nyuuzyou/wikis huggingface dataset:
Condition for Nu Function to be 1

== Theorem ==
Let:
:$n = \ds \prod_{i \mathop = 1}^s p_i^{m_i}$
where $p_1, p_2, \ldots, p_s$ are distinct primes.
Then:
:$(1): \quad m_1, m_2, \ldots, m_s = 1$, that is, $n$ is square-free
:$(2): \quad \forall i, j \in \set {1, 2, \ldots, s}: p_i \not \equiv 1 \pmod {p_j}$
:
:every group $G$ of order $n$ is cyclic and so $\map \nu n = 1$.
== Proof ==
== Examples ==
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.4$
Category:Condition for Nu Function to be 1
Category:Nu Function
Category:Finite Cyclic Groups
Condition for Open Extension Space to be First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a first-countable space $T$ is.
== Proof ==
Let $T = \struct {S, \tau}$ be a first-countable space.
Then every point in $S$ has a countable local basis.
Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition.
So if $x$ has a countable local basis in $T$, then it has one in $T^*_{\bar p}$ as well.
Finally we note that $p$ is by definition in exactly one open set of $T^*_{\bar p}$, that is, $S^*_p$, and thus (trivially) has a countable local basis in $T^*_{\bar p}$.
Then every point in $S^*_p$ has a countable local basis in $T^*_{\bar p}$.
So if $T$ is a first-countable space, then $T^*_{\bar p}$ is also a first-countable space.
Now suppose $T = \struct {S, \tau}$ is not a first-countable space.
Then $\exists x \in S$ such that $x$ has no countable local basis in $T$.
As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $x$ has no countable local basis in $T^*_{\bar p}$ either.
So if $T $ is not a first-countable space, then neither is $T^*_{\bar p}$ a first-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:First-Countable Spaces
Category:Open Extension Topology
Condition for Open Extension Space to be Second-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a second-countable space $T$ is.
== Proof ==
Let $T = \struct {S, \tau}$ be a second-countable space.
Then $\tau$ has a countable basis.
Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition.
So if $\tau$ has a countable basis in $T$, then $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has one as well.
So if $T$ is a second-countable space, then $T^*_{\bar p}$ is also a second-countable space.
Now suppose $T = \struct {S, \tau}$ is not a second-countable space.
Then $\tau$ has no countable basis in $T$.
As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has no countable basis either.
So if $T$ is not a second-countable space, then neither is $T^*_{\bar p}$ a second-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Second-Countable Spaces
Category:Open Extension Topology
Condition for Open Extension Space to be Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a separable space $T$ is.
== Proof ==
Let $T = \struct {S, \tau}$ be a separable space.
Then there exists a countable subset $H \subseteq S$ which is everywhere dense in $T$.
That is, $H^- = S$ where $H^-$ is the closure of $H$ in $S$.
So in $T_{\bar p}^*$, $S\subseteq H^-=S^-\subseteq S^*_p$.
Hence $H^- = S$ or $H^- = S \cup \set p = S^*_p$.
From Topological Closure is Closed we deduce that $H^-$ is closed in $T_{\bar p}^*$.
From the definition of open extension topology, the closed sets of $T_{\bar p}^*$ are $S^*_p \setminus U$ where $U \in \tau$.
Because $\set p \notin \tau$, $S$ is not closed.
Thus it must be that $H^- = S^*_p$.
Hence $H$ is a countable subset $H \subseteq S^*_p$ which is everywhere dense in $T^*_{\bar p}$.
 $T^*_{\bar p} = \struct {S^*_p, \tau}$ is separable, but that $T = \struct {S, \tau}$ is not separable.
Let $H \subseteq S^*_p$ be a countable subset which is dense in $T^*_{\bar p}$.
If $H \subseteq S$, then $H^- = S$ in $T$, but this cannot be since $T$ is not separable.
Then $p \in H$.
Consider the subset $V = H \setminus \set p$.
