nyuuzyou/wikis huggingface dataset:
Existence of Dedekind Completion

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Then there exists a Dedekind completion of $S$.
That is, there exists a Dedekind complete ordered set $\tilde S$ and an order embedding $\phi: S \to \tilde S$ such that:
:For all Dedekind complete ordered sets $X$, and for all order embeddings $f: S \to X$, there exists an order embedding $\tilde f: \tilde S \to X$ such that:
::$\tilde f \circ \phi = f$
== Proof ==
For all subsets $I \subseteq S$, define:
:$\map {\operatorname U} I = \left\{{x \in S: x}\right.$ is an upper bound for $\left.{I}\right\}$
:$\map {\operatorname L} I = \left\{{x \in S: x}\right.$ is a lower bound for $\left.{I}\right\}$
Note that, for all $I, J \subseteq S$:
:$I \subseteq \map {\operatorname {LU} } I$
:$I \subseteq J \implies \map {\operatorname {LU} } I \subseteq \map {\operatorname {LU} } J$
:$\map {\operatorname {ULU} } I = \map {\operatorname U} I$
All of the above follow directly from the relevant definitions.
Let:
:$\tilde S = \left\{{I \subseteq S: I = \map {\operatorname {LU} } I}\right.$, $I$ is non-empty and bounded above$\left.{}\right\}$
We have that $\struct {\tilde S, \subseteq}$ is an ordered set.
Let $A \subseteq \tilde S$ be non-empty and bounded above.
By Union is Smallest Superset: General Result, $\ds \bigcup A$ is non-empty and bounded above.
It follows that:
:$\ds {\operatorname{LU} } {\bigcup A} \in \tilde S$
From Set is Subset of Union: General Result, it follows that $\ds \map {\operatorname{LU} } {\bigcup A}$ is an upper bound for $A$.
If $I \in \tilde S$ is an upper bound for $A$, then, by Union is Smallest Superset: General Result:
:$\ds \bigcup A \subseteq I$
It follows that:
:$\ds \map {\operatorname{LU} } {\bigcup A} \subseteq \map {\operatorname{LU} } I = I$
Hence:
:$\ds \operatorname{LU} {\bigcup A} = \sup A$
By definition, $\struct {\tilde S, \subseteq}$ is Dedekind complete.
Let $\phi: S \to \tilde S$ be the order embedding defined as:
:$\forall x \in S: \map \phi x = x^\preceq$
where $x^\preceq$ denotes the weak lower closure of $x$ (in $S$).
Let $\struct {X, \preccurlyeq}$ be a Dedekind complete ordered set, and let $f: S \to X$ be an order embedding.
Let $\tilde f: \tilde S \to X$ be the increasing mapping defined as:
:$\forall I \in \tilde S: \map {\tilde f} I = \sup \map f I$
where $\map f I$ denotes the image of $I$ under $f$.
Then $\tilde f \circ \phi = f$.
Let $I, J \in \tilde S$ such that $\map {\tilde f} I \preccurlyeq \map {\tilde f} J$.
Then $\forall x \in I: \forall u \in \map {\operatorname U} J: \map f x \preccurlyeq \map {\tilde f} J \preccurlyeq \map f u$.
Since $f$ is an order embedding, it follows that $I \subseteq \map {\operatorname {LU} } J = J$.
Hence, $\tilde f$ is an order embedding.
== Also see ==
* Dedekind Completion is Unique up to Isomorphism
Category:Order Theory
Existence of Digital Root

