nyuuzyou/wikis huggingface dataset:
Closure of Infinite Union may not equal Union of Closures/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Consider the real number line $\struct {\R, \tau_d}$ with the usual (Euclidean) topology $\tau_d$.
Let:
:$H_n \subseteq \R: H_n = \closedint {\dfrac 1 n} 1$ for $n \ge 2$
where $\closedint {\dfrac 1 n} 1$ denotes the closed real interval from $\dfrac 1 n$ to $1$.
From Closed Real Interval is Closed Set, $\closedint {\dfrac 1 n} 1$ is a closed set of $\struct {\R, \tau_d}$.
Then from Set is Closed iff Equals Topological Closure:
:$\map \cl {H_n} = H_n$
Also:
:$\ds \bigcup_{n \mathop \ge 2} \map \cl {H_n} = \bigcup_{n \mathop \ge 2} H_n = \hointl 0 1$
However:
:$\ds \map \cl {\bigcup_{n \mathop \ge 2} H_n} = \closedint 0 1$
So:
:$\ds \bigcup_{n \mathop \ge 2} \map \cl {H_n} \ne \map \cl {\bigcup_{n \mathop \ge 2} H_n}$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Example $3.7.19$
Category:Closure of Infinite Union may not equal Union of Closures
Closure of Infinite Union may not equal Union of Closures/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $\struct {\R, \tau_d}$ denote the real number line with the usual (Euclidean) topology $\tau_d$.
Let $\struct {\Q, \tau_d}$ be the rational number space, also under the usual (Euclidean) topology $\tau_d$.
For a rational number $\alpha \in \Q$, let $B_\alpha$ denote the singleton containing $\alpha$.
From Union of Closures of Singleton Rationals is Rational Space:
:$\ds \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \Q$
From Closure of Union of Singleton Rationals is Real Number Line:
:$\ds \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha} = \R$
So it is seen that:
:$\ds \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} \subset \ds \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha}$
but it is not the case that:
:$\ds \bigcup_{\alpha \mathop \in \Q} \map \cl {B_\alpha} = \ds \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha}$
and the result is apparent.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Closure of Infinite Union may not equal Union of Closures
Closure of Integer Reciprocal Space

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n: n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then:
:$A^- = A \cup \set 0$
where $A^-$ denotes the closure of $A$ in $\R$.
== Proof ==
By definition, the closure of $A$ is:
:$A \cup A'$
where $A'$ is the derived set of $A$.
By definition of derived set, $A'$ consists of all the limit points of $A$ in $\R$.
From Zero is Limit Point of Integer Reciprocal Space, the only limit point of $A$ is $0$.
Hence the result:
:$A^- = A \cup \set 0$
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $1 \ \text{(a)}$
Category:Integer Reciprocal Space
Category:Set Closures
Closure of Interior of Closure of Union of Adjacent Open Intervals

== Theorem ==
Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the two adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^{- \circ -} = A^{\circ -} = A^- = \closedint a c$
where:
:$A^\circ$ is the interior of $A$
:$A^-$ is the closure of $A$.
== Proof ==
Then:
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(a)}$
Category:Set Interiors
Category:Set Closures
Category:Union of Adjacent Open Intervals
Closure of Intersection and Symmetric Difference imply Closure of Set Difference

== Theorem ==
Let $\RR$ be a system of sets such that for all $A, B \in \RR$:
:$(1): \quad A \cap B \in \RR$
:$(2): \quad A \symdif B \in \RR$
where $\cap$ denotes set intersection and $\symdif$ denotes set symmetric difference.
Then:
:$\forall A, B \in \RR: A \setminus B \in \RR$
where $\setminus$ denotes set difference.
== Proof ==
Let $A, B \in \RR$.
From Set Difference as Symmetric Difference with Intersection:
:$A \symdif \paren {A \cap B} = A \setminus B$
By hypothesis:
:$A \cap B \in \RR$
and:
:$A \symdif \paren {A \cap B} \in \RR$
and so:
:$A \setminus B \in \RR$
Category:Set Intersection
Category:Set Difference
Category:Symmetric Difference
Category:Set Systems
Closure of Intersection and Symmetric Difference imply Closure of Union