From Topological Closure is Closed we have that $V^-$ is closed in $T_{\bar p}^*$.
From the definition of open extension space, all non-empty closed sets contain the point $p$.
So $V \cup \set p = H \subseteq V^-$. 
From Set Closure is Smallest Closed Set in Topological Space, $S^*_p = H^- \subseteq V^- \implies V$ is dense in $T_{\bar p}^*$.
But because $V \subseteq S$, $V$ is dense in $S$ also.
From Subset of Countably Infinite Set is Countable, $V$ is countable as it is a subset of a countable set.
Thus $T$ is separable, which is a contradiction.
Finally, our assumption is false: $T$ is indeed separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Separable Spaces
Category:Open Extension Topology
Condition for Open Extension Space to be T0 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a $T_0$ (Kolmogorov) space $T$ is.
== Proof ==
By definition:
:$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$
Let $T = \struct {S, \tau}$ be a $T_0$ space.
Let $\forall x, y \in S$ such that $x \ne y$.
First, suppose $x \ne p \ne y$.
Then by definition of $T$ as a $T_0$ space:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
Then as $U \in \tau^*_{\bar p}$, it follows that:
:$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$
or:
:$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$
Now suppose that $y = p$.
Then by definition:
:$\forall U \in \tau^*_{\bar p}: x \in U, p \notin U$
unless $U = S^*_p$.
So we have shown that:
:$\forall x, y \in S^*_p$ such that $x \ne y$, either:
::$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$
:or:
::$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$
and so $T^*_{\bar p}$ is a $T_0$ space.
Now suppose that $T = \struct {S, \tau}$ is not a $T_0$ space.
Let $x, y \in S$ such that:
:$\forall U \in \tau: x \in U \implies y \in U$
:$\forall U \in \tau: y \in U \implies x \in U$
As $U \in \tau^*_{\bar p}$ it follows that:
:$\forall U \in \tau^*_{\bar p}: x \in U \implies y \in U$
:$\forall U \in \tau^*_{\bar p}: y \in U \implies x \in U$
and if $U = S^*_p$ the same applies.
Hence $T^*_{\bar p}$ is not a $T_0$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:T0 Spaces
Category:Open Extension Topology
Condition for Open Extension Space to be T5 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_{\bar p} = \struct {S^*_p, \tau^*_{\bar p} }$ be the open extension space of $T$.
Then $T^*_{\bar p}$ is a $T_5$ space $T$ is a $T_5$ space.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:T5 Spaces
Category:Open Extension Topology
Condition for Operation to be Left Distributive over Constant Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ be the constant operation for some $c \in S$.
Then:
:$\circ$ is left distributive over $\sqbrk c$
:
:$\forall x \in S: x \circ c = c$
== Proof ==
=== Sufficient Condition ===
Let $\circ$ be left distributive over $\sqbrk c$.
That is:
:$\forall x \in S: x \circ c = c$
=== Necessary Condition ===
Let:
:$\forall x \in S: x \circ c = c$
Then:
That is, $\circ$ is left distributive over $\sqbrk c$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.24$
Category:Constant Operation
Category:Distributive Operations
Condition for Operation to be Right Distributive over Constant Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ be the constant operation for some $c \in S$.
Then:
:$\circ$ is right distributive over $\sqbrk c$
:
:$\forall x \in S: c \circ x = c$
== Proof ==
=== Sufficient Condition ===
Let $\circ$ be right distributive over $\sqbrk c$.
That is:
:$\forall x \in S: c \circ x = c$
=== Necessary Condition ===
Let:
:$\forall x \in S: c \circ x = c$
Then:
That is, $\circ$ is right distributive over $\sqbrk c$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.24$
Category:Constant Operation
Category:Distributive Operations
Condition for Ordered Set of All Mappings to be Total Ordering

