nyuuzyou/wikis huggingface dataset:
Open Real Interval is Regular Open

== Theorem ==
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\openint a b$ be an open interval of $\R$.
Then $\openint a b$ is regular open in $\struct {\R, \tau_d}$.
== Proof ==
From Open Sets in Real Number Line, $\openint a b$ is open in $\struct {\R, \tau_d}$.
From Closure of Open Real Interval is Closed Real Interval:
:$\openint a b^- = \closedint a b$
where $\openint a b^-$ denotes the closure of $\openint a b$.
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint a b^\circ = \openint a b$
where $\closedint a b^\circ$ denotes the interior of $\closedint a b$.
Hence the result, by definition of regular open.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(c)}$
Category:Open Sets
Category:Regular Open Sets
Category:Real Number Line with Euclidean Topology
Open Real Interval is Subset of Closed Real Interval

== Theorem ==
Let $a, b \in \R$ be real numbers.
Then:
:$\openint a b \subseteq \closedint a b$
where:
:$\openint a b$ is the open interval between $a$ and $b$
:$\closedint a b$ is the closed interval between $a$ and $b$.
== Proof ==
Let $x \in \openint a b$.
Then by definition of open interval:
:$a < x < b$
Thus:
:$a \le x \le b$
and so by definition of closed interval:
:$x \in \closedint a b$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets
Category:Real Intervals
Open Real Interval is not Closed Set

== Theorem ==
Let $\R$ be the real number line with the usual (Euclidean) metric.
Let $I = \openint a b$ be an open real interval.
Then $I$ is not a closed set of $\R$.
=== Corollary ===
== Proof ==
Consider the relative complement of $I$ in $\R$:
:$J = \relcomp \R I = \R \setminus I = \hointl \gets a \cup \hointr b \to$
Let $\epsilon \in \R_{>0}$.
Consider the open $\epsilon$-ball $\map {B_\epsilon} a$.
Whatever the value of $\epsilon$ is, $a + \epsilon$ is not in $\map {B_\epsilon} a$.
So, by definition, $J$ is not an open set of $\R$.
By Relative Complement of Relative Complement, $\relcomp \R J = I$.
By definition, it follows that $I$ is not a closed set of $\R$.
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Open Real Interval is not Closed Set/Corollary

== Theorem ==
Let $\R$ be the real number line with the usual (Euclidean) metric.
Let $a, b \in \R$.
Let:
:$I_a = \openint \gets a$
:$I_b = \openint b \to$
be unbounded open real intervals.
Then neither $I_a$ nor $I_b$ are closed sets of $\R$.
== Proof ==
Consider the relative complement of $I_a$ in $\R$:
:$J = \relcomp \R I = \R \setminus I = \hointr a \to$
Let $\epsilon \in \R_{>0}$.
Consider the open $\epsilon$-ball $\map {B_\epsilon} a$.
Whatever the value of $\epsilon$ is, $a - \epsilon$ is not in $\map {B_\epsilon} a$.
So, by definition, $J$ is not an open set of $\R$.
By Relative Complement of Relative Complement, $\relcomp \R J = I$.
By definition, it follows that $I_a$ is not a closed set of $\R$.
Mutatis mutandis, the same proof shows that neither is $I_b$ a closed set of $\R$.
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Open Real Interval is not Compact

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $I = \openint a b$ be an open real interval.
Then $I$ is not compact.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Closed Real Interval is Compact in Metric Space
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Compact Spaces
Category:Real Intervals
Category:Open Real Interval is not Compact
Open Real Interval is not Compact/Normed Vector Space

== Theorem ==
== Proof ==
We have that $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.
Also, $I$ is bounded by $a$ and $b$.
Consider a sequence $\ds \sequence {x_n}_{n \mathop \in \N}$ with $\ds x_n = a + \frac {b - a} {2n}$.
Let $\epsilon \in \R_{\mathop > 0}$.
Let $\ds N = \frac {b - a} {2 \epsilon}$.
Then for all $n \in \N$ such that $n > N$ we have that:
Hence, $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$.
By Limit of Subsequence equals Limit of Real Sequence, every subsequence of $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$ as well.
However, $a \notin I$.
By definition, $I$ is not compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Open Real Interval is not Compact
Open Real Interval is not Compact/Proof 2

