nyuuzyou/wikis huggingface dataset:
Closure of Subset of Metric Space by Convergent Sequence

== Lemma ==
Let $M$ be a metric space.
Let $C \subseteq M$.
Let $x \in M$.
Let $\map \cl C$ denote the closure of $C$ in $T$.
Then $x \in \map \cl C$ there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$.
== Proof ==
=== Necessary Condition ===
Suppose there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$.
Let $\epsilon > 0$.
Then by definition:
:$\exists N \in \N: \forall n > N: x_n \in \map {B_\epsilon} x$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
Since $\forall n: x_n \in C$, it follows that:
:$\forall \epsilon > 0: \map {B_\epsilon} x \cap C \ne \O$
Let $U$ be an open neighborhood of $x$.
From Open Balls form Basis for Open Sets of Metric Space, there exists $\epsilon> 0$ such that $\map {B_\epsilon} x \subseteq U$.
Since $\map {B_\epsilon} x \cap C \ne \O$, we have $U \cap C \ne \O$. 
Since $U$ was an arbitrary open neighborhood, we have $x \in \map \cl C$.
=== Sufficient Condition ===
Now suppose $x \in \map \cl C$.
From Open Ball is Open Set, $\map {B_\epsilon} x$ is open for each $\epsilon > 0$. 
From the definition of closure, we have that $C \cap \map {B_{1/n} } x \ne \O$ for each $n \in \N$.
Hence we can pick $x_n \in C \cap \map {B_{1/n} } x$.
Then we have $\map d {x_n, x} < 1/n$ and $x_n \in C$ for each $n \in \N$.
Taking $n \to \infty$, we have $x_n \to x$. 
Hence we have found a sequence with terms in $C$ converging to $x$.
== Sources ==
* : $7.2$: Sequential compactness: Lemma $7.2.2$
Category:Metric Spaces
Category:Convergent Sequences
Closure of Subset of Metric Space is Closed

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H^-$ denote the closure of $H$.
Then $H^-$ is a closed set of $M$.
== Proof ==
Let $\overline {\paren {H^-} }$ denote the complement of $H^-$.
Let $x \in \overline {\paren {H^-} }$.
By definition of the closure of $H$:
:$x$ is not a limit point of $H$.
So:
:$\exists \epsilon \in \R_{> 0}: \paren {\map {B_\epsilon} x \setminus \set x} \cap H = \O$
From Intersection with Set Difference is Set Difference with Intersection:
:$\paren {\map {B_\epsilon} x \cap H} \setminus \set x = \O$
From Set Difference with Superset is Empty Set:
:$\map {B_\epsilon} x \cap H \subseteq \set x$
By definition of the closure of $H$:
:$x$ is not an isolated point of $H$.
So:
:$\map {B_\epsilon} x \cap H \ne \set x$
Thus:
:$\map {B_\epsilon} x \cap H = \O$
From Empty Intersection iff Subset of Complement:
:$\map {B_\epsilon} x \subseteq \overline {\paren{H^-} }$
It follows that $\overline {\paren{H^-} }$ is open in $M$.
Thus $H^-$ is closed in $M$ by definition.
Category:Set Closures
Closure of Subset of Metric Space is Intersection of Closed Supersets

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H^-$ denote the closure of $H$.
Then $H^-$ is the intersection of all closed sets of $M$ of which $H$ is a subset.
== Proof ==
Let $\mathbb K$ be the set of all closed sets $K$ of $M$ such that $H \subseteq K$.
From Closure of Subset of Closed Set of Metric Space is Subset:
:$\forall K \in \mathbb K: H^- \subseteq K$
From Intersection is Largest Subset:
:$\ds H^- \subseteq \bigcap \mathbb K$
From Closure of Subset of Metric Space is Closed, $H^-$ is closed.
From Subset of Metric Space is Subset of its Closure, $H \subseteq H^-$.
So $H^-$ is, by definition, a closed set of $M$ which contains $H$.
But $\ds \bigcap \mathbb K$ is the intersection of all closed sets of $M$ which contain $H$.
So from Intersection is Subset it follows that:
:$\ds \bigcap \mathbb K \subseteq H^-$
Finally, we have that:
:$\ds H^- \subseteq \bigcap \mathbb K$
:$\ds \bigcap \mathbb K \subseteq H^-$
So by definition of set equality:
:$\ds H^- = \bigcap \mathbb K$
which is what we needed to prove.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$
Category:Set Closures
Closure of Subspace of Normed Vector Space is Subspace

