nyuuzyou/wikis huggingface dataset:
Collatz Conjecture

== Conjecture ==
:
Let $f: \N \to \N$ be the mapping defined on the natural numbers as follows:
:$\forall n \in \N: \map f n = \begin {cases} n / 2 & : n \text { even} \\ 3 n + 1 & : n \text { odd} \end {cases}$
For any given value of $n$, let the sequence $\sequence {S_k}$ be defined:
:$\forall k \in \N: S_k = \begin {cases} n & : k = 0 \\ \map f {S_{k - 1} } & : k > 0 \end {cases}$
The Collatz conjecture is:
:For all $n \in N$, there exists some $i \in \N$ such that sequence $S_i = 1$.
That is, by repeatedly applying the rule:
:halve it if even, otherwise multiply it by $3$ and add $1$
to any natural number, eventually you will reach $1$.
== Also known as ==
== Failed Proof ==
== Examples ==
== Progress ==
As of $7$th March $2017$, all numbers up to $2^{60}$ have been tested, and all end in $1$.
== Sources ==
* : $27$
* : $27$
Category:Collatz Conjecture
Category:Number Theory
Category:Unproven Hypotheses
Collatz Conjecture/Examples/17

== Example of Collatz Process ==
The following integer sequence is generated from $17$ by the Collatz process:
:$17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$
== Sources ==
* : $27$
* : $27$
Category:Collatz Conjecture
Category:17
Collatz Conjecture/Examples/27

== Example of Collatz Process ==
The following integer sequence is generated from $27$ by the Collatz process:
:$27 \to 82 \to 41 \to 124 \to 62 \to 31 \to 94 \to 47 \to 142 \to 71 \to 214 \to 107 \to 322 \to$
:$161 \to 484 \to 242 \to 121 \to 364 \to 182 \to 91 \to 274 \to 137 \to 412 \to 206 \to 103 \to$
:$310 \to 155 \to 466 \to 233 \to 700 \to 350 \to 175 \to 526 \to 263 \to 790 \to 395 \to 1186 \to$
:$593 \to 1780 \to 890 \to 445 \to 1336 \to 668 \to 334 \to 167 \to 502 \to 251 \to 754 \to 377 \to$
:$1132 \to 566 \to 283 \to 850 \to 425 \to 1276 \to 638 \to 319 \to 958 \to 479 \to 1438 \to 719 \to$
:$2158 \to 1079 \to 3238 \to 1619 \to 4858 \to 2429 \to 7288 \to 3644 \to 1822 \to 911 \to 2734 \to$
:$1367 \to 4102 \to 2051 \to 6154 \to 3077 \to 9232 \to 4616 \to 2308 \to 1154 \to 577 \to 1732 \to$
:$866 \to 433 \to 1300 \to 650 \to 325 \to 976 \to 488 \to 244 \to 122 \to 61 \to 184 \to 92 \to$
:$46 \to 23 \to 70 \to 35 \to 106 \to 53 \to 160 \to 80 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$
== Sources ==
* : $27$
* : $27$
Category:Collatz Conjecture
Category:27
Collatz Conjecture/Failed Proof

