nyuuzyou/wikis huggingface dataset:
Closed Set in Topological Subspace

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $T' = \struct{H, \tau_H}$ be a subspace of $T$ where $H \subseteq S$.
Then $V \subseteq T'$ is closed in $T'$ $V = T' \cap W$ for some $W$ closed in $T$.
=== Corollary ===
== Proof ==
=== Necessary Condition ===
Suppose $V \subseteq T'$ is closed in $T'$.
Then $T' \setminus V$ is open in $T'$ by definition.
So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$.
Then:
Thus $V$ is closed in $T$.
=== Sufficient Condition ===
Conversely, suppose $V = T' \cap W$ where $W$ closed in $T$.
Then $T' \setminus V = T' \setminus \paren {T' \cap W} = T' \cap \paren {T \setminus W}$ which is open in $T'$.
So $V$ is closed in $T'$.
== Also see ==
* Neighborhood in Topological Subspace
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.6$
* : $\S 3$: New Spaces From Old: Subspaces
Category:Closed Sets
Category:Topological Subspaces
Closed Set in Topological Subspace/Corollary

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $T' = \struct{H, \tau_H}$ be a subspace of $T$ where $H \subseteq S$.
Let $H$ be closed in $T$.
Then $V \subseteq H$ is closed in $T'$ $V$ is closed in $T$.
== Proof ==
Let $V \subseteq H$ be closed in $T'$.
Then, from Closed Set in Topological Subspace, $V = H \cap V$ is closed in $T'$.
If $V$ is closed in $T'$ then $V = H \cap W$ where $W$ is closed in $T$.
Since $H$ is closed in $T$, it follows by Topology Defined by Closed Sets that $V$ is closed in $T$.
== Also see ==
* Open Set in Open Subspace
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Corollary $3.7.7$
* : $\S 3$: New Spaces From Old: Subspaces
Category:Closed Sets
Category:Topological Subspaces
Closed Set is F-Sigma Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be a closed set of $T$.
Then $V$ is an $F_\sigma$ set of $T$.
== Proof ==
$V$ is the union of a singleton.
So $V$ is trivially the union of a countable number of closed sets of $T$.
The result follows by definition of $F_\sigma$ set.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:F-Sigma Sets
Category:Closed Sets
Closed Set of Countable Fort Space is G-Delta

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Let $H \subseteq S$ be closed in $T$.
Then $H$ is a $G_\delta$ set.
== Proof ==
By definition of Fort space, $H$ is finite or contains $p$.
Consider the set of sets defined by:
:$\DD = \set {S \setminus \set z: z \notin H}$
Because $H$ is finite, $S \setminus H$ is countably infinite.
From its method of construction, $\DD$ has the same cardinality as $S \setminus H$ and so is countably infinite.
We have that the elements of $\DD$ are all open sets.
Thus, by definition, any set of the form $\ds H \subseteq \bigcap_{V \mathop \in \DD} V$ is a $G_\delta$ set.
Let $r \in H$.
Then:
:$\forall z \in S: z \notin H: r \in S \setminus \set z$
Thus by definition of $\DD$:
:$\ds H \subseteq \bigcap_{V \mathop \in \DD} V$
Let $r \notin H$.
Then:
:$\exists z \in S: r \notin S \setminus \set z \in \DD$
So:
:$\ds \relcomp S H \subseteq \relcomp S {\bigcap_{V \mathop \in \DD} V}$
and so:
:$\ds H = \bigcap_{V \mathop \in \DD} V$
Hence the result.
Category:Fort Spaces
Category:G-Delta Sets
Closed Set of Ultraconnected Space is Ultraconnected

== Theorem ==
Let $T = \struct {S, \tau}$ be an ultraconnected topological space.
Let $F \subset S$ be a closed set in $T$.
Then $F$ is ultraconnected.
== Proof ==
Let $A, B$ be two non-empty closed sets in $\struct {F, \tau}$.
By Closed Set in Topological Subspace, $A, B$ are closed in $T$ as well.
By , $A$ and $B$ are not disjoint.
Since $A$, $B$ are arbitrary, no two non-empty closed sets of $\struct {F, \tau}$ are disjoint.
Hence the result from .
== Also see ==
* Open Set of Irreducible Space is Irreducible
Category:Ultraconnected Spaces
Closed Set of Uncountable Finite Complement Topology is not G-Delta

