nyuuzyou/wikis huggingface dataset:
Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Necessary Condition

== Theorem ==
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.
Let $\sequence {f_n}$ be uniformly convergent on $S$.
Then $\sequence {f_n}$ is uniformly Cauchy on $S$.
== Proof ==
Let $\epsilon \in \R_{> 0}$ be arbitrary.
Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that: 
:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 2$
for all $x \in S$ and $n > N$. 
Then if $x \in S$ and $n, m > N$, we have:
Since $\epsilon$ was arbitrary, $\sequence {f_n}$ is uniformly Cauchy on $S$.
Category:Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent
Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Sufficient Condition

== Theorem ==
Let $S \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.
Let $\sequence {f_n}$ be uniformly Cauchy on $S$. 
Then $\sequence {f_n}$ is uniformly convergent on $S$.
== Proof ==
Let $\epsilon \in \R_{> 0}$ be arbitrary.
As $\sequence {f_n}$ is uniformly Cauchy on $S$, there exists $N \in \N$ such that: 
:$\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$
for all $n, m > N$ and $x \in S$.
Let $x \in S$ be fixed.
Then: 
:$\size {\map {f_n} x - \map {f_m} x} < \epsilon$
for all $n, m > N$. 
As $\epsilon$ was arbitrary, the sequence $\sequence {\map {f_n} x}$ is therefore Cauchy.
By Cauchy's Convergence Criterion on Real Numbers, it follows that $\sequence {\map {f_n} x}$ is convergent.
Define a function $f : S \to \R$ by: 
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
for all $x \in S$. 
We aim to show that $f_n \to f$ uniformly.
For all $x \in S$, we have:
:$\ds \lim_{m \mathop \to \infty} \size {\map {f_n} x - \map {f_m} x} = \size {\map {f_n} x - \map f x}$
We established that for all $\epsilon \in \R_{> 0}$ we can find $N \in \N$ such that: 
:$\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$
for $x \in S$ and $n, m > N$. 
We therefore have: 
:$\size {\map {f_n} x - \map f x} \le \dfrac \epsilon 2 < \epsilon$
for all $x \in S$ and $n > N$.
So $\sequence {f_n}$ converges uniformly to $f$ on $S$.
Category:Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent
Sequence of General Harmonic Numbers Converges for Index Greater than 1

== Theorem ==
Let $\map {H^{\paren r} } n$ denote the general harmonic number:
:$\ds \map {H^{\paren r} } n = \sum_{k \mathop = 1}^n \frac 1 {k^r}$
for $r \in \R_{>0}$.
Let $r > 1$.
Then as $n \to \infty$, $\map {H^{\paren r} } n$ is convergent with an upper bound of $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.
== Proof ==
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: $(4)$
Category:Sequence of General Harmonic Numbers Converges for Index Greater than 1
Category:General Harmonic Numbers
Sequence of General Harmonic Numbers Converges for Index Greater than 1/Proof

== Theorem ==
== Proof ==
Let $m \in \N$ be arbitrary.
Then:
Since $m$ is arbitrary, every partial sum $\harm r n$ is bounded from above by $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.
By the Monotone Convergence Theorem, as $n \to \infty$, $\harm r n$ is convergent with an upper bound of $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $3$
Category:Sequence of General Harmonic Numbers Converges for Index Greater than 1
Sequence of Golden Rectangles

== Theorem ==
A golden rectangle can be divided into a square and another golden rectangle.
That golden rectangle can in turn be divided into a square and another golden rectangle.
The sequence can be continued indefinitely.
:
=== Equiangular Spiral ===
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Golden Mean
Sequence of Golden Rectangles/Equiangular Spiral

== Theorem ==
The points where the vertices of successive squares of this sequence meet can be joined together by an equiangular spiral.
This equiangular spiral can be approximated by quarter circles constructed as shown.
The equiangular spiral tends towards the point of intersection of the diagonals of the golden rectangles.
== Proof ==
== Sources ==
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
* : $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
Category:Golden Mean
Sequence of Imaginary Reciprocals

