nyuuzyou/wikis huggingface dataset:
Set of 2-Dimensional Indefinite Real Orthogonal Matrices is not Compact in Normed Real Square Matrix Vector Space

== Theorem ==
Let $\struct {\R^{2 \times2}, \norm {\, \cdot \,}_\infty}$ be the normed real matrix vector space.
Let $\map O {1, 1} := \set {\mathbf R \in \R^{2 \times2} : \mathbf R^\intercal \mathbf J_{1,1} \mathbf R = \mathbf J_{1,1}}$ be the indefinite orthogonal group of degree $\paren {1, 1}$ over real numbers where:
$\quad \mathbf J_{1, 1} := \begin{bmatrix}
1 & 0\\
0 & -1 \\
\end{bmatrix}$
Then $\map O {1, 1}$ is not a compact set in $\struct {\R^{2 \times 2}, \norm {\, \cdot \,}_\infty}$.
== Proof ==
Let:
$\quad \begin{bmatrix}
\map \cosh t& \map \sinh t\\
\map \sinh t & \map \cosh t \\
\end{bmatrix} := \map {\mathbf R} t$
We have that:
Hence, $\map {\mathbf R} t \in \map O {1, 1}$.
From difference of squares of hyperbolic cosine and sine:
The set of matrix elements constitutes a finite subset of the set of real numbers which is ordered.
By Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements, there is a maximal element.
Consider the supremum norm of $\map {\mathbf R} t$:
Furthermore:
:$\ds \lim_{t \mathop \to \infty} \map \cosh t = \infty$
Hence, $\map O {1, 1}$ is not bounded.
By the Heine-Borel theorem, $\map O {1, 1}$ is not compact in $\struct {\R^{2 \times 2}, \norm {\, \cdot \,}_\infty}$.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Matrix Groups
Category:Compact Spaces
Set of 3 Integers each Divisor of Sum of Other Two

== Theorem ==
There exists exactly one set of distinct coprime positive integers such that each is a divisor of the sum of the other two:
:$\set {1, 2, 3}$
== Proof ==
We note that if $\set {a, b, c}$ is such a set, then $\set {k a, k b, k c}$ satisfy the same properties trivially.
Hence the specification that $\set {a, b, c}$ is a coprime set.
We have that:
:$5 \times 1 = 2 + 3$ so $1 \divides 2 + 3$
:$2 \times 2 = 1 + 3$ so $2 \divides 1 + 3$
:$1 \times 3 = 1 + 2$ so $3 \divides 1 + 2$
It remains to be shown that this is the only such set.
We are to find all the sets $\set {a, b, c}$ such that:
:$a \divides b + c$
:$b \divides a + c$
:$c \divides a + b$
where $a \ne b$, $a \ne c$ and $b \ne c$.
, suppose $a < b < c$.
Since $2 c > a + b > 0$ and $c \divides a + b$, we must have $a + b = c$.
Then it follows that:
:$a \divides \paren {b + a + b}$
:$b \divides \paren {a + a + b}$
which reduces to:
:$a \divides 2 b$
:$b \divides 2 a$
Suppose $b$ is odd.
Then by Euclid's Lemma, we would have $b \divides a$.
By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction.
Thus $b$ is even.
Suppose $a$ is even.
Then $a, b, c$ are all even.
So $\gcd \set {a, b, c} \ne 1$, which is a contradiction.
Therefore it must be the case that $a$ is odd.
Then by Euclid's Lemma, we have:
:$a \divides \dfrac b 2$
and:
:$\dfrac b 2 \divides a$
By Absolute Value of Integer is not less than Divisors, this gives:
:$\dfrac b 2 = a$
Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$.
Hence the set $\set {1, 2, 3}$ is obtained.
== Sources ==
* : $6$
* : $6$
Category:Divisors
Set of 3 Integers each Divisor of Sum of Other Two/Mistake

== Source Work ==
:
:The Dictionary
:: $6$
:
:The Dictionary
:: $6$
== Mistake ==
:$1$, $2$, $3$ is also the only set of $3$ integers such that each divides the sum of the other two.
== Correction ==
The statement should be:
:$1$, $2$, $3$ is also the only coprime set of $3$ distinct integers such that each divides the sum of the other two.
First it is noted that there are two trivial such sets: $\set {1, 1, 1}$ and $\set {1, 1, 2}$.
However, as it specifies set of $3$ integers, and sets by convention contain distinct elements only, the issue is merely one of clarification.
On the other hand, if $\set {a, b ,c}$ is such a set, then $\set {k a, k b, k c}$ is also such a set, which definitely requires the set to be coprime.
== Sources ==
* : $6$
* : $6$
Category:Mistakes/Curious and Interesting Numbers
Set of 5 Triplets whose Sums and Products are Equal

