nyuuzyou/wikis huggingface dataset:
Topological Space induced by Neighborhood Space induced by Topological Space

== Theorem ==
Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\struct {S, \tau'}$ be the topological space induced by $\NN$ on $S$.
Then $\tau = \tau'$.
== Proof ==
Let $U \in \tau$ be an open set of $\struct {S, \tau}$.
By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points.
By definition, $U$ is an open set of $\struct {S, \NN}$.
Thus by definition of neighborhood space induced by $\tau$ on $S$:
:$U \in \NN$
Then, by definition of the topological space induced by $\NN$ on $S$:
:$U \in \tau'$.
Thus $\tau \subseteq \tau'$.
Now let $U \in \tau'$ be an open set of $\struct {S, \tau'}$.
Then, by definition, in the neighborhood space $\struct {S, \NN}$, $U$ is an open set of $\struct {S, \NN}$.
That is, $U$ is a neighborhood in $\struct {S, \NN}$ of each of its points.
But the neighborhoods of $\struct {S, \NN}$ are the neighborhoods of $\struct {S, \tau}$.
Thus by Set is Open iff Neighborhood of all its Points, $U$ is an open set of $\struct {S, \tau}$.
That is:
:$U \in \tau$.
It follows by definition of set equality that $\tau = \tau'$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces
Category:Neighborhood Spaces
Category:Topology
Topological Space is Connected iff any Proper Non-Empty Subset has Non-Empty Boundary

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Then $\struct {X, \tau}$ is connected for each proper non-empty subset $S \subseteq X$, we have $\partial S \ne \O$.
== Proof ==
From Connected iff no Proper Clopen Sets, we have that:
:$\struct {X, \tau}$ is connected there exists no proper non-empty clopen set $S \subseteq X$.
From Set is Clopen iff Boundary is Empty, we have that: 
:$S \subseteq X$ is clopen $\partial S = \O$.
Hence we have:
:$\struct {X, \tau}$ is connected there exists no proper non-empty set $S \subseteq X$ such that $\partial S = \O$.
That is:
:$\struct {X, \tau}$ is connected for each proper non-empty set $S \subseteq X$, we have $\partial S \ne \O$.
Category:Connected Spaces
Category:Set Boundaries
Topological Space is Discrete iff All Points are Isolated

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
:$\tau$ is the discrete topology on $S$ all points in $S$ are isolated points of $T$.
== Proof ==
=== Necessary Condition ===
Let $T = \struct {S, \tau}$ be the discrete space on $S$.
Then by definition $\tau = \powerset S$, that is, $\tau$ is the power set of $S$.
Let $x \in S$.
Then from Set in Discrete Topology is Clopen it follows that $\set x$ is open in $T$.
Thus by definition $x \in S$ is an isolated point of $T$.
=== Sufficient Condition ===
Let $T$ be such that $x \in S$ is an isolated point of $T$ for all $x \in T$.
 $T = \struct {S, \tau}$ is not the discrete space on $S$.
Then from Basis for Discrete Topology it follows that:
:$\BB := \set {\set x: x \in S}$
is not a basis for $T$.
Thus:
:$\exists x \in S: \set x \notin \tau$
Let $U \in \tau$ such that $x \in U$.
Then as $U \ne \set x$ it follows that $\exists y \in U: y \ne x$.
That is, there are no open sets of $T$ which contain only $x$.
So $x$ is not an isolated point.
It follows by Proof by Contradiction that $T$ is the discrete space on $S$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $2$
Category:Isolated Points
Category:Discrete Topology
Topological Space is Open Neighborhood of Subset

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.
Then $S$ is an open neighborhood of $H$.
== Proof ==
From Underlying Set of Topological Space is Clopen, $S$ is open in $T$.
By hypothesis, $H \subseteq S$.
The result follows from Open Superset is Open Neighborhood.
Category:Topology
Category:Neighborhoods
Topological Space is Quasiuniformizable

== Theorem ==
Every topological space is quasiuniformizable.
== Proof ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be defined as:
:$\BB := \set {u_G: u_G = \paren {G \times G} \cup \paren {\paren {S \setminus G} \times G}, G \in \tau}$
Then $\BB$ is a filter sub-basis for a quasiuniformity on $S$ such that $\struct {\struct {S, \UU}, \tau}$ is a quasiuniform space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Uniformities
Category:Uniformities
Topological Space may be Empty

