nyuuzyou/wikis huggingface dataset:
Closure for Finite Collection of Relations and Operations

== Theorem ==
Let $\RR_1, \RR_2, \ldots \RR_n$ be relations.
Let $\SS_1, \SS_2, \ldots \SS_m$ be operations.
Let $T$ be a small class.
Let the image of $\RR_i$ over any small class $x$ be small classes for $1 \le i \le n$.
Let the image of $\SS_i$ over any Cartesian product $x \times x$ be small classes for $1 \le i \le m$.
Then there exists a small class $X$ such that:
:$(1): \quad$ $T \subseteq X$
:$(2): \quad$ Each $\RR_i$ is closed $X$. Each $\SS_i$ is closed $X \times X$.
:$(3): \quad$ $X$ is the smallest small class satisfying conditions $(1)$ and $(2)$ above.
If $Y$ satisfies conditions $(1)$ and $(2)$, then $X \subseteq Y$.
== Proof ==
Let $R \sqbrk x$ denote the image of $x$ under $R$.
Set:
:$\ds \map G x = x \cup \bigcup_{i \mathop = 1}^n \RR_i \sqbrk x \cup \bigcup_{i \mathop = 1}^m \SS_i \sqbrk x$
Using the Principle of Recursive Definition, construct the function $F$ as follows:
:$\map F 0 = T$
:$\map F {n + 1} = \map G {\map F n}$
Define $X$ as follows:
:$\ds X = \bigcup_{n \mathop \in \omega} \map F n$
=== Proof that $T \subseteq X$ ===
=== Proof of Closure ===
Take any $\RR_i$.
Suppose $x \mathrel {\RR_i} y$ and $x \in \map F n$ for some $n$.
Then $y \in \map F {n + 1}$ by definition of image.
Thus $y \in X$ and $R_i$ is closed $X$.
Similarly, take any $\SS_i$.
Suppose $\paren {x \mathrel {S_i} y} = z$ and that $x \in X$ and $y \in X$.
It follows that $x, y \in \map F n$ for some $n$.
So:
:$z \in \map F {n + 1}$
Therefore:
:$z \in X$
=== Proof that $X$ is the Smallest Relational Closure ===
Suppose $T \subseteq Y$ and that $R_i$ and $S_i$ are closed $Y$.
$\map F n \subseteq Y$ shall be proven by induction:
For all $n \in \omega$, let $\map P n$ be the proposition:
:$\map F n \subseteq Y$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\map F 0 \subseteq Y$
which has been proved above.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\map F k \subseteq Y$
Then we need to show:
:$\map F {k + 1} \subseteq Y$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by Principle of Mathematical Induction.
Therefore:
:$\forall n \in \omega: \map F n \subseteq R$
Hence by Indexed Union Subset:
:$T \subseteq Y$
== Sources ==
* : $\S 9.6$
Category:Relational Closures
Closure in Infinite Particular Point Space is not Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Let $A \in \tau_p$ be open in $T$.
Let $A^-$ be the closure of $A$.
Then $A^-$ is not compact.
== Proof ==
From Closure of Open Set of Particular Point Space, we have that $A^- = S$.
The result follows from Infinite Particular Point Space is not Compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $9 \text { - } 10$. Infinite Particular Point Topology: $5$
Category:Compact Spaces
Category:Particular Point Topology
Closure in Weak-* Topology in terms of Annihilators

