideal:
:$x^\preceq$ and $y^\preceq$ are ideals in $\struct {S, \preceq}$
By assumption and definition of mapping preserves the supremum on subset:
:$\map {f^\to} {x^\preceq}$ and $\map {f^\to} {y^\preceq}$ admit suprema in $\struct {T, \precsim}$
and:
:$\map \sup {\map {f^\to} {x^\preceq} } = \map f {\map \sup {x^\preceq} }$
and:
:$\map \sup {\map {f^\to} {y^\preceq} } = \map f {\map \sup {y^\preceq} }$
By Supremum of Lower Closure of Element:
:$\map \sup {x^\preceq} = x$ and $\map \sup {y^\preceq} = y$
By Lower Closure is Increasing:
:$x^\preceq \subseteq y^\preceq$
By Image of Subset under Mapping is Subset of Image:
:$\map {f^\to} {x^\preceq} \subseteq \map {f^\to} {y^\preceq}$
Thus by Supremum of Subset:
:$\map f x \precsim \map f y$
Thus by definition:
:$f$ is increasing.
== Sources ==
* 
* 
Category:Order Theory
Category:Increasing Mappings
Suprema and Infima of Combined Bounded Functions

== Theorem ==
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
=== Bounded Above ===
=== Bounded Below ===
== Warning ==
== Sources ==
* : $\S 7.16 \ (6)$
Category:Real Analysis
Category:Suprema
Category:Infima
Category:Bounded Mappings
Category:Suprema and Infima of Combined Bounded Functions
Suprema and Infima of Combined Bounded Functions/Bounded Above

== Theorem ==
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
Let both $f$ and $g$ be bounded above on $S \subseteq \R$.
Then:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$
where $\ds \map \sup {\map f x}$ is the supremum of $\map f x$.
== Proof ==
First we show that:
:$\ds \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
Next we show that $\ds \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$:
Let:
:$\ds H = \map {\sup_{x \mathop \in S} } {\map f x}$
:$\ds K = \map {\sup_{x \mathop \in S} } {\map g x}$
Then:
:$\forall x \in S: \map f x + \map g x \le H + K$
Hence $H + K$ is an upper bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
== Warning ==
== Sources ==
* : $\S 7.16 \ (6)$
Category:Suprema and Infima of Combined Bounded Functions
Suprema and Infima of Combined Bounded Functions/Bounded Below

== Theorem ==
Let $f$ and $g$ be real functions.
Let $c$ be a constant.
Let both $f$ and $g$ be bounded below on $S \subseteq \R$.
Then:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
where $\ds \map \inf {\map f x}$ is the infimum of $\map f x$.
== Proof ==
First we show that:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
Let $T = \set {\map f x: x \in S}$.
Then:
Next we show that:
:$\ds \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
Let:
:$\ds H = \map {\inf_{x \mathop \in S} } {\map f x}$
:$\ds K = \map {\inf_{x \mathop \in S} } {\map g x}$
Then:
:$\forall x \in S: \map f x + g \map x \ge H + K$
Hence $H + K$ is a lower bound for $\set {\map f x + \map g x: x \in S}$.
The result follows.
== Warning ==
Category:Suprema and Infima of Combined Bounded Functions