== Philosophical Position ==
Notwithstanding the result Empty Set Satisfies Topology Axioms, it is frequently stipulated in the literature that the class of topological spaces does not include the empty set.
This convention is sufficiently commonplace as to be often omitted in published texts, and taken for granted. When it is mentioned, it is usually given as an afterthought.
However, there exists a philosophical position that disallowing the empty topological space is unhelpful, and even harmful.
 adopts this philosophical position, allowing that the underlying set of a given topological space may indeed be empty.
Note, however, that many of the possible properties of a topological space are held vacuously by the empty space.
This position having been taken, it is necessary in many cases to add a condition to a given general statement made about spaces specifically to exclude the empty space from the scope of that statement.
== Also see ==
* Definition:Topological Space
* Definition:Empty Set
* Definition:Empty Topological Space
Category:Philosophical Positions
Category:Empty Topological Space
Topological Space with Generic Point is Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T$ have a generic point $g \in S$.
Then $T$ is path-connected.
== Proof ==
By Path-Connectedness is Equivalence Relation, it suffices to prove that every point is path-connected with $g$.
Let $x \in S$.
Define a path $\gamma: \closedint 0 1 \to S$ by:
:$\map \gamma t = \begin{cases}
x & : t \le \dfrac 1 2 \\
g & : t > \dfrac 1 2
\end{cases}$
We show that $\gamma$ is indeed continuous.
Let $U \subseteq S$ be open and non-empty.
Because $g$ is a generic point, $g \in U$.
If $x \in U$, then its preimage $\gamma^{-1} \sqbrk U = \closedint 0 1$ is open.
If $x \notin U$, then its preimage $\gamma^{-1} \sqbrk U = \hointl {\dfrac 1 2} 1$ is also open.
Thus $x$ and $g$ are path-connected.
Category:Path-Connected Spaces
Topological Space with One Quasicomponent is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which has one quasicomponent.
Then $T$ is connected.
== Proof ==
Let $x \in S$.
By hypothesis, the quasicomponent of $x$ is $S$ itself.
Thus by definition of quasicomponent:
:$\forall y \in S: y \sim x$
where $\sim$ is the relation defined on $T$ as:
:$x \sim y \iff T$ is connected between the two points $x$ and $y$
Let $K = \ds \bigcap_{x \mathop \in U} U: U$ is clopen in $T$.
By Quasicomponent is Intersection of Clopen Sets:
:$\ds \bigcap K = S$
Thus there is no non-empty clopen set of $T$ apart from $S$.
The result follows by definition of connected space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Connectedness Between Two Points
Category:Connected Spaces
Topological Subspace is Topological Space

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $H \subseteq X$ be a non-empty subset of $X$.
Let $\tau_H = \set {U \cap H: U \in \tau}$ be the subspace topology on $H$.
Then the topological subspace $\struct {H, \tau_H}$ is a topological space.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Topological Subspace
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.4$: Subspaces: Definition $3.4.1$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 9$
* : $\S 3$: New Spaces From Old: Subspaces
Category:Topological Subspaces
Category:Topological Subspace is Topological Space
Topological Subspace is Topological Space/Proof 1