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $V \in \tau$ be a closed set of $T$.
Then $V$ is not a $G_\delta$ set.
== Proof ==
Let $V$ be a closed set of $T$.
 $V$ is $G_\delta$ set.
Then by Complement of $G_\delta$ Set is $F_\sigma$ Set:
:$S \setminus V$ is an $F_\sigma$ set.
By definition of closed set, $S \setminus V$ is an open set of $T$.
But by Open Set of Uncountable Finite Complement Topology is not $F_\sigma$:
:$S \setminus V$ is not an $F_\sigma$ set.
It follows by Proof by Contradiction that $V$ is not a $G_\delta$ set.
== Sources ==
* : Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $3$
Category:G-Delta Sets
Category:Finite Complement Topology
Closed Set under Chain Unions with Choice Function is of Type M

== Theorem ==
Let $S$ be a set which is closed under chain unions.
Let there exist a choice function for $S$.
Then $S$ is of type $M$, that is:
:every element of $S$ is a subset of a maximal element of $S$ under the subset relation.
== Proof ==
Let $S$ be closed under chain unions.
Let $C$ be a choice function for $S$.
For $x \in S$, let $x^*$ be defined as the set of all elements $y$ of $S$ such that $x$ is a proper subset of $y$:
:$\forall x \in S: x^* := \set {y \in S: x \subsetneqq y}$
We have that $x^*$ is empty $x$ is a maximal element of $S$ under the subset relation.
Let us define a progressing mapping $g$ on $S$ as follows:
:$\forall x \in S: \map g x = \begin{cases} x & : \text {$x$ is maximal} \\ \map C {x^*} & : \text {otherwise} \end{cases}$
Thus:
:if $x$ is a maximal element of $S$, then $\map g x = x$
:if $x$ is not a maximal element of $S$, then $x$ is a proper subset of $\map g x$.
By Set which is Superinductive under Progressing Mapping has Fixed Point: Corollary:
:for all $b \in S$, $b$ is a subset of some $x \in S$ such that $x = \map g x$.
This element $x$ is then a maximal element of $S$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Theorem $5.3$
Category:Type M Sets
Category:Choice Functions
Category:Closure under Chain Unions
Closed Sets in Indiscrete Topology

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is a closed set of $T$ either $H = S$ or $H = \O$.
== Proof ==
A set $U$ is closed in a topology $\tau$ :
:$\relcomp S U \in \tau$
where $\relcomp S U$ denotes the complement of $U$ in $S$.
That is, the closed sets are the complements of the open sets.
From Open Sets in Indiscrete Topology, in $\tau = \set {\O, S}$, the only open sets are $\O$ and $S$.
Hence the only closed sets in the indiscrete topology on $S$ are:
:$\relcomp S \O = S$ from Relative Complement of Empty Set
and:
:$\relcomp S S = \O$ from Relative Complement with Self is Empty Set
as stated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $2$
Category:Indiscrete Topology
Closed Sets in Noetherian Topological Space

== Theorem ==
Let $T = \struct {X, \tau}$ be a Noetherian topological space.
Let $Y \subseteq X$ be a non-empty closed set of $T$.
Then there exist closed irreducible subsets $Y_1, \ldots, Y_r$ such that:
:$Y = Y_1 \cup \cdots \cup Y_r$
Furthermore, if we require:
:$\forall i, j : i \ne j \implies Y_i \not \subseteq Y_j$
then $Y_1, \ldots, Y_r$ are uniquely determined.
== Proof ==
== Sources ==
* Proposition $1.5$
Category:Noetherian Topological Spaces
Closed Sets of Closed Extension Topology