== Theorem ==
Let $\R R$ be a system of sets such that for all $A, B \in \RR$:
:$(1): \quad A \cap B \in \RR$
:$(2): \quad A \symdif B \in \RR$
where $\cap$ denotes set intersection and $\symdif$ denotes set symmetric difference.
Then:
:$\forall A, B \in \RR: A \cup B \in \RR$
where $\cup$ denotes set union.
== Proof ==
Let $A, B \in \RR$.
From Union as Symmetric Difference with Intersection‎:
:$\paren {A \symdif B} \symdif \paren {A \cap B} = A \cup B$
By hypothesis:
:$A \cap B \in \RR$
and:
:$\paren {A \symdif B} \symdif \paren {A \cap B} \in \RR$
and so:
:$A \cup B \in \RR$
Category:Set Intersection
Category:Set Union
Category:Symmetric Difference
Category:Set Systems
Closure of Intersection is Subset of Intersection of Closures

== Theorem ==
Let $T$ be a topological space.
Let $I$ be an indexing set.
Let $\forall i \in I: H_i \subseteq T$.
Then:
:$\ds \map \cl {\bigcap_I H_i} \subseteq \bigcap_I \map \cl {H_i}$
where $\map \cl {H_i}$ denotes the closure of $H_i$.
== Proof ==
Since $\ds \bigcap_I \map \cl {H_i}$ is an intersection of closed sets, it is closed, from Topology Defined by Closed Sets.
Also, it contains $\ds \bigcap_I H_i$ and so by the main definition of closure also contains $\ds \map \cl {\bigcap_I H_i}$.
== Also see ==
* Closure of Intersection may not equal Intersection of Closures, which shows that equality does not generally hold.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.17$
Category:Set Intersection
Category:Set Closures
Closure of Intersection may not equal Intersection of Closures

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H_1$ and $H_2$ be subsets of $S$.
Let ${H_1}^-$ and ${H_2}^-$ denote the closures of $H_1$ and $H_2$ respectively.
Then it is not necessarily the case that:
:$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$
== Outline of Proof ==
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.17$
* : Problems: Section $1: \ 2$
Category:Set Closures
Category:Set Intersection
Category:Closure of Intersection may not equal Intersection of Closures
Closure of Intersection may not equal Intersection of Closures/Examples/Arbitrary Subsets of Real Numbers

== Examples of Closure of Intersection may not equal Intersection of Closures ==
Let $H$ and $K$ be subsets of the set of real numbers $\R$ defined as:
Let $\map \cl H$ denote the closure of $H$.
Then:
:$H \cap \map \cl K$
:$\map \cl H \cap K$
:$\map \cl H \cap \map \cl K$
:$\map \cl {H \cap K}$
are all different.
== Proof ==
From Closure of Open Real Interval is Closed Real Interval:
Hence by definition of set intersection:
All defined sets, as can be seen, are different.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 24$
Category:Closure of Intersection may not equal Intersection of Closures
Closure of Intersection may not equal Intersection of Closures/Outline

== Theorem ==
== Proof ==
From Closure of Intersection is Subset of Intersection of Closures, it is seen that it is always the case that:
:$\paren {H_1 \cap H_2}^- \subseteq {H_1}^- \cap {H_2}^-$
It remains to be shown that it does not always happen that:
:$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$
The result is demonstrated by Proof by Counterexample.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.17$
Category:Closure of Intersection may not equal Intersection of Closures
Closure of Intersection may not equal Intersection of Closures/Proof 1