== Theorem ==
The following set of $5$ triplets of integers have the property that:
:the sum of the integers in each triplet are equal
and:
:the product of the integers in each triplet are equal:
:$\tuple {6, 480, 495}$, $\tuple {11, 160, 810}$, $\tuple {12, 144, 825}$, $\tuple {20, 81, 880}$, $\tuple {33, 48, 900}$
The sum is $981$, and the product is $1 \, 425 \, 600$.
This is the only known such set of $5$ triplets of integers with this property.
== Proof ==
== Sources ==
* : $981$
* : $981$
Category:Recreational Mathematics
Category:981
Category:1,425,600
Set of 7 Anagrams which are Square

== Theorem ==
The following integers are all anagrams, and all square:
:$1 \, 048 \, 576, 1 \, 056 \, 784, 1 \, 085 \, 764, 5 \, 740 \, 816, 5 \, 764 \, 801, 6 \, 754 \, 801, 7 \, 845 \, 601$
== Proof ==
== Sources ==
* : $1,048,576$
Category:Square Numbers
Set of Affine Mappings on Real Line under Composition forms Group

== Theorem ==
Let $S$ be the set of all real functions $f: \R \to \R$ of the form:
:$\forall x \in \R: \map f x = r x + s$
where $r \in \R_{\ne 0}$ and $s \in \R$
Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.
Then $\struct {S, \circ}$ is a group.
== Proof ==
We note that $S$ is a subset of the set of all real functions on $\R$.
From Set of all Self-Maps under Composition forms Semigroup, we have that $\circ$ is associative.
Consider the real function $I: \R \to \R$ defined as:
:$\forall x \in \R: \map I x = 1 \times x + 0$
We have that:
:$I \in S$
:$I$ is the identity mapping.
So $S$ is not empty.
Then we note:
Let $f, g \in S$ such that:
:$\map f x = r_1 x + s_1$
:$\map g x = r_2 x + s_2$
For all $x \in \R$, we have:
This demonstrates that $\struct {S, \circ}$ is closed.
Let $\phi: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map \phi x = p x + q$
where $p \in \R_{\ne 0}$ and $q \in \R$.
Then:
Thus we define $\psi: \R \to \R$ be the real function defined as:
:$\forall y \in \R: \map \psi y = \dfrac 1 p y + \paren {-\dfrac q p}$
where $\dfrac 1 p \in \R_{\ne 0}$ and $-\dfrac q p \in \R$.
It is seen that $\psi \in S$.
For all $x \in \R$, we have:
Hence:
:$\phi \circ \psi = I$
and so $\psi$ is the inverse of $\phi$.
We have then that $\struct {S, \circ}$ is closed under inversion.
The result follows by the Two-Step Subgroup Test.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{B iv}$
Category:Affine Geometry
Set of All Completely Irreducible Elements is Smallest Order Generating

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.
Then $\map {\operatorname {Irr} } L$ is order generating and
:$\forall X \subseteq S: X$ is order generating $\implies \map {\operatorname {Irr} } L \subseteq X$
where $\map {\operatorname {Irr} } L$ denotes the set of all completely irreducible elements of $L$.
== Proof ==
By Not Preceding implies Exists Completely Irreducible Element in Algebraic Lattice:
:$\forall x, y \in S: y \npreceq x \implies \exists p \in \map {\operatorname {Irr} } L: p \preceq x \land p \npreceq y$
Thus by Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding:
:$\map {\operatorname {Irr} } L$ is order generating.
Let $X \subseteq S$ such that
:$X$ is order generating.
Let $x \in \map {\operatorname {Irr} } L$.
By definition of $\operatorname {Irr}$:
:$x$ is completely irreducible.
Thus by Order Generating Subset Includes Completely Irreducible Elements:
:$x \in X$
== Sources ==
* 
* 
Category:Meet Irreducible Elements
Category:Order Generating Subsets
Set of All Mappings is Small Class