== Theorem ==
Let $X$ be a Banach space. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $N$ be a linear subspace of $X^\ast$. 
Then: 
:$\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$
where:
:$\cl_{w^\ast}$ denotes closure in the weak-$\ast$ topology
:${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$
:$\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$. 
== Proof ==
From Linear Subspace is Subset of Double Annihilator, we have:
:$N \subseteq \paren { {}^\bot N}^\bot$
So, from Set Closure Preserves Set Inclusion:
:$\map {\cl_{w^\ast} } N \subseteq \map {\cl_{w^\ast} } {\paren { {}^\bot N}^\bot}$
From Annihilator of Subspace of Banach Space is Weak-* Closed and Set is Closed iff Equals Topological Closure, we have:
:$\map {\cl_{w^\ast} } N \subseteq \paren { {}^\bot N}^\bot$
It remains to show:
:$\paren { {}^\bot N}^\bot \subseteq \map {\cl_{w^\ast} } N$
From Set Complement inverts Subsets, we can equivalently show:
:$X^\ast \setminus \map {\cl_{w^\ast} } N \subseteq X^\ast \setminus \paren { {}^\bot N}^\bot$
Let:
:$f \in X^\ast \setminus \map {\cl_{w^\ast} } N$
Applying Existence of Non-Zero Continuous Linear Functional vanishing on Proper Closed Subspace of Hausdorff Locally Convex Space to the locally convex space $\struct {X^\ast, w^\ast}$, there exists $\Phi \in \struct {X^\ast, w^\ast}^\ast$ such that $\Phi \ne {\mathbf 0}_{X^\ast}$ and:
:$\map \Phi g = 0$ for each $g \in \map {\cl_{w^\ast} } N$
and:
:$\map \Phi f \ne 0$
From Characterization of Continuity of Linear Functional in Weak-* Topology, there exists $x \in X$ such that $\Phi = x^\wedge$. 
Then: 
:$\map {x^\wedge} g = \map g x = 0$ for all $g \in N$
So $x \in {}^\bot N$.
But we also have $0 \ne \map {x^\wedge} f = \map f x$, so $f$ does not vanish at some point in ${}^\bot N$.
So $f \not \in \paren { {}^\bot N}^\bot$ by the definition of the annihilator of a subspace of $X$. 
We therefore have:
:$X^\ast \setminus \map {\cl_{w^\ast} } N \subseteq X^\ast \setminus \paren { {}^\bot N}^\bot$
hence:
:$\paren { {}^\bot N}^\bot \subseteq \map {\cl_{w^\ast} } N$
from Set Complement inverts Subsets.
We conclude:
:$\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$
Category:Annihilators of Subspaces of Banach Spaces
Category:Annihilators of Subspaces of Normed Dual Spaces
Category:Weak-* Topologies
Closure is Closed

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl: S \to S$ be a closure operator.
Let $x \in S$.
Then $\map \cl x$ is a closed element of $S$ with respect to $\cl$.
=== Power Set ===
When the ordering in question is the subset relation on a power set, the result can be expressed as follows:
== Proof ==
By the definition of closure operator, $\cl$ is idempotent.
Therefore:
:$\map \cl {\map \cl x} = \map \cl x$
It follows by definition that $\map \cl x$ is a closed element.
== Also see ==
* Topological Closure is Closed
Category:Closure Operators
Closure is Closed/Power Set

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\cl: \powerset S \to \powerset S$ be a closure operator.
Let $T \subseteq S$.
Then $\map \cl T$ is a closed set with respect to $\cl$.
== Proof ==
By the definition of closure operator, $\cl$ is idempotent.
Therefore $\map \cl {\map \cl T} = \map \cl T$, so $\map \cl T$ is closed.
Category:Closure Operators
Closure is Smallest Closed Successor

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $f: S \to S$ be a closure operator on $S$.
Let $x \in S$.
Then $\map f x$ is the smallest closed element that succeeds $x$.
== Proof ==
By the definition of closure operator, $f$ is inflationary.
Thus $x \preceq \map f x$.
By definition, $\map f x$ is closed.
So $\map f x$ is a closed element that succeeds $x$.
We will now show that it is the smallest such.
Let $k$ be a closed element of $S$ such that $x \preceq k$.
Since $f$ is a closure operator, it is increasing.
Therefore $\map f x \preceq \map f k$.
Since $k$ is closed, $\map f k = k$.
Thus $\map f x \preceq k$.
Category:Closure Operators
Closure of Absorbing Set is Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A \subseteq X$ be an absorbing set.
Then $A^-$ is absorbing.
== Proof ==
From the definition of closure, we have $A \subseteq A^-$.
From Superset of Absorbing Set is Absorbing, $A^-$ is absorbing.
Category:Absorbing Sets
Closure of Balanced Set in Topological Vector Space is Balanced

