nyuuzyou/wikis huggingface dataset:
Like Electric Charges Repel

== Theorem ==
Let $a$ and $b$ be stationary particles, each carrying an electric charge of $q_a$ and $q_b$ respectively.
Let $q_a$ and $q_b$ be of the same polarity.
That is, let $q_a$ and $q_b$ be like charges.
Then the forces exerted by $a$ on $b$, and by $b$ on $a$, are such as to cause $a$ and $b$ to repel each other.
== Proof ==
By Coulomb's Law of Electrostatics:
:$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$
where:
:$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$
:$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$
:$r$ is the distance between $a$ and $b$.
Let $q_a$ and $q_b$ both be positive.
Then $q_a q_b$ is a positive number multiplied by a positive number. 
Thus $q_a q_b$ is a positive number.
Similarly, let $q_a$ and $q_b$ both be negative.
Then $q_a q_b$ is a negative number multiplied by a negative number.
Thus $q_a q_b$ is similarly a positive number.
So, when $q_a$ and $q_b$ are like charges, $\dfrac {q_a q_b} {r^3}$ is a positive number.
Hence $\mathbf F_{a b}$ is in the same direction as the displacement vector from $a$ to $b$.
That is, the force exerted on $b$ by the electric charge on $a$ is in the direction away from $a$.
The same applies to the force exerted on $a$ by the electric charge on $b$.
That is, the force exerted on $b$ by the electric charge on $a$ is in the direction away from $b$.
The effect of these forces is to cause $a$ and $b$ to push apart, that is, to repel each other.
== Also see ==
* Unlike Electric Charges Attract
== Sources ==
* : Chapter $\text {I}$: Preliminary Survey: $\S 1$. Electrostatics
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge
Category:Electric Charge
Like Unit Vectors are Equal

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be like vector quantities.
Then:
:$\mathbf {\hat a} = \mathbf {\hat b}$
where $\mathbf {\hat a}$ and $\mathbf {\hat b}$ denote the unit vectors in the direction of $\mathbf a$ and $\mathbf b$.
== Proof ==
By definition of like vector quantities, $\mathbf a$ and $\mathbf b$ have the same direction.
By definition of unit vector, $\mathbf {\hat a}$ and $\mathbf {\hat b}$ are both of magnitude $1$.
Hence the result, by Equality of Vector Quantities.
Category:Vectors
Like Vector Quantities are Multiples of Each Other

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be like vector quantities.
Then:
:$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$
where:
:$\size {\mathbf a}$ denotes the magnitude of $\mathbf a$
:$\dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$ denotes the scalar product of $\mathbf b$ by $\dfrac {\size {\mathbf a} } {\size {\mathbf b} }$.
== Proof ==
By the definition of like vector quantities:
:$\mathbf a$ and $\mathbf b$ are like vector quantities they have the same direction.
By definition of unit vector:
:$\dfrac {\mathbf a} {\size {\mathbf a} } = \dfrac {\mathbf b} {\size {\mathbf b} }$
as both are in the same direction, and both have length $1$.
By definition of scalar division:
:$\dfrac 1 {\size {\mathbf a} } \mathbf a = \dfrac 1 {\size {\mathbf b} } \mathbf b$
Hence, multiplying by $\size {\mathbf a}$:
:$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and Subtraction of Vectors: $5$. Multiplication by a number
Category:Scalar Multiplication
Likelihood Function/Examples/Arbitrary Independent Observations

== Example of Likelihood Function ==
Let S$ be a sample of $n$ independent observations of a random variable with a given probability distribution.
The likelihood function $\map {\mathrm L} \theta$ is:
:$\map {\mathrm L} \theta := \map f {x_1, x_2, \ldots, x_n}$
Because of independence:
:$\map {\mathrm L} \theta = \map f {x_1, \theta} \map f {x_2, \theta} \cdots,\map f {x_n, \theta}$
Suppose $\theta_1$ and $\theta_2$ are values of $\theta$.
Suppose that:
:$\map {\mathrm L} {\theta_2} < \map {\mathrm L} {\theta_1}$
This implies that the sample has a smaller value of the joint frequency function if the unknown parameter is $\theta_2$ rather than $\theta_1$.
This in turn means that the sample is less likely to have come from a population where $\theta = \theta_2$ rather than where $\theta = \theta_1$.
This line of reasoning leads to the concept of maximum likelihood estimation.
== Sources ==
* 
* 
Category:Examples of Likelihood Functions
Limit Comparison Test

