nyuuzyou/wikis huggingface dataset:
Closure of Proper Ideal in Unital Banach Algebra is Proper Ideal

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.
Let $I$ be a proper ideal of $A$.
Then $I^-$ is a proper ideal of $A$.
== Proof ==
From Closure of Subspace of Normed Vector Space is Subspace, $I^-$ is a vector subspace of $A$. 
Let $x \in I^-$ and $y \in A$.
We need to show that $x y \in I^-$. 
From the definition of a closed set in a normed vector space, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ valued in $I$ such that:
:$x_n \to x$ 
Since $I$ is an ideal, we have $x_n y \in I$ for each $n \in \N$.
From Product Rule for Sequence in Normed Algebra, we have:
:$x_n y \to x y$
From the definition of closure, we have $x y \in I^-$. 
So $I^-$ is an ideal.
It remains to show that it is proper.
Let $\map G A$ be the group of units of $A$.
From Ideal of Unit is Whole Ring, we have:
:$J \cap \map G A = \O$
From Group of Units in Unital Banach Algebra is Open, $\map G A$ is open.
Hence from Open Set Disjoint from Set is Disjoint from Closure, we have that:
:$J^- \cap \map G A = \O$
From Ideal of Unit is Whole Ring, we conclude that $I^-$ is a proper ideal of $A$.
Category:Banach Algebras
Closure of Range of Compact Linear Transformation is Separable

== Theorem ==
Let $H, K$ be Hilbert spaces.
Let $T \in \map {B_0} {H, K}$ be a compact linear transformation.
Then $\map \cl {\Rng T}$ is separable.
== Proof ==
== Sources ==
* $II.4.5$
Category:Linear Transformations on Hilbert Spaces
Category:Compact Linear Transformations
Closure of Rational Interval is Closed Real Interval

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real numbers under the usual (Euclidean) topology.
Let $\struct {\Q, \tau_d}$ be the rational number space under the same topology.
Let $a, b \in \R$ such that $a < b$.
Let $\Bbb I \subseteq \R$ be an interval of $\R$
Then the closure of the set :
:$\Bbb I \cap \Q$
is the closed real interval $\closedint a b$.
== Proof ==
Let $\Bbb I$ be an open real interval.
From Closure of Real Interval is Closed Real Interval:
:$\Bbb I^- = \closedint a b$
From Closure of Rational Numbers is Real Numbers:
:$\Q^- = \R$
From Closure of Intersection is Subset of Intersection of Closures:
:$\paren{\Bbb I \cap \Q}^- \subseteq \Bbb I^- \cap \Q^-$
From Intersection with Subset is Subset:
:$\Bbb I^- \cap \Q^- = \closedint a b$
and so:
:$\paren {\Bbb I \cap \Q}^- \subseteq \closedint a b$
From Rationals are Everywhere Dense in Topological Space of Reals:
:$\closedint a b \subseteq \paren {\Bbb I \cap \Q}^-$
and the result follows.
Category:Real Intervals
Category:Irrational Number Space
Closure of Rational Numbers is Real Numbers

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\struct {\Q, \tau_d}$ be the rational number space under the same topology.
Then:
:$\Q^- = \R$
where $\Q^-$ denotes the closure of $\Q$.
== Proof ==
From Rationals are Everywhere Dense in Topological Space of Reals, $\Q$ is everywhere dense in $\R$.
It follows by definition of everywhere dense that $\Q^- = \R$.
Category:Rational Number Space
Category:Real Number Line with Euclidean Topology
Category:Set Closures
Closure of Real Interval is Closed Real Interval

== Theorem ==
Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then $I^-$ is the closed real interval $\closedint a b$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.13$
* 
* 
Category:Closure of Real Interval is Closed Real Interval
Category:Real Intervals
Category:Set Closures
Closure of Real Interval is Closed Real Interval/Proof 1