== Theorem ==
Let $S$ be a set.
Let $\struct {T, \preccurlyeq}$ be an ordered set.
Let $\struct {T^S, \preccurlyeq}$ denote the ordered set of all mappings from $S$ to $T$.
Then:
:$\struct {T^S, \preccurlyeq}$ is a totally ordered set
:
:$\card S \le 1$
or:
:$\card T \le 1$
== Proof ==
=== Sufficient Condition ===
Let $\struct {T^S, \preccurlyeq}$ be a totally ordered set.
Let $f, g \in T^S$ be arbitrary.
Let $S$ and $T$ be such that:
Let $f: S \to T$ be defined as:
As $\struct {T^S, \preccurlyeq}$ is a totally ordered set, either $f \preccurlyeq g$ or $g \preccurlyeq f$.
Suppose $f \preccurlyeq g$.
Then:
As $x \ne y$ this shows that $\preccurlyeq$ is not antisymmetric on $T$.
This contradicts the assumption that $\struct {T, \preccurlyeq}$ is an ordered set.
Hence if both $\card T > 1$ and $\card S > 1$ it cannot be the case that $\struct {T^S, \preccurlyeq}$ is a totally ordered set.
Thus by the Rule of Transposition, either $\card T \le 1$ or $\card S \le 1$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.20$
Category:Ordered Sets
Category:Total Orderings
Condition for Pairs of Lines through Origin to be Harmonic Conjugates

== Theorem ==
Consider $4$ lines $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$ through the origin $O$ whose equations embedded in the Cartesian plane are as follows:
Then the pairs of lines $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ are harmonic conjugates :
:$\dfrac {\lambda - \lambda'} {\lambda - \mu'} / \dfrac {\mu - \lambda'} {\mu - \mu'} = -1$
=== Homogeneous Quadratic Equation Form ===
== Proof ==
=== Sufficient Condition ===
Let $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ be harmonic conjugates as asserted.
A straight line in the plane which does not pass through $O$ will either:
:intersect all four of $\LL_1$, $\LL_2$, $\LL_3$ and $\LL_4$
or:
:be parallel to one such straight line and intersect the other three.
Hence, let a straight line be drawn parallel to $\LL_1$ so as to intersect each of $\LL_2$, $\LL_3$, $\LL_4$ at $L'$, $M$ and $M'$.
It would cut $\LL_1$ at $L$, but $L$ is the point at infinity of $\LL_1$.
From Straight Line which cuts Harmonic Pencil forms Harmonic Range, the points of intersection define a harmonic range.
We have that $L'M'$ is parallel to $\LL_1$.
From Harmonic Range with Unity Ratio, the harmonic range defined by $L'M'$ is such that $M$ is the midpoint of $L'M'$.
$L'M'$ can be expressed in slope-intercept form as $y = \lambda x + c$ for some $c \in \R_{\ne 0}$.
Hence the abscissae of the points $L'$, $m$ and $M'$ are respectively:
:$\dfrac c {\lambda' - \lambda}$, $\dfrac c {\mu - \lambda}$, $\dfrac c {\mu' - \lambda}$
Hence the condition that the pairs of lines $\set {\LL_1, \LL_2}$ and $\set {\LL_3, \LL_4}$ are harmonic conjugates is:
:$\dfrac 2 {\mu - \lambda} = \dfrac 1 {\lambda' - \lambda} + \dfrac 1 {\mu' - \lambda}$
which is more usually expressed as:
:$\dfrac {\lambda - \lambda'} {\lambda - \mu'} / \dfrac {\mu - \lambda'} {\mu - \mu'} = -1$
=== Sufficient Condition ===
== Also presented as ==
This result can also be presented as:
:$\dfrac {\paren {\lambda - \lambda'} \paren {\mu - \mu'} } {\paren {\lambda - \mu'} \paren {\mu - \lambda'} } = -1$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $20$. Condition that the two pairs of lines through the origin, $y = \lambda x$, $y = \mu x$ and $y = \lambda' x$, $y = \mu' x$ should be apolar
Category:Harmonic Ranges
Condition for Pairs of Lines through Origin to be Harmonic Conjugates/Homogeneous Quadratic Equation Form