== Theorem ==
Let $S$ and $T$ be small classes.
It follows that the set of all mappings $S^T$ is a small class.
== Proof ==
The set of all mappings $S^T$ is equal to the collection of all mappings $f : S \to T$.
Each of these mappings $f$ is a subset of $S \times T$.
Thus, $S^T \subseteq \powerset {S \times T}$.
Therefore, by Cartesian Product is Small and the axiom of powers, $S^T$ is a small class.
== Sources ==
* : $\S 10.48$
Category:Cardinals
Category:Zermelo-Fraenkel Class Theory
Set of All Mappings of Cartesian Product

== Theorem ==
Let $R$, $S$, and $T$ be sets.
Then:
:$R^{S \times T} \sim \paren {R^S}^T$
where $R^{S \times T}$ denotes the set of all mappings from $S \times T$ to $R$.
== Proof ==
Define the mapping $F: \paren {R^S}^T \to R^{S \times T}$ as follows:
:$\map {\map F f} {x, y} = \map {\paren {\map f x} } y$ for all $x \in S , y \in T$.
Suppose $\map F {f_1} = \map F {f_2}$.
Then $\map {\paren {\map {f_1} x} } y = \map {\paren {\map {f_2} x} } y$ for all $x \in S , y \in T$ by the definition of $F$.
Therefore, $\map {f_1} x = \map {f_2} x$ and $f_1 = f_2$ by Equality of Mappings.
It follows that $F$ is an injection.
Take any $g \in R^{S \times T}$.
Define a function $f$ as $\map {\paren {\map f x} } y = \map g {x, y}$.
It follows that:
Therefore, $F$ is a surjection.
Thus, $F$ is a bijection.
It follows that $ R^{S \times T} \sim \paren {R^S}^T$ by the definition of set equivalence.
== Sources ==
* : $\S 10.50$
Category:Cardinals
Category:Cardinality
Set of All Real Intervals is Semiring of Sets

== Theorem ==
Let $\mathbb S$ be the set of all real intervals.
Then $\mathbb S$ is a semiring of sets, but is not a ring of sets.
== Proof ==
Consider the types of real interval that exist.
In the following, $a, b \in \R$ are real numbers.
There are:
* Closed intervals:
** $\closedint a b = \set {x \in \R: a \le x \le b}$
* Open intervals:
** $\openint a b = \set {x \in \R: a < x < b}$
* Half-open intervals:
** $\hointr a b = \set {x \in \R: a \le x < b}$
** $\hointl a b = \set {x \in \R: a < x \le b}$
* Unbounded closed and unbounded open intervals:
** $\hointr a {+\infty} = \set {x \in \R: a \le x}$
** $\hointl {-\infty} a = \set {x \in \R: x \le a}$
** $\openint a {+\infty} = \set {x \in \R: a < x}$
** $\openint {-\infty} a = \set {x \in \R: x < a}$
** $\openint {-\infty} {+\infty} = \R$
* The empty interval:
** $\openint a a = \set {x \in \R: a < x < a} = \O$
* Singleton intervals:
** $\closedint a a = \set {x \in \R: a \le x \le a} = \set a$
=== Set of Intervals is Not a Ring ===
Consider, for example, an open interval:
: $\openint a b = \set {x \in \R: a < x < b}$
such that $a < b$.
Any subset $\openint c d \subset \openint a b$ such that $a < c, d < b$ is such that:
:$\openint a b \setminus \openint c d = \hointl a c \cup \hointr d b$
But $\hointl a c \cup \hointr d b$ is not in itself a real interval, and therefore is not an element of $\mathbb S$.
Hence $\mathbb S$ is not closed under the operation of set difference and is therefore not a ring of sets.
=== Set of Intervals is a Semiring ===
It is tedious but straightforward to examine each type of interval, and pass it through the same sort of exhaustive examination as follows.
We will take a general half-open interval:
: $A = \hointr a b$
and note that the argument generalizes.
Let $c, d \in \R: a \le c < d \le b$.
Then $C = \openint c d$ is a subset of $\hointr a b$.
There are four cases:
* $a < c, d < b$:
Then:
: $A = \closedint a c \cup \openint c d \cup \hointr d b$
and it can be seen that this is a partition of $A$.
* $a = c, d < b$:
Then:
: $A = \set a \cup \openint a d \cup \hointr d b$
and it can be seen that this is a partition of $A$.
* $a < c, d = b$:
Then:
and it can be seen that this is a partition of $A$.
* $a = c, d = b$:
Then:
and it can be seen that this is a partition of $A$.
The same technique can be used to generate a finite expansion of any interval of $\R$ from any subset of that interval.
Hence $\mathbb S$ is a semiring of sets.
Category:Semirings of Sets
Category:Analysis
Set of All Relations is a Monoid