== Theorem ==
== Proof ==
There are four cases to cover:
:$(1): \quad$ Let $I = \openint a b$.
From Closure of Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
:$(2): \quad$ Let $I = \hointr a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
:$(3): \quad$ Let $I = \hointl a b$.
From Closure of Half-Open Real Interval is Closed Real Interval:
:$I^- = \closedint a b$
:$(4): \quad$ Let $I = \closedint a b$.
From Closed Real Interval is Closed in Real Number Line:
:$I$ is closed in $\R$.
From Set is Closed iff Equals Topological Closure:
:$I^- = \closedint a b$
Thus all cases are covered.
The result follows by Proof by Cases.
Category:Closure of Real Interval is Closed Real Interval
Closure of Real Interval is Closed Real Interval/Proof 2

== Theorem ==
== Proof ==
Let $I$ be one of the intervals as specified in the exposition.
Note that:
:$(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ every open set in $\R$ containing $x$ contains a point in $I$.
:$(2): \quad$ From Union of Open Sets of Metric Space is Open, every open set in $\R$ is a union of open intervals.
Thus we also have that $x \in I^-$ every open interval containing $x$ also contains a point in $I$.
This equivalence will be made use of throughout.
=== Lemma 1 ===
=== Lemma 2 ===
By the two lemmas proven above:
:$\closedint a b = I^-$
Category:Closure of Real Interval is Closed Real Interval
Closure of Real Interval is Closed Real Interval/Proof 2/Lemma 1

== Lemma for Closure of Real Interval is Closed Real Interval ==
Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then:
:$x \in \closedint a b \implies x \in I^-$
== Proof ==
Let $x \in \closedint a b$.
Let $\openint c d$ be an open interval in $\R$ such that $x \in \openint c d$.
We must show that $\openint c d$ contains a point in $I$.
One of the following three possibilities holds:
:$a < x < b$
:$x = a$
:$x = b$
==== Case: $a < x < b$ ====
In this case, $x \in I$ and $x \in \openint c d$.
Therefore $\openint c d$ contains a point in $I$.
==== Case: $x = a$ ====
If $I$ contains $a$, then this means $x \in I$, and the proof is complete.
So, assume that $a \notin I$.
Since $I$ is nonempty but does not contain $a$, we must have $a < b$.
Let $r$ be the minimum of $d$ and $b$, so that $r \le d$ and $r \le b$.
Since $a = x < d$ by choice of $d$ and since $a < b$ by assumption, we must have $a < r$.
Thus, by Real Numbers are Densely Ordered, there exists some $s \in \R$ such that $a < s < r$.
To summarize, we have $c < x = a < s < r$, where $r \le d$ and $r \le b$.
This means that $s$ satisfies both $c < s < d$ and $a < s < b$.
Hence, $s$ is a point in $\openint c d$ which is also in $I$.
The existence of such a point is what we wanted to show.
==== Case: $x = b$ ====
This case is analogous to case when $x = a$.
Here we instead let $l$ be the maximum of $c$ and $a$, and select an $s$ such that $l < s < x = b < d$, where $c \le l$ and $a \le l$.
Category:Closure of Real Interval is Closed Real Interval
Closure of Real Interval is Closed Real Interval/Proof 2/Lemma 2

== Lemma for Closure of Real Interval is Closed Real Interval ==
Let $I$ be a non-empty real interval such that one of these holds:
:$I = \openint a b$
:$I = \hointr a b$
:$I = \hointl a b$
:$I = \closedint a b$
Let $I^-$ denote the closure of $I$.
Then:
:$x \notin \closedint a b \implies x \notin I^-$
== Proof ==
Suppose $x \notin \notin \closedint a b$.
We must find an open interval containing $x$ which does not contain an elements in $I$.
There are two possibilities:
:$x < a$
or:
:$x > b$
==== Case: $x < a$ ====
By Real Numbers are Densely Ordered, there exists $r \in \R$ such that $x < r < a$.
Thus $\openint {x - 1} r$ is an open interval, all of whose elements are less than $a$, and hence not in $I$.
==== Case: $x > b$ ====
This is similarly to the case when $x < a$.
Here instead we pick $r$ such that $b < r < x$, and consider the open interval $\openint r {x + 1}$.
Category:Closure of Real Interval is Closed Real Interval
Closure of Self-Adjoint Subset of Banach *-Algebra is Self-Adjoint

