nyuuzyou/wikis huggingface dataset:
Limit Point of Sequence may only be Adherent Point of Range

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ may be only an adherent point of $A$ and not a limit point of $A$.
== Proof ==
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Let $x \in S$.
Then 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:
:$\sequence {x_n} = \tuple {x, x, x, \ldots}$
From Limit Point of Sequence in Discrete Space not always Limit Point of Open Set:
:$x$ is not a limit point of $U$.
But from Limit Point of Sequence is Adherent Point of Range:
:$x$ is an adherent point of $\set x$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
Category:Limits
Category:Adherent Points
Limit Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence of distinct terms of $S$.
Let $\alpha$ be a limit point of $\sequence {x_n}$.
Then $\alpha$ is also an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
== Proof ==
Let $\alpha$ be an limit point of $\sequence {x_n}$.
From Limit of Sequence is Accumulation Point‎, $\alpha$ is an accumulation point of $\sequence {x_n}$.
From Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range, $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
Category:Limit Points
Category:Omega-Accumulation Points
Limit Point of Set in Complex Plane not Element is Boundary Point

== Theorem ==
Let $S \subseteq \C$ be a subset of the complex plane.
Let $z \in \C$ be a limit point of $S$ such that $z \notin S$.
Then $z$ is a boundary point of $S$.
== Proof ==
Let $z \in \C$ be a limit point of $S$ such that $z \notin S$.
Suppose $z$ is not a boundary point of $S$.
Then there exists an $\epsilon$-neighborhood $\map {N_\epsilon} z$ of $z$ such that either:
:$(1): \quad$ All elements of $\map {N_\epsilon} z$ are in $S$
or
:$(2): \quad$ All elements of $\map {N_\epsilon} z$ are not in $S$.
If $(1)$, then $z \in S$ which is a contradicts $z \notin S$.
If $(2)$, then $z$ is not a limit point of $S$, which also contradicts the conditions on $z$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Point Sets: $126$
Category:Complex Analysis
Limit Point of Set may or may not be Element of Set

== Theorem ==
Let $S$ be a set.
Let $H \subseteq S$ be a subset of $S$.
Let $T = \struct {H, \tau}$ be a topological space on the underlying set $H$.
Let $a \in S$ be a limit point of $T$.
Then $a$ may or may not be an element of $H$.
Whether it is or not depends upon the nature of both $a$ and $T$.
== Proof ==
Consider:
:the open real interval $\openint a b$
:the closed real interval $\closedint a b$.
Both of these are subsets of the set of real numbers $\R$.
From Limit Point Examples: End Points of Real Interval, $a$ is a limit point of both $\openint a b$ and $\closedint a b$.
But $a \in \closedint a b$ while $a \notin \openint a b$.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Definition $3.7.10 \ \text {(a)}$
Category:Limit Points
Limit Point of Subset is Limit Point of Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subset of $S$ such that
:$A \subseteq B$
Let $x$ be a point of $S$.
Then:
:if $x$ is limit point of $A$, then $x$ is limit point of $B$.
== Proof ==
Assume $x$ is limit point of $A$.
By definition of limit point it suffices to prove
:$\forall U \in \tau: x \in U \implies B \cap \paren {U \setminus \set x} \ne \O$
Let $U \in \tau$ such that
:$x \in U$
By definition of limit point:
:$A \cap \paren {U \setminus \set x} \ne \O$
By Set Intersection Preserves Subsets/Corollary:
:$A \cap \paren {U \setminus \set x} \subseteq B \cap \paren {U \setminus \set x}$
Thus:
:$B \cap \paren {U \setminus \set x} \ne \O$
Category:Limit Points
Limit Point of Subset of Metric Space is at Zero Distance

