nyuuzyou/wikis huggingface dataset:
Existence of Arbitrarily Long Aliquot Sequences

== Theorem ==
It is possible to construct an aliquot sequence of arbitrary length which is monotonically increasing.
== Proof ==
Let $p_i$ be the $i$th prime.
Construct the sequence $\sequence {t_i}_{t \mathop = 1}^\infty$ of natural numbers $t_i$ recursively as follows:
:$t_1 = 2$
:$t_{i + 1} = \map \phi {p_i^{t_i + 1} } - 1 = p_i^{t_i} \paren {p_i - 1} - 1$ for $i \ge 1$
We claim that this sequence has the property:
:$(1): \quad p_i^{t_i + 1} \divides \map {\sigma_1} {p_{i + 1}^{t_{i + 1} } }$ for $i \ge 1$
For $i = 1$, $p_1 = 2$ and $t_1 = 2$.
:$t_2 = 2^2 \paren {2 - 1} - 1 = 3$
:$\map {\sigma_1} {p_2^{t_2} } = \map {\sigma_1} {3^3} = 40$
which is a multiple of $2^3 = 8$.
For $i > 1$:
For $i > 1$, $p_{i + 1} - 1 > 2$.
Hence $p_{i + 1} - 1$ is even and composite.
Since $p_i$ is odd and $p_i > \dfrac {p_{i + 1} - 1} 2$ by Bertrand-Chebyshev Theorem:
:$p_{i + 1} - 1$ and $p_i^{t_i + 1}$ are coprime.
By Euclid's Lemma, $p_i^{t_i + 1} \divides \map {\sigma_1} {p_{i + 1}^{t_{i + 1} } }$.
We define for $l \ge 1$: $A_l = \set {m \in \N: p_i^{t_i} \divides m, p_i^{t_i + 1} \nmid m, \forall i: 1 \le i \le l}$.
Denote $\map s n$ the aliquot sum of $n$.
Then for $m \in A_l$ where $l \ge 2$, we claim:
Let $m \in A_l$ where $l \ge 2$.
From definition we have:
Hence $m$ is a multiple of $27 \times 4 = 108$.
Since $108$ is abundant, $m$ is also abundant by Multiple of Abundant Number is Abundant.
$\map s m > m$ follows from definition of abundant numbers.
Write $m = p_1^{t_1} \cdots p_l^{t_l} k$, where $k$ is not divisible by $p_1, \dots, p_l$.
Then:
Since $K p_1 \cdots p_{l - 1}$ is divisible by $p_1, \dots, p_{l - 1}$ while $p_l^{t_l} k$ isn't:
:$p_1, \dots, p_{l - 1} \nmid \paren {K p_1 \cdots p_{l - 1} - p_l^{t_l} k}$
Hence for each $i$ where $1 \le i \le l - 1$:
:$p_i^{t_i} \divides \map s m, p_i^{t_i + 1} \nmid \map s m$
This shows that $\map s m \in A_{l - 1}$.
Hence to create an aliquot sequence of length $l$, one picks an element $m$ from $A_l$.
Applying $(2)$ and $(3)$ repeatedly, we see that:
:$m, \map s m, \map s {\map s m}, \dots, \map {s^l} m$
is a monotonically increasing aliquot sequence of length $l$.
== Historical Note ==
== Sources ==
* 
* : $276$
* : $276$
Category:Aliquot Sequences
Existence of Banach Limits