== Theorem ==
== Proof ==
It suffices to demonstrate this for a particular open interval: we use $\openint 0 1$.
Consider the set of open intervals $\openint {\dfrac 1 n} 1$ for all $n \in \Z_{>1}$.
Each of these is an open set in $\openint 0 1$.
Also:
:$\openint 0 1 = \ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$
Thus $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ is an open cover for $\openint 0 1$.
Note that:
:$\forall x \in \openint 0 1: \exists n \in \Z: n > \dfrac 1 x$
and so:
:$x \in \openint {\dfrac 1 n} 1$
Consider a finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$.
It will be in the form:
:$\set {\openint {\dfrac 1 {n_1} } 1, \openint {\dfrac 1 {n_2} } 1, \ldots, \openint {\dfrac 1 {n_r} } 1}$
But this is just $\openint {\dfrac 1 N} 1$, where $N = \max \set {n_1, n_2, \ldots, n_r}$.
So no finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ can be a cover for $\openint 0 1$.
== Sources ==
* : $5$: Compact spaces: $5.2$: Definition of compactness
* 
Category:Open Real Interval is not Compact
Open Real Intervals are Homeomorphic

== Theorem ==
Let $\R$ be the real number line with the Euclidean topology.
Let $I_1 := \openint a b$ and $I_2 := \openint c d$ be non-empty open real intervals.
Then $I_1$ and $I_2$ are homeomorphic.
== Proof ==
By definition of open real interval, for $I_1$ and $I_2$ to be non-empty it must be the case that $a < b$ and $c < d$.
In particular it is noted that $a \ne b$ and $c \ne d$.
Thus $a - b \ne 0$ and $c - d \ne 0$.
Consider the real function $f: I_1 \to I_2$ defined as:
:$\forall x \in I_1: \map f x = c + \dfrac {\paren {d - c} \paren {x - a} } {b - a}$
Then after some algebra:
:$\forall x \in I_2: \map {f^{-1} } x = a + \dfrac {\paren {b - a} \paren {x - c} } {d - c}$
Both of these are defined as $a - b \ne 0$ and $c - d \ne 0$.
By the Combination Theorem for Continuous Real Functions, both $f$ and $f^{-1}$ are continuous on the open real intervals on which they are defined.
Hence the result by definition of homeomorphic.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Exercise $2$
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(a)}$
Category:Examples of Homeomorphisms
Open Reciprocal-N Balls form Neighborhood Basis in Real Number Line

== Theorem ==
Let $\R$ denote the real number line with the usual (Euclidean) metric.
Let $a \in \R$ be a point in $\R$.
Let $\BB_a$ be defined as:
:$\BB_a := \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N} }$
that is, the set of all open $\epsilon$-balls of $a$ for $\epsilon$ which are reciprocals of integers.
Then $\BB_a$ is a basis for the neighborhood system of $a$.
== Proof ==
Let $N$ be a neighborhood of $a$ in $M$.
Then by definition:
:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$.
From Open Ball in Real Number Line is Open Interval:
:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
From Between two Real Numbers exists Rational Number:
:$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$
Let $\epsilon$ be expressed in canonical form:
:$\epsilon = \dfrac p q$
Let $\epsilon = \dfrac 1 q$
Then $\epsilon \le \epsilon < \epsilon'$
and so:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
From Open Real Interval is Open Ball
:$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$
is the open $\epsilon$-ball at $a$.
From Subset Relation is Transitive:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq N$
From Open Ball is Neighborhood of all Points Inside, $\openint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$.
Hence there exists a neighborhood $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$.
Hence the result by definition of basis for the neighborhood system of $a$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $4 \ \text{iii)}$
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Open Rectangles Closed under Intersection