== Theorem ==
Consider the two homogeneous quadratic equations:
each representing two straight lines through the origin.
Then the two straight lines represented by $(\text E1)$ are harmonic conjugates of the two straight lines represented by $(\text E2)$ :
:$a_1 b_2 + a_2 b_1 - 2 h_1 h_2 = 0$
== Proof ==
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $(\text E1)$ and $(\text E2)$ represent straight lines through the origin :
Let the two straight lines represented by $(\text E1)$ be defined by the equations:
Let the two straight lines represented by $(\text E2)$ be defined by the equations:
Then we can write the Condition for Pairs of Lines through Origin to be Harmonic Conjugates as:
:$(1): \quad 2 \paren {\lambda_1 \mu_1 + \lambda_2 \mu_2} = \paren {\lambda_1 + \mu_1} \paren {\lambda_2 + \mu_2}$
We can express $a_1 x^2 + 2 h_1 x y + b_1 y^2 = 0$ as:
:$b_1 \paren {y - \lambda_1 x} \paren {y - \mu_1 x} = 0$
and $a_2 x^2 + 2 h_2 x y + b_2 y^2 = 0$ as:
:$b_2 \paren {y - \lambda_2 x} \paren {y - \mu_2 x} = 0$
from which we obtain via Sum of Roots of Quadratic Equation and Product of Roots of Quadratic Equation:
Therefore $(1)$ reduces to:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $21$. Condition that the two pairs of lines $\ldots$ should be apolar
Category:Harmonic Ranges
Condition for Partition between Invertible and Non-Invertible Elements to induce Congruence Relation on Monoid/Counterexample

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$ such that $\struct {S, \circ}$ is specifically not a group.
Let $\struct {H, \circ}$ be the subgroup of $\struct {S, \circ}$ consisting of its invertible elements.
Let $N$ be the set of non-invertible elements of $\struct {S, \circ}$.
Let $\RR$ be the equivalence relation induced by the partition $\set {N \mid H}$.
Let $\circ$ be such that it is neither a cancellable operation nor a commutative operation.
Then it is not necessarily the case that either:
:$\RR$ is a congruence relation on $\circ$
or:
:the quotient structure $\struct {S / \RR, \circ_\RR}$ is isomorphic to $\struct {\Z_2, \times_2}$, the multiplicative monoid of integers modulo $2$.
== Proof ==
Consider the set of natural numbers $\N$.
Let $\N^\N$ denote the set of all mappings from $\N$ to $\N$.
Now consider the algebraic structure $\struct {\N^\N, \circ}$, where $\circ$ in this context denotes composition of mappings.
From Set of all Self-Maps under Composition forms Monoid, $\struct {\N^\N, \circ}$ is a monoid.
From Composition of Mappings is not Commutative, $\circ$ is not a commutative operation.
Let $g: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map g x = \begin {cases} x - 1 & : x > 0 \\ 0 & : x = 0 \end {cases}$
$g$ is not injective, as:
:$\map g 0 = \map g 1 = 0$
From Injection iff Left Cancellable, $g$ is not cancellable.
Thus we have that $\circ$ is not a cancellable operation.
So we have a monoid $\struct {\N^\N, \circ}$ whose operation is neither cancellable nor commutative.
From Existence of One-Sided Inverses on Natural Numbers whose Composition is Identity Mapping, there exist mappings $f: \N \to \N$ and $g: \N \to \N$ such that:
:$g \circ f = I_\N$
such that:
:$g$ is surjective but not injective
:$f$ is injective but not surjective.
By definition of a bijection, neither $f$ nor $g$ are invertible elements of $\struct {\N^\N, \circ}$ since they are not bijective.
Hence we have $f, g \in N$ while $g \circ f = I_\N \in H$.
Also note that $f \circ f$ is not surjective, so $f \circ f \in N$.
By definition of $\RR$:
:$x \mathrel \RR y \iff \set {x, y} \subseteq N \lor \set {x, y} \subseteq H$
We now have:
:$g \mathrel \RR f \land f \mathrel \RR f$
:$\neg \paren {\paren {g \circ f} \mathrel \RR \paren {f \circ f}}$
thus $\RR$ is not a congruence relation, and the quotient structure $\struct {S / \RR, \circ_\RR}$ is not well-defined.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.17 \ \text {(b)}$
Category:Condition for Partition between Invertible and Non-Invertible Elements to induce Congruence Relation on Monoid
Condition for Planes to be Parallel