== Theorem ==
Let $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$ be the normed vector space of bounded sequences on $\R$. 
Let $\struct {\paren {\map {\ell^\infty} \R}^\ast, \norm \cdot_{\paren {\ell^\infty}^\ast} }$ be the normed dual space of $\struct {\map {\ell^\infty} \R, \norm \cdot_\infty}$.
Then there exists a Banach limit $L \in \paren {\map {\ell^\infty} \R}^\ast$.
== Proof ==
Let $\map c \R$ be the set of the convergent real sequences.
Then:
:$\map c \R \subseteq \map {\ell^\infty} \R$
is a linear subspace in view of:
:Convergent Real Sequence is Bounded
:Linear Combination of Convergent Sequences in Topological Vector Space is Convergent 
Define a mapping $f_0 : \map c \R \to \R$ by:
:$\ds \map {f_0} {\sequence {x_n} } := \lim _{n \mathop \to \infty} x_n$
=== Lemma 1 ===
$f_0 \in {\map c \R}^\ast$ such that:
:$\norm {f_0}_{ {\map c \R}^\ast} = 1$
and:
:$\map {f_0} {\mathbf 1} = 1$
=== Proof of Lemma 1 ===
First, $f_0$ is linear, since Infinite Limit Operator is Linear Mapping.
Furthermore, $f_0$ is bounded with:
:$\norm {f_0}_{ {\map c \R}^\ast} = 1$
since:
:$\ds \size {\map {f_0} {\sequence {x_n} } } = \size { \lim _{n \mathop \to \infty} x_n } \le \sup_n \size {x_n} = \norm {\sequence {x_n} }_\infty$
and:
:$\map {f_0} {\mathbf 1} = 1 = \norm {\mathbf 1}_\infty$
By Hahn-Banach Theorem there is an extension $f \in \paren {\ell^\infty}^\ast$ of $f_0$ such that:
:$(a):\quad \norm f_{\paren {\ell^\infty}^\ast} = \norm {f_0}_{ {\map c \R}^\ast} = 1$
Observe that by Lemma 1:
:$(b):\quad \map f {\mathbf 1} = \map {f_0} {\mathbf 1} = 1$
Define a mapping $L : \map {\ell^\infty} \R \to \R$ by:
:$\ds \map L {\sequence {x_n} } := \map f {\sequence {x_0, \frac {x_0 + x_1} 2, \frac {x_0 + x_1 + x_2} 3, \ldots } }$
=== Lemma 2 ===
$L \in \paren {\map {\ell^\infty} \R}^\ast$ such that:
:$\norm L_{\paren {\ell^\infty}^\ast} = 1$
and:
:$\map L {\mathbf 1} = 1$
=== Proof of Lemma 2 ===
$L$ is linear, since the mapping:
:$\ds \sequence {x_n} \mapsto \frac { x_0 + \cdots + x_k } {k+1}$
is linear for each $k \in \N$.
Moreover, observe:
and:
:$\map L {\mathbf 1} = \map f {\mathbf 1} = 1 = \norm {\mathbf 1}_\infty$
by $(b)$.
We check that $L$ is a Banach limit.
First, we check that for all $x = \sequence {x_n} \in \map {\ell^\infty} \R$:
:$\map L x \ge 0$
if $x_n \ge 0$ for all $n \in \N$.
Observe that:
:$(c):\quad \ds \forall n \in \N : \size {x_n - \frac {\norm x_{\map {\ell^\infty} \R} } 2} \le \frac {\norm x_{\map {\ell^\infty} \R} } 2$
since:
:$x_n \in \closedint 0 {\norm x_{\map {\ell^\infty} \R} }$
Thus, by Lemma 2:
That is:
:$0 \le \map L {\sequence {x_n} }$
Secondly:
Finally, by Lemma 2:
:$\map L {\mathbf 1} = 1$
Category:Banach Limits
Existence of Biconnected Set without Dispersion Point

== Theorem ==
There exists at least one example of a biconnected set which does not have a dispersion point.
== Proof ==
Let $T$ be a Miller's biconnected set.
From Miller's Biconnected Set is Biconnected, $T$ is a biconnected set.
From Miller's Biconnected Set is has no Dispersion Point, $T$ does not have a dispersion point.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Biconnectedness and Continua
Category:Biconnected Sets
Category:Dispersion Points
Existence of Bijection between Coproducts of two Sets