== Theorem ==
The set of all relations $\Bbb E = \set {\RR: \RR \subseteq S \times S}$ on a set $S$ forms a monoid in which the operation is composition of relations.
The only invertible elements of $\Bbb E$ are permutations.
If $\RR$ is invertible, its inverse is $\RR^{-1}$.
== Proof ==
=== Closure ===
A relation followed by another relation is another relation, from the definition of composition of relations.
As the domain and codomain of two relations are the same, then:
:$\forall \RR_1, \RR_2 \subseteq S \times S: \RR_1 \circ \RR_2 \subseteq S \times S$
Therefore composition of relations on a set is closed.
=== Associativity ===
Composition of Relations is Associative.
=== Identity ===
From Identity Mapping is Left Identity and Identity Mapping is Right Identity we have that:
:$\forall \RR \subseteq S \times S: \RR \circ I_S = \RR = I_S \circ \RR$
Hence the identity mapping is the identity element of this set of relations.
It is closed and associative, so it is a semigroup. It has an identity element, so it is a monoid.
=== Inverses ===
Now if $\RR \in \Bbb E$ were to be invertible, it would need to satisfy:
:$\RR^{-1} \circ \RR = I_S$
and:
:$\RR \circ \RR^{-1} = I_S$
From Inverse Relation is Left and Right Inverse iff Bijection this can only happen if $\RR$ is a bijection on $S$, that is, a permutation, and its inverse is then $\RR^{-1}$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(g)}$
Category:Relation Theory
Category:Examples of Monoids
Set of Associating Elements forms Subsemigroup of Magma

== Theorem ==
Let $S$ be a set.
Let $\oplus$ be an operation on $S$ such that $\struct {S, \oplus}$ is a magma.
Let $T \subseteq S$ be the subset of $S$ defined as:
:$T = \set {x \in S: \paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z} }$
Suppose $T \ne \O$.
Then $\struct {T, \oplus {\restriction_T} }$ is a subsemigroup of $\struct {S, \oplus}$.
== Proof ==
Taking the semigroup axioms in turn:
=== ===
Because $T$ consists only of elements $x$ such that $\paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z}$, it follows that $\oplus$ is associative on $T$.
That is, $\struct {T, \oplus {\restriction_T} }$ is associative.
=== ===
Let $x, y \in T$.
Let $a, b \in T$.
Then:
Thus $x \oplus y \in T$ and so $\struct {T, \oplus {\restriction_T} }$ is closed.
The semigroup axioms are thus seen to be fulfilled, and so $\struct {T, \oplus {\restriction_T} }$ is a semigroup.
The result follows by definition of subsemigroup.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.8$
Category:Subsemigroups
Category:Associativity
Set of Cardinality not Greater than Cardinality of Finite Set is Finite

== Theorem ==
Let $X, Y$ be sets such that
:$\card X \le \card Y$
and
:$Y$ is finite,
where $\card X$ denotes the cardinality of $X$.
Then $X$ is finite.
== Proof ==
By Finite iff Cardinality Less than Aleph Zero:
:$\card Y < \aleph_0$
Then
:$\card X < \aleph_0$
Thus by Finite iff Cardinality Less than Aleph Zero:
:$X$ is a finite set.
== Sources ==
* 
Category:Cardinals
Set of Chains is Closed under Chain Unions under Subset Relation

== Theorem ==
Let $\struct {P, \le}$ be an ordered set.
Let $\map {\operatorname {ch} } P$ be the set of chains in $P$.
Then $\map {\operatorname {ch} } P$ is closed under chain unions.
That is, if $\CC$ is a chain in $\map {\operatorname {ch} } P$ with respect to the subset ordering, the union $\ds \bigcup \CC$ of $\CC$ is also a chain in $P$.
== Proof ==
Let $\ds D = \bigcup \CC$. 
Observe that any $C \in \CC$ is a chain in $P$, hence $C \subseteq P$.
Therefore, $D \subseteq P$, by Set Union Preserves Subsets.
Now suppose that $a, b \in D$.
By definition of union, there exist $A, B \in \CC$ such that $a \in A$ and $b \in B$. 
Since $\CC$ is a chain in $\struct {\map {\operatorname {ch} } P, \subseteq}$, either $A \subseteq B$ or $B \subseteq A$.
Hence either $a \in B$ or $b \in A$.
Both $A$ and $B$ are chains in $P$, and thence $a \le b$ or $b \le a$.
Therefore, $D$ is totally ordered by $\le$.
That is, it is a chain in $P$.
Category:Order Theory
Category:Closure under Chain Unions
Category:Subset Relation
Set of Closed Elements wrt Closure Operator under Subset Operation is Complete Lattice