== Theorem ==
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$.
Let $\ds \frac {a_n} {b_n} \to l$ as $n \to \infty$ where $l \in \R_{>0}$.
Then the series $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ are either both convergent or both divergent.
== Proof ==
Let $\ds \sum_{n \mathop = 1}^\infty b_n$ be convergent.
Then by Terms in Convergent Series Converge to Zero, $\sequence {b_n}$ converges to zero.
A Convergent Sequence is Bounded.
So it follows that:
:$\exists H: \forall n \in \N_{>0}: a_n \le H b_n$
Thus, by the corollary to the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
Since $l > 0$, from Sequence Converges to Within Half Limit:
:$\exists N: \forall n > N: a_n > \dfrac 1 2 l b_n$
Hence the convergence of $\ds \sum_{n \mathop = 1}^\infty a_n$ implies the convergence of $\ds \sum_{n \mathop = 1}^\infty b_n$.
== Sources ==
* : $\S 6.26 \ (2)$
Category:Convergence Tests
Category:Series
Limit Inferior/Examples

== Examples of Limits Inferior ==
=== Sequence of Reciprocals ===
=== Divergent Sequence $\paren {-1}^n$ ===
=== Farey Sequence ===
=== Limit Inferior of $\paren {-1}^n \paren {1 + \dfrac 1 n}$ ===
Category:Examples of Limits Inferior
Limit Inferior/Examples/(-1)^n

== Example of Limit Inferior ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{>0}: a_n = \paren {-1}^n$
The limit inferior of $\sequence {a_n}$ is given by:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = -1$
== Proof ==
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
From Divergent Sequence may be Bounded, $\sequence {a_n}$ is bounded above by $1$ and bounded below by $-1$.
We have the subsequences:
:$(1): \quad \sequence {a_{n_r} }$ where $\sequence {n_r}$ is the integer sequence defined as $n_r = 2 r$
:$(2): \quad \sequence {a_{n_s} }$ where $\sequence {n_s}$ is the integer sequence defined as $n_s = 2 s + 1$.
We have that:
:$\sequence {a_{n_r} }$ is the sequence $1, 1, 1, 1, \ldots$
:$\sequence {a_{n_s} }$ is the sequence $-1, -1, -1, -1, \ldots$
and so:
:$\ds \map {\lim_{n \mathop \to \infty} } {a_{n_r} } = 1$
:$\ds \map {\lim_{n \mathop \to \infty} } {a_{n_s} } = -1$
Hence by definition of limit inferior:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = -1$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.14$: Example $\text {(ii)}$
Category:Examples of Limits Inferior
Limit Inferior/Examples/(-1)^n (1 + n^-1)

== Example of Limit Inferior ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{>0}: a_n = \paren {-1}^n \paren {1 + \dfrac 1 n}$
The limit inferior of $\sequence {a_n}$ is given by:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = -1$
This is not the same as:
:$\ds \inf_{n \mathop \ge 1} {\paren {-1}^n \paren {1 + \dfrac 1 n} }$
== Proof ==
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
We have that:
and:
:$\sequence {a_n}$ contains no subsequence which converges to a limit different from $-1$ or $1$.
Hence:
:$L = \set {-1, 1}$
and it follows that:
:$\ds \map {\liminf_{n \mathop \to \infty} } {\paren {-1}^n \paren {1 + \dfrac 1 n} } = 0$
But note that:
:$\ds \sup_{n \mathop \ge 1} \paren {-1}^n \paren {1 + \dfrac 1 n}$
occurs when $n = 1$, at which point:
:$\paren {-1}^n \paren {1 + \dfrac 1 n} = \paren {-1}^1 \paren {1 + \dfrac 1 1} = -2$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: Exercise $\S 5.15 \ (1) \ \text {(ii)}$
Category:Examples of Limits Inferior
Limit Inferior/Examples/Farey Sequence