== Theorem ==
Let $S_1$ and $S_2$ be sets.
Let $\struct {C, i_1, i_2}$ and $\struct {D, j_1, j_2}$ be two coproducts on $S_1$ and $S_2$.
Then there exists a unique bijection $\theta: D \to C$ such that:
:$\theta \circ j_i = i_1$
:$\theta \circ j_2 = i_2$
== Proof ==
Let $X$ be an arbitrary set.
Let $f_1: S_1 \to X$ and $f_2: S_2 \to X$ be arbitrary mappings.
Let $h_C: C \to X$ be the unique mapping such that:
:$h_C \circ i_1 = f_1$
:$h_C \circ i_2 = f_2$
Let $h_D: D \to X$ be the unique mapping such that:
:$h_D \circ j_1 = f_1$
:$h_D \circ j_2 = f_2$
The existence and uniqueness of $h_C$ and $h_D$ follow from the fact of $\struct {C, i_1, i_2}$ and $\struct {D, j_1, j_2}$ both being coproducts of $S_1$ and $S_2$.
We have that:
:$h_C \circ i_1 = f_1 = h_D \circ j_1$
:$h_C \circ i_2 = f_2 = h_D \circ j_2$
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $18$
Category:Set Coproducts
Existence of Canonical Form of Rational Number

== Theorem ==
Let $r \in \Q$.
Then:
:$\exists p \in \Z, q \in \Z_{>0}: r = \dfrac p q, p \perp q$
That is, every rational number can be expressed in its canonical form.
== Proof ==
We have that the set of rational numbers is the field of quotients of the set of integers.
From Divided by Positive Element of Field of Quotients:
:$\exists s \in \Z, t \in \Z_{>0}: r = \dfrac s t$
Now if $s \perp t$, our task is complete.
Otherwise, let:
:$\gcd \set {s, t} = d$
where $\gcd \set {s, t}$ denotes the greatest common divisor of $s$ and $t$.
Let $s = p d, t = q d$.
We have that $t, d \in \Z_{>0}$
Therefore $q \in \Z_{>0}$ also.
From Integers Divided by GCD are Coprime:
:$p \perp q$
Also:
Thus:
:$r = \dfrac p q$
where $p \perp q$ and $q \in \Z_{>0}$.
== Also see ==
* Canonical Form of Rational Number is Unique
== Sources ==
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
Category:Canonical Form of Rational Number
Existence of Cartesian Axes for which Products of Inertia are Zero

== Theorem ==
Let $B$ be a rigid body which is rotating in space about some axis $\LL$.
Let $B$ be embedded in Cartesian $3$-space whose origin lies on $\LL$.
Then there exists a set of coordinate axes $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$ such that the products of inertia of $B$ about $\mathbf O x$, $\mathbf O y$ and $\mathbf O z$ are zero.
== Proof ==
== Sources ==
* 
* 
Category:Products of Inertia
Existence of Chebyshev Polynomials of the First Kind

== Theorem ==
There exists a Chebyshev polynomial of the first kind for all natural numbers $n$.
== Proof ==
For $n = 0$:
:$\map {T_0} x = 1$, $T_0 \in \Bbb P$
For $n = 1$:
:$\map {T_1} x = x$, $T_1 \in \Bbb P$
Assume $\map {T_n} {\cos \theta} = \map \cos {n \theta}$.
:$\map {T_{n+1} } x = 2x \map {T_n} x - \map {T_{n - 1} } x$, $T_{n+1} \in \Bbb P$
By the Second Principle of Mathematical Induction, $T_n \in \Bbb P$ for all natural numbers $n$.
Category:Chebyshev Polynomials
Existence of Chebyshev Polynomials of the Second Kind

== Theorem ==
There exists a Chebyshev polynomial of the second kind for all natural numbers $n$.
== Proof ==
For $n = 0$:
:$\map {U_0} x = 1$, $U_0 \in \Bbb P$
For $n = 1$:
:$\map {U_1} x = 2x$, $U_1 \in \Bbb P$
Assume $\map {U_n} {\cos \theta}\sin \theta = \map \sin { \paren {n+1} \theta }$.
:$\map {U_{n+1} } x = 2x \map {U_n} x - \map {U_{n - 1} } x$, $U_{n+1} \in \Bbb P$
By the Second Principle of Mathematical Induction, $U_n \in \Bbb P$ for all natural numbers $n$.
Category:Chebyshev Polynomials
Existence of Class Complement