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $S \subseteq A$ be self-adjoint.
Then the closure $S^-$ of $S$ is self-adjoint.
== Proof ==
Let $x \in S^-$. 
From Point in Closure of Subset of Metric Space iff Limit of Sequence, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ with $x_n \to x$. 
Since $S$ is self-adjoint, we have $x_n^\ast \in S$. 
From Star of Convergent Sequence in Banach *-Algebra Converges, we have $x_n^\ast \to x^\ast$. 
From Point in Closure of Subset of Metric Space iff Limit of Sequence, we have $x^\ast \in S^-$. 
Hence if $x \in S^-$, we have $x^\ast \in S^-$. 
So $S^-$ is self-adjoint.
Category:Banach *-Algebras
Closure of Set of Condensation Points equals Itself

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:$\paren {A^0}^- = A^0$
where
:$A^0$ denotes the set of condensation points of $A$
:$A^-$ denotes the closure of $A$
== Proof ==
By Set is Subset of its Topological Closure:
:$A^0 \subseteq \paren {A^0}^-$
To prove the equality by definition of set equality it suffices to show the inclusion:
:$\paren {A^0}^- \subseteq A^0$
Let $x \in \paren {A^0}^-$.
We will prove that
:$(1): \quad \forall U \in \tau: x \in U \implies A \cap U$ is uncountable
Let $U$ be an open subset of $S$ such that
:$x \in U$
By Condition for Point being in Closure:
:$A^0 \cap U \ne \O$
By definition of empty set:
:$\exists y: y \in A^0 \cap U$
By definition of intersection:
:$y \in A^0 \land y \in U$
By definition of set of condensation points:
:$y$ is condensation point of $A$
Then by definition of condensation point:
:$A \cap U$ is uncountable
We will prove that
:$x$ is limit point of $A$
Let $U$ be an open subset of $S$ such that
:$x \in U$
By $(1)$:
:$A \cap U$ is uncountable
:$A \cap \paren {U \setminus \set x}$ is uncountable
Thus:
:$A \cap \paren {U \setminus \set x} \ne \O$
Then by definition and $(1)$:
:$x$ is condensation point of $A$
Thus by definition of set of condensation points:
:$x \in A^0$
== Sources ==
* 
* 
Category:Condensation Points
Category:Set Closures
Closure of Singleton is Lower Closure of Element in Scott Topological Lattice

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a up-complete topological lattice with the Scott topology.
Let $x \in S$.
Then:
:$\set x^- = x^\preceq$
where
:$\set x^-$ denotes the topological closure of $\set x$
:$x^\preceq$ denotes the lower closure of $x$.
== Proof ==
By Lower Closure of Element is Closed under Directed Suprema:
:$x^\preceq$ is closed under directed suprema.
By Lower Closure of Singleton:
:$\set x^\preceq = x^\preceq$
By Lower Closure is Lower Section:
:$x^\preceq$ is a lower section.
By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
:$x^\preceq$ is closed.
By definitions of lower closure of element and reflexivity:
:$x \in x^\preceq$
By Singleton of Element is Subset:
:$\set x \subseteq x^\preceq$
We will prove that:
:for every a closed subset $C$ of $S$: $\set x \subseteq C \implies x^\preceq \subseteq C$
Let $C$ be a closed subset of $S$ such that:
:$\set x \subseteq C$
By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
:$C$ is a lower section.
Let $y \in x^\preceq$.
By definition of lower closure of element:
:$y \preceq x$
By definitions of subset and singleton:
:$x \in C$
Thus by definition of lower section:
:$y \in C$
Thus by definition of topological closure:
:$\set x^- = x^\preceq$
== Sources ==
* 
* 
Category:Topological Order Theory
Closure of Subalgebra in Normed Algebra is Subalgebra

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.
Let $B$ be a subalgebra of $A$.
Then $B^-$ is a subalgebra of $A$.
== Proof ==
From Closure of Subspace of Normed Vector Space is Subspace, $B^-$ is a vector subspace of $A$. 
Now let $x, y \in B^-$.
From the definition of a closed set in a normed vector space, there exists sequences $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ valued in $B$ such that:
:$x_n \to x$ and $y_n \to y$.
From Product Rule for Sequence in Normed Algebra, we have:
:$x_n y_n \to x y$
From the definition of closure, we have $x y \in B^-$.
So $B^-$ is a vector subspace of $A$ that is closed under multiplication.
Hence $B^-$ is a subalgebra of $A$.
Category:Normed Algebras
Closure of Subgroup is Group