== Theorem ==
== Proof ==
We verify the open set axioms for $\tau_H$ to be a topology on $H$.
=== ===
Let $\AA \subseteq \tau_H$.
It is to be shown that:
:$\ds \bigcup \AA \in \tau_H$
Define:
:$\ds \AA' = \set {V \in \tau: V \cap H \subseteq \bigcup \AA} \subseteq \tau$
Let:
:$\ds U = \bigcup \AA'$
By the definition of a topology, we have $U \in \tau$.
Then, from Intersection Distributes over Union and by Union is Smallest Superset: Family of Sets:
:$\ds U \cap H = \bigcup_{V \mathop \in \AA'} \paren {V \cap H} \subseteq \bigcup \AA$
By the definition of $\tau_H$ and by Set is Subset of Union: General Result, we have:
:$\ds \forall S \in \AA: \exists V \in \tau: S = V \cap H \subseteq \bigcup \AA$
That is:
:$\forall S \in \AA: \exists V \in \AA': S = V \cap H$
By Set is Subset of Union: General Result, we have:
:$\ds \forall V \in \AA': V \subseteq U$
Since intersection preserves subsets, it follows that:
:$\forall S \in \AA: S \subseteq U \cap H$
By Union is Smallest Superset: General Result, we conclude that:
:$\ds \bigcup \AA \subseteq U \cap H$
Hence, by definition of set equality:
:$\ds \bigcup \AA = U \cap H \in \tau_H$
=== ===
Let $A, B \in \tau_H$.
Let $U, V \in \tau$ be such that $A = U \cap H$ and $B = V \cap H$.
By the definition of a topology, we have $U \cap V \in \tau$.
From Set Intersection is Self-Distributive:
:$A \cap B = \paren {U \cap V} \cap H \in \tau_H$
=== ===
By the definition of a topology, we have $X \in \tau$.
By Intersection with Subset is Subset, it follows that $H = X \cap H \in \tau_H$.
All the open set axioms are fulfilled, and the result follows.
== Also see ==
* Definition:Topological Subspace
Category:Topological Subspace is Topological Space
Topological Sum is Coproduct in Category of Topological Spaces

== Theorem ==
Let $\mathbf{Top}$ be the category of topological spaces.
Let $X$ and $Y$ be topological spaces, and let $X \sqcup Y$ be their topological sum.
Then $X \sqcup Y$ is the coproduct of $X$ and $Y$ in $\mathbf{Top}$.
== Proof ==
== Sources ==
* : $\S 3.2$: Example $3.6$
Category:Category of Topological Spaces
Category:Topological Sum
Topological Vector Space as Union of Dilations of Open Neighborhood of Origin

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$. 
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
Let $\sequence {r_n}_{n \mathop \in \N}$ be a sequence of positive real numbers such that $r_n \to \infty$. 
Then: 
:$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$
where $r_n V$ denotes the dilation of $V$ by $r_n$.
== Proof ==
Clearly: 
:$\ds \bigcup_{n \mathop = 1}^\infty r_n V \subseteq X$
We show that: 
:$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$
Let $x \in X$. 
From the definition of a topological vector space, the mapping $K \times X \to X$ defined by $\tuple {\lambda, x} \mapsto \lambda x$ is continuous.
From Horizontal Section of Continuous Function is Continuous, it follows that the $x$-horizontal section with $\lambda \mapsto \lambda x$ is continuous.
Hence, the set:
:$U = \set {\lambda \in \Bbb F : \lambda x \in V}$ 
is open.
Since $0 x = {\mathbf 0}_X$ and $V$ is an open neighborhood of ${\mathbf 0}_X$, we have: 
:$U$ is an open neighborhood of $0 \in \Bbb F$.
So, there exists $\epsilon > 0$ such that if $\cmod \lambda < \epsilon$, we have $\lambda x \in U$.
From the definition of a real sequence tending to $+\infty$, there exists $N \in \N$ such that: 
:$\ds r_n > \frac 1 \epsilon$ for $n \ge N$. 
Then: 
:$\ds 0 < \frac 1 {r_N} < \epsilon$
so that: 
:$\dfrac 1 {r_N} x \in V$
giving:
:$x \in r_N V$
So, we have: 
:$\ds x \in \bigcup_{n \mathop = 1}^\infty r_n V$
Since $x \in X$ was arbitrary we have: 
:$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$
giving:
:$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$
== Sources ==
* : $1.15$: Theorem
Category:Topological Vector Spaces
Topological Vector Space is Hausdorff iff T1