== Theorem ==
== Proof ==
=== Outline ===
=== Counterexample ===
Let $\struct {\R, \tau}$ be the real number line under the usual (Euclidean) topology.
Let $\Q$ denote the set of rational numbers.
Let $\R \setminus \Q$ denote the set of irrational numbers.
From Closure of Intersection of Rationals and Irrationals is Empty Set:
:$\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$
From Intersection of Closures of Rationals and Irrationals is Reals:
:$\Q^- \cap \paren {\R \setminus \Q}^- = \R$
The result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30 \text { - } 31$. The Rational and Irrational Numbers: $3$
Category:Closure of Intersection may not equal Intersection of Closures
Closure of Intersection may not equal Intersection of Closures/Proof 2

== Theorem ==
== Proof ==
=== Outline ===
=== Counterexample ===
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $H_1 = \openint 0 {\dfrac 1 2}$ and $H_2 = \openint {\dfrac 1 2} 1$.
By inspection it can be seen that:
:$H_1 \cap H_2 = \O$
Thus from Closure of Empty Set is Empty Set:
:$\paren {H_1 \cap H_2}^- = \O$
From Closure of Open Real Interval is Closed Real Interval:
:$H_1 = \closedint 0 {\dfrac 1 2}, H_2 = \closedint {\dfrac 1 2} 1$
Thus:
:${H_1}^- \cap {H_2}^- = \set {\dfrac 1 2}$
So $\paren {H_1 \cap H_2}^- \ne {H_1}^- \cap {H_2}^-$
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(d)}$
Category:Closure of Intersection may not equal Intersection of Closures
Closure of Intersection of Rationals and Irrationals is Empty Set

== Theorem ==
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\Q$ be the set of rational numbers.
Then:
:$\paren {\Q \cap \paren {\R \setminus \Q} }^- = \O$
where:
:$\R \setminus \Q$ denotes the set of irrational numbers
:$\paren {\Q \cap \paren {\R \setminus \Q} }^-$ denotes the closure of $\Q \cap \paren {\R \setminus \Q}$.
== Proof ==
From Set Difference Intersection with Second Set is Empty Set:
:$\Q \cap \paren {\R \setminus \Q} = \O$
By Empty Set is Closed in Topological Space, $\O$ is closed in $\R$.
From Closed Set Equals its Closure:
:$\O^- = \O$
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30 \text { - } 31$. The Rational and Irrational Numbers: $3$
Category:Rational Numbers
Category:Irrational Numbers
Category:Real Number Line with Euclidean Topology
Category:Set Closures
Closure of Irrational Interval is Closed Real Interval

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real numbers under the usual (Euclidean) topology.
Let $\struct{\R \setminus \Q, \tau_d}$ be the irrational 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:
:$\Bbb I \cap \paren {\R \setminus \Q}$
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 Irrational Numbers is Real Numbers:
:$\paren {\R \setminus \Q}^- = \R$
From Closure of Intersection is Subset of Intersection of Closures:
:$\paren{\Bbb I \cap \paren{\R \setminus \Q}}^- \subseteq \Bbb I^- \cap \paren {\R \setminus \Q}^-$
From Intersection with Subset is Subset:
:$\Bbb I^- \cap \paren {\R \setminus \Q}^- = \closedint a b$
and so:
:$\paren {\Bbb I \cap \paren {\R \setminus \Q} }^- \subseteq \closedint a b$
From Irrationals are Everywhere Dense in Topological Space of Reals:
:$\closedint a b \subseteq \paren {\Bbb I \cap \paren {\R \setminus \Q} }^-$
and the result follows.
Category:Real Intervals
Category:Irrational Number Space
Category:Set Closures
Closure of Irrational Numbers is Real Numbers

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

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $Y \subseteq S$ be a subset of $S$ which is irreducible in $T$.
Then its closure $Y^-$ in $T$ is also irreducible in $T$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Closure of Connected Set is Connected
Category:Closure of Irreducible Subspace is Irreducible
Category:Irreducible Spaces
Category:Set Closures
Closure of Irreducible Subspace is Irreducible/Proof 1