== Example of Limit Inferior ==
Consider the Farey sequence:
:$\sequence {a_n} = \dfrac 1 2, \dfrac 1 3, \dfrac 2 3, \dfrac 1 4, \dfrac 2 4, \dfrac 3 4, \dfrac 1 5, \dfrac 2 5, \dfrac 3 5, \dfrac 4 5, \dfrac 1 6, \ldots$
The limit inferior of $\sequence {a_n}$ is given by:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = 0$
== Proof ==
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
From the definition of $F$ we have that:
:$\forall n \in \N_{>0}: 0 < a_n < 1$
From Lower and Upper Bounds for Sequences we have that:
:$L \subseteq \closedint 0 1$
Consider the subsequences:
:$(1): \quad \sequence {a_{n_r} } = \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \dfrac 1 5 \to 0$ as $n \to \infty$
:$(2): \quad \sequence {a_{n_s} } = \dfrac 1 2, \dfrac 2 3, \dfrac 3 4, \dfrac 4 5 \to 1$ as $n \to \infty$
Hence by definition of limit inferior:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = 0$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.14$: Example $\text {(iii)}$
Category:Examples of Limits Inferior
Limit Inferior/Examples/Sequence of Reciprocals

== Example of Limit Inferior ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{>0}: a_n = \dfrac 1 n$
The limit inferior of $\sequence {a_n}$ is given by:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = 0$
== Proof ==
From Sequence of Reciprocals is Null Sequence, $\sequence {a_n}$ is convergent:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
By Limit of Subsequence equals Limit of Real Sequence, all such subsequences have limit $0$.
 
Hence by definition of limit inferior:
:$\ds \map {\liminf_{n \mathop \to \infty} } {a_n} = \inf \set 0 = 0$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.14$: Example $\text {(i)}$
Category:Examples of Limits Inferior
Limit Inferior of Inclusion Net is Supremum of Directed Subset

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be an up-complete lattice.
Let $D \subseteq S$ be a directed subset of $S$.
Let $\struct {D, \preceq'}$ be a directed ordered subset of $L$.
Let $i_D: D \to S$, the inclusion mapping, be a net in $S$.
Then $\liminf i_D = \sup D$
== Proof ==
We will prove that:
:(lemma): $\forall j \in D: \map {\inf_L} {\map {\preceq'} j} = j$
Let $j \in D$.
By definitions of image of element and upper closure of element:
:$\map {\preceq'} j = j^{\succeq'}$
By Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure:
:$j^{\succeq'} = D \cap j^\succeq$
By Intersection is Subset:
:$j^{\succeq'} \subseteq j^\succeq$
By Infimum of Subset and Infimum of Upper Closure of Element:
:$j \preceq \map {\inf_L} {\map {\preceq'} j}$
By definition of reflexivity:
:$j \preceq' j$
By definition of image of element:
:$j \in \map {\preceq'} j$
By definitions of infimum and lower bound:
:$\map {\inf_L} {\map {\preceq'} j} \preceq j$
Thus by definition of antisymmetry:
:$\map {\inf_L} {\map {\preceq'} j} = j$
Thus
== Sources ==
* 
Category:Inclusion Mappings
Category:Limits Inferior of Nets
Limit Inferior of Norm of Weakly Convergent Sequence is Bounded Below by Norm of Weak Limit

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $x \in X$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ converging weakly to $x$. 
Then, we have: 
:$\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$
== Proof ==
From Existence of Support Functional, there exists $f \in X^\ast$ such that: 
:$\map f x = \norm x$
and:
:$\norm f_{X^\ast} = 1$
Since:
:$x_n \weakconv x$
we have: 
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f x$ 
That is:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \norm x$
From Absolute Value of Limit, we have: 
:$\ds \lim_{n \mathop \to \infty} \cmod {\map f {x_n} } = \norm x$ 
From Convergence of Limsup and Liminf, we have: 
:$\ds \liminf_{n \mathop \to \infty} \cmod {\map f {x_n} } = \norm x$
From Fundamental Property of Norm on Bounded Linear Functional, we have: 
:$\cmod {\map f {x_n} } \le \norm f_{X^\ast} \norm {x_n}$
Since:
:$\norm f_{X^\ast} = 1$
we have: 
:$\cmod {\map f {x_n} } \le \norm {x_n}$
So, from Inequality Rule for Real Sequences: 
:$\ds \liminf_{n \mathop \to \infty} \cmod {\map f {x_n} } \le \liminf_{n \mathop \to \infty} \norm {x_n}$
That is: 
:$\ds \norm x \le \liminf_{n \mathop \to \infty} \norm {x_n}$
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Limit Inferior of Repetition Net