== Theorem ==
Let $X$ be a class.
Then there is a unique class $Z$ such that, for every set $u$:
:$u \in Z \iff u \notin X$
== Proof ==
Let $X$ be arbitrary.
By Axiom $\text B 3$, there is a class $Z$ such that:
:$\forall u: u \in Z \iff u \notin X$
This satisfies the existence portion of the theorem.
Let $Z'$ be a class such that:
:$\forall u: u \in Z' \iff u \notin X$
Then, for every set $u$:
:$u \in Z \iff u \notin X \iff u \in Z'$
Therefore, by the Axiom of Extension:
:$Z = Z'$
satisfying the uniqueness portion of the theorem.
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Class Intersection

== Theorem ==
For any classes $X, Y$, the intersection $X \cap Y$ exists and is unique.
That is:
:$\forall X, Y: \exists! Z: \forall u: u \in Z \iff u \in X \land u \in Y$
where $X \cap Y := Z$.
== Proof ==
Let $X, Y$ be arbitrary.
By Axiom $\text B 2$, there is some class $Z$ such that:
:$\forall u: u \in Z \iff u \in X \land u \in Y$
satisfying the existence portion of the theorem.
Now, let some $Z'$ satisfy:
:$\forall u: u \in Z' \iff u \in X \land u \in Y$
For each set $u$:
:$u \in Z \iff u \in X \land u \in Y \iff u \in Z'$
and thus:
:$u \in Z \iff u \in Z'$
Thus, by Axiom of Extension:
:$Z = Z'$
satisfying the uniqueness portion of the theorem.
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Class Union

== Theorem ==
For any classes $X, Y$, the union $X \cup Y$ exists and is unique.
That is:
:$\forall X, Y: \exists! Z: \forall u: u \in Z \iff u \in X \lor u \in Y$
where $X \cup Y := Z$.
== Proof ==
By Existence of Class Complement, there exist unique classes $\overline X, \overline Y$ such that:
:$u \in \overline X \iff u \notin X$
:$u \in \overline Y \iff u \notin Y$
for all sets $u$.
Then, by Existence of Class Intersection, there is a unique class $\overline X \cap \overline Y$ such that:
:$u \in \overline X \cap \overline Y \iff u \in \overline X \land u \in \overline Y$
By the equivalences above:
:$u \in \overline X \cap \overline Y \iff u \notin X \land u \notin Y$
Finally, by Existence of Class Complement, there is a unique class $\overline {\overline X \cap \overline Y}$ such that:
:$u \in \overline {\overline X \cap \overline Y} \iff u \notin \overline X \cap \overline Y$
By the equivalence above:
:$u \in \overline {\overline X \cap \overline Y} \iff \neg \paren {u \notin X \land u \notin Y}$
But then, by De Morgan's Laws:
:$u \in \overline {\overline X \cap \overline Y} \iff u \in X \lor u \in Y$
Therefore, $Z := \overline {\overline X \cap \overline Y}$ exists and is unique.
Category:Class Theory
Category:Von Neumann-Bernays-Gödel Set Theory
Existence of Closed Geodesics

== Theorem ==
Let $\struct {M, g}$ be a compact connected Riemannian manifold. 
Every nontrivial free homotopy class in $M$ is represented by a closed geodesic that has minimum length among all admissible loops in the given free homotopy class.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Closed Geodesics
Category:Homotopy Theory
Category:Riemannian Manifolds
Existence of Compact Hausdorff Space which is not T5

== Theorem ==
There exists at least one example of a compact $T_2$ (Hausdorff) space which is not a $T_5$ space.
== Proof ==
Let $T$ be the Tychonoff plank.
From Tychonoff Plank is Compact, $T$ is a compact topological space.
From Tychonoff Plank is Hausdorff, $T$ is a $T_2$ (Hausdorff) space.
From Tychonoff Plank is not Completely Normal, $T$ is not a completely normal space.
From $T_2$ Space is $T_1$ Space, $T$ is also a $T_1$ (Fréchet) space.
As $T$ is a $T_1$ space, it follows by definition of completely normal space that $T$ is not a $T_5$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Category:Hausdorff Spaces
Category:Compact Spaces
Category:T5 Spaces
Existence of Compact Space which Satisfies No Separation Axioms