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $Y \subseteq X$ be a subspace of $X$.
Let $Y^-$ be the closure of $Y$.
Then $Y^- \subseteq X$ is also a subspace of $X$.
== Proof ==
Suppose $y \in Y^-$.
Then there is a sequence $\ds \sequence {y_n}_{n \mathop \in \N} \in Y$ which converges to $y$.
Suppose $y \in Y$ and $y$ is a limit point.
Then one can define a constant sequence:
:$\sequence {y_n}_{n \mathop \in \N} = y$
=== Closed under restriction of vector addition ===
Let $x, y \in Y^-$.
Let $\sequence {x_n}_{n \mathop \in \N}, \sequence {y_n}_{n \mathop \in \N}$ be sequences in $Y$.
Suppose:
:$\ds \lim_{n \mathop \to \infty} x_n = x$
:$\ds \lim_{n \mathop \to \infty} y_n = y$
Since $Y$ is a subspace:
:$\forall n \in \N : x_n + y_n \in Y \subseteq Y^-$
Furthermore:
:$\ds \lim_{n \mathop \to \infty} \paren {x_n + y_n} = x + y$
$Y^-$ is closed and contains its limit points.
Hence:
:$x + y \in Y^-$
=== Closed under restriction of scalar multiplication ===
Let $\alpha \in \Bbb K$, where $\Bbb K$ is a field.
Let $y \in Y^-$.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence in $Y$.
Suppose:
:$\ds \lim_{n \mathop \to \infty} y_n = y$
Since $Y$ is a subspace:
:$\forall n \in \N : \alpha \cdot y_n \in Y \subseteq Y^-$
Furthermore:
:$\ds \lim_{n \mathop \to \infty} \alpha \cdot y_n = \alpha \cdot y$
$Y^-$ is closed and contains its limit points.
Hence:
:$\alpha \cdot y \in Y^-$
=== Nonemptiness ===
$Y$ is a subspace.
Hence:
:$0 \in Y \subseteq Y^-$
Thus $Y^-$ contains at least one element and is non-empty.
Hence $Y^-$ is a closed subspace
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Set Closures
Closure of Symmetric Subset of Normed Vector Space is Symmetric

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $A \subseteq X$ be symmetric.
Then the topological closure of $A$ is symmetric.
== Proof ==
Let $A^-$ be the topological closure of $A$.
Let $a \in A^-$.
Then from Point in Closure of Subset of Metric Space iff Limit of Sequence, we have: 
:there exists a sequence $\sequence {a_n}_{n \mathop \in \N}$ in $A$ such that $a_n \to a$. 
Since $A$ is symmetric, we have: 
:$-a_n \in A$ for each $n \in \N$.
We also have, from Multiple Rule for Sequences in Normed Vector Space:
:$-a_n \to -a$ 
So we have: 
:$-a \in A^-$
Category:Set Closures
Category:Symmetric Subsets of Vector Spaces
Closure of Topological Closure equals Closure

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
Then:
:$\map \cl {\map \cl H} = \map \cl H$
where $\cl$ denotes the closure of $H$.
== Proof ==
It follows directly from Set is Subset of its Topological Closure that:
:$\map \cl H \subseteq \map \cl {\map \cl H}$
Let $x \in \map \cl {\map \cl H}$.
Then from Condition for Point being in Closure, any $U$ which is open in $T$ such that $x \in U$ contains some $y \in \map \cl H$.
If we consider $U$ as an open set containing $y$, it follows that:
:$U \cap H \ne \O$
Hence $x \in \map \cl H$.
Hence by definition of subset:
:$\map \cl {\map \cl H} \subseteq \map \cl H$
Hence the result by definition of set equality.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.15 \ \text{(c)}$
Category:Set Closures
Closure of Union and Complement imply Closure of Set Difference

== Theorem ==
Let $\RR$ be a system of sets on a universe $\mathbb U$ such that for all $A, B \in \RR$:
:$(1): \quad A \cup B \in \RR$
:$(2): \quad \map \complement A \in \RR$
where $\cup$ denotes set union and $\complement$ denotes complement (relative to $\mathbb U$).
Then:
:$\forall A, B \in \RR: A \setminus B \in \RR$
where $\setminus$ denotes set difference.
== Proof ==
Let $A, B \in \RR$.
As both set union and complement are closed in $\RR$ the result follows.
== Sources ==
* : $\S 1.1$: Rings and Algebras
Category:Set Union
Category:Set Difference
Category:Set Complement
Category:Set Systems
Closure of Union contains Union of Closures