== Conjecture ==
== Failed Proof ==
 that for some $n \in N$, there does NOT exist any $i \in \N$ such that the sequence $S_i = 1$.
From the Well-Ordering Principle, this hypothetical sequence $\sequence {S_k}$ must have a smallest element.
Let $a = S_k$ be the smallest element of this hypothetical sequence.
We know that $a$ can not be even because:
which contradicts our assertion that $a$ was the smallest element in the sequence.
We also know that $a \not \equiv 1\pmod 4$ because:
Because $S_{k + 3} < S_k$, this contradicts our assertion that $a$ was the smallest element in the sequence.
When $a \equiv 3 \pmod 4$, we obtain
:but we can go no further because the parity of $S_{k + 4}$ depends on $r$.
The results here are inconclusive.
We now have that if $a = S_k$ is the smallest element of our hypothetical sequence, then $a$ must reside in the residue class of $3$ (modulo $4$).
Let us now see if we can improve our filter for $a$ by removing additional residue classes from contention.
Inspecting the set of residue classes modulo $16$, we focus our attention on $3$ (modulo $16$), $7$ (modulo $16$), $11$ (modulo $16$) and $15$ (modulo $16$).
We know that $a \not \equiv 3 \pmod {16}$ because:
Because $S_{k + 6} < S_k$, this contradicts our assertion that $a$ was the smallest element in the sequence.
Just like $3$ (modulo $4$) shown above, the results for $7$ (modulo $16$), $11$ (modulo $16$) and $15$ (modulo $16$) are inconclusive.
We now have the following:
If a counterexample to the Collatz Conjecture exists, then such a counterexample would require a sequence with a smallest element.
This smallest element could only reside in the following residue classes:
(modulo $4$): $3$ 
(modulo $16$): $7$, $11$, $15$
(modulo $64$): $7$, $15$, $27$, $31$, $39$, $47$, $59$, $63$
(modulo $256$): $27$, $31$, $47$, $63$, $71$, $91$, $103$, $111$, $127$, $155$, $159$, $167$, $191$, $207$, $223$, $231$, $239$, $251$, $255$
As we can see, (modulo $4^n$), there are $\approx 2^{1.5 \paren{n - 1} }$ residue classes where $a$ could reside.
This means we have a Cantor set-esque filter which allows our hypothetical smallest element $a$ to reside in an ever dwindling proportion of the possible residue classes (that is, $\dfrac {2^ {1.5 \paren {n - 1} } } {4^n}$).
Just like the Cantor set and just like prime numbers, at increasing values of $n$, we get an ever increasing set of something (what remains of the Cantor set, prime numbers and in this case, candidate residue classes where $a$ could reside).
However, the proportion of the item of interest relative to all possible values heads towards zero.
And we've proved nothing.
Math is hard.
Category:Collatz Conjecture
Collection of All Ordered Sets is not Set

== Theorem ==
Let $\mathrm {OS}$ denote the collection of all ordered sets.
Then $\mathrm {OS}$ is not a set.
== Proof ==
Let $C$ be the collection of all singletons:
:$\set {x: \exists y: x = \set y}$
Define a mapping $\map f {\set y} = \RR$ where $\RR$ is a reflexive relation on $\set y$.
By Reflexive Relation on Singleton is Well-Ordering, $\RR$ is an ordering.
Thus:
:$f: C \to \mathrm {OS}$
By Equality of Ordered Pairs:
:$\map f {y_1} = \map f {y_2} \implies y_1 = y_2$
Thus $f$ is an injection.
By definition of cardinality, $C$ is the collection of all sets which are equivalent to $\set \O$.
By Collection of Sets Equivalent to Set Containing Empty Set is Proper Class, $C$ is proper.
Thus, by Injection from Proper Class to Class, $\mathrm {OS}$ is proper.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $25$
Category:Ordered Sets
Collection of Sets Equivalent to Set Containing Empty Set is Proper Class

== Theorem ==
Let $S = \set \O$ be the singleton whose element is the empty set.
Let $C$ be the collection of all sets which are equivalent to $S$.
$C$ is a proper class.
== Proof ==
By definition of cardinality, $C$ is the collection of all singletons:
:$\set {x: \exists y: x = \set y}$
Define a class mapping $f: C \to V$, where $V$ is the universal class, such that $\map f {\set x} = x$.
This is a mapping on the domain $C$, as all elements of $C$ are singletons.
Take an arbitrary $y \in V$.
Then by definition of $C$:
:$\set y \in C$
and by definition of $f$: 
:$\map f {\set y} = y$
Thus:
:$\exists x \in A: \map f x = y$
and $f$ is a surjection.
From Universal Class is Proper, $V$ is proper.
It follows from Surjection from Class to Proper Class that $C$ is proper.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $3$: Cardinality: Exercise $2$
Category:Class Theory
Category:Singletons
Category:Empty Set
Column Equivalence is Equivalence Relation