== Theorem ==
== Proof ==
By definition, $Y$ is an irreducible subset of $S$ in $T$ the subspace $\struct {Y, \tau_Y}$ is an irreducible topological space.
That is, such that two arbitrary non-empty open sets of $\struct {Y, \tau_Y}$ are not disjoint.
The open sets of $T$ in $Y^-$ are the same as the open sets of $\struct {Y, \tau_Y}$.
Hence by definition of irreducible space, any two open sets in $Y^-$ are not disjoint in $Y^-$.
That is $Y^-$ is also irreducible.
More generally, we can also show that if $Y^-$ is irreducible for a subset $Y \subseteq S$, then $Y$ is also irreducible in $T$. 
 $Y$ is not irreducible.
Then there exist two proper subsets $Y_1$, $Y_2$ of $Y$ which are closed in $Y$ such that $Y = Y_1 \cup Y_2$.
Then:
:$Y^- = {Y_1}^- \cup {Y_2}^-$
which contradicts the assumption.
Category:Closure of Irreducible Subspace is Irreducible
Closure of Irreducible Subspace is Irreducible/Proof 2

== Theorem ==
== Proof ==
In view of , it suffices to show that for all closed sets $A_1$ and $A_2$ in $T$:
:$Y^- \subseteq A_1 \cup A_2 \implies \exists i_0 \in \set {1, 2} : Y^- \subseteq A_{i_0}$
To this end, let $A_1$ and $A_2$ be closed sets in $T$ such that:
:$Y^- \subseteq A_1 \cup A_2$
Then, in particular:
:$Y \subseteq A_1 \cup A_2$
Since $Y \subseteq S$ is an irreducible subset:
:$\exists i_0 \in \set {1, 2} : Y \subseteq A_{i_0}$
By Closure of Subset of Closed Set of Topological Space is Subset:
:$Y^- \subseteq A_{i_0}$
Category:Closure of Irreducible Subspace is Irreducible
Closure of Irreducible Subspace is Irreducible/Proof 3

== Theorem ==
== Proof ==
Observe that for each closed set $V$ in $T$:
:$(1): \quad V \subsetneqq Y^- \implies V \cap Y \subsetneqq Y$
Indeed:
 $Y^-$ is not irreducible.
That is, there exist $V_1, V_2 \subsetneqq Y^-$, closed in $\struct {Y^-, \tau_{Y^-} }$, such that:
:$Y^- = V_1 \cup V_2$
$V_1$ and $V_2$ are also closed in $T$, since $Y^-$ is closed in $T$,
Then, from $(1)$:
:$V_1 \cap Y \subsetneqq Y$
and:
:$V_2 \cap Y \subsetneqq Y$
Therefore $V_1 \cap Y$ and $V_2 \cap Y$ are proper subsets of $Y$ such that:
:$Y = \paren {V_1 \cap Y} \cup \paren {V_2 \cap Y}$
Furtheremore, $V_1 \cap Y$ and $V_2 \cap Y$ are closed in $\struct {Y, \tau_Y}$.
This contradicts the fact that $Y$ is irreducible.
Category:Closure of Irreducible Subspace is Irreducible
Closure of Linear Subspace of Topological Vector Space is Linear Subspace

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$.
Let $Y$ be a linear subspace of $X$.
Then the closure $Y^-$ of $Y$ is a linear subspace of $X$.
== Proof ==
Since $Y \subseteq Y^-$, we have that $Y^-$ is non-empty.
We now use the One-Step Vector Subspace Test.
We show that for each $\lambda \in K$ and $u, v \in Y^-$ we have: 
:$u + \lambda v \in Y^-$
That is, we want to show that: 
:$Y^- + \lambda Y^- \subseteq Y^-$
for each $\lambda \in K$. 
Let $\lambda \in K$.
Then: 
Since $Y$ is a linear subspace, for each $u, v \in Y$ we have: 
:$u + \lambda v \in Y$
So:
:$Y + \lambda Y \subseteq Y$
From Topological Closure of Subset is Subset of Topological Closure, we therefore have: 
:$\paren {Y + \lambda Y}^- \subseteq Y^-$
so that: 
:$Y^- + \lambda Y^- \subseteq Y^-$
for each $\lambda \in K$ as required.
== Sources ==
* : $1.13$: Theorem
Category:Topological Vector Spaces
Category:Set Closures
Closure of Non-Empty Bounded Subset of Metric Space is Bounded