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ denotes the power set of $S$.
Then the union of the closures of the elements of $\mathbb H$ is a subset of the closure of the union of $\mathbb H$:
:$\ds \bigcup_{H \mathop \in \mathbb H} \map \cl H \subseteq \map \cl {\bigcup_{H \mathop \in \mathbb H} H}$
== Proof ==
Let $\ds K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\ds L = \bigcup_{H \mathop \in \mathbb H} H$.
We have:
:$\forall H \in \mathbb H: H \subseteq L$
so from Topological Closure of Subset is Subset of Topological Closure:
:$\map \cl H \subseteq \map \cl L$
It follows from Union is Smallest Superset: General Result that:
:$K \subseteq \map \cl L$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
* : Problems: Section $1: \ 1$
Category:Set Union
Category:Set Closures
Closure of Union of Adjacent Open Intervals

== Theorem ==
Let $a, b, c \in R$ where $a < b < c$.
Let $A$ be the union of the two adjacent open intervals:
:$A := \openint a b \cup \openint b c$
Then:
:$A^- = \closedint a c$
where:
:$A^-$ is the closure of $A$.
== Proof ==
Category:Set Closures
Category:Union of Adjacent Open Intervals
Closure of Union of Singleton Rationals is Real Number Line

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the usual (Euclidean) topology $\tau_d$.
Let $B_\alpha$ be the singleton containing the rational number $\alpha$.
Then the closure in the set of real numbers $\R$ of the union of all $B_\alpha$ is $\R$ itself:
:$\ds \map \cl {\bigcup_{\alpha \mathop \in \Q} B_\alpha} = \R$
== Proof ==
By definition:
:$B_\alpha = \set \alpha$
Thus:
:$\ds \bigcup_{\alpha \mathop \in \Q} B_\alpha = \Q$
The result follows from Closure of Rational Numbers is Real Numbers.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $1$
Category:Rational Number Space
Category:Real Number Line with Euclidean Topology
Category:Singletons
Closure of product equals product of closures

== Theorem ==
Let $\set {X_i}_{i \mathop \in I}$ be an arbitrary family of topological spaces.
Let:
:$A_i \subset X_i$
for each $i \in I$.
Let $X = \ds \prod_{i \mathop \in I} X_i$ have the product topology.
Then:
:$\ds \prod_{i \mathop \in I} \overline{A_i} = \overline {\prod_{i \mathop \in I} A_i}$
== Proof ==
We proceed by double inclusion.
Suppose:
:$\ds \family {x_i}_{i \mathop \in I} \in \prod_{i \mathop \in I} \overline {A_i}$
This means:
:$x_i \in \overline {A_i}$ 
for each $i \in I$. 
Let $U$ be an open set containing $\family {x_i}_{i \mathop \in I}$. 
Since:
:$x_i \in \overline {A_i}$
we can choose:
:$y_i \in U_i \cap A_i$ 
and then:
:$\ds \family {y_i}_{i \mathop \in I} \in U \cap \prod_{i \mathop \in I} A_i$
But $U$ was arbitrary, hence:
:$\ds \family {x_i}_{i \mathop \in I} \in \overline {\prod_{i \mathop \in I} A_i}$
Conversely, suppose:
:$\ds \family {x_i}_{i \mathop \in I} \in \overline {\prod_{i \mathop \in I} A_i}$
We want to show that:
:$x_j \in \overline {A_j}$ for all $j$
Fixed $j$, we take $V_j$ to be open set in $X_j$ such that:
:$x_j \in V_j$
Since $\pi_j^{-1} \sqbrk {V_j}$ is open in $\ds \prod_{i \mathop \in I} X_i$, we can pick:
:$\ds \family {y_i}_{i \mathop \in I} \in \prod_{i \mathop \in I} A_i$
Then:
:$y_j \in V_j \cap A_j$
so it follows that:
:$x_j \in \overline {A_j}$
namely:
:$\family {x_i}_{i \mathop \in I} \in \prod_{i \mathop \in I} \overline{A_i}$
Closure of von Neumann-Bounded Subset of Topological Vector Space is von Neumann-Bounded