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be a limit point of $H$.
Let $\map d {x, H}$ denote the distance between $x$ and $H$:
:$\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$
Then:
:$\map d {x, H} = 0$
== Proof ==
Let $x$ be a limit point of $H$.
 $\map d {x, H} \ne 0$.
By definition of metric, that means:
:$\map d {x, H} > 0$
Then:
:$\exists \epsilon \in \R_{>0}: \forall y \in H: \map d {x, y} > \epsilon$
That is:
:$\forall y \in H: y \notin \map {B_\epsilon} x \setminus \set x$
where $\map {B_\epsilon} x \setminus \set x$ denotes the deleted $\epsilon$-neighborhood of $x$.
That is:
:$\paren {\map {B_\epsilon} x \setminus \set x} \cap H = \O$
Hence, by definition, $x$ is not a limit point of $H$.
From this contradiction it follows that it cannot be the case that $\map d {x, H} \ne 0$
That is:
:$\map d {x, H} = 0$
== Warning ==
The converse is not necessarily true.
It may be the case that $H$ has no limit points.
The corresponding result is in fact:
If $\map d {x, H} = 0$, then either:
:$x$ is a limit point of $H$
or:
:$x \in H$
This is demonstrated in Point not in Subset of Metric Space iff Distance from Elements is Greater than Zero.
Category:Limit Points
Category:Distance Function
Limit Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1

== Theorem ==
Let $\sequence {a_n}$ denote the sequence defined as:
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $S$ denote the set of terms of $\sequence {a_n}$ considered as a subset of $\struct {\R, \tau_d}$.
Then $0$ is a limit point of $S$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
Hence $\dfrac 1 m \in \openint {-\epsilon} \epsilon$.
Hence the result by definition of limit point of $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $3$
Category:Sequences
Limit Points in Closed Extension Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Let $x \in S$.
Then $x$ is a limit point of $p$.
=== Limit Points in Subset ===
== Proof ==
Every open set of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau^*_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a point, $x$ is a limit point of $p$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:Limit Points in Closed Extension Space
Limit Points in Closed Extension Space/Subset

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Let $U \subseteq S^*_p$ such that $p \in U$.
Let $x \in S$.
Then $x$ is a limit point of $U$.
== Proof ==
Every open set of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau^*_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a set, $x$ is a limit point of $U$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:Limit Points in Closed Extension Space
Limit Points in Excluded Point Space

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Let $x \in S$ such that $x \ne p$.
Then $p$ is the only limit point of $x$.
Similarly, let $U \subseteq S$.
Then $p$ is the only limit point of $U$.
== Proof ==
Let $U \subseteq S$.
Let $x \in S$ such that $x \ne p$.
From:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Limit Points in Open Extension Space
it follows that:
:$p$ is a limit point of $U$
:$p$ is a limit point of $x$.
Now suppose $y \in S$ such that $y \ne p$.
We have by definition of excluded point space that $\set y$ is open in $T$.
So there is no $z \in \set y: z \ne y, z \in U$. 
Hence $y$ can not be a limit point of $U$.
Similarly $x \notin \set y$.
So $y$ can not be a limit point of $x$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $5$
Category:Examples of Limit Points
Category:Excluded Point Topology
Limit Points in Fort Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space.
Let $x \in S$ such that $x \ne p$.
Then $p$ is the only limit point of $x$.
== Proof ==
From , we have $\relcomp S {\set x} \in \tau_p$.
For any $y \ne x$, $y \in \relcomp S {\set x}$.
Therefore $\relcomp s {\set x}$ is an open neighborhood of $y$.
From we also have $x \notin \relcomp S {\set x}$.
Hence $y$ is not a limit point of $x$.
By , $x$ cannot be a limit point of $x$.
Therefore $x$ has no limit points in $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $23 \text { - } 24$. Fort Space: $7$
Category:Examples of Limit Points
Category:Fort Spaces
Limit Points in Open Extension 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 $T$.
Let $x \in S$.
Then $p$ is a limit point of $x$.
=== Limit Points of Subset ===
== Proof ==
Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition.
So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$.
So by definition of the limit point of a point, $p$ is a limit point of $x$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Examples of Limit Points
Category:Open Extension Topology
Category:Limit Points in Open Extension Space
Limit Points in Open Extension Space/Subset