== Theorem ==
Let $G$ be a topological group.
Let $H\leq G$ be a subgroup.
Let $\overline H$ denote its closure.
Then $\overline H$ is a subgroup of $G$.
== Proof ==
We use the One-Step Subgroup Test.
Because $H \subset \overline H$, $\overline H$ is non-empty.
Let $a, b \in \overline H$.
Let $U$ be a neighborhood of $a b^{-1}$.
Let the mapping $f: G\times G \to G$ be defined as:
:$\map f {x, y} = x y^{-1}$
By definition of topological group, $f$ is continuous.
By definition of product space, there exist neighborhoods $A, B$ of $a, b$ respectively such that:
:$A \times B \subset f^{-1} \sqbrk U$
By assumption, there exist $x \in A \cap H$ and $y \in B \cap H$.
Then:
:$x y^{-1} \subset U \cap H$
Because $U$ was arbitrary:
:$a b^{-1} \in \overline H$
By One-Step Subgroup Test, $\overline H$ is a subgroup of $G$.
Category:Topological Groups
Closure of Subset Contains Parallel Elements

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $\sigma: \powerset S \to \powerset S$ denote the closure operator of $M$.
Let $A \subseteq S$.
Let $x, y \in S$.
If $x \in \map \sigma A$ and $y$ is parallel to $x$ then:
:$y \in \map \sigma A$
== Proof ==
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Let $x \in \map \sigma A$
Let $y$ be parallel to $x$.
By the definitions of the closure operator and depends depends:
:$\map \rho A = \map \rho {A \cup \set x}$
and:
:$y \in \map \sigma A$ $\map \rho A = \map \rho {A \cup \set y}$
From Rank Function is Increasing:
:$\map \rho {A \cup \set x} = \map \rho A \le \map \rho {A \cup \set y} \le \map \rho {A \cup \set {x, y} }$
To show $\map \rho A = \map \rho {A \cup \set y}$ it is sufficient to show:
:$\map \rho {A \cup \set {x, y} } = \map \rho {A \cup \set x}$
From Rank Function is Increasing:
:$\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x, y} }$
By the definition of parallel:
:$\set x$ is independent
From Independent Subset is Contained in Maximal Independent Subset:
:$\exists X \in \mathscr I : \set x \subseteq X \subseteq A \cup \set x : \size X = \map \rho {A \cup \set x}$
From Subset Relation is Transitive:
:$X \subseteq A \cup \set {x, y}$
From Independent Subset is Contained in Maximal Independent Subset:
:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A \cup \set {x, y} : \size Y = \map \rho {A \cup \set {x, y} }$
By definition of a subset:
:$x \in Y$
By definition of parallel:
:$\set {x, y}$ is dependent
From Superset of Dependent Set is Dependent:
:$\set {x, y} \nsubseteq Y$
Then:
:$y \notin Y$
So:
:$Y \subseteq A \cup \set x$
By definition of the rank function:
:$\size Y \le \map \rho {A \cup \set x}$
As $\size Y = \map \rho {A \cup \set {x, y} }$ then:
:$\map \rho {A \cup \set {x, y} } \le \map \rho {A \cup \set x}$
As $\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x, y} }$ then:
:$\map \rho {A \cup \set {x, y} } = \map \rho {A \cup \set x}$
The result follows.
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Closure
Category:Matroid Parallelism
Closure of Subset contains Loop

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x$ be a loop of $M$.
Let $A \subseteq S$.
Then:
:$x \in \map \sigma A$
where $\map \sigma A$ denotes the closure of $A$.
== Proof ==
By definition of the closure of $A$:
:$x \in \map \sigma A$ $x \sim A$
where $\sim$ is the depends relation on $M$.
By definition of the depends relation:
:$x \sim A$ $\map \rho {A \cup \set x} = \map \rho A$
where $\rho$ is the rank function on $M$.
So it remains to show that:
:$\map \rho {A \cup \set x} = \map \rho A$
By definition of the rank function:
:$\map \rho {A \cup \set x} = \max \set {\size X : X \subseteq A \cup \set x \land X \in \mathscr I}$
From Max Operation Equals an Operand:
:$\exists X \in \mathscr I : X \subseteq A \cup \set x \land \size X = \map \rho {A \cup \set x}$
From the contrapositive statement of Set is Dependent if Contains Loop:
:$x \notin X$
Now:
From Max Operation Yields Supremum of Operands:
:$\size X \le \max \set {\size Y : Y \subseteq A \land Y \in \mathscr I} = \map \rho A$
From Rank Function is Increasing:
:$\map \rho A \le \map \rho {A \cup \set x} = \size X$
Thus:
:$\map \rho A = \size X = \map \rho {A \cup \set x}$
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Closure
Category:Matroid Loops
Closure of Subset in Subspace

