nyuuzyou/wikis huggingface dataset:
Dirichlet Series Convergence Lemma/General

== Theorem ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty a_n e^{-\map {\lambda_n} s}$ be a general Dirichlet series.
Let $\map f s$ converge at $s_0 = \sigma_0 + i t_0$.
Then $\map f s$ converge for all $s = \sigma + i t$ where $\sigma > \sigma_0$.
== Proof ==
Let $s = \sigma + i t$
Let $s_0 \in \C$ be such that $\map f {s_0}$ converges. 
Let $\map S {m, n} = \ds \sum_{k \mathop = n}^m a_k e^{-\lambda_k s_0}$
We may create a new Dirichlet series that converges at 0 by writing:
Thus it suffices to show $\map g s$ converges for $\sigma > 0$.
By Cauchy's Convergence Criterion, it suffices to show that for all $\epsilon>0$ there exists an $N$ such that for all $m, n > N$: 
:$\ds \cmod {\sum_{k \mathop = n}^m a_n e^{-\lambda_k s_0} e^{-\lambda_k s} } < \epsilon$
By Abel's Lemma: Formulation 2 we may write:
Because $\map S {m, n}$ is the difference of partial sums of a convergent, and thus cauchy, sequence, its modulus, $\cmod {\map S {m, n} }$, is bounded, say by $Q$.
Thus we have:
We see that:
Thus we have:
Because $\lambda_n$ tends to infinity, both summands tend to $0$ as $n$ goes to $\infty$ if $\sigma > 0$.
Thus we can pick $N$ large enough such that both terms are less than $\dfrac \epsilon 2$ for $n, m > N$, giving us the desired result.
Category:General Dirichlet Series
Dirichlet Series Convergence Lemma/Lemma

== Lemma to Dirichlet Series Convergence Lemma ==
Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac{a_n} {n^s}$ be a Dirichlet series.
Suppose that for some $s_0 = \sigma_0 + i t_0 \in \C$, $\map f {s_0}$ has bounded partial sums:
:$(1): \quad \ds \size {\sum_{n \mathop = 1}^N a_n n^{-s_0} } \le M$
for some $M \in \R$ and all $N \ge 1$.
Then for every $s = \sigma + i t \in \C$ with $\sigma > \sigma_0$:
:$\ds \size {\sum_{n \mathop = m}^N a_n n^{-s} } \le 2 M m^{\sigma_0 - \sigma} \paren {1 + \frac {\size {s - s_0} } {\sigma - \sigma_0} }$
== Proof ==
From the Summation by Parts formula:
:$\ds \sum_{n \mathop = m}^N f_n g_n = f_N G_N - f_m G_{m-1} - \sum_{n \mathop = m}^{N - 1} \map {G_n} {f_{n + 1} - f_n}$
Let:
:$g_n = a_n n^{-s_0}$
:$f_n = n^{s_0 - s}$
For $N \ge 1$, the quantities $G_N$ are the partial sums $(1)$.
Thus $G_N \le M$ for all $N \ge 1$.
We have:
Finally, because $N \ge m$ and $\sigma_0 - \sigma < 0$, we have:
:$N^{\sigma_0 - \sigma} + m^{\sigma_0 - \sigma} \le 2 m^{\sigma_0 - \sigma}$
Therefore:
Hence the result.
== Sources ==
* : $\S 11.6$: Lemma $2$, Theorem $11.8$
Category:Dirichlet Series
Dirichlet Series is Analytic

== Theorem ==
Let $(a_n)$ be sequence of complex numbers.
Let:
:$\ds \map f z = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^z}$
be the associated Dirichlet Series, which is defined at the points where the series converges.
Then $f$ is analytic in every open set such that the sum converges in the set.
== Proof ==
Category:Dirichlet Series
Dirichlet Series of Convolution of Arithmetic Functions

== Theorem ==
Let $f, g: \N \to \C$ be arithmetic functions.
Let $h = f * g$ be their Dirichlet convolution.
Let $F, G, H$ be their Dirichlet series.
Let $s$ be a complex number such that $\map F s$ and $\map G s$ converge absolutely.
Then $\map H s$ converges absolutely to $\map F s \times \map G s$.
== Proof ==
== Also see ==
* Dirichlet Series of Inverse of Arithmetic Function
* Upper Bound for Abscissa of Absolute Convergence of Product of Dirichlet Series
Category:Dirichlet Series
Dirichlet Series of Inverse of Arithmetic Function