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $N = \struct {\N, \le}$ be a directed ordered set.
Let $a, b \in S$.
Let $f = \sequence {c_i}_{i \mathop \in \N} = \tuple {a, b, a, b, \dots}: \N \to S$ be a net.
Then $\liminf \sequence {c_i}_{i \mathop \in \N} = a \wedge b$
== Proof ==
We will prove that
:(lemma): $\forall j \in \N: f \sqbrk {\le \paren j} = \set {a, b}$
Let $j \in \N$.
Let $x \in S$.
Assume:
:$x \in f \sqbrk {\le \paren j}$
By definition of image of set:
:$\exists i \in \le \paren j: x = \map f i$
By definition of $f$:
:$x = a$ or $x = b$
Thus by definition of unordered tuple:
:$x \in \set {a, b}$
Assume:
:$x \in \set {a, b}$
We have:
:$j \le j$ and $j \le j+1$
By definition of image of element:
:$j, j + 1 \in \le \paren j$
By definition of $f$:
:$\set {\map f j, \map f {j + 1} } = \set {a, b}$
By definition of image of set:
:$\map f j, \map f {j + 1} \in f \sqbrk {\le \paren j}$
Thus by definition of unordered tuple:
$x \in f \sqbrk {\le \paren j}$
Thus by definition of set equality:
:$f \sqbrk {\le \paren j} = \set {a, b}$
Thus:
== Sources ==
* 
Category:Mapping Theory
Category:Limits Inferior of Nets
Limit Inferior of Restriction Net is Supremum of Image of Directed Subset

== Theorem ==
Let $L = \struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be up-complete lattices.
Let $f: S \to T$ be an increasing mapping.
Let $D \subseteq S$ be a directed subset of $S$.
Let $\struct {D, \preceq'}$ be a directed ordered subset of $L$.
Let $f \restriction D: D \to T$, the restriction of mapping, be a net in $T$.
Then $\map \liminf {f \restriction D} = \map \sup {f \sqbrk D}$
== Proof ==
We will prove that
:(lemma): $\forall j \in D: \map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } = \map f j$
Let $j \in D$.
By definitions of image of element and upper closure of element:
:$\map {\preceq'} j = j^{\succeq'}$
By Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure:
:$j^{\succeq'} = D \cap j^{\succeq_1}$
By Intersection is Subset:
:$j^{\succeq'} \subseteq j^{\succeq_1}$
By Image of Subset under Mapping is Subset of Image:
:$\paren {f \restriction D} \sqbrk {j^{\succeq'} } \subseteq \paren {f \restriction D} \sqbrk {j^{\succeq_1} }$
By Infimum of Subset and Infimum of Image of Upper Closure of Element under Increasing Mapping:
:$\map f j \preceq_2 \map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} }$
By definition of reflexivity:
:$j \preceq' j$
By definition of image of element:
:$j \in \map {\preceq'} j$
By definition of image of set:
:$\map f j \in \paren {f \restriction D} \sqbrk {\map {\preceq'} j}$
By definitions of infimum and lower bound:
:$\map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } \preceq_2 \map f j$
Thus by definition of antisymmetry:
:$\map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } = \map f j$
Thus
== Sources ==
* 
Category:Restrictions
Category:Limits Inferior of Nets
Limit Ordinal Equals its Union