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $B \subseteq X$ be a balanced set.
Then the closure $B^-$ of $B$ is balanced.
== Proof ==
Let $\lambda \in \Bbb F$ have $\cmod \lambda \le 1$. 
Then, we have: 
:$\lambda B \subseteq B$ 
So, from Topological Closure of Subset is Subset of Topological Closure we have: 
:$\paren {\lambda B}^- \subseteq B^-$
From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have $\paren {\lambda B}^- = \lambda B^-$ and so: 
:$\lambda B^- \subseteq B^-$
So $B^-$ is balanced.
== Sources ==
* : $1.13$: Theorem
Category:Set Closures
Category:Balanced Sets
Category:Topological Vector Spaces
Closure of Cartesian Product is Product of Closures

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
:$\map \cl {H \times K} = \map \cl H \times \map \cl K$
where $\map \cl H$, for example, denotes the closure of $H$.
=== Indexed Family ===
== Proof ==
Consider the relative complements of $H$ and $K$ in $T_1$ and $T_2$ respectively:
:$\overline H = \relcomp {S_1} H$
:$\overline K = \relcomp {S_2} K$
Then from Interior of Cartesian Product is Product of Interiors:
:$\Int {\overline H \times \overline K} = \Int {\overline H} \times \Int {\overline K}$
From Complement of Interior equals Closure of Complement:
:$\map \cl H = \Int {\overline H}$
:$\map \cl K = \Int {\overline K}$
:$\map \cl {H \times K} = \Int {\overline H \times \overline K}$
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 35 \ \text {(ii)}$
Category:Set Closures
Category:Product Topology
Closure of Complement of Closure is Regular Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of $T$.
Let $A^-$ denote the closure of $A$ in $T$.
Let $A'$ denote the complement of $A$ in $S$: $A' = S \setminus A$.
Then $A^{-'-}$ is regular closed.
== Proof ==
Let $A^\circ$ denote the interior of $A$.
From Set is Closed iff Equals Topological Closure, $A^-$ is closed in $T$.
Since $A^-$ is a closed set, $A^{-'}$ is open.
Therefore:
:$\forall x \in A^{-'}: \exists \epsilon \in \R_{>0}$
Additionally, because $A^{-'}$ is the complement of $A^-$ relative to $S$, we have:
:$\exists \epsilon \in \R_{>0}: \map {V_\epsilon} x \cap A^- = \O \iff x \in A^{-'}$
Therefore, for $y: y \in A^{-'-}$ and $ y \notin A^{- '}$ we know that:
:$\forall \epsilon \in \R_{>0}: \map {V_\epsilon} y \cap A^- \ne \O$
meaning that $y$ is a limit point of $A^-$.
Therefore:
:$y \notin A^{-'-\circ}$
Now:
:$x \in A^{-'-\circ} \iff x \in A^{-'}$
In conclusion, this means that:
:$A^{-'-\circ-} = A^{-'-},$ verifying the definition of $A^{-'-}$ being regular closed. 
Category:Regular Closed Sets
Closure of Connected Set is Connected

== Theorem ==
Let $T$ be a topological space.
Let $H$ be a connected set of $T$.
Let $H^-$ denote the closure of $H$ in $T$. 
Then $H^-$ is connected in $T$.
== Proof ==
By Set is Subset of Itself, the result follows by setting $K = H^-$ in Set between Connected Set and Closure is Connected.
== Also see ==
* Closure of Irreducible Subspace is Irreducible
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Set Closures
Category:Connected Sets
Closure of Convex Set in Topological Vector Space is Convex