== Theorem ==
Consider the subset $S$ of the complex plane defined as:
:$S := \set {\dfrac i n : n \in \Z_{>0} }$
That is:
:$S := \set {i, \dfrac i 2, \dfrac i 3, \dfrac i 4, \ldots}$
where $i$ is the imaginary unit.
Then $S$ has the following properties:
=== Boundedness ===
=== Limit Points ===
=== Closedness ===
=== Boundary Points ===
=== Interior ===
=== Openness ===
=== Connectedness ===
=== Not an Open Region ===
=== Closure ===
=== Complement ===
=== Countability ===
=== Not Compact ===
=== Closure is Compact ===
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45$
Category:Examples of Convergent Complex Sequences
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Boundary Points

== Theorem ==
Every point of $S$, along with the point $z = 0$, is a boundary point of $S$.
== Proof ==
Consider the point $z = \dfrac i n \in S$.
Let $\delta \in \R_{>0}$.
Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$.
Then $\map {N_\delta} z$ contains at least one point of $S$ ($i / n$ itself) as well as points which are not in $S$.
Hence, by definition, $z$ is a boundary point of $S$.
Let $z = 0$.
Similarly, let $\map {N_\delta} z$, 
Let $\delta \in \R_{>0}$.
Let $\map {N_\delta} z$ be the $\delta$-neighborhood of $z$.
Let $n \in \N$ such that $n > \dfrac 1 \delta$.
Then $\cmod {\dfrac i n} < \delta$ and so:
: $\dfrac i n \in \map {N_\delta} z$
Thus every $\delta$-neighborhood of $z = 0$ also contains points of $S$ and points not in $S$.
Hence, by definition, $z = 0$ is also a boundary point of $S$.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(d)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Boundedness

== Theorem ==
The set $S$ is bounded in $\C$.
== Proof ==
Let $z \in S$.
Then, for example:
:$\cmod z \le 2$
That is, $S$ is contained entirely within a circle of radius $2$ whose center is at the origin.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(a)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Closedness

== Theorem ==
The set $S$ is not closed.
== Proof ==
From Sequence of Imaginary Reciprocals: Limit Points, $S$ has one limit point $z = 0$.
But:
:$\nexists n \in \N: \dfrac i n = 0$
so $0 \notin S$.
As $S$ does not contain (all) its limit point(s), it follows by definition that $S$ is not closed.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(c)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Closure

== Theorem ==
The closure of $S$ is the set:
:$\set {0, i, \dfrac i 2, \dfrac i 3, \dfrac i 3, \ldots}$
== Proof ==
By definition, the closure of $S$ is the set $S$ along with all its limit points.
Sequence of Imaginary Reciprocals: Limit Points, $S$ has one limit point $z = 0$.
Hence the result.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(h)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Closure is Compact

== Theorem ==
The closure $S^-$ of the set $S$ is compact.
== Proof ==
From Topological Closure is Closed, $S^-$ is closed.
From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$.
It follows trivially that $S^-$ is also bounded in $\C$.
Hence the result by definition of compact.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(l)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Complement

== Theorem ==
The complement of $S$ in $\C$ is the set:
:$\C \setminus \set {i, \dfrac i 2, \dfrac i 3, \dfrac i 3, \ldots}$
== Proof ==
By definition of set complement.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(i)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Connectedness

== Theorem ==
$S$ is not connected.
== Proof ==
Let $z_1 \in S$ and $z_2 \in S$ be joined by a polygonal path $P$.
Then there are points of $P$ which are not in $S$.
Hence, by definition, $S$ is not connected.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(f)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Countability

== Theorem ==
The set $S$ is countably infinite.
== Proof ==
Let $\phi: \N \to S$ be the mapping defined as:
:$\forall n \in \N: \map \phi n = \dfrac i n$
$\phi$ is a bijection.
Hence the result by definition of countably infinite.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(j)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Interior