== Theorem ==
Let $P: \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be a plane in $\R^3$.
Then the plane $P'$ is parallel to $P$ there is a $\gamma' \in \R$ such that:
:$P' = \set {\tuple {x_1, x_2, x_3} \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'}$
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Let $P' \ne P$ be a plane given by the equation:
:$\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'$
 we have a point:
:$\mathbf x = \tuple {x_1, x_2, x_3} \in P \cap P'$
Then, as $\mathbf x \in P$, it also satisfies:
:$\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$
It follows that $\gamma = \gamma'$, so $P = P'$.
This contradiction shows that $P \cap P' = \O$, that is, $P$ and $P'$ are parallel.
The remaining case is when $P' = P$.
By definition, $P$ is parallel to itself. 
The result follows.
== Also see ==
* Condition for Straight Lines in Plane to be Parallel
== Sources ==
* : Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations
Category:Planes
Category:Solid Analytic Geometry
Condition for Point being in Closure

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $H^-$ denote the closure of $H$ in $T$.
Let $x \in S$.
Then $x \in H^-$ every open neighborhood of $x$ contains a point in $H$.
=== Metric Space ===
=== Topological Vector Space ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.12$
Category:Set Closures
Category:Condition for Point being in Closure
Condition for Point being in Closure/Metric Space

== Theorem ==
Let $M = \struct {S, d}$ be a metric space.
Let $H \subseteq S$.
Let $\map \cl H$ denote the closure of $H$ in $M$.
Let $x \in S$.
Then $x \in \map \cl H$ :
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.
== Proof ==
By definition of closure of $H$ in $M$:
:$\map \cl H = H^i \cup H'$
where:
:$H^i$ denotes the set of isolated points of $H$
:$H'$ denotes the set of limit points of $H$.
=== Sufficient Condition ===
Let $x \in \map \cl H$.
Suppose $x \in H^i$.
By definition of isolated point:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set a$
But also by definition of isolated point, we do at least have that:
:$x \in H$
and from Center is Element of Open Ball, we have that:
:$\forall \epsilon \in \R_{>0}: x \in \map {B_\epsilon} x$
and so:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$
Suppose $x \in H'$:
By definition of limit point:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} x \setminus \set x} \cap H \ne \O$
and so:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$
So in both cases:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$
=== Necessary Condition ===
Let $x$ be such that:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$
To show that $x \in \map \cl H$, it is sufficient to show that either:
:$x \in H^i$
or that:
:$x \in H'$
There are $2$ cases to consider.
$(1): \quad$ Suppose that:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$
Then $x$ is by definition an isolated point of $H$.
That is:
:$x \in H^i$
$(2): \quad$ Suppose that:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \set x$
Let $\epsilon$ be arbitrary.
As $\map {B_\epsilon} x \cap H \ne \O$, it must follow that:
:$\exists y \in \map {B_\epsilon} x \cap H: y \ne x$
and so:
:$y \in \paren {\map {B_\epsilon} x \setminus \set x} \cap H$
Hence:
:$\paren {\map {B_\epsilon} x \setminus \set x} \cap H \ne \O$
Then $x$ is by definition a limit point of $H$.
That is:
:$x \in H'$
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.12$
Category:Set Closures
Category:Condition for Point being in Closure
Condition for Point being in Closure/Proof 1