== Theorem ==
Let $\paren {\openint {\mathbf a} {\mathbf b} }$ and $\paren {\openint {\mathbf c} {\mathbf d} }$ be open $n$-rectangles.
Then $\paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} }$ is also an open $n$-rectangle.
== Proof ==
From Cartesian Product of Intersections: General Case, we have:
:$\ds \paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} } = \prod_{i \mathop = 1}^n \openint {a_i} {b_i} \cap \openint {c_i} {d_i}$
Therefore, it suffices to show that the intersection of two open intervals is again an open interval.
Now let $x \in \openint {a_i} {b_i} \cap \openint {c_i} {d_i}$.
Then $x$ is subject to:
:$x > a_i$ and $x > c_i$, that is: $x > \max \set {a_i, c_i}$
:$x < b_i$ and $x < d_i$, that is: $x < \min \set {b_i, d_i}$
and we see that these conditions are satisfied :
:$x \in \openint {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
Thus, we conclude:
:$\openint {a_i} {b_i} \cap \openint {c_i} {d_i} = \openint {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
showing that indeed the intersection is an open interval.
Combining this with the above reasoning, it follows that indeed:
:$\paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} }$
is an open $n$-rectangle.
Category:Analysis
Open Set/Complex Analysis/Examples/Open Unit Circle

== Example of Open Set in the context of Complex Analysis ==
Let $S$ be the subset of the complex plane defined as:
:$\cmod z < 1$
where $\cmod z$ denotes the complex modulus of $z$.
Then $S$ is open.
== Proof ==
By definition, $S$ is open $S$ consists only of interior points.
Let $z_1 \in S$.
Then $\cmod {z_1} < 1$.
Let $\epsilon \in \R: \epsilon < 1 - \cmod {z_1}$.
Then:
:$\map {N_\epsilon} {z_1} \subseteq S$
and so $z_1$ is an interior points of $S$.
As $z_1$ is arbitrary, the result follows.
== Sources ==
* : $1$: Complex Numbers: Point Sets: $6.$
Category:Examples of Closed Sets (Complex Analysis)
Open Set Characterization of Denseness

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $S \subseteq X$.
Then $S$ is (everywhere) dense in $X$ every non-empty ($\tau$-)open set contains an element of $S$.
=== Analytic Basis ===
=== Open Ball ===
== Proof ==
=== Necessary Condition ===
Let $S$ be everywhere dense in $X$.
Let $U$ be open and non-empty.
Then $U$ has an element $x$.
Since $S$ is everywhere dense in $X$, $x \in S^-$, the closure of $S$.
By Equivalence of Definitions of Adherent Point, every open neighborhood of $x$ contains an element of $S$.
Thus in particular, $U$ contains an element of $S$.
=== Sufficient Condition ===
Suppose that every non-empty open set in $X$ contains an element of $S$.
Let $x \in X$.
Let $U$ be an open neighborhood of $x$.
Then $U$ contains an element $s$ of $S$.
As this holds for all open neighborhoods of $x$, Equivalence of Definitions of Adherent Point shows that $x \in S^-$.
As this holds for all $x \in X$, $S$ is everywhere dense in $X$.
Category:Denseness
Open Set Characterization of Denseness/Analytic Basis

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $S \subseteq X$.
Let $\BB \subseteq \tau$ be an analytic basis for $\tau$.
Then $S$ is (everywhere) dense in $X$ every non-empty open set of $\BB$ contains an element of $S$.
== Proof ==
=== Necessary Condition ===
Let $S$ be everywhere dense in $X$.
By Open Set Characterization of Denseness then every non-empty open set contains an element of $S$.
Every non-empty set of an analytic basis is an open set by definition.
Hence every non-empty set of non-empty open set of $\BB$ contains an element of $S$.
=== Sufficient Condition ===
Suppose that every non-empty open set of $\BB$ contains an element of $S$.
Let $U$ be any non-empty open set.
Let $x \in U$.
By definition of an analytic basis then:
:$\exists B \in \BB: x \in B \subseteq U$
By assumption $B$ contains an element $s$ of $S$.
Hence $s \in U$.
Since $U$ was an arbitrary non-empty open set then every non-empty open set contains an element of $S$
By Open Set Characterization of Denseness, $S$ is everywhere dense in $X$.
Category:Denseness
Open Set Characterization of Denseness/Open Ball