== 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 the closed sets of $T^*_p$ (apart from $S^*_p$) are the closed sets of $T$.
This explains why $\tau^*_p$ is called the closed extension topology of $\tau$.
== Proof ==
By definition:
:$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \O$
Let $V \subseteq S^*_p$ be closed in $T^*_p$.
Then $S^*_p \setminus V$ is open in $T^*_p$.
Then $\struct {S^*_p \setminus V} \setminus \set p$ is open in $T$.
From Set Difference with Union we have:
:$\struct {S^*_p \setminus V} \setminus \set p = S^*_p \setminus \struct {V \cup \set p}$
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $12$. Closed Extension Topology: $20$
Category:Closed Extension Topology
Closed Sets of Either-Or Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then the closed sets of $T$ are:
:$\O$
:$S$
:$\set {-1}$
:$\set 1$
:$\set {-1, 1}$
:Any subset of $\closedint {-1} 1$ containing $\set 0$ as a subset.
== Proof ==
From Open and Closed Sets in Topological Space we have that $\O$ and $S$ are closed sets trivially.
From the definition of closed set, we have:
:$U$ is open in $T$ $S \setminus U$ is closed in $T$
:$U$ is closed in $T$ $S \setminus U$ is open in $T$
where $S \setminus U$ denotes the relative complement of $U$ in $S$.
Now we have that:
:$\openint {-1} 1$ is open in $T$ so $S \setminus \openint {-1} 1 = \set {-1, 1}$ is closed in $T$.
:Both $\hointr {-1} 1 \supseteq \openint {-1} 1$ and $\hointr {-1} 1 \supseteq \openint {-1} 1$ therefore are open in $T$, so $\set {-1}$ and $\set 1$ are closed in $T$.
Apart from $S$ itself, there are no other subsets of $S = \closedint {-1} 1$ which have $\openint {-1} 1$ as a subset.
Finally, let $U \in \tau: \set 0 \nsubseteq U$.
Then by the definition of relative complement $\set 0 \subseteq S \setminus U$.
Now suppose $\set 0 \subseteq V$.
Then again by the definition of relative complement $\set 0 \nsubseteq S \setminus V$.
So $S \setminus V$ is open in $T$ and so $V$ is closed in $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology
Category:Either-Or Topology
Closed Sets of Fortissimo Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Then $H \subseteq S$ is closed in $T$ :
:$p \in H$
or
:$H$ is countable
or both.
== Proof ==
By definition of a Fortissimo space, $U \subseteq S$ is open in $T$ :
:$p \in \relcomp S U$
or
:$\relcomp S U$ is countable
or both.
The result follows from the definition of closed set.
Category:Closed Sets
Category:Fortissimo Space
Closed Sets of Right Order Space on Real Numbers

== Theorem ==
Let $T = \struct {\R, \tau}$ be the right order space on $\R$.
Then $H \subseteq S$ is closed in $T$ :
:$H = \O$ or $\R$
or
:$H = \hointl {-\infty} a$ for some $a \in \R$.
== Proof ==
By definition of the right order space on $\R$, $U \subseteq S$ is open in $T$ :
:$U = \O$ or $\R$
or
:$U = \openint a \infty$ for some $a \in \R$.
Note that:
:$\R \setminus \O = \R$
:$\R \setminus \R = \O$
:$\R \setminus \openint a \infty = \hointl {-\infty} a$
The result follows from the definition of closed set.
Category:Closed Sets
Category:Right Order Topology
Closed Statement/Examples/Arbitrary Example 1

== Example of Open Statement ==
Consider the statement expressed symbolically as:
:$\exists x: \exists y: x \text { is the son of $y$}$
This is a closed statement, as both the variables $x$ and $y$ appears as bound variables.
Whether it is true or false depends upon the ranges of $x$ and $y$.
== Sources ==
* 
* 
Category:Examples of Open Statements
Closed Statement/Examples/Arbitrary True Statement

== Example of Closed Statement ==
The statement:
:$\exists x \in \Z: x^2 + 2 = 11$
is a closed statement that is true, as it is satisfied by the integers $x = 3$ and $x = -3$.
== Sources ==
* : $\S 1$: Some mathematical language: Variables and quantifiers
Category:Examples of Closed Statements
Closed Subgroups of P-adic Integers