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $H \subseteq S$ be an arbitrary subset of $S$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace on $H$.
Let $A \subseteq H$ be an arbitrary subset of $H$.
Then:
:$\map {\cl_H} A = H \cap \map \cl A$
where:
:$\map {\cl_H} A$ denotes the closure of $A$ in $T_H$
:$\map \cl A$ denotes the closure of $A$ in $T$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 28 \ \text {(i)}$
* : $\S 3$: New Spaces From Old: Subspaces
Category:Set Closures
Category:Topological Subspaces
Closure of Subset in Subspace/Corollary 1

== Corollary to Closure of Subset in Subspace ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H$ be a subset of $S$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace on $H$.
Let $K \subseteq S$.
Let $\map {\cl_T} K$ denote the closure of $K$ in $T$.
Let $\map {\cl_H} {K \cap H}$ denote the closure of $K \cap H$ in $T_H$.
Then:
:$\map {\cl_H} {K \cap H} \subseteq \map {\cl_T} K \cap H$
== Proof ==
Category:Set Closures
Category:Topological Subspaces
Closure of Subset in Subspace/Corollary 2

== Corollary to Closure of Subset in Subspace ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H$ be a subset of $S$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace on $H$.
Let $A$ be a subset of $H$.
Let $H$ be closed in $T$.
Then:
:$\map {\cl_H} A = \map \cl A$
where:
:$\map {\cl_H} A$ denotes the closure of $A$ in $T_H$
:$\map \cl A$ denotes the closure of $A$ in $T$.
== Proof ==
From Closure of Subset in Subspace:
:$\map {\cl_H} A = \map \cl A \cap H$
From Intersection is Subset:
:$\map {\cl_H} A \subseteq \map \cl A$
From Topological Closure is Closed:
:$\map \cl A$ is closed in $T$
From Intersection of Closed Sets is Closed in Topological Space:
:$\map {\cl_H} A$ is closed in $T$
From Set is Subset of its Topological Closure:
:$A \subseteq \map {\cl_H} A$
From Set Closure is Smallest Closed Set in Topological Space:
:$\map \cl A \subseteq \map {\cl_H} A$
By definition of set eqality:
:$\map {\cl_H} A = \map \cl A$
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 28 \ \text {(ii)}$
Category:Set Closures
Category:Topological Subspaces
Closure of Subset of Closed Set of Metric Space is Subset

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $F$ be a closed set of $M$.
Let $H \subseteq F$ be a subset of $F$.
Let $H^-$ denote the closure of $H$.
Then $H^- \subseteq F$.
== Proof 1 ==
== Proof 2 ==
Category:Set Closures
Category:Closure of Subset of Closed Set of Metric Space is Subset
Closure of Subset of Closed Set of Metric Space is Subset/Proof 1

== Theorem ==
== Proof ==
From Metric Induces Topology, the topology $\tau$ induced by the metric $d$ is a topology on $M$.
From Metric Closure and Topological Closure of Subset are Equivalent, it is sufficient to show that the topological closure of $H$ is contained in $F$.
From Set is Closed in Metric Space iff Closed in Induced Topological Space:
:$F$ is closed in the topological space $\struct{A, \tau}$ induced by the metric $d$.
We have:
Category:Closure of Subset of Closed Set of Metric Space is Subset
Closure of Subset of Closed Set of Metric Space is Subset/Proof 2