== Theorem ==
No point of $S$ is an interior point.
== Proof ==
From Sequence of Imaginary Reciprocals: Boundary Points, every $z \in S$ is a boundary point of $S$.
Thus no $z \in S$ is an interior point.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(d)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Limit Points

== Theorem ==
The set $S$ has exactly one limit point, and that is $z = 0$.
== Proof ==
We have that $S$ is countably infinite and also bounded.
Hence by the Bolzano-Weierstrass Theorem $S$ has at least one limit point.
Let $\epsilon \in \R_{>0}$.
Let $\map {N_\epsilon} z$ be the $\epsilon$-neighborhood of $z$.
Let $n \in \N$ such that $n > \dfrac 1 \epsilon$.
Then $\cmod {\dfrac i n} < \epsilon$ and so:
: $\dfrac i n \in \map {N_\epsilon} z$
where $\map {N_\epsilon} {z_0}$ denotes the $\epsilon$-neighborhood of $z$.
Hence, by definition, $z$ is a limit point of $S$.
Now let $z_1 \in \C$ such that $z_1 \ne 0$.
Let $z \in S$ such that $z \ne z_1$.
Then:
:$\cmod {z_1 - z} > 0$
Let $\epsilon \in \R_{>0}$ such that $\epsilon < \cmod {z_1 - z}$.
Then:
:$z \notin \map {N_\epsilon} {z_1}$
By making $\epsilon$ smaller than the minimum $\cmod {z_1 - z}$ for $z \in S$, it is seen that:
:$\forall z \in S: z \ne z_1 \implies z \notin \map {N_\epsilon} {z_1}$
and so $z_1$ is not a limit point of $S$.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(b)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Not Compact

== Theorem ==
The set $S$ is not compact.
== Proof ==
From Sequence of Imaginary Reciprocals: Boundedness, $S$ is bounded in $\C$.
But from Sequence of Imaginary Reciprocals: Closedness, $S$ is not closed.
Hence the result by definition of compact.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(k)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Not an Open Region

== Theorem ==
$S$ is not an open region.
== Proof ==
From Sequence of Imaginary Reciprocals: Openness, $S$ is not an open set.
From Sequence of Imaginary Reciprocals: Connectedness, $S$ is not connected.
Hence, by definition, $S$ is not an open region of $\C$.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(g)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Imaginary Reciprocals/Openness

== Theorem ==
$S$ is not an open set.
== Proof ==
From Sequence of Imaginary Reciprocals: Interior, no $z \in S$ is an interior point.
Hence $S$ cannot be open.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Point Sets: $45 \ \text {(e)}$
Category:Sequence of Imaginary Reciprocals
Sequence of Implications of Connectedness Properties

== Theorem ==
Let $P_1$ and $P_2$ be connectedness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
 		 		Ultraconnected 	
 		 		$\Big\Downarrow$	
Arc-Connected		$\implies$		Path-Connected 	
 		 		$\Big\Downarrow$	
Irreducible 		$\implies$		Connected 	
== Proof ==
The relevant justifications are listed as follows:
:Ultraconnected Space is Path-Connected
:Arc-Connected Space is Path-Connected
:Path-Connected Space is Connected
:Irreducible Space is Connected
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Connected Spaces
Sequence of Implications of Disconnectedness Properties