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Then the closed subgroups of $\Z_p$ are the principal ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N : p^k \Z_p$
== Proof ==
From Metric Space is Hausdorff:
:$\Z_p$ is a Hausdorff space
From Finite Subspace of Hausdorff Space is Closed:
:$\set 0$ is closed
From Cosets Form Local Basis of P-adic Number:
:$\forall k \in \N : p^k \Z_p = 0 + p^k \Z_p$ is closed
Hence the ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N : p^k \Z_p$
are closed subgroups.
It remains to show that an arbitrary closed subgroups is indeed one of these ideals.
Let $H$ be a closed subgroup of the additive group of $\Z_p$.
From Correspondence between Abelian Groups and Z-Modules:
:$\Z H \subseteq H$
Let $a \in \Z_p$.
Let $h \in H$.
From Integers are Dense in P-adic Integers, there exists a sequence $\sequence {a_n}$:
:$\forall n \in \N: a_n \in Z$
:$\ds \lim_{n \mathop \to \infty} a_n = a$
From Multiple Rule for Sequences in Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} a_n h = a h$
We have:
:$a_n h \in \Z H \subseteq H$
From Subset of Metric Space contains Limits of Sequences iff Closed:
:$a h = \ds \lim_{n \mathop \to \infty} a_n h \in H$
Since $a \in \Z_p$ and $h \in H$ were arbitrary, it follows that:
:$\Z_p H \subseteq H$
By definition of ideal:
:$H$ is an ideal of $\Z_p$
From Ideals of P-adic Integers, $H$ is one of the principal ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N : p^k \Z_p$
Category:P-adic Integers
Closed Subset is Upper Section in Lower Topology

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a transitive relational structure with lower topology.
Let $A \subseteq S$ such that
:$A$ is closed.
Then $A$ is an upper section of $S$.
== Proof ==
By definition of closed set:
:$S \setminus A$ is open.
By Open Subset is Lower Section in Lower Topology:
:$S \setminus A$ is a lower section.
Thus by Complement of Lower Section is Upper Section and Relative Complement of Relative Complement:
:$A$ is an upper section of $S$.
== Sources ==
* 
Category:Topological Order Theory
Closed Subset of Irreducible Space with Same Krull Dimension is Itself

== Definition ==
Let $X$ be an irreducible topological space.
Let $Y \subseteq X$ be a closed subset.
Suppose:
:$\map \dim Y = \map \dim X < + \infty$
where $\dim$ denotes the Krull dimension.
Then:
:$Y = X$
== Proof ==
Let $n = \map \dim Y$.
Then there exists a chain of closed irreducible sets of $Y$:
:$A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n$
If $Y \subsetneq X$, then:
:$A_0 \subsetneq A_1 \subsetneq \cdots \subsetneq A_n \subsetneq X$
would be a chain of closed irreducible sets of $X$, which implies that:
:$n + 1 \le \map \dim X$
which is a contradiction.
Thus $Y = X$.
Category:Irreducible Spaces
Closed Subset of Real Number Line is G-Delta

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $H \subseteq \R$ be a closed subset of $\R$.
Then $H$ is a $G_\delta$ set.
== Proof ==
We have:
:Real Number Line is Metric Space
:Closed Set in Metric Space is $G_\delta$
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $5$
Category:Real Number Line with Euclidean Topology
Category:G-Delta Sets
Closed Subset of Real Numbers with Lower Bound contains Infimum

== Theorem ==
Consider the real number line as a metric space under the usual metric.
Let $A \subseteq \R$ such that $A$ is closed in $\R$ and $A \ne \O$.
Let $A$ be bounded below.
Then $A$ contains its infimum.
== Proof ==
From Infimum of Bounded Below Set of Reals is in Closure:
:$\inf A \in \map \cl A$
From Set is Closed iff Equals Topological Closure:
:$A = \map \cl A$
Therefore $\inf A \in A$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $5$
Category:Bounded Below Sets of Real Numbers
Category:Real Analysis
Closed Subsets of Symmetry Group of Square