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\Bbb F$.
Let $C \subseteq X$ be convex.
Then the closure $C^-$ of $C$ is convex.
== Proof ==
Let $t \in \closedint 0 1$.
Since $C$ is convex, we have: 
:$t C + \paren {1 - t} C \subseteq C$ 
We show that: 
:$t C^- + \paren {1 - t} C^- \subseteq C^-$
We have: 
== Also see ==
* Closure of Convex Subset in Normed Vector Space is Convex
== Sources ==
* : $1.13$: Theorem
Category:Convex Sets (Vector Spaces)
Category:Topological Vector Spaces
Category:Set Closures
Closure of Convex Subset in Normed Vector Space is Convex

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $C \subseteq X$ be a convex subset of $X$.
Let $C^-$ be the closure of $C$.
Then $C^- \subseteq X$ is also a convex subset of $X$.
== Proof ==
Let $x, y \in C^-$.
Suppose $x, y$ are limit points.
Then there are sequences $\sequence {x_n}_{n \mathop \in \N}, \sequence {x_n}_{n \mathop \in \N}$ in $C$, such that:
:$\ds \lim_{n \mathop \to \infty} x_n = x$
:$\ds \lim_{n \mathop \to \infty} x_y = y$
Let $\alpha \in \closedint 0 1$.
Then:
Since $C$ is convex:
:$\ds \forall n \in \N : \paren {1 - \alpha} x_n + \alpha y_n \in C \subseteq C^-$
$C^-$ is closed and contains its limit points.
Hence:
:$\paren {1 - \alpha} x + \alpha y \in C^-$
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Set Closures
Category:Convex Sets (Vector Spaces)
Category:Normed Vector Spaces
Closure of Dense-in-itself is Dense-in-itself in T1 Space

== Theorem ==
Let $T$ be a $T_1$ topological space.
Let $A \subseteq T$. 
Let $A$ be dense-in-itself.
Then the closure $A^-$ of $A$ is also dense-in-itself.
== Proof ==
Let $A$ be dense-in-itself.
Then by Dense-in-itself iff Subset of Derivative:
:$(1): \quad A \subseteq A'$
where $A'$ denotes the derivative of $A$.
By Derivative of Derivative is Subset of Derivative in $T_1$ Space:
:$(2): \quad A \subseteq A'$
By Dense-in-itself iff Subset of Derivative it is sufficient to proof that:
:$A^- \subseteq \left({A^-}\right)'$
Thus:
== Sources ==
* 
* 
Category:Denseness
Category:Set Closures
Category:T1 Spaces
Closure of Densely-Defined Bounded Linear Operator

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\tuple {\map D T, T}$ be a densely-defined linear operator that is bounded.
Let $\widetilde T : \HH \to \HH$ be the bounded linear operator extending $T$, given by Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain.
Then $\tuple {\map D T, T}$ is closable with closure $\tuple {\HH, \widetilde T}$.
== Proof ==
From Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph and Closable Densely-Defined Linear Operator has Smallest Closed Extension, to show that the closure of $\tuple {\map D T, T}$ is $\tuple {\HH, \widetilde T}$, it is enough to show that:
:$\map \cl {\map \GG T} = \map \GG {\widetilde T}$
where the closure is understood in the direct product norm and $\GG$ denotes the graph of a linear operator.
Let $\tuple {x, y} \in \map \cl {\map \GG T}$.
Then there exists a convergent sequence $\sequence {\tuple {x_n, T x_n} }_{y \mathop \in \N}$ valued in $\map \GG T$ such that:
:$\tuple {x_n, T x_n} \to \tuple {x, y}$
From Convergence in Direct Product Norm, we have:
:$x_n \to x$ as $n \to \infty$
and:
:$\widetilde T x_n = T x_n \to y$ as $n \to \infty$.
We cannot conclude that $T x_n \to T x$ since we do not necessarily have $x \in \map D T$. 
However, from the boundedness and global definition of $\widetilde T$ we do obtain $\widetilde T x_n \to \widetilde T x$ as $n \to \infty$.
From Convergent Sequence in Normed Vector Space has Unique Limit, we have $\widetilde T x = y$. 
Hence we have:
:$\tuple {x, y} = \tuple {x, \widetilde T x} \in \map \GG {\widetilde T}$
So:
:$\map \cl {\map \GG T} \subseteq \map \GG {\widetilde T}$. 
Conversely let $\tuple {x, \widetilde T x} \in \map \GG {\widetilde T}$.
Since $\map D T$ is everywhere dense in $\HH$, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ such that $x_n \to x$.
Then we have $T x_n = \widetilde T x_n \to \widetilde T x$. 
Hence from Convergence in Direct Product Norm we have:
:$\tuple {x_n, T x_n} \to \tuple {x, \widetilde T x}$
Hence $\tuple {x, \widetilde T x}$ is the limit of a sequence in $\map \GG T$.
Hence:
:$\map \GG {\widetilde T} \subseteq \map \cl {\map \GG T}$
We conclude:
:$\map \cl {\map \GG T} = \map \GG {\widetilde T}$
Hence by Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph and Closable Densely-Defined Linear Operator has Smallest Closed Extension, $\tuple {\map D T, T}$ is closable with closure $\tuple {\HH, \widetilde T}$.
Category:Closable Densely-Defined Linear Operators
Closure of Derivative is Derivative in T1 Space