== Theorem ==
Column equivalence is an equivalence relation.
== Proof ==
In the following, $\mathbf A$, $\mathbf B$ and $\mathbf C$ denote arbitrary matrices in a given matrix space $\map \MM {m, n}$ for $m, n \in \Z{>0}$.
We check in turn each of the conditions for equivalence:
=== Reflexive ===
Let $\kappa_i$ denote an arbitrary column of $\mathbf A$.
Let $e$ denote the elementary column operation $\kappa_i \to 1 \kappa_i$ applied to $\mathbf A$.
Then trivially:
:$\map e {\mathbf A} = \mathbf A$
and so $\mathbf A$ is trivially column equivalent to itself.
So column equivalence has been shown to be reflexive.
=== Symmetric ===
Let $\mathbf A$ be column equivalent to $\mathbf B$.
Let $\Gamma$ be the column operation that transforms $\mathbf A$ into $\mathbf B$.
From Column Operation has Inverse there exists a column operation $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$.
Thus $\mathbf B$ is column equivalent to $\mathbf A$.
So column equivalence has been shown to be symmetric.
=== Transitive ===
Let $\mathbf A$ be column equivalent to $\mathbf B$, and let $\mathbf B$ be column equivalent to $\mathbf C$.
Let $\Gamma_1$ be the column operation that transforms $\mathbf A$ into $\mathbf B$.
Let $\Gamma_2$ be the column operation that transforms $\mathbf B$ into $\mathbf C$.
From Sequence of Column Operations is Column Operation, the application of $\mathbf C$ is column equivalent to $\mathbf A$.
So column equivalence has been shown to be transitive.
Column equivalence has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Also see ==
* Row Equivalence is Equivalence Relation
Category:Examples of Equivalence Relations
Category:Column Operations
Column Operation has Inverse

== Theorem ==
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix. 
Let $\Gamma$ be a column operation which transforms $\mathbf A$ to a new matrix $\mathbf B \in \map \MM {m, n}$.
Then there exists another column operation $\Gamma'$ which transforms $\mathbf B$ back to $\mathbf A$.
== Proof ==
Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the finite sequence of elementary column operations that compose $\Gamma$.
Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding finite sequence of the elementary column matrices.
From Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices, we have:
:$\mathbf A \mathbf K = \mathbf B$
where $\mathbf K$ is the product of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
:$\mathbf K = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k$
By Elementary Column Matrix is Nonsingular, each of $\mathbf E_i$ is nonsingular.
By Product of Matrices is Nonsingular iff Matrices are Nonsingular, it follows that $\mathbf K$ is likewise nonsingular.
Thus $\mathbf K$ has an inverse $\mathbf K^{-1}$.
Hence:
We have:
From Elementary Column Matrix for Inverse of Elementary Column Operation is Inverse, each of ${\mathbf E_i}^{-1}$ is the elementary column matrix corresponding to the inverse $e'_i$ of the corresponding elementary column operation $e_i$.
Let $\Gamma'$ be the column operation composed of the finite sequence of elementary column operations $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$.
Thus $\Gamma'$ is a column operation which transforms $\mathbf B$ into $\mathbf A$.
Hence the result.
== Also see ==
* Row Operation has Inverse
Category:Column Operations
Column Rank of Matrix equals Rank of Matrix

== Theorem ==
Let $\mathbf A$ be a matrix.
The column rank of $\mathbf A$ is equal to the rank of $\mathbf A$.
== Proof ==
The rank of $\mathbf A$ is defined as the dimension of the column space of $\mathbf A$.
Thus the result follows by definition.
== Sources ==
* 
Category:Column Rank
Category:Rank of Matrix
Column Rank of Matrix equals Row Rank

== Theorem ==
Let $\mathbf A$ be a matrix.
The column rank of $\mathbf A$ is equal to the row rank of $\mathbf A$.
== Proof ==
=== Proof Outline and Definitions ===
Recall:
:The row rank of $\mathbf A$ is defined as the dimension of the row space.
:The column rank of $\mathbf A$ is defined as the dimension of the column space.
, we define the rank of $\mathbf A$ to be the column rank of $\mathbf A$.
That is, the columns of a matrix are the rows of its transpose and the rows of a matrix are the columns of its transpose.
Thus, Row Rank of Transpose is Column Rank and of course Column Rank of Transpose is Row Rank:
:$\map r {\mathbf A} = \map c {\mathbf A^\intercal}$
:$\map c {\mathbf A} = \map r {\mathbf A^\intercal}$
where $\map r {\mathbf A}$ and $\map c {\mathbf A}$ denote the row rank and column rank respectively.
Hence this result is equivalent to:
:$\map c {\mathbf A} = \map c {\mathbf A^\intercal}$
where $\mathbf A^\intercal$ denotes the transpose of $\mathbf A$.
== Proof using Orthogonality ==
== Proof using Rank Factorization ==
== Also see ==
* Definition:Rank of Matrix: what this number is called
== Sources ==
* 
* 
* 
* 
Category:Column Rank of Matrix equals Row Rank
Category:Column Rank
Category:Row Rank
Column Rank of Matrix equals Row Rank/Proof using Orthogonality