== 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 $T$.
Let $U \subseteq S^*_p$.
Then $p$ is a limit point of $U$.
== Proof ==
Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition.
So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$.
So by definition of the limit point of a set, $p$ is a limit point of $U$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $9$
Category:Limit Points in Open Extension Space
Limit Points in Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a limit point of $p$.
=== Limit Points in Subset ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $2$
Category:Examples of Limit Points
Category:Particular Point Topology
Category:Limit Points in Particular Point Space
Limit Points in Particular Point Space/Proof 1

== Theorem ==
== Proof ==
Let $U \in \tau$ be an open set of $T$.
Then by definition of $T$:
:$p \in U$
That is, $U$ contains (at least) one point of $S$ which is distinct from $x$.
As $U$ is arbitrary, it follows that all open set of $T$ have this property.
The result follows by definition of the limit point of a point.
Category:Limit Points in Particular Point Space
Limit Points in Particular Point Space/Subset

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a limit point of $U$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $3$
Category:Limit Points in Particular Point Space
Limit Points in Particular Point Space/Subset/Proof 1

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
== Proof ==
Every open set of $T = \struct {S, \tau_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a set, $x$ is a limit point of $U$.
Category:Limit Points in Particular Point Space
Limit Points in T1 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which satisfies the $T_1$ (Fréchet) axiom.
Let $H \subset S$ be any subset of $S$.
Let $x \in H$.
Then $x$ is a limit point of $H$ every neighborhood of $x$ contains infinitely many points of $H$.
== Proof ==
=== Necessary Condition ===
Suppose every neighborhood of $x$ contains infinitely many points of $H$.
Let $N$ be an open neighborhood of $x$.
Then $H \cap N$ is infinite, and so is $H \cap \paren {N \setminus \set x}$.
Since $N$ is arbitrary, $x$ is a limit point of $H$ by definition.
=== Sufficient Condition ===
We prove the contrapositive:
Suppose there exists some neighborhood $V$ of $x$ containing only finitely many points of $H$.
Then there exists some open set $N \subseteq V$ such that $x \in N$.
By Set Intersection Preserves Subsets, $N \cap H \subseteq V \cap H$.
By Subset of Finite Set is Finite, $N \cap H$ is finite.
For each $y \in H \cap N$ and $y \ne x$, since $T$ is $T_1$:
:$\exists U_y \in \tau: x \in U_y, y \notin U_y$
Consider the set $\ds U = N \cap \bigcap_{y \mathop \in H \cap \paren {N \setminus \set x} } U_y$.
This is a finite intersection of open sets.
By General Intersection Property of Topological Space, $U \in \tau$.
By , since $x \in N$ and $x \in U_y$ for all $y \in H \cap \paren {N \setminus \set x}$, we have $x \in U$.
Therefore $U$ is an open neighborhood of $x$.
 $H \cap \paren {U \setminus \set x} \ne \O$.
Let $z \in H \cap \paren {U \setminus \set x}$.
Then:
But from definition of $U_z$: 
:$z \in H \cap \paren {N \setminus \set x} \implies z \notin U_z$
which is a contradiction.
This shows that $H \cap \paren {U \setminus \set x} = \O$.
By , $x$ is not a limit point of $H$.
Thus the contrapositive is proved.
Category:Examples of Limit Points
Category:T1 Spaces
Limit Points in Uncountable Fort Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an uncountable Fort space.
Let $U \subseteq S$ be a countably infinite subset of $S$.
Then $p$ is the only limit point of $U$.
== Proof ==
Suppose $y \in S, y \ne p$.
We have by definition of Fort space that $\set y$ is open in $T$.
So there is no $z \in \set y: z \ne y, z \in U$. 
Hence $y$ can not be a limit point of $U$.