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $\tau_d$ be the topology induced by the metric $d$.
Let $S \subseteq X$.
Then $S$ is (everywhere) dense in $\struct {X, \tau_d}$ every open ball contains an element of $S$.
== Proof ==
By Open Balls form Basis for Open Sets of Metric Space, the set of open balls are an analytic basis for the topology $\tau_d$.
By Analytic Basis Characterization of Denseness then:
:$S$ is (everywhere) dense in $\struct {X, \tau_d}$ every open ball contains an element of $S$.
Category:Denseness
Open Set Disjoint from Set is Disjoint from Closure

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$. 
Let $B$ be an open set of $T$ disjoint from $A$. 
Then:
: $A^- \cap B = \O$
where $A^-$ denotes the closure of $A$.
 
== Proof ==
Since $B \in \tau$, it follows by definition that $S \setminus B$ is closed.
Thus:
Category:Set Closures
Category:Open Sets
Open Set Less One Point is Open

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $U \subseteq M$ be an open set of $M$.
Let $\alpha \in U$.
Then $U \setminus \set \alpha$ is open in $M$.
=== Corollary ===
== Proof ==
Let $x \in U \setminus \set \alpha$.
Let $\delta = \map d {x, \alpha}$.
Let $\map {B_\epsilon} x \subseteq U$ be an open $\epsilon$-ball of $x$ in $U$.
Let $\zeta = \min \set {\epsilon, \delta}$.
Then:
:$\map {B_\epsilon} x \subseteq U \setminus \set \alpha$
The result follows.
Category:Open Sets
Open Set Less One Point is Open/Corollary

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $U \subseteq M$ be an open set of $M$.
Let $S = \set {\alpha_1, \alpha_2, \ldots, \alpha_n} \subseteq U$ be a finite set of points in $U$.
Then $U \setminus S$ is open in $M$.
== Proof ==
Follows directly from Open Set Less One Point is Open and Finite Intersection of Open Sets of Metric Space is Open.
Let:
:$U_1 = U \setminus \set {\alpha_1}, U_2 = U \setminus \set {\alpha_2}, \ldots, U_n = U \setminus \set {\alpha_n}$
From the above, $U_1, U_2, \ldots, U_n$ are all open in $M$.
From Finite Intersection of Open Sets of Metric Space is Open $\ds \bigcap_{i \mathop = 1}^n U_i$ is open.
Category:Open Sets
Open Set Not in System of Open Neighborhoods Iff Subset of Complement of Singleton Closure

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $x \in S$.
Let $\map \UU x$ denote the system of open neighborhoods of $x$.
Let $\set x^-$ denote the topological closure of $\set x$.
Then for all $U \in \tau$:
:$U \notin \map \UU x$ $U \subseteq S \setminus \set x^-$
== Proof ==
First note that by definition of closed set:
:$S \setminus U$ is a closed set
We have:
Category:Systems of Open Neighborhoods
Open Set in Open Subspace

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $T_U = \struct{U, \tau_U}$ be a subspace of $T$ where $U \subseteq X$ is open.
Let $V \subseteq U$ be a subset.
Then $V$ is open in $T_U$ $V$ is open in $T$.
== Proof ==
=== Necessary Condition ===
Let $V$ be open in $T$.
By Intersection with Subset is Subset, $V\cap U = V$.
By definition of topological subspace, $V$ is open in $T_U$.
=== Sufficient Condition ===
Let $V$ be open in $T_U$.
By definition of topological subspace, there exists an open subset $W\subseteq X$ such that $V=U\cap W$.
Because $U$ and $W$ are open in $T$, $V=U\cap W$ is open in $T$.
== Also see ==
* Closed Set in Closed Subspace
== Sources ==
* : $\S 3$: New Spaces From Old: Subspaces
Category:Topology
Category:Topological Subspaces
Open Set in Partition Topology is Component