== Theorem ==
Let $n \in \N$ be a natural number.
Let $b \in \N$ such that $b \ge 2$ also be a natural number.
Let $n$ be expressed in base $b$.
Then the digital root base $b$ exists for $n$.
== Proof ==
By definition, the digital root base $b$ for $n$ is the single digit resulting from:
:adding up the digits in $n$, and expressing the result in base $b$
:adding up the digits in that result, and again expressing the result in base $b$
:repeating until down to one digit.
Let $n = d_1 + b d_2 + \dotsb + b^{m - 1} d_m$ where, for all $i$, $0 \le d_i < b$.
Let $\map S n$ be the digit sum of $n$.
Then:
:$\map S n = d_1 + d_2 + \dotsb + d_m$
Thus:
:$\map S n < n$
unless $d_2, d_3, \dotsb, d_m = 0$
in which case $n$ is a one digit number.
Similarly: 
:$\map S {\map S n} < \map S n$
Every time the digit sum is taken, the result is at least one less than the previous digit sum.
As $n$ is finite, it will take a finite number of steps to reduce the result to a one digit number.
Hence the result.
Category:Number Bases
Category:Digital Roots
Existence of Disjoint Well-Ordered Sets Isomorphic to Ordinals

== Theorem ==
Let $\alpha$ and $\beta$ be ordinals.
Then there exist two well-ordered sets $a$ and $b$ such that:
:$a$ and $b$ are order isomorphic to $\alpha$ and $\beta$ respectively
:$a$ and $b$ are disjoint.
== Proof ==
Let $a$ and $b$ be defined as:
Then let $a$ and $b$ be ordered by their first coordinate.
The result follows.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 6$ Ordinal arithmetic
Category:Ordinals
Category:Well-Orderings
Existence of Disjunctive Normal Form of Statement

== Theorem ==
Any propositional formula can be expressed in disjunctive normal form (DNF).
== Proof ==
A propositional variable is already trivially in disjunctive normal form (DNF).
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:
:$P_1 \lor P_2 \lor \cdots \lor P_n$
where $P_1, P_2, \ldots, P_n$ are either:
:literals
or:
:statements of the form $\paren {Q_1 \land Q_2 \land \ldots \land Q_n}$
If all the $Q_1, \ldots, Q_n$ are literals we have finished.
Otherwise they will be of the form $Q_j = \paren {R_1 \lor R_2 \lor \ldots \lor R_m}$
If the latter is the case, then use Conjunction Distributes over Disjunction to convert:
:$Q_1 \land Q_2 \land \ldots \land \paren {R_1 \lor R_2 \lor \ldots \lor R_m} \ldots \land Q_n$
into:
It is taken for granted that Conjunction is Associative and Conjunction is Commutative.
It can be seen then that each of the
:$\paren {Q_1 \land Q_2 \land \ldots \land Q_n \land R_k}$
are terms in the DNF expression required.
If any terms are still not in the correct format, then use the above operation until they are.
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Exercises, Group $\text{III}: 43$
* 
Category:Disjunctive Normal Form
Existence of Distance Functional

== Theorem ==
Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a normed vector space over $\mathbb F$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $X$. 
Let $Y$ be a proper closed linear subspace of $X$. 
Let $x \in X \setminus Y$.
Let:
:$d = \map {\operatorname {dist} } {x, Y}$
where $\map {\operatorname {dist} } {x, Y}$ denotes the distance between $x$ and $Y$.
Then there exists $f \in X^\ast$ such that:
:$(1): \quad$ $\norm f_{X^\ast} = 1$
:$(2): \quad$ $\map f y = 0$ for each $y \in Y$
:$(3): \quad$ $\map f x = d$.
That is: 
:there exists a distance functional for $x$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* $20.2$: The Distance Functional
Category:Normed Dual Spaces
Category:Existence of Distance Functional
Existence of Distance Functional/Proof 2

== Theorem ==
== Proof ==
Consider the normed quotient vector space $X/Y$ with quotient mapping $\pi$.
From Kernel of Quotient Mapping, we have $\map \pi x \ne 0$. 
So, from Existence of Support Functional, there exists $f \in \paren {X/Y}^\ast$ such that: 
:$\norm f_{\paren {X/Y}^\ast} = 1$
and:
:$\map f {\map \pi x} = \norm {\map \pi x}_{X/Y}$
From the definition of the quotient norm, we have: 
:$\norm {\map \pi x}_{X/Y} = \map {\operatorname {dist} } {x, Y}$
From Normed Dual Space of Normed Quotient Vector Space is Isometrically Isomorphic to Annihilator, $g = f \circ \pi \in X^\ast$ and: 
:$\norm g_{X^\ast} = \norm f_{\paren {X/Y}^\ast} = 1$ 
with: 
:$\map g x = \map {\operatorname {dist} } {x, Y}$
So $g$ is a linear functional satisfying our requirements.
Category:Existence of Distance Functional
Existence of Divisor with Remainder between 2b and 3b