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
:$(1): \quad \struct {X, \tau}$ is Hausdorff
:$(2): \quad \struct {X, \tau}$ is a $T_1$ space
:$(3): \quad$ for each $x \in X \setminus \set { {\mathbf 0}_X}$, there exists an open neighborhood $U_x$ of ${\mathbf 0}_X$ such that $x \not \in U_x$.
== Proof ==
=== $(1)$ implies $(2)$ ===
This is precisely $T_2$ Space is $T_1$ Space.
=== $(2)$ implies $(3)$ ===
Suppose that $\struct {X, \tau}$ is a $T_1$ space.
Then, for each $x, y \in X$ with $x \ne y$, there exists an open neighborhood $U_{x, y}$ such that:
:$x \in U_{x, y}$
and:
:$y \not \in U_{x, y}$.
Let:
:$x \in X \setminus \set { {\mathbf 0}_X}$
Let:
:$y = {\mathbf 0}_X$
Then, for each $x, {\mathbf 0}_X \in X$ with $x \ne {\mathbf 0}_X$, there exists an open neighborhood $X \setminus \set { {\mathbf 0}_X}$ such that:
:$x \in X \setminus \set { {\mathbf 0}_X}$ 
and:
:${\mathbf 0}_X \not \in X \setminus \set { {\mathbf 0}_X}$
Hence the result.
=== $(3)$ implies $(1)$ ===
Suppose that:
:for each $x \in X \setminus \set { {\mathbf 0}_X}$, there exists an open neighborhood $U_x$ of ${\mathbf 0}_X$ such that $x \not \in U_x$.
Let $x, y \in X$ be such that:
:$x \ne y$
Then:
:$x - y \ne {\mathbf 0}_X$
Consider:
:$U_{x - y}$
From Open Neighborhood of Point in Topological Vector Space contains Sum of Open Neighborhoods: Corollary 1, there exists a symmetric set $V_{x, y}$ such that:
:$V_{x, y} + V_{x, y} \subseteq U_{x - y}$
By definition of symmetric subset of vector space:
:$V_{x, y} = -V_{x, y}$
we have:
:$V_{x, y} - V_{x, y} \subseteq U_{x - y}$
For brevity, let:
:$V = V_{x, y}$
We show that:
:$\paren {x + V} \cap \paren {y + V} = \O$
 that:
:$\paren {x + V} \cap \paren {y + V} \ne \O$
Let:
:$z \in \paren {x + V} \cap \paren {y + V}$
Then there exists $u, v \in V$ such that:
:$x + u = z$
and:
:$y + v = z$. 
Then we have:
:$x - y = \paren {z - u} - \paren {z - v} = v - u$. 
We have:
:$v - u \in V - V$
and so:
:$v - u \in U_{x - y}$
But:
:$x - y \not \in U_{x - y}$
so this is a contradiction.
Hence:
:$\paren {x + V} \cap \paren {y + V} = \O$
From Translation of Open Set in Topological Vector Space is Open, $x + V$ is an open neighborhood of $x$ and $y + V$ is an open neighborhood of $y$. 
Hence $\struct {X, \tau}$ is Hausdorff.
Category:Hausdorff Spaces
Category:T1 Spaces
Category:Topological Vector Spaces
Topological Vector Space over Connected Topological Field is Connected

== Theorem ==
Let $K$ be a connected topological field.
Let $X$ be a topological vector space over $K$. 
Then $X$ is connected.
== Proof ==
From the definition of a topological vector space, the mapping $\circ_X : K \times X \to X$ defined by:
:$\map {\circ_X} {\lambda, x} = \lambda x$
for $\tuple {\lambda, x} \in K \times X$ is continuous.
Let $x \in X$. 
From Horizontal Section of Continuous Function is Continuous, the mapping $c_x : K \to X$ defined by:
:$\map {c_x} \lambda = \lambda x$
for $\lambda \in K$ is continuous.
Since $K$ is connected, we have that:
:$c_x \sqbrk K$ is connected
from Continuous Image of Connected Space is Connected. 
That is:
:$K x$ is connected.
Since $x \in K x$ for each $x \in X$, we have:
:$\ds X = \bigcup_{x \in X} K x$
Since $0_K \circ x = {\mathbf 0}_X$ for each $x \in X$, we have that:
:$\ds {\mathbf 0}_X \in \bigcap_{x \in X} K x$
From Union of Connected Sets with Common Point is Connected, we have that:
:$\ds \bigcup_{x \in X} K x$ is connected.
Hence $X$ is connected.
Category:Topological Vector Spaces
Category:Connected Spaces
Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$ admitting a weak topology $w$. 
Then $\struct {X, w}$ is a topological vector space. 
=== Corollary ===
== Proof ==
Follows from combining: 
:Vector Addition is Continuous in Weak Topology
:Scalar Multiplication is Continuous in Weak Topology
Category:Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology
Category:Weak Topologies on Topological Vector Spaces
Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology/Corollary