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $S \subseteq A$ be bounded in $M$.
Then:
:$\map \cl S$ is also bounded in $M$.
where $\map \cl S$ denotes the closure of $S$ in $M$.
== Proof ==
By definition of bounded:
:$S$ is bounded :
:$\exists K \in \R: \forall x, y \in M': \map {d_B} {x, y} \le K$
That is, such that $S$ has a diameter $K$.
From Diameter of Closure of Subset is Diameter of Subset, if $S$ has a diameter $K$, then so does $\map \cl S$.
That is, $\map \cl S$ is also bounded.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 32$
Category:Metric Spaces
Category:Set Closures
Category:Boundedness
Closure of Open 1-Ball in Standard Discrete Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be the standard discrete metric space on a set $A$.
Let $x \in A$.
Let $\map {B_1} x$ be the open $1$-ball of $x$ in $M$.
Then:
:$\map \cl {\map {B_1} x} = \set x$
while:
:$\set {y \in A: \map d {x, y} \le 1} = A$
== Proof ==
By definition of the standard discrete metric:
:$\map d {x, y} = \begin {cases}
0 & : x = y \\ 
1 & : x \ne y
\end {cases}$
That is:
:$\forall \tuple {x, y} \in A: \map d {x, y} \le 1$
Thus:
:$\set {y \in A: \map d {x, y} \le 1} = A$
From Open Ball in Standard Discrete Metric Space:
:$\map {B_1} x = \set x$
Let $y \in A: y \ne x$.
By definition of closure:
:$\map \cl {\map {B_1} x} = \paren {\map {B_1} x}^i \cup \paren {\map {B_1} x}'$
where:
:$\paren {\map {B_1} x}^i$ denotes the set of isolated points of $\map {B_1} x$
:$\paren {\map {B_1} x}'$ denotes the set of limit points of $\map {B_1} x$.
But:
:from Point in Standard Discrete Metric Space is Isolated, all points in $\map {B_1} x$ are isolated:
::$\paren {\map {B_1} x}^i = \map {B_1} x$
:from Set in Standard Discrete Metric Space has no Limit Points, $\map {B_1} x$ has no limit points:
::$\paren {\map {B_1} x}' = \O$
Hence:
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Example $3.7.22 \ \text {(a)}$
Category:Open Balls
Closure of Open Ball in Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M$.
Let $y \in \map \cl {\map {B_\epsilon} x}$, where $\cl$ denotes the closure of $\map {B_\epsilon} x$.
Then $\map d {x, y} \le \epsilon$.
== Proof ==
Suppose $\map d {x, y} > \epsilon$.
Then $\map {B_{\map d {x, y} - \epsilon} } y$ is an open set containing $y$ and not meeting $\map {B_\epsilon} x$.
Hence:
:$y \notin \map \cl {\map {B_\epsilon} x}$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.23$
Category:Open Balls
Closure of Open Ball may not equal Closed Ball of Same Radius

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$ of radius $\epsilon$ for some $\epsilon \in \R_{>0}$.
Let $\map { {B_\epsilon}^-} x$ be the closed $\epsilon$-ball of $x$ of radius $\epsilon$.
Then it is not necessarily the case that:
:$\map \cl {\map {B_\epsilon} x} = \map { {B_\epsilon}^-} x$
where $\cl$ denotes closure.
== Proof ==
Proof by Counterexample:
Let $M = \struct {A, d}$ be the standard discrete metric space on a set $A$.
From Closure of Open $1$-Ball in Standard Discrete Metric Space we have that:
:$\map \cl {\map {B_1} x} = \set x$
but:
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$
Category:Open Balls
Closure of Open Real Interval is Closed Real Interval