== Theorem ==
== Proof ==
Let $x \in H^-$.
From Point in Closure of Subset of Metric Space iff Limit of Sequence
:there exists a sequence $\sequence {a_n}$ of points of $H$ which converges to the limit $x$.
By assumption:
:$\sequence {a_n}$ is also a sequence of points of $F$
From Subset of Metric Space contains Limits of Sequences iff Closed:
:$x \in F$
Thus it has been shown:
:$H^- \subseteq F$
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$
Category:Closure of Subset of Closed Set of Metric Space is Subset
Closure of Subset of Closed Set of Topological Space is Subset

== Theorem ==
Let $T$ = $\struct {S, \tau}$ be a topological space.
Let $F$ be a closed set of $T$.
Let $H \subseteq F$ be a subset of $F$.
Let $H^-$ denote the closure of $H$.
Then $H^- \subseteq F$.
== Proof 1 ==
== Proof 2 ==
Category:Set Closures
Category:Closure of Subset of Closed Set of Topological Space is Subset
Closure of Subset of Closed Set of Topological Space is Subset/Proof 2

== Theorem ==
== Proof ==
Let $x \notin F$.
Then $x$ is in the open set $S \setminus F$.
From Set Difference with Subset is Superset of Set Difference:
:$S \setminus F \subseteq S \setminus H$
From Subsets of Disjoint Sets are Disjoint:
:$S \setminus F \cap H = \O$
From Set is Open iff Neighborhood of all its Points:
:$S \setminus F$ is an neighborhood of $x$
By definition of the closure of $H$:
:$x \notin H^-$
We have shown that:
:$S \setminus F \subseteq S \setminus H^-$
From Set Difference with Subset is Superset of Set Difference:
:$H^- \subseteq F$
== Sources ==
* : $\S 4.6$: Closure, Interior, Boundary: Lemma $4.3$
Category:Closure of Subset of Closed Set of Topological Space is Subset
Closure of Subset of Double Pointed Topological Space

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Let $X \subseteq S \times D$ be a subset of $S \times D$.
Then the closure of $X$ in $\tau$ is:
:$\map \cl X = \map \cl {\map {\pr_1} X} \times D$
where $\pr_1$ denotes the first projection on $S \times D$.
== Proof ==
By Closed Sets of Double Pointed Topology, $\map \cl {\map {\pr_1} X} \times D$ is closed in $\tau$.
Furthermore, for $\tuple {s, d} \in X$, one has:
:$s \in \map {\pr_1} X \subseteq \map \cl {\map {\pr_1} X}$
by definition of closure.
Since also $d \in D$, we conclude that:
:$X \subseteq \map \cl {\map {\pr_1} X} \times D$
By $(3)$ of Equivalence of Definitions of Topological Closure, it now suffices to prove that:
:$\map \cl {\map {\pr_1} X} \times D \subseteq C$
for any closed set $C$ with $X \subseteq C$.
By Closed Sets of Double Pointed Topology, for some $C'$ closed in $\tau$, we have:
:$C = C' \times D$
so that $C' = \map {\pr_1} C$.
By Image Preserves Subsets, it follows that:
:$\map {\pr_1} X \subseteq C'$
and by $(3)$ of Equivalence of Definitions of Topological Closure, this means:
:$\map \cl {\map {\pr_1} X} \subseteq C'$
whence by Cartesian Product of Subsets:
:$\map \cl {\map {\pr_1} X} \times D \subseteq C' \times D = C$
Hence the result.
== Also see ==
* Interior of Subset of Double Pointed Topological Space
Category:Double Pointed Topologies
Category:Set Closures
Closure of Subset of Indiscrete Space

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\O \subsetneq H \subseteq S$ (that is, let $H$ be a non-empty subset of $T$).
Then:
:$H^- = H^{- \circ} = H^{- \circ -} = S$
where:
:$H^\circ$ denotes the interior of $H$
:$H^-$ denotes the closure of $H$.
== Proof ==
From Limit Points of Indiscrete Space, every point in $S$ is a limit point of $H$.
So from the definition of closure, $H \ne \O \implies H^-= S$.
From the open set axioms, $S$ is open in $T$.
From Interior of Open Set, $S^\circ = S$.
From Underlying Set of Topological Space is Clopen, $S$ is closed in $T$.
From Closed Set Equals its Closure, $S^- = S$.
The result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $6$
Category:Set Closures
Category:Indiscrete Topology