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\Bbb F$.
Let $E$ be a von Neumann-bounded subset of $X$.
Then the closure $E^-$ of $E$ is von Neumann-bounded.
== Proof ==
Let $V$ be an open neighborhood of ${\mathbf 0}_X$.
From Disjoint Compact Set and Closed Set in Topological Vector Space separated by Open Neighborhood: Corollary, there exists an open neighborhood $W$ of ${\mathbf 0}_X$ such that: 
:$W^- \subseteq V$
Since $E$ is von Neumann-bounded, there exists $s > 0$ such that: 
:$E \subseteq t W$ for $t \in \Bbb F$ with $t > s$.
From Topological Closure of Subset is Subset of Topological Closure, we have: 
:$E^- \subseteq \paren {t W}^-$ for $t \in \Bbb F$ with $t > s$.
From Dilation of Closure of Set in Topological Vector Space is Closure of Dilation we have: 
:$\paren {t W}^- = t W^-$
Then we have: 
:$E^- \subseteq t W^- \subseteq t V$ for $t \in \Bbb F$ with $t > s$.
Since $V$ was arbitrary, we have that $E^-$ is von Neumann-bounded.
== Sources ==
* : $1.13$: Theorem
Category:Set Closures
Category:Von Neumann-Bounded Subsets of Topological Vector Spaces
Closure under Chain Unions with Choice Function implies Elements with no Immediate Extension

== Theorem ==
Let $S$ be a set of sets which:
:is closed under chain unions
:has a choice function $C$ for its union $\bigcup S$.
Let $b \in S$.
Then $b$ is the subset of an element of $S$ which has no immediate extension in $S$.
== Proof ==
Let the hypothesis be assumed.
Let $A = \bigcup S$
Let $x \in S$ be arbitrary.
Let $\map E x$ be the set of all elements $a \in A$ such that $x \cup \set a$ is an immediate extension of $x$.
Note that:
:$\map E x \subseteq \bigcup S$
Suppose $x$ has an immediate extension in $S$.
Then:
:$\map E x$ is non-empty
:because $\map E x \subseteq \bigcup S$, $\map E x$ is in the domain of the choice function $C$.
Let us define a progressing mapping $g: S \to S$ as follows:
:$\forall x \in S: \map g x = \begin {cases}
x & : \text {$x$ has no immediate extension in $S$} \\ \\ 
x \cup \set {\map C {\map E x} } & : \text {$x$ does have an immediate extension in $S$}
\end{cases}$
Thus $g$ is a progressing mapping whose fixed points are those elements of $S$ that have no immediate extensions in $S$.
Hence the criteria hold for to be applied.
That tells us that an arbitrary $b \in S$ is the subset of a fixed point $x$ of $g$.
Such an element, by definition of $g$, has no immediate extension in $S$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles: Theorem $6.2$
Category:Closure under Chain Unions
Category:Choice Functions
Closures of Elements of Locally Finite Set is Locally Finite

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\AA$ be a locally finite set of subsets of $T$.
Then:
:$\set{A^- : A \in \AA}$ is locally finite in $T$
where $A^-$ denotes the closure of $A$ in $T$.
== Proof ==
Let $\BB = \set{A^- : A \in \AA}$.
Let $x \in S$.
By definition of locally finite set of subsets:
:$\exists U_x \in \tau : x \in U_x :$ the set $\AA_x = \set{A \in \AA : A \cap U_x \ne \O}$ is finite
Let $\BB_x = \set{A^- \in \BB : A^- \cap U_x \ne \O}$
Let $A^- \in \BB_x$.
From Open Set Disjoint from Set is Disjoint from Closure:
:$U_x \cap A \ne \O$
Hence:
:$A \in \AA_x$
It follows that $\BB_x$ is finite.
Since $x$ was arbitrary, we have shown that:
:$\forall x \in S : \exists U_x \in \tau : x \in U_x :$ the set $\BB_x = \set{A^- \in \BB : A^- \cap U_x \ne \O}$ is finite
Hence $\BB$ is locally finite set of subsets by definition.
Category:Locally Finite Sets of Subsets
Cluster Analysis/Examples/Human Statistics

== Examples from Cluster Analysis ==
When analysing a human population, various statistics such as:
* age
* income
* ownership of home or car
* time spent out of the house each week
are likely to show evidence for several distinct clusters, each corresponding to a category such as:
* unemployed
* pensioners
* students
and so on.
== Sources ==
* 
* 
Category:Examples from Cluster Analysis
Cluster Point of Ultrafilter is Unique

== Theorem ==
Let $S$ be a set.
Let $\FF$ be an ultrafilter on $S$.
Let $x \in S$ be a cluster point of $\FF$.
Then there is no point $y \in S: y \ne x$ such that $y$ is also a cluster point of $\FF$.
== Proof ==
Let $x, y$ both be cluster points of an ultrafilter $\FF$ on a set $S$.
Then by definition:
:$\forall U \in \FF: \set {x, y} \subseteq U$
But then we can create a finer filter $\FF \cup \set {\set x}$ on $S$.
So $\FF$ is not an ultrafilter.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Filters
Category:Filter Theory
Co-Countable Measure is Measure

