nyuuzyou/wikis huggingface dataset:
Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable

== Theorem ==
Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$. 
Let $w^\ast$ be the weak-$\ast$ topology on $B^-_{X^\ast}$.
Then $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable.
== Proof ==
Let $\sequence {x_n}_{n \in \N}$ be a countable dense subset of $X$.
Let $\sequence {x_n^\wedge }_{n \in \N}$ be the evaluation maps associated with $\sequence {x_n}_{n \in \N}$.
We show that $\sequence {x_n^\wedge }_{n \in \N}$ separates points.
Suppose that: 
:$\map {x_n^\wedge } f = \map {x_n^\wedge } g$ 
for each $f, g \in X^\ast$.
Then: 
:$\map f {x_n} = \map g {x_n}$ 
for each $n \in \N$.
Since $f$ and $g$ are continuous and $\sequence {x_n}_{n \in \N}$ is dense in $X$, we have that $\sequence {x_n^\wedge }_{n \in \N}$ separates points as desired.
Now let $\sigma$ be the initial topology generated by $\sequence {x_n^\wedge \restriction_{B^-_{X^\ast} } }_{n \in \N}$
From Initial Topology Generated by Countable Family of Functions Separating Points is Metrizable, $\sigma$ is metrizable. 
Note that the weak-$\ast$ topology on $B^-_{X^\ast}$ is generated by $\set {y^\wedge \restriction_{B^-_{X^\ast} } : y \in X}$ from Subspace Topology on Initial Topology is Initial Topology on Restrictions.
So $\sigma \subseteq w^\ast$. 
Hence the identity mapping $\iota : \struct {B^-_{X^\ast}, w^\ast} \to \struct {B^-_{X^\ast}, \sigma}$ is continuous.
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact.
Since $\sigma$ is metrizable, it is Hausdorff from Metric Space is Hausdorff.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $\iota$ is a homeomorphism.
Hence $\sigma = w^\ast$, and $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable.
== Also see == 
* Weak-* Metrizability of Closed Unit Ball in Normed Dual of Normed Vector Space implies Original Space is Separable shows that the converse of this result is also true.
== Sources ==
* : Proposition $3.24$
Category:Weak-* Topologies
Category:Metrizable Topologies
Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Separable

== Theorem ==
Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.
Let $w^\ast$ be the weak-$\ast$ topology on $B^-_{X^\ast}$.
Then $\struct {B^-_{X^\ast}, w^\ast}$ is separable. 
== Proof ==
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact. 
From Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable. 
From Compact Metric Space is Separable, we have that $\struct {B^-_{X^\ast}, w^\ast}$ is separable.
Category:Weak-* Topologies
Category:Separable Spaces
Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Sequentially Compact

== Theorem ==
Let $X$ be a separable normed vector space.
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.
Let $w^\ast$ be the weak-$\ast$ topology on $B^-_{X^\ast}$.
Then $\struct {B^-_{X^\ast}, w^\ast}$ is sequentially compact. 
== Proof ==
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact. 
From Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable, $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable. 
From Compactness and Sequential Compactness are Equivalent in Metric Spaces, $\struct {B^-_{X^\ast}, w^\ast}$ is sequentially compact.
Category:Sequentially Compact Spaces
Category:Weak-* Topologies
Closed Unit Ball in Normed Vector Space is Weakly Closed

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.
Let $B^-$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,} }$.
Then $B^-$ is weakly closed.
== Proof ==
From Closed Unit Ball is Convex Set, $B^-$ is convex.
From Closed Ball is Closed, $B^-$ is $\norm {\, \cdot \,}$-closed. 
From Mazur's Theorem: Corollary, we can conclude that $B^-$ is weakly closed.
Category:Weakly Closed Sets
Category:Closed Balls
Category:Normed Vector Spaces
Closed Unit Ball is Convex Set

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map {B_1^-} 0$ be the closed unit ball in $X$.
Then $\map {B_1^-} 0$ is convex.
== Proof ==
Let $x, y \in \map {B_1^-} 0$.
Let $\alpha \in \closedint 0 1$ be arbitrary.
Then:
Therefore, $\paren {1 - \alpha}x + \alpha y \in \map {B_1^-} 0$.
By definition, $\map {B_1^-} 0$ is convex.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $1$
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
* $3.1$: Norms
Category:Vector Spaces
Category:Closed Balls
Category:Convex Sets (Vector Spaces)
Closed Unit Interval is Homeomorphic to Letter L

== Theorem ==
Let $\R$ be the real number line under the Euclidean metric.
Let $\Bbb I := \closedint 0 1$ be the closed unit interval.
Let $\mathsf L \subseteq \R^2$ denote the letter $L$:
:$\mathsf L := \closedint 0 1 \times \set 0 \cup \set 0 \times \closedint 0 1$
Then $\Bbb I$ and $\mathsf L$ are homeomorphic.
== Proof ==
:
Consider the mapping $f: \Bbb I \to \mathsf L$ defined as:
:$\forall x \in \Bbb I: \map f x = \begin {cases} \tuple {0, 1 - 2 x} & : x \in \closedint 0 {\dfrac 1 2} \\ \tuple {2 x - 1, 0} & : x \in \closedint {\dfrac 1 2} 1 \end {cases}$
It is seen that:
:$f \closedint 0 {\dfrac 1 2} = \set 0 \times \closedint 0 1$
and:
:$f \closedint {\dfrac 1 2} 1 = \closedint 0 1 \times \set 0$
such that:
:$\map f {\dfrac 1 2} = \tuple {0, 0}$
By construction, $f$ is a bijection.
Its continuity follows from the Combination Theorem for Continuous Real Functions.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(c)}$
Category:Examples of Homeomorphisms
Category:Letter L
Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets

== Theorem ==
Let $\mathbb I = \closedint 0 1$ be the closed unit interval.
Then $\mathbb I$ cannot be expressed as the union of a countably infinite set of pairwise disjoint closed sets.
== Proof ==
 $\ds \mathbb I = \bigcup_{i \mathop = 1}^\infty C_i$ where $\set {C_i}$ is a set of pairwise disjoint closed sets.
Let:
:$\ds B = \bigcup \partial C_i = \mathbb I \setminus \bigcup C_i^\circ$
where $\partial C_i$ is the boundary of $C_i$ and $C_i^\circ$ is the interior of $C_i$.
Let $J \subseteq \mathbb I$ be a subinterval of $\mathbb I$.
Then $J$ is non-meager, and so some $C_k$ is dense in some open interval $L \subseteq J$.
Since $C_k$ is closed we have that $L \subseteq C_k^\circ$ and so $L \cap B = \O$.
So $J$ contains an open subset $L$ disjoint from $B$.
So $B$ is nowhere dense in $\mathbb I$.
So every open interval $U$ containing some $x \in \partial C_j$ must intersect $B \setminus \partial C_j$.
This is because $U$ is an open neighborhood of $x$ and therefore contains some point $u \in \mathbb I \setminus C_j$, say $u \in C_m$.
Then if $U \cap B \cap \partial C_m = \O$ it follows that $C_m^\circ \cap U$ is a non-empty open set in $U$. which is also relatively closed.
Now $B$ is itself non-meager, since it is a closed subset of $\mathbb I$.
Thus some $\partial C_k$ is dense in some non-empty open set $U \cap B$, for some open interval $U$ in $\mathbb I$.
Since $\partial C_k$ is closed, this means $\partial C_k \cap U = B \cap U$.
But this is impossible, since if $U \cap C_k \ne \O$ then $U \cap \paren {B \setminus \partial C_k} \ne \O$.
This contradiction shows that $\mathbb I$ cannot be expressed as $\ds \bigcup_{i \mathop = 1}^\infty C_i$.
== Sources ==
* : Notes: Part $\text {II}$: Examples: $18$
Category:Closed Sets
Category:Topology
Closed and Bounded Subset of Normed Vector Space is not necessarily Compact

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,} }$ be a normed vector space.
Let $K \subset X$ be a closed and bounded subset.
Then $K$ is not necessarily compact.
== Proof 1 ==
== Proof 2 ==
Category:Normed Vector Spaces
Category:Compact Spaces
Category:Closed and Bounded Subset of Normed Vector Space is not necessarily Compact
Closed and Bounded Subset of Normed Vector Space is not necessarily Compact/Proof 1

== Theorem ==
== Proof ==
Let $\struct {\ell^2, \norm {\, \cdot \,}}_2$ be the 2-sequence space.
Let $K$ be a closed unit ball in $\struct {\ell^2, \norm {\, \cdot \,}}_2$:
:$K := \set {\mathbf x \in \ell^2 : \norm {\mathbf x}_2 \le 1}$
$K$ is bounded and closed.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a sequence such that:
:$\mathbf x_n := \tuple {0, \ldots 0, 1, 0, \ldots}$
where $1$ is in the $n$-th component of $\mathbf x_n$.
Then:
:$\forall n \in \N : \mathbf x_n \in K \subset \ell^2$
However:
:$\forall n, m \in \N : n \ne m : \norm {\mathbf x_n - \mathbf x_m}_2 = \sqrt 2$
In other words, $\sequence {\mathbf x_n}$ is not a Cauchy sequence.
By Convergent Subsequence of Cauchy Sequence, $\sequence {\mathbf x_n}$ has no convergent subsequence.
Therefore, a closed and bounded space $K$ has a sequence $\sequence {\mathbf x_n}_{n \mathop \in \N}$ which contains a non-convergent subsequence.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Closed and Bounded Subset of Normed Vector Space is not necessarily Compact
Closed and Bounded Subset of Normed Vector Space is not necessarily Compact/Proof 2

== Theorem ==
== Proof ==
Let $\struct {\CC \closedint 0 1, \norm {\, \cdot \,}_\infty}$ be the normed vector space of continuous on closed interval real-valued functions with supremum norm.
Consider a closed unit ball with the center at $0$ at $\struct {\CC, \norm {\, \cdot \,}_\infty}$:
:$K := \set {x \in \CC \closedint 0 1 : \norm x_\infty \le 1}$.
We have that $K$ is bounded and closed.
Consider the following sequence of graphs:
:$\map {x_n} t = \begin{cases}
2^{n + 2} t & : \ds \sum_{i \mathop = 1}^n \frac 1 {2^i} \le t < \sum_{i \mathop = 1}^n \frac 1 {2^i} + \frac 1 {2^{n + 2}} \\
\ds 1 - 2^{n + 2} \paren {t - \paren {\sum_{i \mathop = 1}^n \frac 1 {2^i} + \frac 1 {2^{n + 2}}}} & : \ds \sum_{i \mathop = 1}^n \frac 1 {2^i} + \frac 1 {2^{n + 2}} \le t \le \sum_{i \mathop = 1}^{n + 1} \frac 1 {2^i} \\
0 & : \text {otherwise}
\end{cases}$
where $n \in \N$.
We have that:
:$\forall n,m \in \N : n \ne m \implies \norm {x_n - x_m}_\infty = 1$
Hence, $\sequence {x_n}_{n \mathop \in \N}$ is not a Cauchy sequence.
We have that every convergent sequence is a Cauchy sequence.
Hence, $\sequence {x_n}_{n \mathop \in \N}$ is not convergent.
Therefore, any subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is also divergent.
Thus, a closed unit ball, which is closed and bounded, is not compact in $\struct {\CC \closedint 0 1, \norm {\, \cdot \,}_\infty}$.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Closed and Bounded Subset of Normed Vector Space is not necessarily Compact
Closed and Bounded Subspace is not necessarily Compact

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $M_H = \struct {H, d_H}$ be the subspace of $M$ induced by $d$.
Let $H$ be closed and bounded.
Then it is not necessarily the case that $M_H$ is compact.
== Proof ==
Proof by Counterexample:
Let $A = \openint 0 1$ be the open unit interval.
From Open Real Interval is not Compact, $\openint 0 1$ is not a compact space.
Let $H = \openint 0 1$.
Then $H \subseteq A$.
From Underlying Set of Topological Space is Clopen, $\openint 0 1$ is both closed and open in $\openint 0 1$.
Thus $H$ is closed and bounded but not compact.
== Sources ==
* : $5$: Compact spaces: $5.7$: Compact Subspaces of $\R^n$: Remark $5.7.2$
Category:Compact Spaces
Closure (Metric Space)/Examples/Union of Disjoint Closed Real Intervals

== Examples of Closure in the context of a Metric Space ==
Let $\R$ be the real number line under the usual (Euclidean) metric.
Let $M$ be the subspace of $\R$ defined as:
:$M = \closedint 0 1 \cup \closedint 2 3$
Let $\map {B_1} 1$ denote the open $1$-ball of $1$ in $M$.
Let $\map { {B_1}^-} 1$ denote the closed $1$-ball of $1$ in $M$.
Then:
:$\map \cl {\map {B_1} 1} = \closedint 0 1$
while:
:$\map { {B_1}^-} 1 = \closedint 0 1 \cup \set 2$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Example $3.7.22 \ \text {(b)}$
Category:Examples of Set Closures
Closure (Topology)/Examples/Open Interval in Open Unbounded Interval

== Example of Closure in the context of Topology ==
Let $S$ be the open real interval:
:$S = \openint a \to$
Let $H$ be the open real interval:
:$H = \openint a b$
Then the closure of $H$ in $S$ is:
:$H^- = \hointl a b$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.14 \ \text {(a)}$
Category:Examples of Set Closures
Closure (Topology)/Examples/Open Interval under Discrete Topology

== Example of Closure in the context of Topology ==
Let $\T = \struct {\R, \tau_d}$ denote the topological space formed from the set of real numbers $\R$ together with the discrete topology $\tau_d$.
Let $H$ be the open real interval:
:$H = \openint a b$
Then the closure of $H$ in $T$ is:
:$H^- = \openint a b$
== Proof ==
From Set in Discrete Topology is Clopen, $H = \openint a b$ is both open and closed.
From Set is Closed iff Equals Topological Closure it follows that $H^- = H = \openint a b$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.14 \ \text {(b)}$
Category:Examples of Set Closures
Closure (Topology)/Examples/Union of Singleton with Open Interval

== Example of Closure in the context of Topology ==
Let $\R$ be the set of real numbers.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
:$H = \set 0 \cup \openint 1 2$
Then the closure of $H$ in $\R$ is:
:$H^- = \set 0 \cup \closedint 1 2$
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.13$
Category:Examples of Set Closures
Closure Condition for Hausdorff Space

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Then $\struct {S, \tau}$ is a Hausdorff space :
:For all $x, y \in S$ such that $x \ne y$, there exists an open set $U$ such that $x \in U$ and $y \notin U^-$, where $U^-$ is the closure of $U$.
== Proof ==
=== Necessary Condition ===
Let $\struct {S, \tau}$ be a Hausdorff space.
Let $x, y \in S$ with $x \ne y$.
Then by the definition of Hausdorff space there exist open sets $U, V \subseteq S$ such that:
:$x \in U$
:$y \in V$
:$U \cap V = \O$
By Empty Intersection iff Subset of Complement, $U \subseteq S \setminus V$.
By definition, $S \setminus V$ is closed.
Thus by definition of closure:
:$U^- \subseteq S \setminus V$
Since $y \in V$, by definition of relative complement it follows that $y \notin S \setminus V$.
By definition of subset, $y \notin U^-$.
=== Sufficient Condition ===
Suppose that for each $x, y \in S$ such that $x \ne y$ there exists an open set $U$ such that $x \in U$ and $y \notin U^-$.
Let $x, y \in S$ with $x \ne y$.
Let $U$ be as described.
Let $V = S \setminus U^-$.
Then $y \in V$ by the definition of set difference.
Since $U \subseteq U^-$, it follows that $U \cap V = \O$.
As such $U$ and $V$ exist for all such $x$ and $y$, it follows that $\struct {S, \tau}$ is a Hausdorff space.
Category:Hausdorff Spaces
Closure Equals Union with Derivative

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Then:
:$A^- = A \cup A'$
where
:$A'$ denotes the derivative of $A$
:$A^-$ denotes the closure of $A$.
== Proof ==
=== Closure Subset of Union ===
It is to be proved that:
:$A^- \subseteq A \cup A'$
Let $x \in A^-$.
In the case where $x \in A$ then $x \in A \cup A'$ by definition of set union.
Let:
:$(1): \quad x \notin A$
From Characterization of Derivative by Open Sets, to prove $x \in A'$ it is enough to show that:
:for every open set $U$ of $T$:
::if $x \in U$
::then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.
Let $U$ be an open set of $T$.
Let $x \in U$.
Then by Condition for Point being in Closure:
:$A \cap U \ne \O$
Then by definition of empty set:
:$\exists y \in S: y \in A \cap U$
By definition of set intersection:
:$y \in A$ and $y \in U$
But as $x \notin A$ it follows by definition of set intersection that:
:$x \notin A \cap U$
So by $(1)$:
:$x \ne y$
Thus $y$ fulfils the conditions of the hypothesis, and so:
:$x \in A'$
Hence by definition of set union:
:$x \in A \cup A'$
Thus in all cases:
:$x \in A^- \implies x \in A \cup A'$
and so:
:$A^- \subseteq A \cup A'$
=== Union Subset of Closure ===
It is to be proved that:
:$A \cup A' \subseteq A^-$
By Set is Subset of its Topological Closure:
:$A \subseteq A^-$
By Derivative is Included in Closure:
:$A' \subseteq A^-$
 
Hence by Union of Subsets is Subset:
:$A \cup A' \subseteq A^-$
Hence by definition of set equality:
:$A^- = A \cup A'$
== Sources ==
* 
* 
Category:Set Derivatives
Category:Set Closures
Closure Operator Preserves Directed Suprema iff Image of Closure Operator Inherits Directed Suprema

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $h: S \to S$ be a closure operator on $L$.
Then $h$ preserves directed suprema
 $\struct {h \sqbrk S, \precsim}$ inherits directed suprema.
where
:$h \sqbrk S$ denotes the image of $h$,
:$\mathord\precsim = \mathord\preceq \cap \paren {h \sqbrk S \times h \sqbrk S}$
== Proof ==
By Operator Generated by Image of Closure Operator is Closure Operator:
:$\map {\operatorname{operator} } {\struct {h \sqbrk S, \precsim} } = h$
where $\map {\operatorname{operator} } {\struct {h \sqbrk S, \precsim} }$ denotes the operator generated by $\struct {h \sqbrk S, \precsim}$
Hence the result holds by Operator Generated by Closure System Preserves Directed Suprema iff Closure System Inherits Directed Suprema.
== Sources ==
* 
* 
Category:Closure Operators
Closure Operator does not Change Infimum of Subset of Image

== Theorem ==
Let $L = \left({S, \preceq}\right)$ be an ordered set.
Let $c: S \to S$ be a closure operator on $L$.
Let $X$ be a subset of $c\left[{S}\right]$ such that
:$X$ admits an infimum,
where $c\left[{S}\right]$ denotes the image of $c$.
Then $\inf X = c\left({\inf X}\right)$
== Proof ==
We will prove that
:$c\left({\inf X}\right)$ is lower bound for $X$.
Let $x \in X$.
By definition of subset:
:$x \in c\left[{S}\right]$
By definition of image of mapping:
:$\exists y \in S: x = c\left({y}\right)$
By definition of closure operator/idempotent:
:$x = c\left({x}\right)$
By definition of infimum:
:$\inf X$ is lower bound for $X$.
By definition of lower bound:
:$\inf X \preceq x$
Thus by definition of closure operator/increasing:
:$c\left({\inf X}\right) \preceq x$
By definition of infimum:
:$c\left({\inf X}\right) \preceq \inf X$
By definition of closure operator/inflationary:
:$c\left({\inf X}\right) \succeq \inf X$
Thus by definition of antisymmetry:
:$c\left({\inf X}\right) = \inf X$
== Sources ==
* 
* 
Category:Closure Operators
Closure Operator from Closed Elements

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $C \subseteq S$.
Suppose that $C$ is a subset of $S$ with the property that every element of $S$ has a smallest successor in $C$.
Let $\cl: S \to S$ be defined as follows:
For $x \in S$:
:$\map \cl x = \map \min {C \cap x^\succeq}$
where $x^\succeq$ is the upper closure of $x$.
That is, let $\map \cl x$ be the smallest successor of $x$ in $C$.
Then:
:$\cl$ is a closure operator on $S$
:The closed elements of $\cl$ are precisely the elements of $C$.
== Proof ==
=== Inflationary ===
$x$ is a lower bound of $x^\succeq$.
Hence by Lower Bound for Subset, $x$ is also a lower bound of $C \cap x^\succeq$.
By the definition of smallest element, $x \preceq \map \cl x$.
=== Order-Preserving ===
Suppose that $x \preceq y$.
Then:
:$C \cap y^\succeq \subseteq C \cap x^\succeq$
By Smallest Element of Subset:
:$\map \cl x \preceq \map \cl y$
=== Idempotent ===
Let $x \in S$.
For each $x \in S$:
:$\map \cl x = \map \min {C \cap x^\succeq}$
Thus:
:$\map \cl x \in \paren {C \cap x^\succeq} \subseteq C$
That is to say, $\map \cl x$ is its own smallest successor in $C$.
Thus:
:$\map \cl x = \map \cl {\map \cl x}$
When $x \in C$, $x$ is the minimum of $C \cap x^\succeq$
Hence, elements of $C$ are closed elements with respect to $\cl$.
Suppose that $x$ is closed with respect to $\cl$.
Then:
:$x = \map \min {C \cap x^\succeq}$
so in particular:
:$x \in C$
Category:Closure Operators
Category:Order Theory
Closure Operator from Closed Sets

== Theorem ==
Let $S$ be a set.
Let $\CC$ be a set of subsets of $S$.
Let $\CC$ be closed under arbitrary intersections:
:$\forall \KK \in \powerset \CC: \bigcap \KK \in \CC$
where $\bigcap \O$ is taken to be $S$.
Define $\cl: \powerset S \to \CC$ by letting:
:$\map \cl T = \bigcap \set {C \in \CC: T \subseteq C}$
Then $\cl$ is a closure operator whose closed sets are the elements of $\CC$.
== Proof ==
First we will show that $\cl$ is a closure operator.
=== Inflationary ===
Let $T \subseteq S$.
By Set Intersection Preserves Subsets/General Result/Corollary, $T \subseteq \map \cl T$.
Since this holds for all such $T$, $\cl$ is inflationary.
=== Increasing ===
Let $T \subseteq U \subseteq S$.
Let $\TT$ and $\UU$ be the sets of elements of $\CC$ containing $T$ and $U$, respectively.
Since Subset Relation is Transitive, every set containing $U$ contains $T$, so $\UU \subseteq \TT$.
By Intersection is Decreasing, $\bigcap \TT \subseteq \bigcap \UU$.
Thus $\map \cl T \subseteq \map \cl U$.
=== Idempotent ===
Let $T \subseteq S$.
By the premise, the intersection of a subset of $\CC$ is in $\CC$.
Thus in particular $\map \cl T \in \CC$.
Therefore:
:$\map \cl {\map \cl T} \subseteq \map \cl T$
Since $\cl$ is inflationary:
:$\map \cl T \subseteq \map \cl {\map \cl T}$
By definition of set equality:
:$\map \cl {\map \cl T} = \map \cl T$
Since this holds for all $T \subseteq S$, $\cl$ is idempotent.
Finally, we need to show that the elements of $\CC$ are the closed sets with respect to $\cl$.
If $C \in \CC$, then since $\cl$ is inflationary:
:$C \subseteq \map \cl C$
But since $C \subseteq C$, $\map \cl C \subseteq C$.
Thus by definition of set equality:
:$\map \cl C = C$
so $C$ is closed with respect to $\cl$.
Suppose instead that $C$ is closed with respect to $\cl$.
Then $\map \cl C = C$.
Since $\CC$ is closed under intersections, $C \in \CC$. 
Category:Closure Operators