== Corollary ==
Let $K$ be a Hausdorff topological field.
Let $X$ be a topological vector space over $K$ admitting a weak topology $w$. 
Then $\struct {X, w}$ is a Hausdorff topological vector space. 
== Proof ==
From Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology, $\struct {X, w}$ is a topological vector space.
From Weak Topology on Topological Vector Space over Hausdorff Topological Field is Hausdorff, $\struct {X, w}$ is a Hausdorff topological vector space.
Category:Topological Vector Space over Topological Field remains Topological Vector Space with Weak Topology
Category:Hausdorff Topological Vector Spaces
Topologically Distinguishable Points are Distinct

== Theorem ==
Let $T = \struct {X, \tau}$ be a topological space.
Let $x, y \in X$ be topologically distinguishable.
Then the singleton sets $\set x$ and $\set y$ are disjoint and so:
:$x \ne y$
== Proof ==
Let $x$ and $y$ be topologically distinguishable.
Then either:
:$\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x$
or:
:$\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y$
 $x = y$.
Then:
:$\map \neg {\exists U \in \tau: x \in U \subseteq N_x \subseteq X: y \notin N_x}$
and:
:$\map \neg {\exists V \in \tau: y \in V \subseteq N_y \subseteq X: x \notin N_y}$
Hence the result by Proof by Contradiction:
:$x \ne y$
and so by Singleton Equality:
:$\set x \ne \set y$
== Sources ==
* 
Category:Topological Distinguishability
Topologically Equivalent Metrics induce Equal Topologies

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be topologically equivalent.
Let $\tau_1$ and $\tau_2$ denote the topologies on $A$ induced by $d_1$ and $d_2$ respectively.
Then $\tau_1$ and $\tau_2$ are equal.
== Proof ==
Let $d_1$ and $d_2$ be topologically equivalent by hypothesis.
By definition of topological equivalence:
$d_1$ and $d_2$ are topologically equivalent :
:$U \subseteq A$ is $d_1$-open $U \subseteq A$ is $d_2$-open.
By definition of the induced topology:
:The topology on the metric space $M = \struct {A, d}$ induced by (the metric) $d$ is defined as the set $\tau$ of all open sets of $M$.
Hence it follows that:
:$U \subseteq A$ is open in the topological space $\struct {A, \tau_1}$
:
:$U \subseteq A$ is open in the topological space $\struct {A, \tau_2}$
That is:
:$U \in \tau_1 \iff U \in \tau_2$
and the result follows.
Category:Topologically Equivalent Metrics
Topologies are not necessarily Comparable by Coarseness

== Theorem ==
Let $S$ be a set with at least $2$ elements.
Let $\mathbb T$ be the set of all topologies on $S$.
For two topologies $\tau_a, \tau_b \in \mathbb T$, let $\tau_a \le \tau_b$ denote that $\tau_a$ is coarser than $\tau_b$.
Then there exist $\tau_1, \tau_2 \in \mathbb T$ such that neither:
:$\tau_1 \le \tau_2$
nor:
:$\tau_2 \le \tau_1$
That is, there are always topologies on $S$ which are non-comparable.
== Proof ==
Let $a, b \in S$.
Let:
:$\tau_a$ be the particular point topology with respect to $a$ on $S$
:$\tau_b$ be the particular point topology with respect to $b$ on $S$
From Homeomorphic Non-Comparable Particular Point Topologies:
:neither $\tau_a \le \tau_b$ nor $\tau_b \le \tau_a$.
Hence the result.
== Also see ==
* Coarseness Relation on Topologies is Partial Ordering
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 3$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:Topology
Topologies induced by Usual Metric and Scaled Euclidean Metric on Positive Integers are Homeomorphic