== Theorem ==
Let $S$ be a set.
Let $\cl$ be a closure operator on the power set $\powerset S$ of $S$.
Let $\mathscr C$ be the set of all subsets $T$ of $S$ such that:
:$\map \cl T = T$
Then the algebraic structure $\struct {\mathscr C, \subseteq}$ forms a complete lattice.
== Proof ==
Recall the closure axioms:
First we note that from we have that:
:$\map \cl S = S$
and so:
:$S \in \mathscr C$
Let $\AA \subseteq \mathscr C$.
Thus $\AA$ is a set of subsets $T$ of $S$ for all of which $\map \cl T = T$.
From Intersection of Closed Sets is Closed:
:$\ds \cap \AA \in \mathscr C$
The result follows from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice,
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.25$
Category:Examples of Complete Lattices
Category:Closure Operators
Set of Closed Subsets of Power Structure of Entropic Structure is Closed

== Theorem ==
Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure.
Let $\struct {\powerset S, \odot_\PP}$ be the power structure of $\struct {S, \odot}$.
Let $\TT$ be the set of all submagmas of $\struct {S, \odot}$.
Then the algebraic structure $\struct {\TT, \odot_\PP}$ is a submagma of $\struct {\powerset S, \odot_\PP}$ which is itself an entropic structure.
== Proof ==
Recall the definition of subset product:
:$A \odot_\PP B = \set {a \odot b: \tuple {a, b} \in A \times B}$
First we show that:
:$\forall A, B \in \TT: A \odot_\PP B \in \TT$
Let $A$ and $B$ be arbitrary elements of $\TT$.
Let $a$ and $c$ be arbitrary elements of $A$.
Let $b$ and $d$ be arbitrary elements of $B$.
Then we have:
Then:
That is:
:$A \odot_\PP B$ is closed in $S$.
As $A$ and $B$ are arbitrary, it follows that $\struct {\TT, \odot_\PP}$ is closed in $\powerset S$.
Then we need to show that:
:$\forall A, B, C, D \in \TT: \paren {A \odot_\PP B} \odot_\PP \paren {C \odot_\PP D} = \paren {A \odot_\PP C} \odot_\PP \paren {B \odot_\PP D}$
demonstrating the entropic nature of $\struct {\TT, \odot_\PP}$.
We have:
and the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(f)}$
Category:Entropic Structures
Category:Power Structures
Set of Codes for URM Programs is Primitive Recursive

== Theorem ==
Let $\operatorname{Prog}$ be the set of all code numbers of URM programs.
Then $\operatorname{Prog}$ is a primitive recursive set.
== Proof ==
A natural number $n$ codes a URM program it codes a sequence of positive integers which are the code numbers of URM instructions.
Suppose $n$ codes such a sequence.
Then $\map \len n$ is the number of terms in this sequence, where $\map \len n$ is the length of $n$.
Also, for $1 \le j \le \map \len n$, $\paren n_j$ is the exponent of the $j$th prime in the prime decomposition of $n$.
So $\paren n_j$ is the $j$th number in the sequence coded by $n$.
So:
:$n \in \operatorname {Prog}$ $\map {\chi_{\operatorname {Seq} } } n = 1$ and $\paren n_j \in \operatorname{Instr}$ for $1 \le j \le \map \len n$
where $\chi_{\operatorname {Seq} }$ is the characteristic function of the set of code numbers of finite sequences of positive integers.
Now $\paren n_j \in \operatorname{Instr} \iff \map {\chi_{\operatorname {Instr} } } {\paren n_j} = 1$.
So $\paren n_j \in \operatorname{Instr}$ for $1 \le j \le \map \len n$ $\map {\chi_{\operatorname {Instr} } } {\paren n_j} = 1$ for $1 \le j \le \map \len n$.
This is the case :
:$\ds \prod_{j \mathop = 1}^{\map \len n} = 1$
Thus $n \in \operatorname{Prog}$ :
:$\ds \map {\chi_{\operatorname {Seq} } } n \times \prod_{j \mathop = 1}^{\map \len n} = 1$
Now we define the function $g: \N^2 \to \N$ by:
:$\map g {n, z} = \begin{cases}
1 & : z = 0 \\
\ds \prod_{j \mathop = 1}^z \map {\chi_{\operatorname {Instr} } } {\paren n_j} & : \text {otherwise}
\end{cases}$
We use $g$ to obtain the characteristic function of the set $\operatorname {Prog}$:
:$\map {\chi_{\operatorname {Prog} } } n = \map {\chi_{\operatorname {Seq} } } n \map g {n, \map \len n}$
(We need to introduce $g$ to ensure $\chi_{\operatorname{Prog}}$ is defined if $\map \len n = 0$.)
Now we have that:
:$\paren n_j$ is primitive recursive;
:$\operatorname {Instr}$ is primitive recursive.
So $g$ is therefore primitive recursive, as it is obtained by substitution from these.
Therefore $\chi_{\operatorname {Prog}}$ is primitive recursive as it is obtained by substitution from:
:the primitive recursive function $\operatorname {mult}$
:the primitive recursive function $\chi_{\operatorname {Seq} }$
:the primitive recursive function $g$
:the primitive recursive function $\len$.
Hence $\operatorname {Prog}$ is a primitive recursive set.
Category:URM Programs
Category:Primitive Recursive Functions
Set of Column Matrices under Entrywise Addition forms Vector Space