== Theorem ==
There exists at least one example of a compact space for which none of the Tychonoff separation axioms are satisfied.
== Proof ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $D = \struct {\set {0, 1}, \vartheta}$ be the indiscrete topology on two points.
Let $T \times D$ be the double pointed topology on $T$.
From Double Pointed Finite Complement Topology is Compact, $T \times D$ is compact.
From Double Pointed Finite Complement Topology fulfils no Separation Axioms, $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness
Category:Separation Axioms
Category:Compact Spaces
Existence of Compact Space which is not Sequentially Compact

== Theorem ==
There exists at least one example of a compact topological space which is not also a sequentially compact space.
== Proof ==
Let $T = \mathbb I^{\mathbb I}$ be the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology.
From Uncountable Cartesian Product of Closed Unit Interval is Compact, $T$ is a compact space.
From Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Sequentially Compact Spaces
Category:Compact Spaces
Existence of Complementary Subspace

== Theorem ==
Let $X$ be a vector space.
Let $N \subseteq X$ be a subspace.
Then $N$ has a complementary subspace.
That is, there exists a subspace $Y \subseteq X$ such that:
:$X = N \oplus Y$
== Proof ==
Let $S$ be the set of all subspaces $V \subseteq X$ such that:
:$N \cap V = \set 0$
By Zorn's Lemma, $\struct {S, \subseteq}$ has a maximal element $Y$.
We claim:
:$X = N \oplus Y$
Indeed, if there would be an $x \in X \setminus \paren {N \oplus Y}$, then:
:$Y \subsetneq \map \span {Y \cup \set x} \in S$
where $\span$ denotes the linear span.
This contradicts the maximality of $Y$.
== Sources ==
* : $2.2$: Index of a Linear Map
Category:Linear Algebra
Existence of Completely Hausdorff Space which is not Regular

== Theorem ==
There exists at least one example of a topological space which is a completely Hausdorff space, but is not also a regular space.
== Proof ==
Let $T$ be a half-disc space.
From Half-Disc Space is Completely Hausdorff, $T$ is a completely Hausdorff space.
From Half-Disc Space is not Regular, $T$ is not a regular space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Completely Hausdorff Spaces
Category:Regular Spaces
Existence of Completely Normal Space which is not Perfectly Normal

== Theorem ==
There exists at least one example of a completely normal topological space which is not perfectly normal.
== Proof ==
Let $T$ be an uncountable Fort space.
From Fort Space is Completely Normal, $T$ is a completely normal space.
From Uncountable Fort Space is not Perfectly Normal, $T$ is not a perfectly normal space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:Completely Normal Spaces
Category:Perfectly Normal Spaces
Existence of Completely Normal Space whose Product Space is Not Normal

== Theorem ==
There exists at least one example of a completely normal topological space $T$ such that the product space $T \times T$ is not a normal topological space.
== Proof ==
Let $T$ be the Sorgenfrey line.
Let $T' = T \times T$ be Sorgenfrey's half-open square topology.
From Sorgenfrey Line satisfies all Separation Axioms, $T$ is a completely normal space.
From Sorgenfrey's Half-Open Square Topology is Not Normal, $T'$ is not a normal space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Separation Axioms
Category:Normal Spaces
Category:Completely Normal Spaces
Existence of Conjunctive Normal Form of Statement

== Theorem ==
Any propositional formula can be expressed in conjunctive normal form (CNF).
== Proof ==
A propositional variable is already trivially in conjunctive normal form (CNF).
So we consider the general propositional formula $S$.
First we convert to negation normal form (NNF).
This is always possible, by Existence of Negation Normal Form of Statement.
Now $S$ will be of the form:
:$P_1 \land P_2 \land \cdots \land P_n$
where $P_1, P_2, \ldots, P_n$ are either:
:literals
or:
:statements of the form $\paren {Q_1 \lor Q_2 \lor \ldots \lor Q_n}$.
If all the $Q_1, \ldots, Q_n$ are literals we have finished.
Otherwise they will be of the form:
:$Q_j = \paren {R_1 \land R_2 \land \ldots \land R_m}$
If the latter is the case, then use the Disjunction Distributes over Conjunction to convert:
:$Q_1 \lor Q_2 \lor \ldots \lor \paren {R_1 \land R_2 \land \ldots \land R_m} \ldots \lor Q_n$
into:
It is taken for granted that Disjunction is Associative and Disjunction is Commutative.
It can be seen then that each of the
:$\paren {Q_1 \lor Q_2 \lor \ldots \lor Q_n \lor R_k}$
are terms in the CNF expression required.
If any terms are still not in the correct format, then use the above operation until they are.
== Sources ==
* : $\S 3.7$: Decision Procedures and Normal Forms: Theorem $2$
* : $\S 4.8$: Completeness: Theorem $6$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Exercises, Group $\text{III}: 42$
* 
Category:Propositional Logic
Existence of Connected Non-T1 Scattered Space