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
 
Let $\openint a b$ be an open interval of $\R$.
Then the closure of $\openint a b$ is the closed interval $\closedint a b$.
== Proof ==
From Limit Points of Open Real Interval, the limit points of $\openint a b$ consist of:
:the points $\openint a b$ itself
and
:the points $a$ and $b$.
By definition, the closure of $\openint a b$ is the union of $\openint a b$ and its limit points.
Hence the result, by definition of the closed interval $\closedint a b$.
Category:Open Sets
Category:Set Closures
Category:Real Number Line with Euclidean Topology
Closure of Open Set of 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 $T$.
Then $U^- = S^*_p$ where $U^-$ denotes the closure of $U$ in $T^*_p$.
== Proof ==
By definition, $\forall U \in \tau^*_p, u \ne \O: p \in U$.
From Limit Points in Closed Extension Space, every point in $S^*_p$ is a limit point of $p$.
So by definition of closure, every point in $S^*_p$ is in $U^-$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $21$
Category:Open Sets
Category:Set Closures
Category:Closed Extension Topology
Closure of Open Set of Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \in \tau_p$ be open in $T$ such that $U \ne \O$.
Then:
:$U^- = S$
where $U^-$ denotes the closure of $U$.
== Proof ==
Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Closure of Open Set of Closed Extension Space
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $2$
Category:Open Sets
Category:Set Closures
Category:Particular Point Topology
Closure of Pointwise Operation on Algebraic Structure

== Theorem ==
Let $S$ be a set such that $S \ne \O$.
Let $\struct {T, \circ}$ be an algebraic structure.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $f, g \in T^S$, that is, let $f: S \to T$ and $g: S \to T$ be mappings.
Let $\oplus: T^S \to T^S$ be the pointwise operation on $T^S$ induced by $\circ$.
Then $\oplus$ is closed on $T^S$ $\struct {T, \circ}$ is closed.
== Proof ==
=== Necessary Condition ===
Let $\struct {T, \circ}$ be closed.
Let $x \in S$ be arbitrary.
Then:
So $\oplus$ is closed on $T^S$.
=== Sufficient Condition ===
Let $\oplus$ is closed on $T^S$.
 $\struct {T, \circ}$ is not closed.
Then:
:$(1): \quad \exists s, t \in T: s \circ t \notin T$
By definition, $T^S$ is the set of all mappings from $S$ \to $T$.
As $S \ne \O$ it follows that $\exists x \in S$.
Thus, let $x \in S$ be arbitrary.
Let $f, g \in T^S$ such that:
:$(2): \quad \map f x = s, \map g x = t$
Then:
That is, $\oplus$ is not closed on $T^S$.
From that contradiction it follows that $\struct {T, \circ}$ is closed.
Category:Pointwise Operations
Closure of Preimage under Continuous Mapping is not necessarily Preimage of Closure

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H \subseteq S_1$ be a subset of $S_1$.
Let $\map \cl H$ denote the closure of $H$.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then it is not necessarily the case that:
:$f^{-1} \sqbrk {\map \cl H} = \map \cl {f^{-1} \sqbrk H}$
== Proof ==
Proof by Counterexample:
Let $\R$ be the real numbers under the usual (Euclidean) topology.
Let $f: \R \to \R$ be the real function:
:$\forall x \in \R: \map f x = \begin {cases} -1 & : x \le -1 \\ x & : -1 \le x \le 1 \\ 1 & : x \ge 1 \end {cases}$
It is accepted that $f$ is a continuous mapping.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
As can be seen:
:$f^{-1} \sqbrk {\map \cl H} \ne \map \cl {f^{-1} \sqbrk H}$
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 27 \ \text {(ii)}$
Category:Set Closures
Category:Continuous Mappings