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a $T_1$ topological space.
Let $A$ be a subset of $S$.
Then
:$A'^- = A'$
where
:$A'$ denotes the derivative of $A$
:$A^-$ denotes the closure of $A$.
== Proof ==
So:
:$A'^- \subseteq A'$
Then by definition of closure:
:$A' \subseteq A'^-$
Hence the result by definition of set equality.
== Sources ==
* 
* 
Category:Set Derivatives
Category:Set Closures
Category:T1 Spaces
Closure of Empty Set is Empty Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then the closure of the empty set $\O$ in $T$ is $\O$.
== Proof ==
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$.
The result follows from Closed Set equals its Closure.
Category:Set Closures
Category:Empty Set
Closure of Finite Union equals Union of Closures

== Theorem ==
Let $T$ be a topological space.
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \ldots, n}: H_i \subseteq T$
Then:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} = \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
where $\map \cl {H_i}$ denotes the closure of $H_i$.
== Proof ==
From Closure of Union contains Union of Closures we have that:
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \subseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
We need now to show that: 
:$\ds \bigcup_{i \mathop = 1}^n \map \cl {H_i} \supseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
Let $\ds K = \bigcup_{i \mathop = 1}^n \map \cl {H_i}$ and $\ds H = \bigcup_{i \mathop = 1}^n H_i$.
From Topological Closure is Closed, all of $\map \cl {H_i}$ are closed.
So $K$ is the union of a finite number of closed sets.
So $K$ is itself closed, from Topology Defined by Closed Sets.
From Set is Subset of its Topological Closure:
:$\forall i \in \closedint 1 n: H_i \subseteq \map \cl {H_i}$
and so $H \subseteq K$.
So from Topological Closure of Subset is Subset of Topological Closure:
:$\map \cl H \subseteq \map \cl K$
The result follows.
== Also see ==
* Closure of Infinite Union may not equal Union of Closures
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.18$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Union
Category:Set Closures
Closure of Hadamard Product

== Theorem ==
Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is closed on $\map {\MM_S} {m, n}$ $\cdot$ is closed on $\struct {S, \cdot}$.
== Proof ==
=== Necessary Condition ===
Let the operation $\cdot$ be closed on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of $\map {\MM_S} {m, n}$.
Let $\sqbrk c_{m n} = \sqbrk a_{m n} \cdot \sqbrk b_{m n}$.
Then:
:$\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} = a_{i j} \circ b_{i j}$
Thus:
:$\struct {S, \cdot}$ is closed $c_{i j} \in S$.
From the definition of Hadamard product, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$.
Thus it follows that:
:$\sqbrk c_{m n} \in \map {\MM_S} {m, n}$
Thus $\struct {\map {\MM_S} {m, n}, \circ}$, as it is defined, is closed.
=== Sufficient Condition ===
Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not closed on $\struct {S, \cdot}$.
Then there exists $a$ and $b$ such that:
:$a \cdot b \notin S$
Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that:
:$a_{i j} = a$, $b_{i j} = b$
where:
:$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$
:$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$
Then:
:$a_{i j} \cdot b_{i j} \notin S$
 