== Theorem ==
There exists at least one example of a connected topological space which is not a $T_1$ (Fréchet) space, which is also a scattered space.
== Proof ==
Let $T$ be a divisor space.
From Divisor Space is Connected, $T$ is a connected space.
From Divisor Space is not $T_1$, $T$ is not a $T_1$ (Fréchet) space.
From Divisor Space is Scattered, $T$ is a scattered space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T1 Spaces
Category:Connected Spaces
Category:Scattered Spaces
Existence of Connected Punctiform Space

== Theorem ==
There exists at least one example of a connected topological space which is also punctiform.
== Proof ==
Let $T$ be Cantor's leaky tent.
From Cantor's Leaky Tent is Connected, $T$ is a connected topological space.
From Cantor's Leaky Tent is Punctiform, $T$ is a punctiform space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Biconnectedness and Continua
Category:Connected Spaces
Category:Punctiform Spaces
Existence of Connected Space which is Totally Pathwise Disconnected

== Theorem ==
There exists at least one example of a topological space which is both connected and totally pathwise disconnected.
== Proof ==
Let $T$ be Gustin's sequence space.
From Gustin's Sequence Space is Connected, $T$ is a connected space.
From Gustin's Sequence Space is Totally Pathwise Disconnected, $T$ is a totally pathwise disconnected space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Connected Spaces
Category:Totally Pathwise Disconnected Spaces
Existence of Coprime Amicable Pair

== Open Question ==
It is not known whether it is possible for the numbers in an amicable pair to be coprime.
All known amicable pairs share at least one common prime factor.
== Progress ==
If there do exist an amicable pair whose numbers are coprime, then, according to , their product exceeds $10^{67}$.
Such numbers cannot be created by Thabit's Rule, or any other similar construction for that matter.
== Sources ==
* : $220$
* : $220$
Category:Amicable Pairs
Category:Open Questions
Existence of Countably Compact Space which is not Sequentially Compact

== Theorem ==
There exists at least one example of a countably compact topological space which is not also a sequentially compact space.
== Proof ==
Let $T := \hointr 0 \Omega \times \mathbb I^{\mathbb I}$ be the Cartesian product of $\hointr 0 \Omega$ under the interval topology, with the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology.
From Product of Uncountable Ordinal Interval Space with Uncountable Cartesian Product of Closed Unit Interval is Countably Compact, $T$ is a countably compact space.
From Product of Uncountable Ordinal Interval Space with Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Sequentially Compact Spaces
Category:Countably Compact Spaces
Existence of Cuboid with Integer Edges, Face Diagonals and Space Diagonal

== Open Question ==
It is not known whether there exists a cuboid:
: whose edges
: the diagonals of whose faces
: the space diagonal
are all integers.
== Also see ==
* Cuboid with Integer Edges and Face Diagonals
== Sources ==
* : $44$
* : $44$
Category:Solid Geometry
Category:Open Questions
Existence of Cyclic Group of Order n

== Theorem ==
Let $n \in \Z_{>0}$.
Then there exists a cyclic group of order $n$ which is unique up to isomorphism.
== Proof ==
Existence follows from Integers Modulo m under Addition form Cyclic Group.
Uniqueness follows from Cyclic Groups of Same Order are Isomorphic.
== Sources ==
* : $\S 1.7$: Theorem $7$
Category:Finite Cyclic Groups