Suppose $p \in V \in \tau_p$ for some $V \subseteq S$.
From , $\relcomp S V$ is finite.
Then:
Thus $\relcomp S {V \setminus \set p}$ is also finite.
Since $U$ is countably infinite:
:$U \nsubseteq \relcomp S {V \setminus \set p}$.
From Empty Intersection iff Subset of Complement:
:$U \cap \paren {V \setminus \set p} \ne \O$.
Since $V$ is arbitrary, every open neighborhood of $p$ has non-empty intersection with $U$.
Thus $p$ is the only limit point of $U$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $24$. Uncountable Fort Space: $4$
Category:Examples of Limit Points
Category:Fort Spaces
Limit Points of Countable Complement Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement space.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then every point of $S$ is a limit point of $H$.
Let $H \subseteq S$ be an countable or finite subset of $S$.
Then $H$ contains all its limit points.
== Proof ==
Let $U \in \tau$ be any open set of $T$.
Let $H \subseteq S$ be an uncountable.
From Uncountable Subset of Countable Complement Space Intersects Open Sets, we have that $U \cap H$ is uncountable $H$ is uncountable.
Let $x \in S$.
Then every open set $U$ in $T$ such that $x \in U$ also contains an uncountable number of points of $H$ other than $x$.
Thus, by definition, $x$ is a limit point of $H$.
Now let $H$ be a countable subset of $S$.
By definition $\relcomp S H$ is open in $T$ and so $H$ is closed.
So by Closed Set Equals its Closure we have by definition that if $x$ is a limit point of $H$, then $x \in H$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $3$
Category:Examples of Limit Points
Category:Countable Complement Topology
Limit Points of Either-Or Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Let $H \subseteq S$ be any subset of $S$.
Then no element of $S$ can be a limit point of $H$ except $0$.
== Proof ==
Let $x \in S$ such that $x \ne 0$.
Then, as $0 \notin \set x$, we have by definition of the either-or topology that $x$ is open in $T$.
So whatever $H$ is, $\set x$ never contains any points of $H$ which are different from $x$.
So $x$ can not be a limit point of $H$.
However, every open set of $T$ which contains $0$ also contains the interval $\openint {-1} 1$ and so can contain points of $H$.
It follows that the only non-empty sets of $T$ for which $0$ is not a limit point are $\set {-1}$, $\set 0$ and $\set 1$ and their unions.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $5$
Category:Examples of Limit Points
Category:Either-Or Topology
Limit Points of Indiscrete Space

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space consisting of at least two points.
Let $H$ be a subset of $T$ such that $H \ne \O$.
Then every point of $T$ is a limit point of $H$.
== Proof ==
By definition, $x \in \struct {S, \tau}$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
Here, of course, there is only one open set that contains any points at all, and that is $S$.
So as $S$ contains more than one point, it follows that every point of $T$ is a limit point of $H$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $4$
Category:Examples of Limit Points
Category:Indiscrete Topology
Limit Points of Infinite Subset of Finite Complement Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement space.
Let $H \subseteq S$ be an infinite subset of $S$.
Then every point of $S$ is a limit point of $H$.
== Proof ==
Let $U \in \tau$ be any open set of $T$.
From Infinite Subset of Finite Complement Space Intersects Open Sets, we have that $U \cap H$ is infinite $H$ is infinite.
Let $x \in S$.
Then every open set $U$ in $T$ such that $x \in U$ also contains an infinite number of points of $H$ other than $x$.
Thus, by definition, $x$ is a limit point of $H$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $1$
Category:Examples of Limit Points
Category:Finite Complement Topology
Limit Points of Open Real Interval