== Theorem ==
For every pair of integers $a, b$ where $b > 0$, there exist unique integers $q$ and $r$ where $2 b \le r < 3 b$ such that:
:$a = q b + r$
== Proof ==
From the Division Theorem, we have that:
:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
So, with a view to where this is going, let $q$ and $r$ be renamed such that $a = q' b + r'$ with $0 \le r' < b$.
Then let $q' = q + 2$.
We have:
Setting $r = 2 b + r'$, it follows that:
:$2 b \le r < 3 b$
and the result follows.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $1$
Category:Division Theorem
Existence of Domain of Class Relation

== Theorem ==
Let $X$ be a class.
Then there is a unique class $Z$ such that:
:$\forall u: u \in Z \iff \exists v: \tuple {u, v} \in X$
where $\tuple {\cdot, \cdot}$ represents the Kuratowski formalization of ordered pairs.
That is, $Z$ is the class of all $z$ which appear as the first coordinate of an ordered pair in $X$.
== Proof ==
Let $X$ be arbitrary.
By Axiom $\text B 4$, there is a class $Z$ such that:
:$\forall u: u \in Z \iff \exists v: \tuple {u, v} \in X$
This satisfies the existence portion of the theorem.
Let $Z'$ be a class such that:
:$\forall u: u \in Z' \iff \exists v: \tuple {u, v} \in X$
Then, for every set $u$:
:$u \in Z \iff \paren {\exists v: \tuple {u, v} \in X} \iff u \in Z'$
Therefore, 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 Dyadic Rational between two Rationals

== Theorem ==
Let $a$ and $b$ be rational numbers such that $a < b$.
Then there exist integers $m$ and $r$ such that:
:$a < \dfrac m {2^r} < b$
That is, there exists a dyadic rational between any pair of rational numbers.
== Proof ==
As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
Thus:
:$\dfrac 1 {b - a} \in \R$
By the Axiom of Archimedes:
:$\exists r \in \N: r > \dfrac 1 {b - a}$
Notice that $2^r > r$.
Thus we also have:
:$2^r > \dfrac 1 {b - a}$
Let $M := \set {x \in \Z: x > a 2^r}$.
By Set of Integers Bounded Below has Smallest Element, there exists $m \in \Z$ such that $m$ is the smallest element of $M$.
That is:
:$m > a 2^r$
and, by definition of smallest element:
:$m - 1 \le a 2^r$
As $2^r > \dfrac 1 {b - a}$, it follows from Ordering of Reciprocals that:
:$\dfrac 1 {2^r} < b - a$
Thus:
Thus we have shown that $a < \dfrac m {2^r} < b$.
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Exercise $2$
Category:Rational Numbers
Existence of Euler-Mascheroni Constant

== Theorem ==
The real sequence:
:$\ds \sequence {\sum_{k \mathop = 1}^n \frac 1 k - \ln n}$
converges to a limit.
This limit is known as the Euler-Mascheroni constant.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
* 
Category:Existence of Euler-Mascheroni Constant
Category:Euler-Mascheroni Constant
Existence of Euler-Mascheroni Constant/Proof 1

== Theorem ==
== Proof ==
Let $f: \R \setminus \set 0 \to \R: \map f x = \dfrac 1 x$.
Clearly $f$ is continuous and positive on $\hointr 1 {+\infty}$.
From Reciprocal Sequence is Strictly Decreasing, $f$ is decreasing on $\hointr 1 {+\infty}$.
Therefore the conditions of the Cauchy Integral Test hold.
Thus the sequence $\sequence {\Delta_n}$ defined as:
:$\ds \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$
is decreasing and bounded below by zero.
But from the definition of the natural logarithm:
:$\ds \int_1^n \frac {\d x} x = \ln n$
Hence the result.
== Sources ==
* : $\S 14.3 \ (5)$
Category:Existence of Euler-Mascheroni Constant
Existence of Euler-Mascheroni Constant/Proof 2