== Theorem ==
Let $f : \N \to\C$ be an arithmetic function.
Let $g : \N \to \C$ be an Dirichlet inverse of $f$.
Let $F, G$ be their Dirichlet series.
Let $s \in \C$ such that both $\map F s$ and $\map G s$ converge absolutely.
Then $\map F s \cdot \map G s = 1$.
== Proof ==
Let $\varepsilon$ be the identity arithmetic function.
By Dirichlet Series of Identity Arithmetic Function, $\varepsilon$ has Dirichlet series $\map E s = 1$.
By Dirichlet Series of Convolution of Arithmetic Functions, $\map F s \map G s = 1$.
Category:Dirichlet Series
Discontinuity (Real Analysis)/Examples/Example 1

== Example of Discontinuity in the context of Real Analysis ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac 1 {1 - x}$
Then $f$ has a discontinuity at $x = 1$, as $f$ is not defined there.
== Sources ==
* 
* 
* 
Category:Examples of Discontinuities (Real Analysis)
Discontinuity (Real Analysis)/Examples/Example 2

== Example of Discontinuity in the context of Real Analysis ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} \dfrac 1 k & : 0 \le x \le k \\ 0 & : \text {otherwise} \end {cases}$
Then $f$ has a discontinuity at $x = 0$ and $x = k$
== Sources ==
* 
Category:Examples of Discontinuities (Real Analysis)
Discontinuity (Real Analysis)/Examples/Example 3

== Example of Discontinuity in the context of Real Analysis ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac 1 {x^2 - 4}$
Then $f$ has a discontinuity at $x = -2$ and $x = 2$.
These are infinite discontinuities.
:
== Sources ==
* 
Category:Examples of Discontinuities (Real Analysis)
Discontinuity of Monotonic Function is Jump Discontinuity

== Theorem ==
Let $X$ be an open set of $\R$.
Let $f: X \to Y$ be a monotone real function.
Then $f$ is discontinuous at a point $c \in X$ $c$ is a jump discontinuity of $f$.
== Proof ==
The backwards implication follows directly from definition of a jump discontinuity.
For the forwards implication, we prove the contrapositive.
 suppose $f$ is increasing.
If $f$ is decreasing, note that $-f$ is increasing and we can simply replace $f$ by $-f$ in the following analysis.
Suppose that $c \in X$ is not a jump discontinuity of $f$.
By definition of jump discontinuity, it is not the case that:
:$\ds \map f {c^-} = \lim_{x \mathop \to c^-} \map f x$ and $\ds \map f {c^+} = \lim_{x \mathop \to c^+} \map f x$ exist and are not equal.
By definition of an open set:
:there exists $\epsilon \in \R_{>0}$ such that $\openint {c - \epsilon} {c + \epsilon} \subseteq X$.
By Limit of Monotone Real Function/Increasing/Corollary, since $c \in \openint {c - \epsilon} {c + \epsilon}$:
:$\map f {c^-}$ and $\map f {c^+}$ both exist;
:$\map f {c^-} \le \map f c \le \map f {c^+}$
Hence by Modus Ponendo Tollens:
:$\map f {c^-} = \map f {c^+}$
Therefore we have:
:$\map f {c^-} = \map f c = \map f {c^+}$
By Limit iff Limits from Left and Right:
:$\map f x \to \map f c$ as $x \to c$
Hence $f$ is continuous at $c$ by definition.
Thus we have shown the contrapositive of the forwards implication.
Category:Monotone Real Functions
Category:Real Analysis
Category:Proofs by Contraposition
Discrepancy between Gregorian Year and Tropical Year

== Theorem ==
The Gregorian year and the tropical year differ such that the Gregorian calendar becomes $1$ day further out approximately every $3319$ years.
== Proof ==
By definition, the length $Y_T$ of the tropical year is defined as $\approx 365 \cdotp 24219 \, 878$ days.
By definition of the Gregorian year:
:$Y_G = \begin{cases} 366 \, \text {days} & : 400 \divides y \\
365 \, \text {days} & : 400 \nmid y \text { and } 100 \divides y\\
366 \, \text {days} & : 100 \nmid y \text { and } 4 \divides y\\
365 \, \text {days} & : 4 \nmid y \end{cases}$
where:
:$Y_G$ denotes the length of the Gregorian year in days
:$y$ denotes the number of the year
:$4 \divides y$ denotes that $y$ is divisible by $4$
:$4 \nmid y$ denotes that $y$ is not divisible by $4$.
Thus in every $400$ years, there are:
:$300$ years which are not divisible by $4$, which have $365$ days
:$1$ year which is divisible by $400$, which has $366$ days
:$3$ years which are divisible by $100$ but not $400$, which have $365$ days
:The remaining $96$ years which are divisible by $4$ but not $100$, which have $366$ days.
Thus:
Thus the mean Gregorian year is $\dfrac 1 {3318 \cdotp 95}$ days longer than the tropical year.
This means that after approximately $3319$ Gregorian years, the Gregorian calendar starts one day later relative to the tropical year.
== Sources ==
* : $365 \cdotp 2422$
* : $365 \cdotp 24219 \, 878$
Category:Calendars
Discrepancy between Julian Year and Tropical Year

== Theorem ==
The Julian year and the tropical year differ such that the Julian calendar becomes $1$ day further out approximately every $128$ years.
== Proof ==
By definition, the length $Y_T$ of the tropical year is defined as $\approx 365 \cdotp 24219 \, 878$ days
By definition of the Julian year:
:$Y_J = \begin{cases} 366 \, \text {days} & : 4 \divides y \\
365 \, \text {days} & : 4 \nmid y \end{cases}$
where:
:$Y_J$ denotes the length of the Julian year in days
:$y$ denotes the number of the year
:$4 \divides y$ denotes that $y$ is divisible by $4$
:$4 \nmid y$ denotes that $y$ is not divisible by $4$.
Thus:
Thus the mean Julian year is $\dfrac 1 {128 \cdotp 18}$ days longer than the tropical year.
This means that after approximately $128$ Julian years, the Julian calendar starts one day later relative to the tropical year.
== Sources ==
* : $365 \cdotp 2422$
* : $365 \cdotp 24219 \, 878$
Category:Calendars
Discrete Category is Order Category

== Theorem ==
Let $\map {\mathbf{Dis} } S$ be a discrete category.
Then $\map {\mathbf{Dis} } S$ is also an order category.
== Proof ==
We have, for any morphism $a \to b$ in $\map {\mathbf{Dis} } S$ that $a = b$.
Thus we see that $\map {\mathbf{Dis} } S$ will be an order category :
:$\forall a, b \in S: a \preceq b \iff a = b$
holds for some ordering $\preceq$ on $S$.
The trivial ordering on $S$ accomplishes this.
Hence the result.
== Sources ==
* : $\S 1.4.12$
Category:Discrete Categories
Category:Order Categories
Discrete Category on Set is Discrete Category

== Theorem ==
Let $S$ be a set.
Let $\map {\mathbf {Dis} } S$ be the discrete category on $S$.
Then $\map {\mathbf {Dis} } S$ determines a unique (up to isomorphism discrete category $\map {\mathbf {Dis} } S$ whose objects precisely comprise $S$.
== Proof ==
== Sources ==
* : $\S 1.4.12$
Category:Discrete Categories
Discrete Data/Examples/Colors of Rainbow

== Example of Discrete Data ==
The set $C$ of colors of the rainbow:
:$C = \set {\text {red}, \text {orange}, \text {yellow}, \text {green}, \text {blue}, \text {indigo}, \text {violet} }$
is an example of discrete data.
In such a circumstance where the discrete data are non-numerical, it is usually possible to assign (natural) numbers to each of the elements, for example:
and so the set $C'$ can be considered instead:
:$C' = \set {1, 2, 3, 4, 5, 6, 7}$
== Sources ==
* : Chapter $1$: Discrete and Continuous Variables
Category:Examples of Discrete Data
Discrete Fourier Transform on Abelian Group

== Theorem ==
Let $G$ be a finite abelian group.
Let $G^*$ be the dual group of characters $G \to \C^\times$.
Let $\eta: G \to \C$ be a mapping from $G$ to the set of complex numbers.
Then for all $x \in G$:
:$\ds \map \eta x = \frac 1 {\map \phi q} \sum_{\chi \mathop \in G^*} \innerprod \eta \chi_G \map \chi x$
where:
:$\ds \innerprod \eta \chi_G = \sum_{x \mathop \in G} \map \eta x \map {\overline \chi} x$
== Proof ==
Category:Fourier Analysis
Category:Representation Theory
Discrete Group Acts Continuously iff Acts by Homeomorphisms

== Theorem ==
Let $G$ be a discrete group acting on a topological space $X$.
Then the following are equivalent:
:$G$ acts continuously
:$G$ acts by homeomorphisms
== Proof ==
If $G$ acts continuously, then by Continuous Group Action is by Homeomorphisms, $G$ acts by homeomorphisms
Let $G$ act by homeomorphisms
Let $\phi: G \times X \to X$ denote the group action.
For $g \in G$, denote $\phi_g : X \to X : x \mapsto \map \phi {g, x}$
Let $U \subset X$ be open.
Then:
Since each $\set g \times {\phi _g^{-1} \sqbrk U }$ is open by , $\phi^{-1} \sqbrk U$ is open.
Thus $\phi$ is continuous.
== Also see ==
* Continuous Group Action is by Homeomorphisms
Category:Topological Group Actions
Discrete Normal Subgroup of Connected Group is Contained in Center

== Theorem ==
Let $G$ be a connected topological group.
Let $\map Z G$ be the center of $G$.
Let $N$ be a discrete normal subgroup of $G$.
Then:
:$N \subseteq \map Z G$
== Proof ==
Let $h \in N$.
We shall show $h \in \map Z G$
Now, let:
:$A _h := \set {g \in G : g^{-1} h g = h}$
Then, we need to show:
:$A _h = G$
Let $e \in G$ denote the identity.
Since $e \in A _h$, we have $A _h \ne \O$.
Thus, it suffices to show that $A _h$ is clopen in view of .
As $N$ is a discrete subgroup, there is an open set $U \subseteq G$ such that:
:$U \cap N = \set e$
Multiplying the above equality from left, we also have:
:$\forall h \in N : h U \cap N = \set h$
Define $\phi _h : G \to N$ by:
:$\map {\phi _h} g := g^{-1} h g$
Then $\phi _h$ is continuous.
Thus both:
and
are open.
That is, $A _h$ is clopen.
Category:Topological Groups
Discrete Random Variable is Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ fulfils the condition:
:$\forall x \in \R: \set {\omega \in \Omega: \map X \omega \le x} \in \Sigma$
That is, $X$ fulfils the condition for it to be a random variable.
== Proof ==
Let $X$ be a discrete random variable.
Then by definition:
:$\forall x \in \R: \set {\omega \in \Omega: \map X \omega = x} \in \Sigma$
But see that:
:$\ds \set {\omega \in \Omega: \map X \omega \le x} = \bigcup_{\substack {y \mathop \in \Omega_X \\ y \mathop \le x} } \set {\omega \in \Omega: \map X \omega = y}$
This is the countable union of events in $\Sigma$.
Hence, as $\Sigma$ is a sigma-algebra, $\set {\omega \in \Omega: \map X \omega \le x} \in \Sigma$ as required.
Category:Probability Theory
Discrete Set/Examples/Natural Numbers

== Example of Discrete Set ==
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
The natural numbers $\N$ form a discrete set within $T$.
== Proof ==
Let $x \in \N$ be an arbitrary natural number.
Let $\map B {x; \dfrac 1 2} \subseteq \R$ be the open ball of radius $\dfrac 1 2$ on $\R$ whose center is $x$.
Then $x$ is the only natural number in $\map B {x; \dfrac 1 2}$.
Hence by definition $x$ is isolated in $\N$.
As $x$ is arbitrary, the result follows.
== Sources ==
* 
* 
Category:Examples of Discrete Sets
Discrete Set/Examples/Rational Numbers

== Example of Discrete Set ==
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
The rational numbers $\Q$ do not form a discrete set within $T$.
== Proof ==
Let $x \in \Q$ be an arbitrary rational number.
Let $\epsilon \in \R_{>0}$ be an arbitrary (strictly) positive rational number.
Let $\map {B_\epsilon} x \subseteq \R$ be the open ball of radius $\epsilon$ on $\R$ whose center is $x$.
That is:
:$\map {B_\epsilon} x := \openint {x - \epsilon} {x + \epsilon}$
is the open interval as described.
Consider the rational number $y = x + \dfrac \epsilon 2$.
It is seen that $y \in \map {B_\epsilon} x$.
As $\epsilon$ is arbitrary, it follows that no such $\map {B_\epsilon} x$ contains only $x$ from $\Q$.
Hence by definition $x$ is not isolated in $\Q$.
As $x$ is arbitrary, the result follows.
== Sources ==
* 
* 
Category:Examples of Discrete Sets
Discrete Set of Subsets is Locally Finite

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a discrete set of subsets of $S$.
Then $\FF$ is a locally finite set of subsets of $S$.
== Proof ==
This follows immediately from the definitions of discrete set of subsets and locally finite set of subsets.
Category:Discrete Sets of Subsets
Category:Locally Finite Sets of Subsets
Discrete Space has Open Locally Finite Cover

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
Then $\CC$ is an open cover of $T$ which is locally finite.
This cover is the finest cover on $S$.
That is, if $\VV$ is a cover of $T$, then $\CC$ is a refinement of $\VV$.
== Proof ==
We have that:
:$\forall x \in S: \exists \set x \in \CC: x \in \set x$
and so $\CC$ is a cover for $S$.
Then from Set in Discrete Topology is Clopen, it follows that $\CC$ is an open cover of $T$.
From Point in Discrete Space is Neighborhood, every point $x \in S$ has a neighborhood $\set x$.
This neighborhood $\set x$ intersects exactly one element of $\CC$, that is: $\set x$ itself.
As $1$ is a finite number, the result follows from definition of locally finite.
Now consider any cover $\VV$ of $S$.
By definition:
:$\forall x \in S: \exists V \in \VV: x \in V$
That is:
:$\forall x \in S: \exists V \in \VV: \set x \subseteq V$
That is, every element of $\CC$ is contained in some element of $\VV$.
Thus by definition, $\CC$ is a refinement of $\VV$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $7$
Category:Covers
Category:Discrete Topology
Discrete Space iff Diagonal Set on Product is Open

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Endow $X \times X$ with the product topology.
Let: 
:$\Delta = \set {\tuple {x, x} : x \in X}$
Then: 
:$\tau = \powerset X$
:
:$\Delta$ is open in $X$.
That is:
:the topology on $X$ is discrete $\Delta$ is open in $X$.
== Proof ==
=== Sufficient Condition ===
Suppose that $\tau = \powerset X$. 
Then for each $x \in X$, the set:
:$\set {\tuple {x, x} }$
is open.
Note that we can write: 
:$\ds \Delta = \bigcup_{x \in X} \set {\tuple {x, x} }$
This is the union of open sets, so from the definition of a topology we have that: 
:$\Delta$ is open.
=== Necessary Condition ===
Suppose that $\Delta$ is open. 
Let $x \in X$ so that $\tuple {x, x} \in \Delta$.
From the definition of the product topology:
:there exists open sets $S, R \subseteq X$ such that $\tuple {x, x} \in S \times R \subseteq \Delta$.
We want to conclude that:
:$S = R = \set x$
Certainly: 
:$\set x \subseteq S$
and:
:$\set x \subseteq R$
Suppose that:
:$S \ne \set x$
then:
:there exists some $y \in S$ with $y \ne x$.
But then: 
:$\tuple {y, x} \in S \times R$
while, since $y \ne x$, we have: 
:$\tuple {y, x} \not\in \Delta$
This contradicts:
:$S \times R \subseteq \Delta$
So we must have: 
:$S = \set x$
So, since $x$ was arbitrary:
:for each $x \in X$, the set $\set x$ is open.
So, from Topology is Discrete iff All Singletons are Open, we have that: 
:the topology on $X$ is discrete.
Category:Discrete Topology
Category:Product Topology
Discrete Space is Compact iff Finite