== Theorem ==
== Proof ==
Let $\map c {\mathbf A}$ denote the column rank of $\mathbf A$.
 
Let $\map r {\mathbf A}$ denote the row rank of $\mathbf A$.
Let $\mathbf A$ be an $m\times n$ matrix whose row rank is $r$.
Therefore, the dimension of the row space of $\mathbf A$ is $r$.
Let $\mathbf x_1, \ldots, \mathbf x_r$ be a basis of the row space of $\mathbf A$.
We claim that the vectors $\mathbf A \mathbf x_1, \ldots, \mathbf A \mathbf x_r$ are linearly independent.
To see why, consider the following linear combination of these vectors with scalar coefficients $c_1, \ldots, c_r$:
:$\mathbf 0 = c_1 \mathbf A \mathbf x_1 + c_2 \mathbf A \mathbf x_2 + \cdots + c_r \mathbf A \mathbf x_r = \mathbf A \paren {c_1 \mathbf x_1 + c_2 \mathbf x_2 + \cdots + c_r \mathbf x_r} = \mathbf A \mathbf v$
where $\mathbf v = c_1 \mathbf x_1 + \cdots + c_r \mathbf x_r$.
We make two observations:
:$(\text a) \mathbf v$ is a linear combination of vectors in the row space of $\mathbf A$, which implies that $\mathbf v$ belongs to the row space of $\mathbf A$
:$(\text b)$ since $\mathbf A \mathbf v = \mathbf 0$, the vector $\mathbf v$ is orthogonal to every row vector of $\mathbf A$ and, hence, is orthogonal to every vector in the row space of $\mathbf A$.
The facts $(\text a)$ and $(\text b)$ together imply that $\mathbf v$ is orthogonal to itself, which proves that $\mathbf v = \mathbf 0$.
Therefore:
:$c_1 \mathbf x_1 + c_2 \mathbf x_2 + \cdots + c_r \mathbf x_r = \mathbf 0$
But recall that $\mathbf x_1, \ldots, \mathbf x_r$ were chosen as a basis of the row space of $\mathbf A$ and so are linearly independent.
This implies that
:$c_1 = \cdots = c_r = 0$
It follows that $\mathbf A \mathbf x_1,\ldots, \mathbf A \mathbf x_r$ are linearly independent.
Now, $\mathbf A \mathbf x_1, \ldots, \mathbf A \mathbf x_r$ are vectors in the column space of $\mathbf A$.
From the above argument, they are also $r$ linearly independent vectors in the column space of $\mathbf A$.
Therefore, the dimension of the column space of $\mathbf A$ (that is, the column rank of $\mathbf A$) must be at least as big as $r$.
This proves that $\map r {\mathbf A} \le \map c {\mathbf A}$, that is, row rank of $\mathbf A$ is no larger than the column rank of $\mathbf A$.
This result applies to any matrix $\mathbf{A}$.
So, we can apply this result to the transpose of $\mathbf A$ to get the reverse inequality:
:$\map r {\mathbf A^\intercal} \le \map c {\mathbf A^\intercal}$
Combining these inequalities, we conclude that row rank equals column rank as follows:
:$\map r {\mathbf A} \le \map c {\mathbf A} = \map r {\mathbf A^\intercal} \le \map c {\mathbf A^\intercal} = \map r {\mathbf A}$
== Sources ==
* 
Category:Column Rank of Matrix equals Row Rank
Column Rank of Matrix equals Row Rank/Proof using Rank Factorization

== Theorem ==
== Proof ==
Let $\map c {\mathbf A}$ denote the column rank of $\mathbf A$.
 