== Corollary to Set of Matrices under Entrywise Addition forms Vector Space ==
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_\GF} {m, 1}$ be set of all column matrices over a field $\GF$.
Let $\struct {\map {\MM_\GF} {m, 1}, +}$ denote the abelian group formed from $\map {\MM_\GF} {m, 1}$ under matrix entrywise addition.
Let $\struct {\map {\MM_\GF} {m, 1}, +, \times}_\GF$ denote the unitary module over $\GF$ where $\times$ denotes the matrix scalar product.
Then $\struct {\map {\MM_\GF} {m, 1}, +, \times}_\GF$ forms a vector space.
== Proof ==
This is an instance of Set of Matrices under Entrywise Addition forms Vector Space with $n = 1$.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $5$
Category:Set of Matrices under Entrywise Addition forms Vector Space
Category:Vector Spaces of Matrices
Category:Column Matrices
Set of Common Divisors of Integers is not Empty

== Theorem ==
Let $a, b \in \Z$ be integers.
Let $S$ be the set of common divisors of $a$ and $b$.
Then $S$ is not empty.
== Proof ==
From One is Common Divisor of Integers:
:$1$ is a common divisor of $a$ and $b$.
Thus, whatever $a$ and $b$ are:
:$1 \in S$
The result follows by definition of empty set.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Divisors
Set of Condensation Points is Monotone

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$A \subseteq B \implies {A^0} \subseteq B^0$
where
:$A^0$ denotes the set of condensation points of $A$
== Proof ==
Assume
:$A \subseteq B$
Let $x \in A^0$.
By definition of set of condensation points:
:$x$ is condensation point of $A$
By definition of condensation point:
:$x$ is limit point of $A$ such that $\forall U \in \tau: A \cap U$ is uncountable
Thus by Limit Point of Subset is Limit Point of Set:
:$x$ is limit point of $B$
Let $U \in \tau$.
By definition of condensation point:
:$A \cap U$ is uncountable
By Set Intersection Preserves Subsets/Corollary:
:$A \cap U \subseteq B \cap U$
Thus by Subset of Countable Set is Countable
:$B \cap U$ is uncountable
Then by definition:
:$x$ is condensation point of $B$
Thus by definition of set of condensation points:
:$x \in B^0$
== Sources ==
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Category:Condensation Points
Set of Condensation Points is Subset of Derivative

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:$A^0 \subseteq A'$
where
:$A^0$ denotes the set of condensation points of $A$
:$A'$ denotes the derivative of $A$
== Proof ==
Let $x \in A^0$.
By definition of set of condensation points:
:$x$ is condensation point of $A$
By definition of condensation point:
:$x$ is limit point of $A$
By definition of derived set:
:$x \in A'$
== Sources ==
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Category:Condensation Points
Category:Set Derivatives
Set of Condensation Points of Countable Set is Empty

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:if $A$ is countable,
:then $A^0 = \O$.
=== Lemma ===
== Proof ==
Assume
:$A$ is countable.
:$A^0 \ne \O$
By definition of empty set:
:$\exists x: x \in A^0$
Then by definition of set of condensation points:
:$x$ is a condensation point of $A$.
This contradicts Lemma.
Thus the result follows by Proof by Contradiction.
== Sources ==
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Category:Condensation Points