== Theorem ==
== Proof ==
From the definition of closure, we have that $H^-$ is the union of $H$ and all the limit points of $H$ in $T$.
By definition, an open neighborhood of $x$ in $T$ is an open set of $T$ which contains $x$.
=== Necessary Condition ===
Let $x \in H^-$.
Then either:
:$(1): \quad x \in H$, in which case every open neighborhood of $x$ in $T$ trivially contains a point in $H$ (that is, $x$ itself);
:$(2): \quad x$ is a limit point of $H$ in $T$.
Suppose $(2)$ holds.
Then it follows directly from the definition of limit point that every open neighborhood of $x$ in $T$ contains a point in $H$ other than $x$.
=== Sufficient Condition ===
Suppose that every open neighborhood of $x$ in $T$ contains a point in $H$.
If $x \in H$, then $x$ is in the union of $H$ and all the limit points of $H$ in $T$.
Hence by definition of closure:
:$x \in H^-$
If $x \notin H$ then $x$ must be a limit point of $H$ by definition.
So again, $x$ is in the union of $H$ and all the limit points of $H$ in $T$.
Hence by definition of closure:
:$x \in H^-$
Category:Condition for Point being in Closure
Condition for Point being in Closure/Proof 2

== Theorem ==
== Proof ==
The condition to be proved is equivalent to showing that $x \in H^-$ , for every open neighborhood $U$ of $x$, the intersection $H \cap U$ is non-empty.
For $U \in \tau$, let $U^{\complement}$ denote the relative complement of $U$ in $S$.
By definition, $U^{\complement}$ is closed in $T$
We have that:
Thus:
:$x \in U \iff x \notin H^-$
The result follows from the Rule of Transposition.
Category:Condition for Point being in Closure
Condition for Point being in Closure/Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$. 
Let $A \subseteq X$. 
Let $A^-$ denote the closure of $A$ in $X$.
Let $x \in X$. 
Then $x \in A^-$ :
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$. 
== Proof ==
From the definition of the closure of $A$, $A^-$ is the set of adherent points of $A$.
So $x \in A^-$ for each open neighborhood $U$ of $x$ we have: 
:$U \cap A \ne \O$ 
From Classification of Open Neighborhoods in Topological Vector Space, every such $U$ has the form $U = x + V$ for some open neighborhood $V$ of ${\mathbf 0}_X$, and conversely every set of this form is an open neighborhood of $x$. 
So we have $x \in A^-$ :
:for each open neighborhood $V$ of ${\mathbf 0}_X$ we have $\paren {x + V} \cap A \ne \O$. 
Category:Condition for Point being in Closure
Category:Topological Vector Spaces
Condition for Point to be Center of Circle

== Theorem ==
== Proof ==
Let $ABC$ be a circle, and $D$ a point in it such that $DA = DB = DC$.
Then $D$ can be proved to be the center of circle $ABC$ as follows.
:
Join $AB$ and $BC$, and bisect them at $E$ and $F$.
Join $ED$ and $FD$, and produce them in both directions to $GK$ and $HL$.
We have that $AE = EB$, $AD = BD$ and $ED$ is common.
From Triangle Side-Side-Side Congruence, $\triangle AED = \triangle BED$ and so $\angle AED = \angle BED$.
So from , each of $\angle AED$ and $\angle BED$ is a right angle.
Hence from the porism to Finding Center of Circle, the center of circle $ABC$ is on $GK$.
The same applies to $\triangle BFD = \triangle CFD$, and so by the same construction, the center of circle $ABC$ is also on $HL$.
The straight lines $GK$ and $HL$ have only point $D$ in common.
Hence the result.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles