nyuuzyou/wikis huggingface dataset:
Discrete Space is Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is locally path-connected.
== Proof ==
From Set in Discrete Topology is Clopen, $\set a$ is open in $T$.
From Basis for Discrete Topology, the set:
:$\BB := \set {\set x: x \in S}$
is a basis for $T$.
Let $\set x \in \BB$.
From Point is Path-Connected to Itself, it follows that $\set x$ is path-connected.
Hence $T$ has a basis consisting entirely of path-connected sets.
So by definition $T$ is locally path-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $10$
Category:Locally Path-Connected Spaces
Category:Discrete Topology
Discrete Space is Non-Meager/Proof 1

== Theorem ==
== Proof ==
Let $U \subseteq S$ such that $U \ne \O$.
From Interior Equals Closure of Subset of Discrete Space, we have:
:$U^\circ = U = U^-$
where $U^\circ$ is the interior and $U^-$ the closure of $U$.
So:
:$\paren {U^-}^\circ = U \ne \O$
Thus, by definition, no non-empty subset of $S$ is nowhere dense.
So $S$ can not be the union (countable or otherwise) of nowhere dense subsets.
So by definition $S$ can not be meager.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $9$
Category:Discrete Space is Non-Meager
Discrete Space is Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is paracompact.
== Proof ==
Let $\VV$ be any open cover of $S$.
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 which is locally finite.
This result also shows that $\CC$ is the finest cover on $T$.
So $\CC$ is an open refinement of $\VV$ which is locally finite.
So $T$ is paracompact, by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $7$
Category:Paracompact Spaces
Category:Discrete Topology
Discrete Space is Scattered

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is a scattered space.
== Proof ==
We have that Topological Space is Discrete iff All Points are Isolated.
So, by definition, no subset $H \subseteq S$ of $T$ such that $H \ne \O$ is dense-in-itself.
So, again, by definition, $T$ is scattered.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Scattered Spaces
Category:Discrete Topology
Discrete Space is Separable iff Countable

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a discrete topological space.
Then:
:$T$ is separable $S$ is countable.
== Proof ==
=== Sufficient Condition ===
Immediate from Separable Discrete Space is Countable.
=== Necessary Condition ===
Immediate from Countable Space is Separable.
== Sources ==
* 
* 
Category:Denseness
Category:Separable Spaces
Category:Discrete Topology
Discrete Space is Strongly Locally Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is strongly locally compact.
== Proof ==
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$ of $T$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure in $T$.
From Singleton Set in Discrete Space is Compact, we have that $\set x$ is compact in $T$.
Hence the result by definition of strongly locally compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $7$
Category:Strongly Locally Compact Spaces
Category:Discrete Topology
Discrete Space is Totally Disconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Then $T$ is totally disconnected.
== Proof ==
Follows from:
* Discrete Space is Extremally Disconnected
* Extremally Disconnected Space is Totally Separated
* Totally Separated Space is Totally Disconnected
== Sources ==
* : $6.5$: Components
Category:Totally Disconnected Spaces
Category:Discrete Topology
Discrete Space is Zero Dimensional

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete space.
Then $T$ is zero dimensional.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:Zero Dimensional Spaces
Category:Discrete Topology
Category:Discrete Space is Zero Dimensional
Discrete Space is Zero Dimensional/Proof 2

== Theorem ==
== Proof ==
Let $\BB$ be the set:
:$\BB := \set {\set x: x \in S}$
From Basis for Discrete Topology, $\BB$ is a basis for $T$.
From Set in Discrete Topology is Clopen, all the elements of $\BB$ are both closed and open.
Hence the result, by definition of zero dimensional space
Category:Discrete Space is Zero Dimensional
Discrete Space is not Dense-In-Itself

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Then $T$ is not dense-in-itself.
== Proof ==
By definition, $T$ is dense-in-itself it contains no isolated points.
The result follows from Topological Space is Discrete iff All Points are Isolated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $9$
Category:Discrete Topology
Discrete Space satisfies all Separation Properties

== Theorem ==
Let $T = \struct {S, \powerset S}$ be the discrete topological space on $S$.
Then $T$ fulfils all separation axioms:
:$T$ is a $T_0$ (Kolmogorov) space
:$T$ is a $T_1$ (Fréchet) space
:$T$ is a $T_2$ (Hausdorff) space
:$T$ is a semiregular space
:$T$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space 
:$T$ is a $T_3$ space 
:$T$ is a regular space
:$T$ is an Urysohn space
:$T$ is a $T_{3 \frac 1 2}$ space
:$T$ is a Tychonoff (completely regular) space
:$T$ is a $T_4$ space
:$T$ is a normal space
:$T$ is a $T_5$ space 
:$T$ is a completely normal space
:$T$ is a perfectly $T_4$ space
:$T$ is a perfectly normal space
== Proof ==
We have that a Standard Discrete Metric induces Discrete Topology.
Then we use Metric Space fulfils all Separation Axioms.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $6$
Category:Discrete Topology
Category:Separation Axioms
Discrete Subgroup of Hausdorff Group is Closed

== Theorem ==
Let $G$ be a Hausdorff topological group.
Let $H$ be a discrete subgroup of $G$.
Then $H$ is closed in $G$.
== Proof ==
Let $g \in \overline H$ be in the closure of $H$.
We will show that $g \in H$.
 $g \notin H$.
Let $e$ be the identity of $G$.
Because $H$ is discrete, there exists an open set $U \subset G$ such that $U \cap H = \set e$.
Then $V = U \cap U^{-1}$ is an open neighborhood of $e$ in $G$.
By Right and Left Regular Representations in Topological Group are Homeomorphisms, $g V$ is a neighborhood of $g$ in $G$.
By definition of closure, $g V \cap H \ne \O$.
Let $h \in g V \cap H$.
Because $g \ne h$ and $G$ is Hausdorff, there exists a neighborhood $W$ of $g$ in $G$ with $h \notin W$.
Then $g V \cap W$ is a neighborhood of $g$.
Let $k \in g V \cap W \cap H$.
Then $k \ne h$.
Then
Bu assumption, $k^{-1} h = e$, so $k = h$, which is a contradiction.
Thus, by Proof by Contradiction, $g \in H$.
Thus $\overline H = H$.
Thus $H$ is closed.
== Also see ==
* Topological Group is T1 iff T2
* Discrete Subgroup of Real Numbers is Closed
Category:Topological Groups
Category:Hausdorff Spaces
Discrete Subgroup of Real Numbers is Closed

== Theorem ==
Let $G$ be a subgroup of the additive group of real numbers.
Let $G$ be discrete.
Then $G$ is closed.
== Proof ==
By Subgroup of Real Numbers is Discrete or Dense, there exists $a \in \R$ such that $G = a \Z$.
If $a = 0$, then $G$ is closed.
Let $a > 0$.
Then:
:$\ds \R \setminus G = \bigcup_{z \mathop \in \Z} \openint {a z} {a z + a}$
By Union of Open Sets of Metric Space is Open, $\R\setminus G$ is open.
Thus $G$ is closed.
== Also see ==
* Discrete Subgroup of Hausdorff Group is Closed
Category:Topological Groups
Category:Real Numbers
Discrete Subspace of Fortissimo Space

== Theorem ==
Let $T = \left({S, \tau_p}\right)$ be a Fortissimo space.
Let $T' = \left({S \setminus \left\{{p}\right\}, \tau_p}\right)$ be the topological subspace induced on $T$ by the subset $S \setminus \left\{{p}\right\}$.
Then $T'$ is a discrete topological space.
== Proof ==
By the definition of Fortissimo space, any $A \subset S \setminus \left\{{p}\right\}$ is open in $S$, because $p \notin A$.
Thus by definition of topological subspace, $A \subset S \setminus \left\{{p}\right\}$ is open in $S \setminus \left\{{p}\right\}$.
The result follows by the definition of discrete space.
Category:Fortissimo Space
Category:Discrete Topology
Discrete Time Series by Accumulation/Examples/Batch Process

== Example of Discrete Time Series by Accumulation ==
Consider a manufacturing plant which outputs a product at some variable rate.
The output can yield a discrete time series, obtained by accumulation of this output over a specified time period.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.1$ Time Series and Stochastic Processes: Time series: $2$
Category:Discrete Time Series
Discrete Time Series by Accumulation/Examples/Rainfall

== Example of Discrete Time Series by Accumulation ==
An example of a discrete time series obtained by accumulation of a quantity would be the contents of a collector of rainfall.
At the end of each time period, usually a day, the rainfall is collected from the container and its volume is measured.
The container is emptied again, ready for the rainfall to be collected over the next period.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.1$ Time Series and Stochastic Processes: Time series: $2$
Category:Discrete Time Series
Discrete Time Series by Sampling/Examples/Gas Furnace

== Example of Discrete Time Series by Sampling ==
An example of a discrete time series obtained by sampling a continuous time series would be the input and output of a gas furnace sampled at intervals of $10$ seconds.
== Sources ==
* :
::Part $\text {I}$: Stochastic Models and their Forecasting:
:::$2$: Autocorrelation Function and Spectrum of Stationary Processes:
::::$2.1$ Autocorrelation Properties of Stationary Models:
:::::$2.1.1$ Time Series and Stochastic Processes: Time series: $1$
Category:Discrete Time Series
Discrete Topology is Finest Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
:$\tau$ is the finest topology on $S$.
== Proof ==
Let $\phi$ be any topology on $S$.
Let $U \in \phi$.
Then, by the definition of topology, $U \subseteq S$.
Then, by the definition of discrete topological space, $U \in \tau$.
Hence by definition of subset, $\phi \subseteq \tau$.
Hence by definition of finer topology, $\tau$ is finer than $\phi$.
== Also see ==
* Indiscrete Topology is Coarsest Topology
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $1$
* 
* : Chapter $\S 2.3$: The normed space $\map {CL} {X, Y}$. Strong and weak operator topologies on $\map {CL} {X, Y}$
Category:Discrete Topology
Discrete Topology is Topology

== Theorem ==
Let $S$ be a set.
Let $\tau$ be the discrete topology on $S$.
:$\tau$ is a topology on $S$. 
== Proof ==
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Then by definition $\tau = \powerset S$, that is, is the power set of $S$.
We confirm the criteria for $T$ to be a topology:
:$(1): \quad$ By definition of power set, $\O \in \powerset S$ and $S \in \powerset S$.
:$(2): \quad$ From Power Set with Union is Monoid, $\powerset S$ is closed under set union.
:$(3): \quad$ From Power Set with Intersection is Monoid, $\powerset S$ is closed under set intersection.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.4$
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology
* : $\S 2$: Topological Spaces: Topologies
Category:Discrete Topology
Discrete Uniform Distribution/Examples/Casting of Die

== Example of Discrete Uniform Distribution ==
Consider the exercise of casting a fair die.
There are $6$ equally probable outcomes: $1$, $2$, $3$, $4$, $5$ and $6$, each with a probability of $\dfrac 1 6$.
Hence this is modelled by a discrete uniform distribution with parameter $6$.
== Sources ==
* 
* 
Category:Examples of Discrete Uniform Distributions
Discrete Uniform Distribution/Examples/Coin Toss

== Example of Discrete Uniform Distribution ==
Consider the exercise of coin-tossing.
There are $2$ possible outcomes: $\mathrm H$eads or $\mathrm T$ails, each with a probability of $\dfrac 1 2$.
Hence this is modelled by a discrete uniform distribution with parameter $2$.
== Sources ==
* 
* 
Category:Examples of Discrete Uniform Distributions
Discrete Uniform Distribution/Examples/Random Digits

== Example of Discrete Uniform Distribution ==
Consider the exercise of selecting a random digit.
There are $10$ equally probable outcomes in the integer interval $\closedint 0 9$, each with a probability of $\dfrac 1 {10}$.
Hence this is modelled by a discrete uniform distribution with parameter $10$.
== Sources ==
* 
* 
Category:Examples of Discrete Uniform Distributions
Discrete Uniform Distribution gives rise to Probability Measure

== Theorem ==
Let $\EE$ be an experiment.
Let the probability space $\struct {\Omega, \Sigma, \Pr}$ be defined as:
:$\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$
:$\Sigma = \powerset \Omega$
:$\forall A \in \Sigma: \map \Pr A = \dfrac 1 n \card A$
where:
:$\powerset \Omega$ denotes the power set of $\Omega$
:$\card A$ denotes the cardinality of $A$.
Then $\Pr$ is a probability measure on $\struct {\Omega, \Sigma}$.
== Proof ==
From Power Set of Sample Space is Event Space we have that $\Sigma$ is an event space.
We check the axioms defining a probability measure:
Axiom $\text I$ is seen to be satisfied by the observation that the cardinality of a set is never negative.
Hence $\map \Pr A \ge 0$.
Then we have:
Axiom $\text {II}$ is thus seen to be satisfied.
Let $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$ where $k = \card A$.
Then by Union of Set of Singletons:
:$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$
Hence:
Hence Axiom $\text {III}$ is thus seen to be satisfied.
All axioms are seen to be satisfied.
Hence the result.
== Sources ==
* : $1$: Events and probabilities: $1.3$: Probabilities: Example $11$
Category:Discrete Uniform Distribution
Discrete Uniformity generates Discrete Topology

== Theorem ==
Let $S$ be a set.
Let $\UU$ be the discrete uniformity on $S$.
Then the topology generated by $\UU$ is the discrete topology on $S$.
The diagonal relation $\Delta_S$ generates the basis for this discrete topology.
== Proof ==
From the construction, let $\tau \subseteq \powerset S$ be a subset of the power set of $S$, created from $\UU$ by:
:$\tau := \set {\map u x: u \in \UU, x \in X}$
where:
:$\forall x \in X: \map u x = \set {y: \tuple {x, y} \in u}$
We need to show that $\tau$ is the discrete topology.
Consider $\Delta_S \in \UU$.
Let $x \in S$.
Then the set:
:$U_x := \set {y: \tuple {x, y} \in \Delta_S}$
We have that:
$\forall x \in S: U_x = \set x \in \UU$
Thus we have:
:$\BB := \set {U_x: x \in S} = \set {\set x: x \in S}$
From Basis for Discrete Topology, we have that $\BB$ is a basis for the discrete topology on $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $11$
Category:Discrete Uniformities
Category:Discrete Topology
Discrete Uniformity is Uniformity

== Theorem ==
Let $S$ be a set.
Let $\UU$ be the discrete uniformity on $S$
Then $\UU$ is indeed a uniformity.
== Proof ==
We examine the uniformity axioms in turn:
;$\text U 1$: $\forall u \in \UU: \Delta_S \subseteq u$
This follows by definition of the discrete uniformity:
:$\UU := \set {u \subseteq S \times S: \Delta_S \subseteq u}$
;$\text U 2$: $\forall u, v \in \UU: u \cap v \in \UU$
We have that $\forall u, v \subseteq S \times S: \Delta_S \subseteq u, \Delta_S \subseteq v$
So $\Delta_S \subseteq u \cap v$ from Intersection is Largest Subset.
So $u \cap v \in \UU$, and $\text U 2$ holds.
;$\text U 3$: $u \in \UU, u \subseteq v \subseteq S \times S \implies v \in \UU$
Let $u \subseteq v \subseteq S \times S$.
We have that $\Delta_S \subseteq u$ and so $\Delta_S \subseteq v$ from Subset Relation is Transitive.
So $v \in \UU$ by definition of $\UU$, and $\text U 3$ holds.
;$\text U 4$: $\forall u \in \UU: \exists v \in \UU: v \circ v \subseteq u$ where $\circ$ is defined as:
::$u \circ v := \set {\tuple {x, z}: \exists y \in S: \tuple {x, y} \in v, \tuple {y, z} \in u}$
We have by definition of $\UU$ that $\Delta_S \in \UU$.
We note from Diagonal Relation is Equivalence that $\Delta_S$ is an equivalence relation and so by definition transitive.
Therefore from Relation contains Composite with Self iff Transitive we have that $\Delta_S \circ \Delta_S \subseteq \Delta_S$.
So:
:$\forall u \in \UU: \exists \Delta_S \in \UU: \Delta_S \circ \Delta_S \subseteq \Delta_S \subseteq u$
and so $\text U 4$ holds.
;$\text U 5$: $\forall u \in \UU: \exists u^{-1} \in \UU$ where $u^{-1}$ is defined as:
::$u^{-1} := \set {\tuple {y, x}: \tuple {x, y} \in u}$
Let $u \in \UU$, so $\Delta_S \subseteq u \subseteq S \times S$.
Thus:
:$\forall x \in u: \tuple {x, x} \in u \implies \tuple {x, x} \in u^{-1}$
and so $\Delta_S \subseteq u^{-1}$
Thus $u^{-1} \in \UU$.
All of the uniformity axioms are satisfied, and $\UU$ is demonstrated to be a uniformity.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $11$
Category:Discrete Uniformities
Discrete Variable/Examples/Number of Children

== Example of Discrete Variable ==
The number $N$ of children in any given family can take one of the values in the set $\N = \set {0, 1, 2, \ldots}$
However, $N$ cannot take the value of a number which is not a natural number, for example, $2 \cdotp 5$ or $3 \cdotp 842$.
That is, $N$ is a discrete variable.
== Sources ==
* : Chapter $1$: Discrete and Continuous Variables: Example 1.
Category:Examples of Discrete Variables
Discriminant is Invariant for Isometry of Conic Section

== Theorem ==
Let a conic section $\KK$ be expressed as a quadratic form in $2$ variables:
:$a x^2 + b x y + c y^2 = r$
The discriminant:
:$b^2 - 4 a c$
is an invariant under translations and rotations of the coordinate axes.
== Proof ==
== Sources ==
* 
* 
* 
* 
Category:Conic Sections
Category:Isometries (Euclidean Geometry)
Category:Invariants
Discriminant of Polynomial/Examples/Cubic