== Theorem ==
Let $\lambda$ be a limit ordinal.
Then:
:$\lambda = \bigcup \lambda$
where $\bigcup \lambda$ denotes the union of $\lambda$.
== Proof ==
From Union of Ordinal is Subset of Itself:
:$\bigcup \lambda \subseteq \lambda$
Suppose $x \in \lambda$.
By Successor of Ordinal Smaller than Limit Ordinal is also Smaller:
:$x^+ < \lambda$
and so:
:$x \in x^+$ and $x^+ \in \lambda$
from which:
:$x \in \bigcup \lambda$
That is:
:$\lambda \subseteq \bigcup \lambda$
Hence by set equality:
:$\lambda = \bigcup \lambda$
== Also presented as ==
This can also be presented in the form:
:$\lambda = \ds \bigcup_{\alpha \mathop \in \lambda} \alpha$
which by definition of union can be seen to be equivalent to $\lambda = \bigcup \lambda$.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.20 \ (3)$
* 
Category:Set Union
Category:Limit Ordinals
Limit Ordinals Closed under Ordinal Exponentiation

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $y$ be a limit ordinal.
Let $x^y$ denote ordinal exponentiation.
Then:
:If $x > 1$, then $x^y$ is a limit ordinal.
:If $x \ne \O$, then $y^x$ is a limit ordinal.
== Proof ==
Suppose $x > 1$.
Suppose also that $x^y$ is the successor of some ordinal $w$.
By definition of ordinal multiplication:
:$\ds x^y = \bigcup_{z \mathop \in y} x^z$ 
Then:
But $z^+ \in y$ by Successor of Ordinal Smaller than Limit Ordinal is also Smaller.
So $w^+ \in x^y$ and $w^+ \in w^+$.
This creates a membership loop and thus is a contradiction by No Membership Loops.
For the second part, since $x$ is not the empty set it follows that $x = z^+$ for some $z$ or that $x$ is a limit ordinal.
If $x$ is a limit ordinal, then $y^x$ is a limit ordinal by the first part.
If $x$ is the successor of $z$, then:
:$y^x = y^w × y$ by the definition of ordinal exponentiation.
Then, $y^x$ is a limit ordinal by Limit Ordinals Preserved Under Ordinal Multiplication.
== Sources ==
* : $\S 8.39$
Category:Ordinal Arithmetic
Limit Ordinals Preserved Under Ordinal Addition

== Theorem ==
Let $x$ and $y$ be ordinals such that $x$ is a limit ordinal.
Then $y + x$ is a limit ordinal.
That is, letting $K_{II}$ denote the class of all limit ordinals:
:$\forall x \in K_{II}: y + x \in K_{II}$
== Proof ==
The result is now obtained by Proof by Contradiction:
Assume that $y + x = z^+$.
But $w \in x \implies w^+ \in x$ by Successor of Ordinal Smaller than Limit Ordinal is also Smaller.
But $z^+ \in z^+$ is clearly a membership loop, and therefore, our assumption must be wrong.
By disjunctive syllogism:
:$y + x \in K_{II}$
== Sources ==
* : $\S 8.11$
Category:Ordinal Arithmetic
Limit Ordinals Preserved Under Ordinal Multiplication

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x$ be non-empty.
Let $y$ be a limit ordinal.
It follows that the ordinal product $\left({x \times y}\right)$ is a limit ordinal.
== Proof ==
$y$ is a limit ordinal and thus is nonzero, by definition.
$x$ and $y$ are both nonzero.
So by Ordinals have No Zero Divisors:
:$x \times y \ne 0$
So by definition of limit ordinal:
:$x \times y \in K_{II} \lor \exists z \in \On: x \times y = z^+$
Suppose that $x \times y = z^+$ for some ordinal $z$.
It follows that:
But by Subset is Right Compatible with Ordinal Addition:
:$\paren {x \times w} + 1 \subseteq \paren {x \times w} + x$
Therefore:
But by Successor of Ordinal Smaller than Limit Ordinal is also Smaller:
:$w^+ \in y$
Therefore:
:$z^+ \in x \times y$
contradicting the fact that $z^+ = x \times y$.
Thus:
:$z^+ \ne x \times y$
and:
:$x \times y \in K_{II}$
== Sources ==
* : $\S 8.23$
Category:Ordinal Arithmetic
Limit Point/Examples/End Points of Real Interval

== Examples of Limit Points ==
The real number $a$ is a limit point of both the open real interval $\openint a b$ as well as of the closed real interval $\closedint a b$.
It is noted that $a \in \closedint a b$ but $a \notin \openint a b$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.10 \ \text {(a)}$
Category:Examples of Limit Points
Limit Point/Examples/Union of Singleton with Open Real Interval