== Theorem ==
Let $P_1$ and $P_2$ be disconnectedness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
Regular		$\impliedby$		Zero Dimensional and $T_0$		$\impliedby$		Discrete Space 		$\implies$ 		$\implies$ 		$\implies$ 		Scattered and $T_1$	
 		 		$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		 		 		$\Big\Downarrow$ 	
 		 		$\Big\Downarrow$ 		 		Extremally Disconnected 		 		 		 		$\Big\Downarrow$ 	
 		 		$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		 		 		$\Big\Downarrow$ 	
 		 		$\implies$ 		$\implies$ 		Totally Separated 		$\implies$ 		Urysohn 		 		$\Big\Downarrow$ 	
 		 		 		 		$\Big\Downarrow$ 		 		 		 		$\Big\Downarrow$ 	
 		 		$T_1$ 		$\impliedby$		Totally Disconnected 		$\impliedby$		$\impliedby$		$\impliedby$		$\impliedby$ 	
 		 		 		 		$\Big\Downarrow$ 		 		 		 		 	
 		 		 		 		Totally Pathwise Disconnected		 		 		 		 	
== Proof ==
The relevant justifications are listed as follows:
* Discrete Space is $T_0$ and Discrete Space is Zero Dimensional.
* Discrete Space is $T_1$ and Discrete Space is Scattered.
* Zero Dimensional Space is T3, and by definition of a regular space as being both $T_3$ and $T_0$.
* Zero Dimensional $T_0$ Space is Totally Separated.
* Discrete Space is Extremally Disconnected.
* Extremally Disconnected Space is Totally Separated.
* Totally Separated Space is Completely Hausdorff and Urysohn.
* Totally Separated Space is Totally Disconnected.
* Scattered $T_1$ Space is Totally Disconnected.
* Totally Disconnected Space is Totally Pathwise Disconnected.
* Totally Disconnected Space is $T_1$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Disconnected Spaces
Sequence of Implications of Global Compactness Properties

== Theorem ==
Let $P_1$ and $P_2$ be compactness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
 		 		Sequentially Compact 		 		 	
 		 		$\Big\Downarrow$ 		 		 	
Compact 		$\implies$		Countably Compact 		$\implies$		Pseudocompact	
$\Big\Downarrow$		 		$\Big\Downarrow$ 		 		 	
$\sigma$-Compact		 		Weakly Countably Compact		 		 	
$\Big\Downarrow$		 		 		 		 	
Lindelöf Space 		 		 		 		 	
== Proof ==
The relevant justifications are listed as follows:
* Sequentially Compact Space is Countably Compact.
* Compact Space is Countably Compact.
* Compact Space is $\sigma$-Compact.
* $\sigma$-Compact Space is Lindelöf.
* Countably Compact Space is Weakly Countably Compact.
* Countably Compact Space is Pseudocompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
Category:Compact Spaces
Sequence of Implications of Local Compactness Properties

== Theorem ==
Let $P_1$ and $P_2$ be compactness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
Compact 		$\implies$		Strongly Locally Compact		 		 	
$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		 	
Weakly $\sigma$-Locally Compact		$\implies$		Weakly Locally Compact 		$\Longleftarrow$		Locally Compact	
$\Big\Downarrow$ 		 		 		 		 	
$\sigma$-Compact 		 		 		 		 	
$\Big\Downarrow$ 		 		 		 		 	
Lindelöf Space 		 		 		 		 	
== Proof ==
The relevant justifications are listed as follows:
* Compact Space is Strongly Locally Compact.
* Strongly Locally Compact Space is Weakly Locally Compact.
* Locally Compact Space is Weakly Locally Compact.
* Compact Space is Weakly $\sigma$-Locally Compact.
* A weakly $\sigma$-locally compact is both weakly locally compact and $\sigma$-compact by definition.
* $\sigma$-Compact Space is Lindelöf.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Localized Compactness Properties
Category:Compact Spaces
Sequence of Implications of Metric Space Compactness Properties