== Theorem ==
Recall the symmetry group of the square:
=== Symmetry Group of Square ===
The subsets of $\SS$ which are closed under composition of mappings are:
:$\O$
:$\set e$
:$\set {e, r^2}$
:$\set {e, t_x}$
:$\set {e, t_y}$
:$\set {e, t_{AC} }$
:$\set {e, t_{BD} }$
:$\set {e, r, r^2, r^3}$
:$\set {e, r^2, t_x, t_y}$
:$\set {e, r^2, t_{AC}, t_{BD} }$
:$\SS$
== Proof ==
Recall that a submagma of an algebraic structure $\SS$ is a subsets of $\SS$ which is closed.
Let $\XX$ be the set of all submagmas of $\SS$.
From Empty Set is Submagma of Magma:
:$\O \in \XX$
From Magma is Submagma of Itself:
:$\SS \in \XX$
From Idempotent Magma Element forms Singleton Submagma:
:$\set e \in \XX$
Let us refer to the Cayley table:
=== Cayley Table of Symmetry Group of Square ===
Taking each of the elements of $\SS$ in order:
Thus we have:
Next note by inspection that:
:$\set {e, r^2, t_x, t_y} \in \XX$
and:
:$\set {e, r^2, t_{AC}, t_{BD} } \in \XX$
Finally note by inspection that:
:any closed subset of $\SS$ which contains both $r$ and any of the reflections contains all the elements of $\SS$
:any closed subset of $\SS$ which contains both $r^3$ and any of the reflections contains all the elements of $\SS$.
Thus there are no more proper subsets of $\SS$ which are submagmas of $\SS$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.5$
Category:Symmetry Groups
Category:Symmetry Group of Square
Closed Subspace of Banach Space forms Banach Space

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $Y$ be a closed linear subspace of $X$.
Let $\norm \cdot_Y$ be the restriction of $\norm \cdot_X$ to $Y$. 
Then $\struct {Y, \norm \cdot_Y}$ is a Banach space.
== Proof ==
From Restriction of Norm on Vector Space to Subspace is Norm, $\struct {Y, \norm \cdot_Y}$ is a normed vector space.
We now show that $\struct {Y, \norm \cdot_Y}$ is complete.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {Y, \norm \cdot_Y}$. 
Then, for each $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\norm {y_n - y_m}_Y < \epsilon$ 
for all $n, m \ge N$.
Since $\norm \cdot_Y$ is the restriction of $\norm \cdot_X$ to $Y$, we have: 
:$\norm {y_n - y_m}_Y = \norm {y_n - y_m}_X$
so:
:$\norm {y_n - y_m}_X < \epsilon$
So $\sequence {y_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {X, \norm \cdot_X}$.
Since $\struct {X, \norm \cdot_X}$ is Banach, we have that $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in X$.
Since $Y$ is closed in $\struct {X, \norm \cdot_X}$, we have that $Y$ contains all its limit points.
So $y \in Y$. 
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we therefore have: 
:$\norm {y_n - y}_X \to 0$
with $y_n - y \in Y$ for each $n \in \N$. 
Since $\norm \cdot_Y$ is the restriction of $\norm \cdot_X$ to $Y$, we have: 
:$\norm {y_n - y}_Y \to 0$
So $\sequence {y_n}_{n \mathop \in \N}$ converges to $y$ in $\struct {Y, \norm \cdot_Y}$. 
Since $\sequence {y_n}_{n \mathop \in \N}$ was arbitrary, we have that all Cauchy sequence in $\struct {Y, \norm \cdot_Y}$ converge.
== Also see ==
* Subspace of Complete Metric Space is Closed iff Complete
Category:Banach Spaces
Closed Subspace of Compact Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a compact topological space.
Let $C = \struct {H, \tau_H}$ be a subspace of $T$.
Let $C$ be closed in $T$.
Then $\struct {H, \tau}$ is compact.
That is, the property of being compact is weakly hereditary.
== Proof ==
Let $T$ be a compact space.
Let $C$ be a closed subspace of $T$.
Let $\UU$ be an open cover of $C$.
We have that $C$ is closed in $T$.
It follows by definition of closed that $T \setminus C$ is open in $T$.
So if we add $T \setminus C$ to $\UU$, we see that $\UU \cup \set {T \setminus C}$ is also an open cover of $T$.
As $T$ is compact, there exists a finite subcover of $\UU \cup \set {T \setminus C}$, say $\VV = \set {U_1, U_2, \ldots, U_r}$.
This covers $C$ by the fact that it covers $T$.
Suppose $T \setminus C$ is an element of $\VV$.
Then $T \setminus C$ may be removed from $\VV$, and the rest of $\VV$ still covers $C$.
Thus we have a finite subcover of $\UU$ which covers $C$.
Hence $C$ is compact.
== Also see ==
* Intersection of Closed Set with Compact Subspace is Compact
== Sources ==
* : $2.35$
* : $5$: Compact spaces: $5.6$: Compactness and Constructions: Proposition $5.6.1$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:Closed Sets
Category:Compact Spaces
Closed Subspace of Lindelöf Space is Lindelöf Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a Lindelöf space.
Let $C = \struct {H, \tau_H}$ be a subspace of $T$.
Let $C$ be closed in $T$.
Then $\struct {H, \tau}$ is Lindelöf space.
That is, the property of being Lindelöf is weakly hereditary.
== Proof ==
Let $T$ be a Lindelöf space.
Let $C$ be a closed subspace of $T$.
Let $\UU$ be an open cover of $H$.
We have that $H$ is closed in $T$.
It follows by definition of closed that $H \setminus C$ is open in $T$.
So if we add $S \setminus H$ to $\UU$, we see that:
:$\UU \cup \set {S \setminus H}$ 
is also an open cover of $S$.
As $T$ is compact, there exists a countable subcover of $\UU \cup \set {S \setminus H}$, that is:
:$\VV = \set {U_1, U_2, \ldots}$
This covers $H$ by the fact that it covers $S$.
Suppose $S \setminus H$ is an element of $\VV$.
Then $S \setminus H$ may be removed from $\VV$, and the rest of $\VV$ still covers $H$.
Thus we have a countable subcover of $\UU$ which covers $H$.
Hence $C$ is a Lindelöf space.
== Also see ==
* Closed Subspace of Compact Space is Compact
== Sources ==
* : Chapter $4$: Separation and Countability: $\S16$: Countability Properties: Theorem $16.6(\text b)$
Category:Closed Sets
Category:Compact Spaces
Closed Topologist's Sine Curve is Connected