== Theorem ==
Let $T = \struct {S, \tau}$ be a partition topological space.
Then each of its open sets are components of $T$.
== Proof ==
Let the partition $\PP$ be a basis of $T$.
From Open Set in Partition Topology is also Closed, open sets are in fact clopen.
So the elements of $\PP$ are clopen.
The result follows from the definition of components.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology: $1$
Category:Partition Topology
Open Set in Quotient Topological Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
Let $N$ be a linear subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\tau_N$ be the quotient topology on $X/N$.
Let $U \subseteq X/N$.
Then we have $U \in \tau_N$ there exists $V \in \tau$ such that $U = \pi \sqbrk V$. 
== Proof ==
=== Necessary Condition ===
Suppose that $U \in \tau_N$.
From the definition of the quotient topology, $\pi^{-1} \sqbrk U \in \tau$. 
Let $V = \pi^{-1} \sqbrk U$. 
Since $\pi$ is surjective, we have $\pi \sqbrk {\pi^{-1} \sqbrk U} = U$ and so $U = \pi \sqbrk V$.
Hence we have shown that there exists $V \in \tau$ such that $U = \pi \sqbrk V$. 
=== Sufficient Condition ===
Suppose that there exists $V \in \tau$ such that $U = \pi \sqbrk V$.
From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, we have that $\pi$ is an open mapping.
So $U \in \tau_N$ as the image of an open set under $\pi$.
Category:Quotient Topological Vector Spaces
Open Set is G-Delta Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ be an open set of $T$.
Then $U$ is a $G_\delta$ set of $T$.
== Proof ==
$U$ is the intersection of a singleton.
So $U$ is trivially the intersection of a countable number of open sets of $T$.
The result follows by definition of $G_\delta$ set.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:G-Delta Sets
Category:Open Sets
Open Set is Union of Elements of Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $B$ be a basis of $T$.
Let $V$ be an open subset of $S$.
Then $V = \bigcup \set {G \in B: G \subseteq V}$
== Proof ==
Let $x$ be arbitrary.
We will prove that:
:$x \in V \implies \exists Y \in \set {G \in B: G \subseteq V}: x \in Y$
Assume that:
:$x \in V$
By definition of basis:
:$\exists F \subseteq B: V = \bigcup F$
By definition of union:
:$\exists Y \in F: x \in Y$
By Set is Subset of Union/General Result:
:$Y \subseteq V$
Thus by definition of subset:
:$Y \in \set {G \in B: G \subseteq V}$
Thus:
:$x \in Y$
Assume that:
:$\exists Y \in \set {G \in B: G \subseteq V}: x \in Y$
By assumption:
:$Y \subseteq V$
Thus by definition of subset:
:$x \in V$
Hence by definition of union:
:$V = \bigcup \set {G \in B: G \subseteq V}$
== Sources ==
* 
Category:Topological Bases
Open Set may not be Open Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $U \subseteq M$ be an open set of $M$.
Then it is not necessarily the case that $U$ is an open ball of some $x \in A$.
== Proof ==
Consider the Euclidean space $\R^2$.
Let:
:$U \subseteq \R^2: U = \set {\tuple {x_1, x_2}: a < x_1 < b, c < x_2 < d}$
Let $x = \tuple {x_1, x_2} \in U$.
Then $\map {B_\epsilon} x \subseteq U$ when $\epsilon = \min \set {x_1 - a, b - x_1, x_2 - c, d - x_2}$:
:
So by definition, $U$ is open in $M$.
However, $U$ is not an open ball.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces
Category:Open Sets (Metric Spaces)
Category:Open Balls
Open Set minus Closed Set is Open