== Theorem ==
Let $P_1$ and $P_2$ be compactness properties and let:
:$P_1 \implies P_2$
mean:
:If a metric space $M$ satsifies property $P_1$, then $M$ also satisfies property $P_2$.
Then the following sequence of implications holds:
Sequentially Compact 		$\implies$		Weakly $\sigma$-Locally Compact		$\implies$		Weakly Locally Compact 	
$\Big\Updownarrow$ 		 		$\Big\Downarrow$ 		 		$\Big\Updownarrow$ 	
Countably Compact 		 		$\sigma$-Compact 		 		Strongly Locally Compact	
$\Big\Updownarrow$ 		 		$\Big\Downarrow$ 		 		 	
Compact 		 		Separable 		 		 	
$\Big\Updownarrow$ 		 		$\Big\Updownarrow$ 		 		 	
Weakly Countably Compact		 		Lindelöf 		 		 	
 		 		$\Big\Updownarrow$ 		 		 	
 		 		Second-Countable 		 		 	
== Proof ==
The relevant justifications are listed as follows:
:Metric Space is Compact iff Countably Compact.
:Metric Space is Countably Compact iff Sequentially Compact.
:Metric Space is Weakly Countably Compact iff Countably Compact.
:Compact Space is Weakly $\sigma$-Locally Compact.
:By definition, a weakly $\sigma$-locally compact space is both weakly locally compact and $\sigma$-compact.
:$\sigma$-Compact Space is Lindelöf.
:Metric Space is Lindelöf iff Second-Countable.
:Metric Space is Separable iff Second-Countable.
:Metric Space is Weakly Locally Compact iff Strongly Locally Compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Compact Spaces
Sequence of Implications of Paracompactness Properties

== Theorem ==
Let $P_1$ and $P_2$ be paracompactness properties and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
 		 		Compact 		$\implies$		Countably Compact 	
 		 		$\Big\Downarrow$		 		$\Big\Downarrow$ 	
Fully Normal 		$\implies$		Paracompact 		$\implies$		Countably Paracompact	
$\Big\Downarrow$		 		$\Big\Downarrow$		 		$\Big\Downarrow$ 	
Fully $T_4$ 		 		Metacompact 		$\implies$		Countably Metacompact	
$\Big\Downarrow$		 		 		 		 	
$T_4$ 		 		 		 		 	
== Proof ==
The relevant justifications are listed as follows:
* Compact Space is Countably Compact.
* Compact Space is Paracompact.
* Fully Normal Space is Paracompact.
* Countably Compact Space is Countably Paracompact.
* Paracompact Space is Countably Paracompact.
* Fully Normal Space is Fully $T_4$ by definition.
* Fully $T_4$ Space is $T_4$.
* Paracompact Space is Metacompact.
* Metacompact Space is Countably Metacompact.
* Countably Paracompact Space is Countably Metacompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Compact Spaces
Sequence of Implications of Separation Axioms

== Theorem ==
Let $P_1$ and $P_2$ be separation axioms and let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies separation axiom $P_1$, then $T$ also satisfies separation axiom $P_2$.
Then the following sequence of separation axioms holds:
 		 		Perfectly Normal 		$\implies$		Perfectly $T_4$ 		$\implies$		$T_4$	
 		 		$\Big\Downarrow$ 		 		 		 		 	
 		 		Completely Normal 		$\implies$		$T_5$ 		 		 	
 		 		$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		 	
 		 		Normal 		$\implies$		$T_4$ 		 		 	
 		 		$\Big\Downarrow$ 		 		 		 		 	
$T_{3 \frac 1 2}$		$\impliedby$		Completely Regular (Tychonoff)		$\implies$		Urysohn 		 		 	
$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		 	
$T_3$ 		$\impliedby$		Regular 		$\implies$		$T_{2 \frac 1 2}$ (Completely Hausdorff)		 		 	
 		 		$\Big\Downarrow$ 		 		$\Big\Downarrow$ 		 		 	
 		 		Semiregular 		$\implies$		$T_2$ (Hausdorff) 		 		 	
 		 		 		 		$\Big\Downarrow$ 		 		 	
 		 		 		 		$T_1$ (Fréchet) 		 		 	
 		 		 		 		$\Big\Downarrow$ 		 		 	
 		 		 		 		$T_0$ (Kolmogorov) 		 		 	
== Proof ==
The relevant justifications are listed as follows:
* Perfectly Normal implies Perfectly $T_4$ implies $T_4$ by definition.
* Perfectly Normal Space is Completely Normal Space.
* Completely Normal implies $T_5$ by definition.
* Completely Normal Space is Normal Space.
* $T_5$ space is $T_4$ space.
* Normal implies $T_4$ by definition.
* Normal Space is Tychonoff (Completely Regular) Space.
* Completely Regular (Tychonoff) implies $T_{3 \frac 1 2}$ by definition.
* Completely Regular (Tychonoff) Space is Regular Space.
* $T_{3 \frac 1 2}$ Space is $T_3$ Space.
* Regular implies $T_3$ by definition.
* Completely Regular (Tychonoff) Space is Urysohn Space.
* Urysohn Space is Completely Hausdorff Space.
* Regular Space is Completely Hausdorff Space.
* Regular Space is Semiregular Space.
* Completely Hausdorff Space is $T_2$ (Hausdorff) Space.
* Semiregular implies $T_2$ (Hausdorff) by definition.
* $T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space.
* $T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Additional Separation Properties
Category:Separation Axioms
Sequence of Integers defining Abelian Group