== Theorem ==
Let $G$ be the graph of the function $y = \map \sin {\dfrac 1 x}$ for $x > 0$.
Let $J$ be the line segment joining the points $\tuple {0, -1}$ and $\tuple {0, 1}$ in $\R^2$.
Then $G \cup J$ is connected.
== Proof ==
Since the open interval $\openint 0 \infty$ is connected, then so is $G$ by Continuous Image of Connected Space is Connected.
It is enough, from Set between Connected Set and Closure is Connected, to show that $J \subseteq \map \cl G$.
Let $p \in J$, say, $\tuple {0, y}$ where $-1 \le y \le 1$.
We need to show that:
:$\forall \epsilon > 0: \map {N_\epsilon} p \cap G \ne \O$
where $\map {N_\epsilon} p$ is the $\epsilon$-neighborhood of $p$.
Let us choose $n \in \N: \dfrac 1 {2 n \pi} < \epsilon$.
From Sine of Half-Integer Multiple of Pi:
:$\map \sin {\dfrac {\paren {4 n + 1} \pi} 2} = 1$
and:
:$\map \sin {\dfrac {\paren {4 n + 3} \pi} 2} = -1$
So by the Intermediate Value Theorem, $\map \sin {\dfrac 1 x}$ takes every value between $-1$ and $1$ in the closed interval $\closedint {\dfrac 2 {\paren {4 n + 3} \pi} } {\dfrac 2 {\paren {4 n + 1} \pi} }$.
In particular, $\map \sin {\dfrac 1 {x_0} } = y$ for some $x_0$ in this interval.
The distance between the points $\tuple {0, y}$ and $\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } = \tuple {x_0, y}$ is $x_0 < \epsilon$.
So:
:$\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } \in \map {N_\epsilon} p \cap G$
as required.
== Sources ==
* : $6.2$: Connectedness: Example $6.2.19$
Category:Topology
Category:Sine Function
Category:Reciprocals
Category:Connected Sets
Closed Unit Ball in Normed Dual Space is Weak-* Closed

== Theorem ==
Let $X$ be a normed vector space. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $B^-_{X^\ast}$ be the closed unit ball in $X^\ast$. 
Then we have that $B^-_{X^\ast}$ is weak-$\ast$ closed.
== Proof ==
From Weak-* Topology is Hausdorff, $\struct {X^\ast, w^\ast}$ is Hausdorff.
From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact. 
From Compact Subspace of Hausdorff Space is Closed, it follows that $B^-_{X^\ast}$ is weak-$\ast$ closed.
Category:Weak-* Topologies
Category:Normed Dual Spaces