== Theorem ==
Let $\Z_{>0}$ be the set of (strictly) positive integers.
Let $d: \Z_{>0} \times \Z_{>0} \to \R$ be the usual (Euclidean) metric on $\Z_{>0}$.
Let $\delta: \Z_{>0} \times \Z_{>0} \to \R$ be the metric on $\Z_{>0}$ defined as:
:$\forall x, y \in \Z_{>0}: \map \delta {x, y} = \dfrac {\size {x - y} } {x y}$
Let $\tau_d$ denote the metric topology for $d$.
Let $\tau_\delta$ denote the metric topology for $\delta$.
Then $\struct {\Z_{>0}, \tau_d}$ and $\struct {\Z_{>0}, \tau_\delta}$ are homeomorphic.
== Proof ==
From Topology induced by Usual Metric on Positive Integers is Discrete‎:
:$\struct {\Z_{>0}, \tau_d}$ is a discrete space.
From Topology induced by Scaled Euclidean Metric on Positive Integers is Discrete‎:
:$\struct {\Z_{>0}, \tau_\delta}$ is a discrete space.
Let $I_{\Z_{>0} }$ be the identity mapping from $\Z_{>0}$ to itself.
From Mapping from Discrete Space is Continuous:
:$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_d} \to \struct {\Z_{>0}, \tau_\delta}$ is continuous
and:
:$I_{\Z_{>0} }: \struct {\Z_{>0}, \tau_\delta} \to \struct {\Z_{>0}, \tau_d}$ is continuous.
Hence the result by definition of homeomorphic.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $10$
Category:Discrete Topology
Category:Homeomorphisms (Topological Spaces)
Topologies on Doubleton

== Theorem ==
Let $S = \set {a, b}$ be a doubleton.
Then there exist $4$ possible different topologies on $S$:
== Proof ==
The power set of $S$ is the set:
:$\powerset S = \set {\O, \set a, \set b, \set {a, b} }$
Because all topologies on $S$ are subsets of $\powerset S$, one of the following must hold:
By definition of a topology, $S$ itself must be an element of the topology.
Thus $\tau_1$ up to $\tau_8$ are not topologies on $S$. 
By Empty Set is Element of Topology, for $\tau$ to be a topology for $S$, it is necessary that $\O \in \tau$. 
Therefore $\tau_9$ up to $\tau_{12}$ are also not topologies on $S$.
By Indiscrete Topology is Topology, $\tau_{13}$ is a topology on $S$.
By Discrete Topology is Topology, $\tau_{16}$ is a topology on $S$.
It is then seen by inspection that $\tau_{14}$ and $\tau_{15}$ are particular point topologies
Indeed, they are Sierpiński topologies.
By Particular Point Topology is Topology, both $\tau_{14}$ and $\tau_{15}$ are topologies.
Hence the result.
The topologies can be assigned arbitrary labels.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Example $4$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 2 \ \text {(i)}$
Category:Doubletons
Category:Examples of Topologies
Topologies on Set form Complete Lattice

== Theorem ==
Let $X$ be a non-empty set.
Let $\LL$ be the set of topologies on $X$.
Then $\struct {\LL, \subseteq}$ is a complete lattice.
== Proof ==
Let $\KK \subseteq \LL$.
Then by Intersection of Topologies is Topology:
:$\bigcap \KK \in \LL$
By Intersection is Largest Subset, $\bigcap \LL$ is the infimum of $\KK$.
Let $\tau$ be the topology generated by the sub-basis $\bigcup \KK$.
Then $\tau \in \LL$ and $\tau$ is the supremum of $\KK$.
We have that each subset of $\LL$ has a supremum and an infimum in $\LL$.
Thus it follows that $\struct {\LL, \subseteq}$ is a complete lattice.
Category:Topology
Category:Complete Lattices
Topologies on Set with More than One Element may not be Homeomorphic

== Theorem ==
Let $S$ be a set which contains at least $2$ elements.
Let $\tau_1$ and $\tau_2$ be topologies on $S$.
Then it is not necessarily the case that $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ are homeomorphic.
== Proof ==
Let $\tau_1$ be the indiscrete topology on $S$.
Let $\tau_2$ be the discrete topology on $S$.
Then $\struct {S, \tau_1}$ has $2$ elements: $S$ and $\O$.
Let $a, b \in S$ such that $a \ne b$.
Then $\set a \in \tau_2$ and $\set b \in \tau 2$, as well as $S$ and $\O$.
So there cannot be a bijection between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
Category:Homeomorphisms (Topological Spaces)