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
For $A \subseteq S$ denote by $\relcomp S A$ the relative complement of $A$ in $S$.
Let $U \in \tau$ and $\relcomp S V \in \tau$.
Then:
:$U \setminus V \in \tau$
and:
:$\relcomp S {V \setminus U} \in \tau$
== Proof ==
From Set Difference as Intersection with Relative Complement:
:$U \setminus V = U \cap \relcomp S V$
Since $\tau$ is a topology:
:$U, \relcomp S V \in \tau \implies U \cap \relcomp S V \in \tau \implies U \setminus V \in \tau$
The other statement follows mutatis mutandis.
== Sources ==
* 
* 
Category:Topology
Category:Open Sets
Category:Closed Sets
Open Set of Irreducible Space is Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be an irreducible topological space.
Let $U$ be a non-empty open set of $T$.
Then $U$ is irreducible in its induced subspace topology.
== Proof ==
Let $T = \struct {S, \tau}$ be an irreducible topological space.
Let $U$ be a non-empty open set of $T$.
 $U$ is not irreducible in $T$.
Then $U = V_1 \cup V_2$ for some closed sets $V_1$ and $V_2$ of $\struct {U, \tau_U}$.
By definition of subspace topology:
:$V_1 = U \cap W_1$
and:
:$V_2 = U \cap W_2$
for some closed sets $W_1$ and $W_2$ of $T$.
Because $W_1 = S \implies V_1 = U \cap W_1 = U$, it follows that:
:$W_1 \ne S$
This is a contradiction, because $V_1$ is a proper subset of $U$.
Now $U \ne \O$ implies that $S \setminus U$ is a proper subset of $S$ which is a closed set of $T$.
Let $W_3 := W_2 \cup \paren {S \setminus U}$.
Then $W_3$ is a closed set of $T$.
Also, $W_3 \ne S$, because otherwise $V_2 = U \cap W_2 = U \cap W_3 = U$.
Thus $S = W_1 \cup W_3$ shows that $T$ is not irreducible.
The result follows by Proof by Contradiction.
== Also see ==
* Space is Irreducible iff Open Sets are Connected
* Closed Set of Ultraconnected Space is Ultraconnected
Category:Irreducible Spaces
Open Set of Uncountable Finite Complement Topology is not F-Sigma

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $U \in \tau$ be an open set of $T$.
Then $U$ is not an $F_\sigma$ set.
== Proof ==
Let $U$ be an open set of $T$.
As $S$ is uncountable, then so is $U$.
By the definition of a finite complement topology, all closed sets of $T$ are finite.
From Countable Union of Countable Sets is Countable, $U$ can not be expressed as the union of a countable number of closed sets.
So by definition $U$ is not an $F_\sigma$ set.
== Sources ==
* : Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $3$
Category:F-Sigma Sets
Category:Finite Complement Topology
Open Sets in Indiscrete Topology

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is an open set of $T$ either $H = S$ or $H = \O$.
== Proof ==
A set $U$ is open in a topology $\tau$ if $U \in \tau$.
In $\tau = \set {\O, S}$, the only open sets are $\O$ and $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $2$
Category:Indiscrete Topology
Open Sets in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $\O$ and $A$ are both open in $M$.
== Proof ==
We have the results:
:Empty Set is Open in Metric Space
:Metric Space is Open in Itself
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.10$
Category:Open Sets (Metric Spaces)
Category:Metric Spaces
Open Sets in Pseudometric Space

== Theorem ==
Let $P = \struct {A, d}$ be a pseudometric space.
Then $\O$ and $A$ are both open in $P$.
== Proof ==
From the definition of open set:
: $\forall y \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} y \subseteq U$
where $U$ is an open set of $P$.
That is:
:One cannot get out of $U$ by moving an arbitrarily small distance from any point in $U$.
Take the case where $U = \O$.
The empty set $\O$ is open by dint of the fact that there are no points $y$ in $\O$ for which the condition is false.
Thus for $\O$:
: $\forall y \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} y \subseteq U$ is vacuously true.
When $U = A$, there are no points in $A$ to which one could get to by leaving $U$, an arbitrarily short distance or no, because there are no points in $A$ that are outside of $U$.
Category:Pseudometric Spaces