== Examples of Limit Points ==
Let $\R$ be the set of real numbers.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
:$H = \set 0 \cup \openint 1 2$
Then $0$ is not a limit point of $H$.
== Proof ==
Consider the open $1$- ball $\map {B_1} 0$ of $0$.
Then the only element of $H$ which is in $\map {B_1} 0$ is $0$ itself.
Hence $0$ is not a limit point of $H$ by definition.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.10 \ \text {(b)}$
Category:Examples of Limit Points
Limit Point iff Superfilter Converges

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on $S$.
Let $x \in S$.
Then $x$ is a limit point of $\FF$ there exists a superfilter $\FF'$ of $\FF$ on $S$ which converges to $x$.
== Proof ==
Let $x$ be a limit point of $\FF$.
Define:
:$\BB := \set {F \cap U : F \in \FF \text{ and } U \text{ is a neighborhood of } x}$
Then $\BB$ is filter basis by definition.
Let $\FF'$ be the corresponding generated filter.
By construction we have $\FF \subseteq \FF'$ and $U \in \FF'$ for every neighborhood $U$ of $x$.
Thus $\FF'$ converges to $x$.
Let there be a filter $\FF'$ on $X$ satisfying $\FF \subseteq \FF'$ which converges to $x$.
Let $U \subseteq X$ be a neighborhood of $x$.
Let $F \in \FF$.
Then:
:$U, F \in \FF'$
and therefore:
: $U \cap F \in \FF'$
Because $\O \notin \FF'$ it follows that:
: $U \cap F \ne \O$
This holds for any neighborhood $U$ of $x$.
Hence $x$ is a limit point of $F$.
Therefore:
:$x \in \overline F$
Because this holds for all $F \in \FF$:
:$\ds x \in \bigcap_{F \mathop \in \FF} \overline F$
and so $x$ is a limit point of $\FF$.
Category:Limit Points
Category:Filter Theory
Limit Point in Metric Space iff Limit Point in Topological Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $T = \struct {A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Then:
:$x \in H$ is a limit point in $M$ $x$ is a limit point in $T$
== Proof ==
From Open Balls form Local Basis for Point of Metric Space, the set:
:$\BB_x = \set{\map {B_\epsilon} x : \epsilon \in \R_{>0}}$
is a local basis of $x$.
From Local Basis Test for Limit Point:
:$x$ is a limit point of $H$ in $T$ $\forall \epsilon \in \R_{>0}: H \cap \map {B_\epsilon} x \setminus \set x \ne \O$
By definition of a limit point in $M$:
:$x$ is a limit point of $H$ in $T$ $x$ is an limit point of $H$ in $M$
Category:Limit Points
Limit Point of Sequence in Discrete Space not always Limit Point of Open Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $U \in \tau$ be an open set of $T$.
Let $\sequence {x_n}$ be a sequence in $U$.
Let $x$ be the limit of $\sequence {x_n}$.
Then $x$ is not always a limit point of $U$.
== Proof ==
Let $x \in S$.
By definition of discrete space:
:$U = \set x$ is an open set of $T$.
Consider the sequence $\sequence {x_n}$ defined as:
:$\forall n \in \N: x_n = x$
That is:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
Thus $x$ is the limit point of $\sequence {x_n}$.
But:
:$U \setminus \set x = \O$
and so $x$ is not a limit point of $U$.
Hence the result.
== Also see ==
* Point in Discrete Space is Adherent Point
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $3$
Category:Discrete Topology
Category:Limit Points
Category:Limits
Limit Point of Sequence is Adherent Point of Range

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\sequence{x_n}$ be a sequence in $S$.
Let $\alpha$ be a limit of $\sequence{x_n}$.
Then $\alpha$ is an adherent point of $\set{x_n: n \in \N}$.
== Proof ==
By definition of Limit of sequence:
:$\forall U \in \tau : \exists N \in \N : \forall n \ge N : x_n \in U$
Hence:
:$\forall U \in \tau : U \cap \set{x_n: n \in \N} \ne \O$.
By definition $\alpha$ is an adherent point of $\set{x_n: n \in \N}$.
Category:Limits
Category:Adherent Points