== Example of Discriminant of Polynomial ==
Let $\alpha_1$, $\alpha_2$ and $\alpha_3$ be the roots of a cubic equation $f$.
The discriminant of $f$ is:
:$\paren {\alpha_1 - \alpha_2}^2 \paren {\alpha_2 - \alpha_3}^2 \paren {\alpha_3 - \alpha_1}^2$
== Sources ==
* 
* 
Category:Examples of Discriminants of Polynomials
Discriminant of Quadratic Equation

== Theorem ==
Let $\QQ$ be the quadratic equation:
:$a x^2 + b x + c$
The discriminant of $\QQ$ is given by:
:$\map \Delta \QQ = \dfrac {b^2 - 4 a c} {a^2}$
== Proof ==
Let $\alpha_1$ and $\alpha_2$ be the roots of $\QQ$.
We have:
== Also presented as ==
== Sources ==
* 
* 
Category:Discriminants of Polynomials
Category:Quadratic Equations
Discriminant of Quadratic Equation/Also presented as

== Discriminant of Quadratic Equation: Also presented as ==
While the discriminant $\Delta$ of a quadratic equation $\QQ: a x^2 + b x + c = 0$ is properly given as:
:$\map \Delta \QQ = \dfrac {b^2 - 4 a c} {a^2}$
it is usual and practically universal that $\map \Delta \QQ$ is given as:
:$\map \Delta \QQ = b^2 - 4 a c$
This is because we are usually interested only in the sign of $\map \Delta \QQ$ and not its actual value.
Indeed, we note that as $a^2 > 0$, multiplying or dividing by $a^2$ has no effect on that sign.
== Sources ==
* 
* 
Category:Discriminants of Polynomials
Category:Quadratic Equations
Dishonest Butler

== Classic Problem ==
A dishonest butler removes $3$ pints of wine from a barrel, replacing them with water.
He repeats this theft twice, removing in total $9$ pints, each time replacing with water.
As a result, the wine in the barrel is half its normal strength, the rest being water.
How much wine was there in the barrel to start with?
== Solution ==
$14.54$ pints.
== Proof ==
Let $x$ pints be the total quantity of wine in the barrel to start with.
After the first theft, there are $x - 3$ pints left.
The concentration of wine is now $\dfrac {x - 3} x$.
So the second theft removes $3 \dfrac {x - 3} x$ pints of wine from the cask.
The concentration of wine is now $\dfrac {x - 3 - 3 \frac {x - 3} x} x$.
So the third theft removes $3 \dfrac {x - 3 - 3 \frac {x - 3} x} x$ pints of wine.
The total removed is now $\dfrac x 2$, and so:
== Historical Note ==
== Sources ==
* : Exchanging the Knights: $104$
Category:Classic Problems
Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood/Corollary

== Corollary ==
Let $F$ be a topological field.
Let $X$ be a topological vector space over $F$.
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
Then there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that: 
:$W^- \subseteq V$
where $W^-$ denotes the closure of $W$.
== Proof ==
From Finite Topological Space is Compact, we have: 
:$\set { {\mathbf 0}_X}$ is compact.
Since $V$ is open, $X \setminus V$ is closed with: 
:$\set { {\mathbf 0}_X} \cap \paren {X \setminus V} = \O$ 
So from Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood, there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that: 
:$\paren {\set { {\mathbf 0}_X} + W} \cap \paren {\paren {X \setminus V} + W} = \O$
That is: 
:$W \cap \paren {\paren {X \setminus V} + W} = \O$
Since $W$ is open: 
:$\paren {X \setminus V} + W$ is open
from Sum of Set and Open Set in Topological Vector Space is Open.
From Open Set Disjoint from Set is Disjoint from Closure: 
:$W^- \cap \paren {\paren {X \setminus V} + W} = \O$
Since ${\mathbf 0}_X \in W$: 
:$X \setminus V \subseteq \paren {X \setminus V} + W$
From Set Intersection Preserves Subsets: 
:$W^- \cap \paren {X \setminus V} = \O$
From Empty Intersection iff Subset of Complement: 
:$W^- \subseteq V$
Category:Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood
Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $V_1$ and $V_2$ be compact sets in $T$.
Then $V_1$ and $V_2$ have disjoint neighborhoods.
=== Lemma ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods
Category:Hausdorff Spaces
Category:Compact Spaces