Let $\map r {\mathbf A}$ denote the row rank of $\mathbf A$.
Let $\mathbf A$ be an $m \times n$ matrix whose column rank is $r$.
Therefore, the dimension of the column space of $\mathbf A$ is $r$.
Let $\mathbf c_1, \ldots, \mathbf c_r$ be any basis for the column space of $\mathbf A$.
Place these vectors as column vectors to form the $m \times r$ matrix $\mathbf C = \begin {bmatrix} \mathbf c_1 & \mathbf c_2 & \cdots & \mathbf c_r \end{bmatrix}$.
Therefore, every column vector of $\mathbf A$ is a linear combination of the columns of $\mathbf C$.
To be precise, if $\mathbf A = \begin {bmatrix} \mathbf a_1 & \mathbf a_2 & \cdots & \mathbf a_n \end {bmatrix}$ is an $m \times n$ matrix with $\mathbf a_j$ as the $j$-th column, then: 
:$\mathbf a_j = f_{1j} \mathbf c_1 + f_{2j} \mathbf c_2 + \cdots + f_{rj} \mathbf c_r \text { for } j = 1, \ldots, n$
where $f_{ij}$ are the scalar coefficients of $\mathbf a_j$ in terms of the basis $\mathbf c_1, \mathbf c_2, \ldots, \mathbf c_r$.
This implies that:
:$\mathbf A = \mathbf C \mathbf F$
where $f_{ij}$ is the $\tuple {i, j}$th element of $\mathbf F$. 
Using transposition of products, we note that $\mathbf A^\intercal = \mathbf F^\intercal \mathbf C^\intercal$.
From the definition of matrix multiplication, this means that each column of $\mathbf A^\intercal$ is a linear combination of the columns of $\mathbf F^\intercal$.
Therefore, the column space of $\mathbf A^\intercal$ is a subspace of the column space of $\mathbf F^\intercal$.
Hence:
:$\map c {\mathbf A^\intercal} \le \map c {\mathbf F^\intercal}$
Now, $\mathbf F^\intercal$ is $n \times r$
Thus there are $r$ columns in $\mathbf F^\intercal$.
Hence:
:$\map c {\mathbf A^\intercal} \le r = \map c {\mathbf A}$
This proves that $\map c {\mathbf A^\intercal} \le \map c {\mathbf A}$.
Now apply the result to the matrix $\mathbf A^\intercal$ to obtain the reverse inequality:
Since $\paren {\mathbf A^\intercal}^\intercal = \mathbf A$, we can write:
:$\map c {\mathbf A} = \map c {\paren {\mathbf A^\intercal}^\intercal} \le \map c {\mathbf A^\intercal}$
This proves $\map c {\mathbf A} \le \map c {\mathbf A^\intercal}$.
We have, therefore, proved:
:$\map c {\mathbf A^\intercal} \le \map c {\mathbf A}$
and the reverse inequality:
:$\map c {\mathbf A} \le \map c {\mathbf A^\intercal}$
Therefore:
:$\map c {\mathbf A} = \map c {\mathbf A^\intercal}$
== Sources ==
* 
Category:Column Rank of Matrix equals Row Rank
Columns of Pascal's Triangle contain Simplicial Polytopic Numbers

== Theorem ==
The columns of Pascal's triangle contain the simplicial polytopic numbers:
: Column $0$: repeated instances of number $1$
: Column $1$: the (strictly) positive integers
: Column $2$: the triangular numbers
: Column $3$: the tetrahedral numbers
: Column $4$: the pentatope numbers
and so on.
== Proof ==
== Sources ==
* : $35$
* : $35$
Category:Pascal's Triangle
Category:Simplicial Polytopic Numbers
Combination Theorem for Bounded Continuous Real-Valued Functions

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded continuous real-valued functions.
Let $\lambda \in \R$.
Then the following results hold.
=== Sum Rule ===
=== Negation Rule ===
=== Difference Rule ===
=== Product Rule ===
=== Multiple Rule ===
=== Absolute Value Rule ===
=== Maximum Rule ===
=== Minimum Rule ===
Category:Combination Theorem for Bounded Continuous Real-Valued Functions 
Category:Continuous Real-Valued Functions
Category:Bounded Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Absolute Value Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded continuous real-valued functions.
Let $\size f : S \to \R$ denote the absolute value of $f$, that is, $\size f$ denotes the mapping defined by:
:$\forall s \in S : \map {\size f} s = \size{\map f s}$
 