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
 
Let $\openint a b$ be an open interval of $\R$.
Then the limit points of $\openint a b$ are:
:all the points in $\openint a b$
as well as:
:the points $a$ and $b$.
== Proof ==
Let $x \in \openint a b$.
Then by definition $a < x < b$.
Let $N_x$ be an open neighborhood of $x$ in $\R$.
So by Open Superset is Open Neighborhood, $N_x$ is an open set of $\R$ containing $x$.
By Open Sets in Real Number Line, $N_x$ is a countable union of pairwise disjoint open intervals.
As $x \in N_x$, it follows that $\exists U \subseteq N_x$ such that $x \in U$ and:
:$\exists p, q \in \R: U = \openint p q$
Let $r = \max \set {a, p}$ and $s = \min \set {b, q}$.
Then as $x > a, x > p$ and $x < b, x < q$ it follows that $x > \max \set {a, p}$ and $x < \min \set {b, q}$.
That is:
:$x \in \openint r s$
From Real Numbers are Densely Ordered:
:$\exists y \in \R: r < y < x$
and so:
:$y \in \openint r s$
By construction, it follows that:
:$y \in \openint p q$
That is, $y$ is an element of $\R$ different from $x$ which is also in $N_x$.
As $N_x$ was arbitrary, it follows that every such open neighborhood of $x$ has the same property.
That is, $x$ is a limit point of $\openint a b$.
Let $N_a$ be an open neighborhood of $a$ in $\R$.
So by Open Superset is Open Neighborhood, $N_a$ is an open set of $\R$ containing $a$.
By Open Sets in Real Number Line, $N_a$ is a countable union of pairwise disjoint open intervals.
As $a \in N_a$, it follows that $\exists U \subseteq N_a$ such that $a \in U$ and:
:$\exists p, q \in \R: U = \openint p q$
Let $s = \min \set {b, q}$.
As $a < b, a < q$ it follows that $a < \min \set {b, q}$.
That is:
:$a \in \openint p s$
From Real Numbers are Densely Ordered:
:$\exists y \in \R: a < y < s$
and so:
:$y \in \openint p s$
By construction, it follows that:
:$y \in \openint p q$
That is, $y$ is an element of $\R$ different from $a$ which is also in $N_a$.
As $N_a$ was arbitrary, it follows that every such open neighborhood of $a$ has the same property.
That is, $a$ is a limit point of $\openint a b$.
By a similar argument it is shown that $a$ is a limit point of $\openint a b$.
Now let:
:$x \notin \openint a b \cup \set {a, b}$
Then either:
:$x < a$
or
:$x > b$
Let $x < a$.
Then by Real Numbers are Densely Ordered:
:$\exists q \in \R: x < q < a$
and so $\exists p \in \R$ such that:
:$x \in \openint p q$
such that:
:$\openint p q \cap \openint a b = \O$
That is, $\openint p q$ is an open set of $\R$ containing $x$ which contains no elements of $\openint a b$.
Hence by definition, $x$ is not a limit point of $\openint a b$.
By a similar argument it is shown that if $x > b$ then $x$ is not a limit point of $\openint a b$.
Category:Examples of Limit Points
Category:Real Intervals
Limit Points of Sequence in Indiscrete Space on Uncountable Set

== Theorem ==
Let $S$ be an uncountable set.
Let $T = \struct {S, \set {\O, S} }$ be the indiscrete topological space on $S$.
Let $\sequence {s_n}$ be a sequence in $T$.
Then every sequence in $T$ has an uncountable number of limits.
== Proof ==
From Sequence in Indiscrete Space converges to Every Point, $\sequence {s_n}$ converges to every point of $S$.
As $S$ is uncountable, the result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $4$
Category:Indiscrete Topology
Limit Superior/Examples/(-1)^n

== Example of Limit Superior ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{>0}: a_n = \paren {-1}^n$
The limit superior of $\sequence {a_n}$ is given by:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 1$
== 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 superior:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 1$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.14$: Example $\text {(ii)}$
Category:Examples of Limits Superior