== Theorem ==
== Proof ==
For $n \in \N_{>0}$ let:
:$\ds \gamma_n := \sum_{k \mathop = 1}^n \frac 1 k - \ln n$
Then:
On the other hand:
Thus by monotone convergence theorem, the sequence $\sequence {\gamma_n}$ converges to a limit in $\R_{\ge 0}$.
Category:Existence of Euler-Mascheroni Constant
Existence of Fibonacci Number Divisible by Number

== Theorem ==
Let $m \in \Z_{\ne 0}$ be an integer.
Then in the first $m^2$ Fibonacci numbers there exists at least one Fibonacci number which is divisible by $m$.
== Proof ==
Consider pairs of Fibonacci numbers $\tuple {F_i, F_{i + 1}}$ modulo $m$.
There are $m^2$ possible pairs of remainders.
Thus by Pigeonhole Principle, among the $\paren {m^2 + 1}$ pairs of $\tuple {F_i, F_{i + 1}}$ with $0 \le i \le m^2$, at least two of them are identical modulo $m$.
That is, there exists $0 \le i < j \le m^2$ such that:
:$\tuple {F_i, F_{i + 1}} \equiv \tuple {F_j, F_{j + 1}} \pmod m$
It is trivial to then show by induction that:
:$F_{i - k} \equiv F_{j - k} \pmod m$
for any integer $k$.
In particular:
:$F_{j - i} \equiv F_{i - i} = F_0 = 0 \pmod m$
so $m \divides F_{j - i}$.
Note that $0 < j - i \le m^2$.
Hence the result.
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Divisors
Existence of First-Countable Space which is not Second-Countable

== Theorem ==
There exists at least one example of a first-countable topological space which is not also a second-countable space.
== Proof ==
Let $T$ be a Sorgenfrey line.
From Sorgenfrey Line is First-Countable, $T$ is a first-countable space.
From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability
Category:First-Countable Spaces
Category:Second-Countable Spaces
Existence of Fraction of Number Smaller than Given

== Theorem ==
Let $a, b \in \R_{>0}$ be two (strictly) positive real numbers, such that $a > b$.
Let the sequence $\sequence {a_n}$ be defined recursively as:
:$a_i = \begin{cases} a & : i = 1 \\
a_{i - 1} - c: \dfrac {a_{i - 1} } 2 < c < a_{i - 1} & : i > 1
\end{cases}$
Then:
: $\exists n \in \N_{>0}: a_n < b$
== Proof ==
:
Let $AB$ and $C$ be two unequal magnitudes such that $AB > C$.
From there exists a multiple of $C$ which is greater than $AB$.
Thus let $DE$ be a multiple of $C$ which is greater than $AB$.
Let $DE$ be divided into parts $DF$, $FG$ and $GE$ (for example) which are equal to $C$.
From $AB$ let $BH$ be subtracted which is greater than half $AB$.
From $AH$ let $HK$ be subtracted which is greater than half $AH$.
Let this process be continued until the divisions of $AB$ are equal in number to the divisions of $DE$.
Let $AK, KH, HB$ be divisions which are equal in number to $DF, FG, GE$.
We have that $DE > AB$.
From $DE$ there has been subtracted $EG$ which is less than half $DE$.
From $AB$ there has been subtracted $BH$ which is greater than half $AB$.
Therefore the remainder $GD$ of $DE$ is greater than the remainder $HA$ of $AB$.
From $GD$ there has been subtracted $GF$ which is half $GD$.
From $HA$ there has been subtracted $HK$ which is greater than half $HA$.
Therefore the remainder $DF$ of $GD$ is greater than the remainder $AK$ of $HA$.
But $DF = C$.
Therefore $C > AK$.
Therefore there is left of the magnitude $AB$ the magnitude $AK$ which is less than the lesser magnitude set out, which is $C$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Existence of Generated C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $S \subseteq A$.
Then there exists a $\text C^\ast$-subalgebra $B \subseteq A$ containing $S$ which is $\subseteq$-minimal among $\text C^\ast$-subalgebras with this property.
== Proof ==
Let $\SS$ be the set of $\text C^\ast$-subalgebras $C \subseteq A$ containing $S$.
We have $A \in \SS$ and hence $\SS \ne \O$.
Let $B = \bigcap \SS$.
From Intersection of C*-Subalgebras is C*-Subalgebra, $B$ is a $\text C^\ast$-algebra.
Further, since $S \subseteq C$ for each $C \in \SS$, we have $S \subseteq B$. 
We now just need to assure ourselves that $B \subseteq A$ is $\subseteq$-minimal in $\SS$.
Let $C \subseteq A$ be a $\text C^\ast$-subalgebra with $C \supseteq B$. 
Then $C \in \SS$.
Hence $B \subseteq C$ from Intersection is Subset.
So $B$ is $\subseteq$-minimal in $\SS$. 
Category:C*-Algebras
Existence of Greatest Common Divisor

== Theorem ==
Let $a, b \in \Z$ be integers such that $a \ne 0$ or $b \ne 0$.
Then the greatest common divisor of $a$ and $b$ exists.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Greatest Common Divisor
Category:Existence of Greatest Common Divisor
Existence of Greatest Common Divisor/Proof 1

== Theorem ==
== Proof ==
=== Proof of Existence ===
This is proved in Greatest Common Divisor is at least $1$.
=== Proof of there being a Largest ===
, suppose $a \ne 0$.
First we note that from Absolute Value of Integer is not less than Divisors:
:$\forall c \in \Z: \forall a \in \Z_{>0}: c \divides a \implies c \le \size c \le \size a$
The same applies for $c \divides b$.
Now we have three different results depending on $a$ and $b$:
So if $a$ and $b$ are both zero, then any $n \in \Z$ divides both, and there is no greatest common divisor.
This is why the proviso that $a \ne 0 \lor b \ne 0$.
So we have proved that common divisors exist and are bounded above.
Therefore, from Set of Integers Bounded Above by Integer has Greatest Element there is always a greatest common divisor.
Category:Existence of Greatest Common Divisor
Existence of Greatest Common Divisor/Proof 2

== Theorem ==
== Proof ==
By definition of greatest common divisor, we aim to show that there exists $c \in \Z_{>0}$ such that:
and:
:$d \divides a, d \divides b \implies d \divides c$
Consider the set $S$:
:$S = \set {s \in \Z_0: \exists x, y \in \Z: s = a x + b y}$
$S$ is not empty, because by setting $x = 1$ and $y = 0$ we have at least that $a \in S$.
From the Well-Ordering Principle, there exists a smallest $c \in S$.
So, by definition, we have $c > 0$ is the smallest such that $c = a x + b y$ for some $x, y \in \Z$.
Let $d$ be such that $d \divides a$ and $d \divides b$.
Then from Common Divisor Divides Integer Combination:
:$d \divides a x + b y$
That is:
:$d \divides c$
We have that:
Mutatis mutandis:
:$c \divides b$
Now suppose $c'$ is such that:
and:
:$d \divides a, d \divides b \implies d \divides c'$
Then we have immediately that:
:$c' \divides c$
and by the same coin:
$c \divides c'$
and so:
:$c = c'$
demonstrating that $c$ is unique.
== Sources ==
* : Introduction $\S 6$: The division process in $I$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: Theorem $1.2$
* : Chapter $3$: The Integers: $\S 11$. Highest Common Factor: Theorem $19$
Category:Existence of Greatest Common Divisor
Existence of Greatest Common Divisor/Proof 3

== Theorem ==
== Proof ==
From Integers form Integral Domain, we have that $\Z$ is an integral domain.
From Euclidean Domain is GCD Domain, $a$ and $b$ have a greatest common divisor $c$.
This proves existence.
From Ring of Integers is Principal Ideal Domain, we have that $\Z$ is a principal ideal domain.
Suppose $c$ and $c'$ are both greatest common divisors of $a$ and $b$.
From Greatest Common Divisors in Principal Ideal Domain are Associates:
:$c \divides c'$
and:
:$c' \divides c$
and the proof is complete.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 28$. Highest Common Factor: Example $54$
Category:Existence of Greatest Common Divisor
Existence of Greatest Common Divisor/Proof 4