Then:
:$\size f$ is a bounded continuous real-valued function
== Proof ==
Follows from:
* Absolute Value Rule for Bounded Real-Valued Function
* Absolute Value Rule for Continuous Real-Valued Function
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Difference Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded contiuous real-valued functions.
Let $f - g : S \to \R$ be the pointwise difference of $f$ and $g$, that is, $f - g$ is the mappping defined by:
:$\forall s \in S : \map {\paren{f - g} } s = \map f s - \map g s$
Then:
:$f - g$ is a bounded coninuous real-valued function
== Proof ==
Follows from:
* Difference Rule for Bounded Real-Valued Functions
* Difference Rule for Continuous Real-Valued Functions
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Maximum Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded continuous real-valued functions.
Let $f \vee g$ denote the pointwise maximum of $f$ and $g$, that is, $f \vee g$ is the mapping defined by:
:$\forall s \in S : \map {\paren{f \vee g} } s = \max \set{\map f s, \map g s}$
Then:
:$f \vee g$ is a bounded continuous real-valued function
== Proof ==
Follows from:
* Maximum Rule for Bounded Real-Valued Functions
* Maximum Rule for Continuous Real-Valued Functions
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Minimum Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded continuous real-valued functions.
Let $f \wedge g$ denote the pointwise minimum of $f$ and $g$, that is, $f \wedge g$ is the mapping defined by:
:$\forall s \in S : \map {\paren{f \wedge g} } s = \min \set{\map f s, \map g s}$
Then:
:$f \wedge g$ is a bounded continuous real-valued function
== Proof ==
Follows from:
* Minimum Rule for Bounded Real-Valued Functions
* Minimum Rule for Continuous Real-Valued Functions
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Multiple Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f :S \to \R$ be bounded contiuous real-valued function.
Let $\lambda \in \R$.
Let $\lambda f : S \to \R$ be the pointwise scalar multiplication of $f$ by $\lambda$, that is, $\lambda f$ is the mappping defined by:
:$\forall s \in S : \map {\paren{\lambda f} } s = \lambda \map f s$
Then:
:$\lambda f$ is a bounded continuous real-valued function
== Proof ==
Follows from:
* Multiple Rule for Bounded Real-Valued Function
* Multiple Rule for Continuous Real-Valued Function
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Negation Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f :S \to \R$ be a bounded continuous real-valued function.
Let $-f : S \to \R$ denote the pointwise negation of $f$, that is, $-f$ denotes the mapping defined by:
:$\forall s \in S : \map {\paren{-f} } s = - \map f s$
Then:
:$-f$ is a bounded continuous real-valued function
== Proof ==
Follows from:
* Negation Rule for Bounded Real-Valued Function
* Negation Rule for Continuous Real-Valued Function
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Product Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded continuous real-valued functions.
Let $f g : S \to \R$ be the pointwise product of $f$ and $g$, that is, $f g$ is the mappping defined by:
:$\forall s \in S : \map {\paren{f g} } s = \map f s \map g s$
Then:
:$f g$ is a bounded continuous real-valued function
== Proof ==
Follows from:
* Product Rule for Bounded Real-Valued Functions
* Product Rule for Continuous Real-Valued Functions
{qed}}
Category:Combination Theorem for Bounded Continuous Real-Valued Functions
Combination Theorem for Bounded Continuous Real-Valued Functions/Sum Rule

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $f, g :S \to \R$ be bounded contiuous real-valued functions.
Let $f + g : S \to \R$ be the pointwise addition of $f$ and $g$, that is, $f + g$ is the mappping defined by:
:$\forall s \in S : \map {\paren{f + g} } s = \map f s + \map g s$
Then:
:$f + g$ is a bounded coninuous real-valued function
== Proof ==
Follows from:
* Sum Rule for Bounded Real-Valued Functions
* Sum Rule for Continuous Real-Valued Functions
Category:Combination Theorem for Bounded Continuous Real-Valued Functions