nyuuzyou/wikis huggingface dataset:
Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Then:
:$\forall m, n \in \struct {S, \circ, \preceq}: m \preceq n \implies \closedint m {n \circ 1} = \closedint m n \cup \set {n \circ 1}$
where $\closedint m n$ is the closed interval between $m$ and $n$.
== Proof ==
Let $m \preceq n$. Then:
Thus:
:$\closedint m {n \circ 1} = \closedint m n \cup \set {n \circ 1}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Corollary $16.4.2$
Category:Naturally Ordered Semigroup
Closed Intervals form Neighborhood Basis in Real Number Line

== Theorem ==
Let $\R$ be the real number line with the usual (Euclidean) metric.
Let $a \in \R$ be a point in $\R$.
Let $\BB_a$ be defined as:
:$\BB_a := \set {\closedint {a - \epsilon} {a + \epsilon}: \epsilon \in \R_{>0} }$
that is, the set of all closed intervals of $\R$ with $a$ as a midpoint.
Then $\BB_a$ is a basis for the neighborhood system of $a$.
== Proof ==
Let $N$ be a neighborhood of $a$ in $M$.
Then by definition:
:$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$
where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$.
From Open Ball in Real Number Line is Open Interval:
:$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$
Let $\epsilon \in \R_{>0}: \epsilon < \epsilon'$.
Then by definition of closed interval and open interval:
:$\closedint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$
From Subset Relation is Transitive:
:$\closedint {a - \epsilon} {a + \epsilon} \subseteq N$
Also by definition of closed interval and open interval:
:$\openint {a - \epsilon} {a + \epsilon} \subseteq \closedint {a - \epsilon} {a + \epsilon}$
From Open Real Interval is Open Ball, $\openint {a - \epsilon} {a + \epsilon}$ is the open $\epsilon$-ball at $a$.
Thus by definition, $\closedint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$.
Hence there exists a neighborhood $\closedint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$.
Hence the result by definition of basis for the neighborhood system of $a$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Exercise $4 \ \text{i)}$
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Closed Linear Subspaces Closed under Intersection

== Theorem ==
Let $V$ be a topological vector space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of closed linear subspaces of $V$.
Then $M := \ds \bigcap_{i \mathop \in I} M_i$ is also a closed linear subspace of $V$.
== Proof ==
By Set of Linear Subspaces is Closed under Intersection, $M$ is a linear subspace of $V$.
By Topology Defined by Closed Sets, the intersection of closed sets is again closed.
As the $M_i$ are all closed, it follows that $M$ is closed.
Hence $M$ is a closed linear subspace of $V$.
Category:Vector Subspaces
Closed Linear Subspaces Closed under Setwise Addition

== Theorem ==
Let $H$ be a Hilbert space.
Let $M, N$ be closed linear subspaces of $H$.
Then $M + N$ is also a closed linear subspace of $H$, where $+$ denotes setwise addition.
== Proof ==
By Linear Subspaces Closed under Setwise Addition, $M + N$ is a linear subspace of $H$.
Now to show that it is closed.
Let $P: H \to H$ denote the orthogonal projection on $M$.
Denote by $I - P$ the complementary projection of $P$.
Define $N' := \set {n - P n: n \in N}$.
$N'$ is a closed linear subspace of $H$.
Observe $m + n = \paren {m + P n} + \paren {n - P n} \in M + N'$; hence, $M + N \subseteq M + N'$. 
By $m + \paren {n - P n} = \paren {m - P n} + n \in M + N$, conclude that $M + N' \subseteq M + N$, hence equality.
Furthermore, $N' \subseteq \map {\operatorname{ran} } {I - P} = \map \ker P$ by Range of Idempotent is Kernel of Complementary Idempotent.
That is, $N' \subseteq M^\perp$ by Properties of Orthogonal Projection, and hence $M \perp N'$.
Denote by $P'$ the orthogonal projection on $N'$.
Suppose now that $h \in M + N'$. Then:
For this last equality, observe $M \perp N'$, hence $M \subseteq N'^\perp$, $N' \subseteq M^\perp$ and $\paren {M + N'}^\perp \subseteq M^\perp \cup N'^\perp$.
The conclusion is that every $h \in M + N'$ can be uniquely decomposed as $P' h + P h$, with $P' h \in N', P h \in M$. 
So suppose there is a sequence $h_n \to h$ in $M + N'$. Then $P h_n$ and $P' h_n$ are sequences in $M, N'$, respectively.
As those are closed linear subspaces of $H$, there are $m \in M, n \in N'$ with $P h_n \to m, P' h_n \to n$.
It follows that $h = m + n \in M + N'$.
That is, $M + N' = M + N$ is a closed linear subspace of $H$.
Category:Hilbert Spaces
Closed Ordinal Space is Compact

== Theorem ==
Let $\Gamma$ be a limit ordinal.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a compact space.
== Proof ==
By definition, $\closedint 0 \Gamma$ is a linearly ordered space.
The result follows from Linearly Ordered Space is Compact iff Complete.
== Sources ==
* : Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $6$
Category:Ordinal Spaces
Category:Compact Spaces
Closed Ordinal Space is Complete Order Space

== Theorem ==
Let $\Gamma$ be a limit ordinal.
Let $\closedint 0 \Gamma$ denote the closed ordinal space on $\Gamma$.
Then $\closedint 0 \Gamma$ is a complete order space.
== Proof ==
Let $H$ be a subset of an ordinal space.
Then $H$ has an infimum: its first element.
Let $H$ be a subset of $\closedint 0 \Gamma$.
Then $H$ has a supremum.
Therefore $\closedint 0 \Gamma$ is a complete order space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $41$. Closed Ordinal Space $[0, \Gamma] \ (\Gamma < \Omega)$: $6$
Category:Ordinal Spaces
Category:Complete Order Topologies
Closed Real Interval/Examples/Example 1

== Examples of Closed Real Intervals ==
Let $I$ be the closed real interval defined as:
:$I := \closedint 1 3$
Then $3 \in I$.
== Proof ==
By definition of open real interval:
:$I = \set {x \in \R: 1 \le x \le 3}$
As $3 \le 3$ it follows that $3 \in I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise: $\S 2.10 \ (1) \ \text{(v)}$
Category:Examples of Intervals
Closed Real Interval is Closed Set

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then $\closedint a b$ is a closed set of $\R$.
== Proof ==
From the corollary to Open Real Interval is Open Set, both $\openint {-\infty} a$ and $\openint b \infty$ are open sets in $M$.
From Union of Open Sets of Metric Space is Open it follows that $\openint {-\infty} a \cup \openint b \infty$ is open in $\R$.
But $\openint {-\infty} a \cup \openint b \infty$ is the relative complement of $\closedint a b$ in $\R$.
The result follows by definition of closed set.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Examples $3.7.2$
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Metric Spaces
Category:Real Intervals
Category:Closed Sets
Closed Real Interval is Closed in Real Number Line

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\closedint a b$ be a closed interval of $\R$.
Then $\closedint a b$ is closed (in the topological sense) in $\struct {\R, \tau_d}$.
== Proof ==
From Open Sets in Real Number Line:
:$U := \openint \gets a \cup \openint b \to$
is an open set in $\struct {\R, \tau_d}$.
Consider the complement relative to $\R$:
:$V := \relcomp \R U = \closedint a b$
By definition, $\relcomp \R U$ is closed (in the topological sense) in $\struct {\R, \tau_d}$.
But by construction:
:$\relcomp \R U = \closedint a b$
Hence the result.
== Sources ==
* 
* 
Category:Closed Sets
Category:Real Number Line with Euclidean Topology
Closed Real Interval is Compact

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $I = \closedint a b$ be a closed real interval.
Then $I$ is compact.
== Topological Space ==
This proves that $I$ is compact in the context of a general topological space:
== Metric Space ==
This proves that $I$ is compact in the context of a metric space:
== Normed Vector Space ==
This proves that $I$ is compact in the context of a normed vector space:
== Also see ==
* Open Real Interval is not Compact
== Sources ==
* : $5$: Compact spaces: $5.2$: Definition of compactness
* 
Category:Compact Spaces
Category:Real Intervals
Category:Closed Real Interval is Compact
Closed Real Interval is Compact/Metric Space

== Theorem ==
== Proof ==
From Closed Real Interval is Closed Set, $I$ is a closed set of $\R$.
From Real Interval is Bounded in Real Numbers, $I$ is bounded in $\R$.
The result follows by definition of compact.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Closed Real Interval is Compact
Closed Real Interval is Compact/Normed Vector Space

== Theorem ==
== Proof ==
We have that $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.
Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $I$.
$\sequence {a_n}$ is bounded by $a$ and $b$.
By Bolzano-Weierstrass theorem, there exists a convergent subsequence $\sequence {a_{n_k}}_{k \in \N}$ with a limit $L$.
Then:
:$\forall k \in \N : a \le a_{n_k} \le b$
Take the limit $k \to \infty$:
:$a \le L \le b$
Hence, $L \in I$.
The result follows by definition of compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Closed Real Interval is Compact
Closed Real Interval is Compact/Topological Space

== Theorem ==
== Proof ==
Let $\UU$ be any open cover of $\closedint a b$.
Let:
:$G = \set {x \in \R: x \ge a, \closedint a x \text{ is covered by a finite subset of } \UU}$
Let points in $G$ be classified as $\text{good}$ (for $\UU$).
Thus if $z$ is $\text{good}$, then $\closedint a z$ has that finite subcover that is to be demonstrated for the whole of $\closedint a b$.
The aim therefore is to show that $b$ is $\text{good}$.
Let $x$ be $\text{good}$, and $a \le y \le x$.
We have that $\closedint a y \subseteq \closedint a x$.
Thus $\closedint a y$ can be covered with any finite subset of $\UU$ that covers $\closedint a x$.
As $x$ be $\text{good}$, at least one such subset is known to exist.
Thus, by definition, $y$ is $\text{good}$.
Now we show that $G \ne \O$.
At the same time we show that $G \supseteq \hointr a {a + \delta}$ for some $\delta > 0$.
As $\UU$ covers $\closedint a x$, it follows that $a$ must belong to some $U \in \UU$.
So, let $U \in \UU$ be some open set such that $a \in U$.
Since $U$ is open, $\hointr a {a + \delta} \subseteq U$ for some $\delta > 0$.
Hence $\closedint a x \subseteq U$ for all $x \in \hointr a {a + \delta}$.
It follows that all these $x$ are $\text{good}$.
So $G \ne \O$, and:
:$(1): \quad G \supseteq \hointr a {a + \delta}$ for some $\delta > 0$
Now the non-empty set $G$ is either bounded above or it is not.
Suppose $G$ is not bounded above.
Then there is some $c$ which is $\text{good}$ such that $c > b$.
From our initial observation that if $x$ is $\text{good}$, and $a \le y \le x$, then $y$ is $\text{good}$, it follows that $b$ is $\text{good}$, and hence the result.
Now suppose $G$ is bounded above.
By the Continuum Property, $G$ admits a supremum in $\R$.
So let $g = \sup G$.
Suppose that $g > b$.
As $g$ is the least upper bound for $G$, there exists some $c$ which is $\text{good}$ such that $c > b$.
Again, from our initial observation, it follows that $b$ is $\text{good}$, and hence the result.
 that $g \le b$.
From $(1)$ above:
:$\hointr a {a + \delta} \subseteq G$ for some $\delta > 0$
and so $g > a$.
Since $g \in \closedint a b$, $g$ must belong to some $U_0 \in \UU$.
Since $U_0$ is open, there exists some open $\epsilon$-ball $\map {B_\epsilon} g$ of $g$ such that $U_0 \supseteq \map {B_\epsilon} g$.
Since $g > a$, we can arrange that $\epsilon < g - a$.
As $g$ is the least upper bound, there must be a $\text{good}$ $c$ such that $c > g - \epsilon$.
This means $\closedint a c$ is covered by a finite subset of $\UU$, say $\set {U_1, U_2, \ldots, U_r}$.
Then $\closedint a {g + \dfrac \epsilon 2}$ is covered by $\set {U_1, U_2, \ldots, U_r, U_0}$.
So $g + \dfrac \epsilon 2$ is $\text{good}$, contradicting the fact that $g$ is an upper bound for $G$.
This contradiction implies that $g > b$, and hence $b \in G$.
Since $\UU$ was arbitrary, we conclude that every open cover of $\closedint a b$ has a finite subcover.
This satisfies the topological definition of compact, thus the proof is complete.
Category:Closed Real Interval is Compact
Closed Real Interval is Neighborhood Except at Endpoints

== Theorem ==
Let $\R$ be the real number line considered as an Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then $\closedint a b$ is a neighborhood of all of its points except $a$ and $b$.
== Proof ==
Let $a < c < b$.
Let $\epsilon < \map \min {b - c, c - a}$.
From the definition of positive it follows that $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.
We have that $c + \epsilon < b$ and $a < c - \epsilon$.
Thus:
:$\map {B_\epsilon} c \subseteq \closedint a b$
It follows that, by definition, $\closedint a b$ is a neighborhood of $c$.
Now consider $a$.
Let $\epsilon \in \R_{>0}$.
We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\closedint a b$.
Similarly, consider $b$.
Let $\epsilon \in \R_{>0}$.
We have that $b < b + \epsilon$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entirely in $\closedint a b$.
It follows that, by definition, $\closedint a b$ is a neighborhood of neither $a$ nor $b$.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Real Intervals
Category:Neighborhoods
Closed Real Interval is Regular Closed

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\closedint a b$ be a closed interval of $\R$.
Then $\closedint a b$ is regular closed in $\struct {\R, \tau_d}$.
== Proof ==
From Closed Real Interval is Closed in Real Number Line, $\closedint a b$ is closed in $\struct {\R, \tau_d}$.
From Interior of Closed Real Interval is Open Real Interval:
:$\closedint a b^\circ = \openint a b$
where $\closedint a b^\circ$ denotes the interior of $\closedint a b$.
From Closure of Open Real Interval is Closed Real Interval:
:$\openint a b^- = \closedint a b$
where $\openint a b^-$ denotes the closure of $\openint a b$.
Hence the result, by definition of regular closed.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $6$
Category:Closed Sets
Category:Regular Closed Sets
Category:Real Number Line with Euclidean Topology
Closed Real Interval is not Open Set

== Theorem ==
Let $\R$ be the real number line considered as a Euclidean space.
Let $\closedint a b \subset \R$ be a closed interval of $\R$.
Then $\closedint a b$ is not an open set of $\R$.
== Proof ==
From Closed Real Interval is Neighborhood Except at Endpoints, $a$ and $b$ have no open $\epsilon$-ball lying entirely in $\closedint a b$.
The result follows by definition of open set.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.9$
Category:Real Intervals
Category:Open Sets
Closed Set/Complex Analysis/Examples/Closed Unit Circle

== Example of Closed Set in the context of Complex Analysis ==
Let $S$ be the subset of the complex plane defined as:
:$\cmod z \le 1$
where $\cmod z$ denotes the complex modulus of $z$.
Then $S$ is closed.
== Proof ==
By definition, $S$ is closed $S$ contains all its limit points.
 $S$ is not closed.
Then $\exists z_1 \in \C: z_1 \notin S$ such that $z_1$ is a limit point of $S$.
We have that $z_1 \notin S$, so $\cmod {z_1} > 1$.
Let $\epsilon \in \R: \epsilon < \cmod {z_1} - 1$.
Then:
:$\map {\N_\epsilon} {z_1} \cap S = \O$
and so $z_1$ is not a limit point of $S$ after all.
Thus there is no such $z_1$.
So by Proof by Contradiction, $S$ is closed.
== Sources ==
* : $1$: Complex Numbers: Point Sets: $3.$
Category:Examples of Closed Sets (Complex Analysis)
Closed Set Measurable in Borel Sigma-Algebra

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\map \BB \tau$ be the associated Borel $\sigma$-algebra.
Let $C$ be a closed set in $\struct {S, \tau}$.
Then $C$ is $\map \BB \tau$-measurable.
== Proof ==
Since $C$ is closed, $S \setminus C$ is open.
By definition of Borel $\sigma$-algebra, $S \setminus C \in \map \BB \tau$.
By axiom $(2)$ for $\sigma$-algebras:
:$S \setminus \paren {S \setminus C} \in \map \BB \tau$
and this set equals $C$ by Set Difference with Set Difference since $C \subseteq S$.
The result follows by definition of measurable set.
Category:Borel Sigma-Algebras
Category:Closed Sets
Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a relational structure with Scott topology
where $\struct {S, \preceq}$ is an up-complete ordered set.
Let $A$ be a subset of $S$.
Then $A$ is closed $A$ is lower and closed under directed suprema.
== Proof ==
=== Sufficient Condition ===
Assume that
:$A$ is closed.
By definition of closed set:
:$\relcomp S A \in \tau$
By definition of Scott topology:
:$\relcomp S A$ is upper and inaccessible by directed suprema.
By Complement of Complement:
:$\relcomp S {\relcomp S A} = A$
Thus by Complement of Upper Section is Lower Section:
:$A$ is a lower section.
Thus by Complement of Inaccessible by Directed Suprema Subset is Closed under Directed Suprema:
:$A$ is closed under directed suprema.
=== Necessary Condition ===
Assume that
$A$ is lower and closed under directed suprema.
By Complement of Lower Section is Upper Section and Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema:
:$\relcomp S A$ is upper and inaccessible by directed suprema.
By definition of Scott topology:
:$\relcomp S A \in \tau$
Thus by definition:
:$A$ is closed set.
== Sources ==
* 
* 
Category:Topological Order Theory
Category:Closed Sets
Closed Set in Coarser Topology is Closed in Finer Topology

== Theorem ==
Let $X$ be a set. 
Let $\tau_1$ and $\tau_2$ be topologies on $X$ such that $\tau_1 \subseteq \tau_2$.
That is, $\tau_1$ is coarser than $\tau_2$. 
Let $C$ be a closed set in $\struct {X, \tau_1}$. 
Then $C$ is closed in $\struct {X, \tau_2}$. 
== Proof ==
Let $C$ be a closed set in $\struct {X, \tau_1}$.
Then by definition of closed set:
:$X \setminus C$ is open in $\struct {X, \tau_1}$.
Hence by definition of open set:
:$X \setminus C \in \tau_1$
By definition of subset:
:$X \setminus C \in \tau_2$
So by definition of open set:
:$X \setminus C$ is open in $\struct {X, \tau_2}$.
So $C$ is closed in $\struct {X, \tau_2}$ by definition.
Category:Closed Sets
Closed Set in Linearly Ordered Space

== Theorem ==
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $C$ be a subset of $X$.
Then $C$ is closed in $X$ for all non-empty subsets $S$ of $C$:
:If $s \in X$ is a supremum or infimum of $S$ in $X$, then $s \in C$.
== Proof ==
=== Necessary Condition ===
Suppose that $C$ be closed.
Let $S$ be a non-empty subset of $C$.
Let $b \in X \setminus C$.
We will show that $b$ is not a supremum of $S$.
If $b$ is not an upper bound of $S$, then by definition $b$ cannot be a supremum of $S$.
Suppose, then, that $b$ is an upper bound of $S$.
Since $C$ is closed and $b \notin C$, there must be an open interval or open ray $U$ containing $b$ that is disjoint from $Y$.
Since $b$ is an upper bound for $S$ and $S$ is non-empty, $U$ cannot be a downward-pointing ray.
Thus $U$ is either an open interval $\openint a q$ or an upward-pointing open ray ${\uparrow}a$.
Then:
:$a \prec b$
Since $b \in U$, and $b$ is an upper bound of $S$, no element strictly succeeding all elements of $U$ can be in $S$.
By the above and the fact that $S \subseteq C$, ${\uparrow}a \cap S = \O$, so $a$ is an upper bound of $S$.
Since $a \prec b$, $b$ is not a supremum of $S$.
A similar argument shows that $b$ is not an infimum.
=== Sufficient Condition ===
Suppose that no non-empty subset of $C$ has a supremum or infimum relative to $X$ that lies in $X \setminus C$.
Let $p \in X \setminus C$.
==== Case 1: $p$ is an upper bound of $C$ ====
Since $C$ is a non-empty subset of $C$, it does not have a supremum in $X \setminus C$.
Thus $p$ is not a supremum of $C$.
Therefore, $C$ has an upper bound $a \in X$ such that:
:$a \prec p$
Thus ${\uparrow_X}a$ contains $p$ and is disjoint from $C$.
So $p$ is not an adherent point of $C$.
==== Case 2: $p$ is a lower bound of $C$ ====
The same approach used for case 1 proves $p$ is not an adherent point of $C$.
==== Case 3: $p$ is neither an upper nor a lower bound of $C$ ====
In this case, $C \cap {\downarrow_X} p$ and $C \cap {\uparrow_X} p$ are non-empty, and their union is $C$.
Thus $p$ is not a supremum of $C \cap {\downarrow_X} p$ and is not an infimum of $C \cap {\uparrow_X}p$.
Thus there are $a, b \in X$ such that $a \prec p \prec b$, $a$ is an upper bound of $C \cap {\downarrow_X}p$, and $b$ is a lower bound of $C \cap {\downarrow_X} p$.
Then $\openint a b$ contains $p$ and is disjoint from $C$.
So $p$ is not an adherent point of $C$.
Since $C$ contains all of its adherent points, it is closed.
== Note ==
Non-empty subsets of $C$ may not have suprema or infima in $X$, but if they have them, they are elements of $C$.
== Also see ==
* Compact Subspace of Linearly Ordered Space
Category:Linearly Ordered Spaces
Closed Set in Metric Space is G-Delta

== Theorem ==
Let $\struct {X, d}$ be a metric space, and let $F \subset X$ be a closed set. 
Then $F$ is a $G_\delta$ set of X.
== Proof ==
Let $n \in \N$.
Let $\ds F_{\frac 1 n} = \bigcup \limits_{x \mathop \in F} \map B {x, \dfrac 1 n}$, where $\map B {x, \dfrac 1 n}$ is the open ball around $x$ with radius $\dfrac 1 n$.
$F_{\frac 1 n}$ is an open set by definition of open ball.
Also, $F_{\frac 1 n}$ contains $F$ as it is the union of open balls around every element of $F$.
$\ds F \subseteq \bigcap \limits_{n \mathop = 1}^{\infty} F_{\frac 1 n}$ because $F \subseteq F_{\frac 1 n}$ for each $n \in \N$.
Note that:
:$\ds \lim_{n \mathop \to \infty} \map B {x, \dfrac 1 n} = \set x$
so the above statement holds even in the limit.
Let $\ds y \in \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$.
Let $\epsilon > 0$ be given.
There exists $n \in \N$ such that $\dfrac 1 n < \epsilon$.
As $y \in \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$, there exists $x \in F$ such that $y \in \map B {x, \dfrac 1 n}$.
Therefore:
:$\map d {x, y} < \dfrac 1 n < \epsilon$
which shows that $y$ is a Limit Point of $F$.
Since $F$ is closed, then $y \in F$.
Therefore:
:$\ds F \supseteq \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$
We conclude that $\ds F = \bigcap \limits_{n \mathop = 1}^\infty F_{\frac 1 n}$, a countable intersection of open sets.
Therefore $F$ is a $G_\delta$ set by definition. 
Category:G-Delta Sets
Category:Metric Spaces
Closed Set in Particular Point Space has no Limit Points

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $H \subsetneq S$ be closed in $T$.
Then $H$ has no limit points.
== Proof ==
Let $H$ be closed in $T$.
Then by definition $p \notin H$.
Let $x \in H$.
By definition, $x$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
Consider the set $U_x := \set {x, p} \subseteq S$.
As $p \in U_x$ we have that $U_x$ is open in $T$.
But there is no $y \in U_x: y \in H, y \ne x$ and so $x$ is not a limit point of $H$.
So $H$ has no limit points.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $9$
Category:Closed Sets
Category:Examples of Limit Points
Category:Particular Point Topology