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is compact $S$ is a finite set, thereby making $\tau$ the finite discrete topology on $S$.
== Proof ==
=== Necessary Condition ===
Let $T$ be a compact discrete space.
 $T$ is infinite.
Let $\CC$ be the cover for $S$ defined as:
:$\CC = \set {\set x: x \in S}$
As $S$ is an infinite set then so is $\CC$.
Let $\CC'$ be a proper subset of $\CC$.
Then:
:$\exists y \in S: \set y \notin \CC'$
and so $\CC'$ is not a cover for $S$.
So by definition $\CC'$ is not a subcover of $\CC$.
So $\CC$ can have no finite subcover.
Hence by definition $T$ can not be compact.
By Proof by Contradiction it follows that if $T$ is compact then $S$ must be finite.
=== Sufficient Condition ===
Let $T$ be a finite discrete space.
Then from Finite Topological Space is Compact it follows that $T$ is a compact space.
== Sources ==
* : $5$: Compact spaces: Exercise $5.10: 2$
Category:Finite Topological Spaces
Category:Compact Spaces
Category:Discrete Topology
Discrete Space is Complete Metric Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is a complete metric space.
== Proof ==
Let $d: S \times S \to \R$ be the standard discrete metric.
From Standard Discrete Metric induces Discrete Topology we have that $\struct {S, d}$ is a metric space whose induced topology is $\tau$.
Consider now a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$.
By the definition of Cauchy sequence:
:$\forall \epsilon > 0: \exists N \in \N$ such that $\forall n, m > N: \map d {x_n, x_m} < \epsilon$
Take $\epsilon = \dfrac 1 2$.
Then:
:$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} < \epsilon = \dfrac 1 2$
From the definition of standard discrete metric:
:$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} = 0 \implies x_n = x_m$
Thus the sequence $\sequence {x_n}_{n \mathop \in \N}$ is constant from some $N \in \N$.
From Eventually Constant Sequence is Convergent:
:$x_n \to x \in S$
Thus every Cauchy sequence in $S$ converges.
The result follows by definition of complete metric space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $9$
Category:Discrete Topology
Category:Complete Metric Spaces
Category:Standard Discrete Metric
Discrete Space is Extremally Disconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is extremally disconnected.
== Proof ==
First we note that as Discrete Space satisfies all Separation Properties, $T$ is a $T_2$ (Hausdorff) space.
Then from Interior Equals Closure of Subset of Discrete Space, it follows directly that the closure of every open set of $T$ is open.
Hence by definition $T$ is extremally disconnected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Extremally Disconnected Spaces
Category:Discrete Topology
Discrete Space is First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is first-countable.
== Proof ==
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$.
From the definition of local basis, it is clear that $\set {\set x}$ is (trivially) a local basis at $x$.
That is, that every open set of $S$ containing $x$ also contains at least one of the sets of $\set {\set x}$.
Equally trivially, we have that $\set {\set x}$ is countable.
Hence the result by definition of first-countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $7$
Category:First-Countable Spaces
Category:Discrete Topology
Discrete Space is Fully Normal

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is fully normal.
== Proof ==
We have that a Discrete Space is fully $T_4$.
Then we note that from Discrete Space satisfies all Separation Properties, a discrete space is a $T_1$ (Fréchet) space.
Therefore, by definition, $T$ is fully normal.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $6$
Category:Fully Normal Spaces
Category:Discrete Topology
Discrete Space is Fully T4

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is fully $T_4$.
== Proof ==
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover:
:$\CC$ is an open cover of $T$ which is locally finite.
Let $x \in S$.
The star of $x$ with respect to $\CC$ is defined as:
:$\ds x^* := \bigcup \set {U \in \CC: x \in U}$
That is, the union of all sets in $\CC$ which contain $x$.
But here the only set of $\CC$ containing $x$ is $\set x$.
Now let $\VV$ be an open cover of $T$.
From the above, it follows that $\CC$ is a cover for $S$ such that:
:$\forall x \in S: \exists U \in \VV: x^* \subseteq U$
where $x^* = \set x$ is the star of $x$ with respect to $\CC$.
That is, $\CC$ is a star refinement of $\VV$.
Therefore, by definition, $T$ is fully $T_4$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $6$
Category:Fully T4 Spaces
Category:Discrete Topology
Discrete Space is Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is locally connected.
== Proof ==
Let $T = \struct {S, \tau}$ be a discrete space.
From Discrete Space is Locally Path-Connected, $T$ is locally path-connected.
From Locally Path-Connected Space is Locally Connected, it follows that $T$ is locally connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $10$
Category:Locally Connected Spaces
Category:Discrete Topology