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $C_n$ be a finite abelian group.
Then $C_n$ is of the form:
:$C_{n_1} \times C_{n_2} \times \cdots \times C_{n_r}$
such that:
:$n = \ds \prod_{k \mathop = 1}^r n_k$
:$\forall k \in \set {2, 3, \ldots, r}: n_k \divides n_{k - 1}$
where $\divides$ denotes divisibility.
== Proof ==
== Examples ==
== Sources ==
* : Chapter $14$: The classification of finite abelian groups
Category:Abelian Groups
Category:Sequence of Integers defining Abelian Group
Sequence of Integers defining Abelian Group/Examples/Order 100

== Examples of Sequences of Integers defining Abelian Groups ==
Let $G$ be an abelian group of order $100$.
From Sequence of Integers defining Abelian Group, $G$ can be expressed in the form:
:$G = C_{n_1} C_{n_2} \cdots C_{n_r}$
The possible sequences $\tuple {n_1, n_2, \ldots n_r}$ of positive integers which can define $G$ are:
== Proof ==
Determined by inspection.
== Sources ==
* : Chapter $14$: The classification of finite abelian groups: Example $14.1$
Category:Sequence of Integers defining Abelian Group
Sequence of Integers whose Factorial minus 1 is Prime

== Sequence ==
The sequence of positive integers $n$ for which $n! - 1$ is prime begins:
:$3, 4, 6, 7, 12, 14, 30, 32, 33, 38, 94, 166, 324, 379, 469, 546, 974, 1963, 3507, 3610, 6917, 21 \, 480, 34 \, 790, 94 \, 550, 103 \, 040, 147 \, 855, 208 \, 003 \ldots$
The sequence of corresponding primes begins:
:$5, 23, 719, 5039, 479 \, 001 \, 599, 87 \, 178 \, 291 \, 199, \ldots$
== Also see ==
* Sequence of Integers whose Factorial plus 1 is Prime
== Sources ==
* 
* : $719$
Category:Factorials
Category:Prime Numbers
Sequence of Integers whose Factorial plus 1 is Prime

== Sequence ==
The sequence of positive integers $n$ for which $n! + 1$ is prime begins:
:$0, 1, 2, 3, 11, 27, 37, 41, 73, 77, 116, 154, 320, 340, 399, 427, 872, 1477, 6380, 26 \, 951, 110 \, 059, 150 \, 209 \ldots$
The sequence of corresponding primes begins:
:$2, 2, 3, 7, 39 \, 916 \, 801, \ldots$
== Also see ==
* Sequence of Integers whose Factorial minus 1 is Prime
== Sources ==
* : $7$
* : $116$
* 
* : $7$
* : $1477$
* : $1477! + 1$
Category:Factorials
Category:Prime Numbers