== Theorem ==
== Proof ==
From the Euclidean Algorithm, we have calculated a sequence $\tuple {r_1, r_2, \ldots r_{n - 2}, r_{n - 1}, r_n}$ such that:
:$b > r_1 > r_2 > \dotsb > r_{n - 2} > r_{n - 1} > r_n = 0$
We have that:
:$r_{n - 1} \divides a$
and:
:$r_{n - 1} \divides b$
Working backwards from the final equation, we see that:
:$r_k \divides r{k - 1}$
for all $k$ such that $1 < k \le n$.
Hence, if $d \divides a$ and $d \divides b$, we can use induction to proceed through the Euclidean Algorithm and see that $d$ divides $r_1, r_2, \ldots, r_{n - 2}, r_{n - 1}$.
Thus we see that $r_{n - 1}$ fulfils the criteria to be the greatest common divisor of $a$ and $b$.
Suppose $c_1$ and $c_2$ are both greatest common divisors of $a$ and $b$.
Then by definition there exist $g, h \in \Z_{>0}$ such that:
:$g c_1 = c_2$
:$h c_2 = c_1$
Hence:
:$c_2 = g h c_2$
and so:
:$g h = 1$
That is:
:$g = h = 1$
and so:
:$c_1 = c_2$
That is, the greatest common divisor of $a$ and $b$ is unique.
== Sources ==
* : $\S 2.2$: Theorem $2.2$
Category:Existence of Greatest Common Divisor
Existence of Group of Finite Order

== Theorem ==
Let $n \in \Z_{>0}$.
Then there exists at least one group whose order is $n$.
== Proof ==
From Existence of Cyclic Group of Order n, there exists a cyclic group whose order is $n$.
In particular, a concrete example of such a group is demonstrated in Roots of Unity under Multiplication form Cyclic Group.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.1$
* : $\S 38.5$ Period of an element
* 
* 
Category:Order of Groups
Existence of Hartogs Number/Proof 1

== Theorem ==
== Proof ==
From Hartogs' lemma there exists an ordinal $\alpha$ such that there is no injection from $\alpha$ to $S$.
We also have that Ordinals are Well-Ordered.
It follows from the definition of well-ordering that there exists a smallest such ordinal.
Hence the result.
Category:Existence of Hartogs Number
Existence of Hausdorff Space which is not Completely Hausdorff

== Theorem ==
There exists at least one example of a topological space which is a $T_2$ (Hausdorff) space, but is not also a completely Hausdorff space.
== Proof ==
Let $T$ be an irrational slope topological space.
From Irrational Slope Space is $T_2$, $T$ is a $T_2$ (Hausdorff) space.
From Irrational Slope Space is not Completely Hausdorff, $T$ is not a completely Hausdorff space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:Separation Axioms
Category:Hausdorff Spaces
Category:Completely Hausdorff Spaces
Existence of Hausdorff Space which is not T3, T4 or T5

== Theorem ==
There exists at least one example of a topological space which is a $T_2$ (Hausdorff) space (and hence also $T_0$ (Kolmogorov) and $T_1$ (Fréchet), but is not a $T_3$ space, a $T_4$ space or a $T_5$ space.
== Proof ==
Let $T$ be an irrational slope topological space.
From Irrational Slope Space is $T_2$, $T$ is a $T_2$ (Hausdorff) space.
Hence from $T_2$ Space is $T_1$ Space and $T_1$ Space is $T_0$ Space, also a $T_0$ (Kolmogorov) space and a $T_1$ (Fréchet) space.
The rest of the result follows from:
:Irrational Slope Space is not $T_3$, $T_4$ or $T_5$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:Separation Axioms
Category:T0 Spaces
Category:T1 Spaces
Category:Hausdorff Spaces