That is:
:$\mathbf A \circ \mathbf B \notin \map {\MM_S} {m, n}$
because (at least) one element of $\mathbf A \circ \mathbf B$ is not an element of $S$.
That is, $\circ$ is not closed on $\map {\MM_S} {m, n}$.
Category:Hadamard Product
Closure of Half-Open Real Interval is Closed Real Interval

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
 
Let $H_1 = \hointl a b$ and $H_2 = \hointr a b$ be half-open intervals of $\R$.
Then the closure of both $H_1$ and $H_2$ in $\R$ are the closed interval $\closedint a b$.
== Proof ==
For an arbitrary $H \subseteq \R$, let $H^-$ denote the closure of $H$ in $\R$.
By the definition of closure:
:$H^-$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H$ as a subset.
Let $A := \openint a b$ be the open interval between $a$ and $b$.
By definition of $H_1$ and $H_2$:
:$A \subseteq H_1$
and:
:$A \subseteq H_2$
From Closure of Open Real Interval is Closed Real Interval:
:$\openint a b^- = \closedint a b$
Thus as $\closedint a b$ is the closure of $\openint a b$, it follows that:
:$\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $\openint a b$.
But $\closedint a b$ also contains as subsets both $H_1$ and $H_2$.
Thus we have:
:$A \subseteq H_1 \subseteq \closedint a b$
and:
:$A \subseteq H_2 \subseteq \closedint a b$
Thus it follows that $\closedint a b$ is the smallest closed set of $\struct {\R, \tau_d}$ containing $H_1$ and $H_2$.
Hence the result.
Category:Real Intervals
Category:Set Closures
Closure of Image under Continuous Mapping is not necessarily Image 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:
:$f \sqbrk {\map \cl H} = \map \cl {f \sqbrk H}$
== Proof ==
Proof by Counterexample:
Let $\R$ be the real numbers under the usual (Euclidean) topology.
Let $f: \R \to \R$ be the (real) hyperbolic tangent function:
:$\forall x \in \R: \map f x = \tanh x$
It is accepted that $f$ is continuous.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
As can be seen:
:$f \sqbrk {\map \cl H} \ne \map \cl {f \sqbrk H}$
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 27 \ \text {(i)}$
Category:Set Closures
Category:Continuous Mappings
Closure of Infinite Subset of Finite Complement Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement space.
Let $H \subseteq S$ be an infinite subset of $S$.
Then $H^- = S$ where $H^-$ is the closure of $S$.
== Proof ==
Let $H$ be an infinite subset of $S$.
From Limit Points of Infinite Subset of Finite Complement Space, every point of $S$ is a limit point of $H$.
Hence the result from the definition of closure.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $1$
Category:Set Closures
Category:Finite Complement Topology
Closure of Infinite Union may not equal Union of Closures

== Theorem ==
Let $T$ be a topological space.
Let $I$ be an infinite indexing set.
Let $\family {H_i}_{i \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $\ds H = \bigcup_{i \mathop \in I} H_i$ be the union of $\family {H_i}_{i \mathop \in I}$.
Then it is not always the case that:
:$\ds \bigcup_{i \mathop \in I} \map \cl {H_i} = \map \cl {\bigcup_{i \mathop \in I} H_i}$
where $\map \cl {H_i}$ denotes the closure of $H_i$.
== Proof 1 ==
== Proof 2 ==
Category:Set Union
Category:Set Closures
Category:Closure of Infinite Union may not equal Union of Closures