== Theorem ==
Let $X$ be an uncountable set.
Let $\Sigma$ be the $\sigma$-algebra of countable sets on $X$.
Then the co-countable measure $\mu$ on $X$ is a measure.
== Proof ==
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
=== Proof of $(1)$ ===
For all $S \in \Sigma$, $\map \mu S$ is $0$ or $1$.
In either case, $\map \mu S \ge 0$.
=== Proof of $(2)$ ===
It is to be shown that (for a sequence $\sequence {S_n}_{n \mathop \in \N}$ of pairwise disjoint sets):
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$
Suppose that at least one $S_n$ is co-countable, say $S_N$.
Since the $S_n$ are pairwise disjoint, it follows that, for all $n \in \N$, $n \ne N$:
:$S_n \subseteq X \setminus S_N$
by Empty Intersection iff Subset of Complement.
From Subset of Countably Infinite Set is Countable, the $S_n$ with $n \ne N$ are all countable, as $S_N$ is co-countable.
Therefore:
:$\map \mu {S_n} = \begin{cases} 1 & : \text {if $n = N$} \\ 0 & : \text {if $n \ne N$} \end {cases}$
and subsequently:
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = 1$
From Superset of Co-Countable Set, $\ds \bigcup_{n \mathop = 1}^\infty S_n$ is co-countable.
Hence:
:$\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = 1$
verifying $(2)$ for $\mu$ in this case.
If on the other hand, all $S_n$ are countable, then for all $n \in \N$:
:$\map \mu {S_n} = 0$
and so:
:$\ds \sum_{n \mathop = 1}^\infty \map \mu {S_n} = 0$
From Countable Union of Countable Sets is Countable:
:$\ds \bigcup_{n \mathop = 1}^\infty S_n$
is also countable, hence:
:$\ds \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = 0$
verifying $(2)$ for $\mu$ in this case as well.
Hence $(2)$ holds for $\mu$, from Proof by Cases.
=== Proof of $(3')$ ===
Note that $\O \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.
By Empty Set is Countable, $\O$ is countable, whence:
:$\map \mu \O = 0$
Having verified the axioms, it follows that $\mu$ is a measure.
Category:Measures
Co-Countable Measure is Probability Measure

== Theorem ==
Let $X$ be an uncountable set.
Let $\AA$ be the $\sigma$-algebra of countable sets on $X$.
Then the co-countable measure $\mu$ on $X$ is a probability measure.
== Proof ==
By Co-Countable Measure is Measure, $\mu$ is a measure.
By Relative Complement with Self is Empty Set, have $\relcomp X X = \O$.
As $\O$ is countable, it follows that $X$ is co-countable.
Hence $\map \mu X = 1$, and so $\mu$ is a probability measure.
Category:Probability Measures
Coarseness Relation on Topologies is Partial Ordering

== Theorem ==
Let $S$ be a set.
Let $\mathbb T$ be the set of all topologies on $S$.
Let $\le$ be the relation on $\mathbb T$ defined as:
:$\forall \tau_1, \tau_2 \in \mathbb T: \tau_1 \le \tau_2$ $\tau_1$ is coarser than $\tau_2$.
Then $\le$ is a partial ordering on $\mathbb T$.
== Proof ==
It follows directly from the definition of being coarser that:
:$\tau_1 \le \tau_2 \iff \tau_1 \subseteq \tau_2$
By Subset Relation is Ordering it thus follows that $\le$ is an ordering.
From Topologies are not necessarily Comparable by Coarseness, it follows that such an ordering is not always total.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:Topology
Category:Partial Orderings
Coarser Between Generator Set and Filter is Generator Set of Filter

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $F$ be a filter on $L$.
Let $G$ be a generator set of $F$.
Let $A$ be a subset of $S$ such that
:$G$ is coarser than $A$ and $A$ is coarser than $F$.
Then $A$ is generator set of $F$.
== Proof ==
By definition of generator set of filter:
:$F = \paren {\map {\operatorname {fininfs} } G}^\succeq$
where
:$\map {\operatorname {fininfs} } G$ denotes the finite infima set of $G$,
:$A^\succeq$ denotes the upper closure of $A$.
By Finite Infima Set of Coarser Subset is Coarser than Finite Infima Set:
:$\map {\operatorname {fininfs} } G$ is coarser than $\map {\operatorname {fininfs} } A$
Thus by Upper Closure of Coarser Subset is Subset of Upper Closure:
:$F \subseteq \paren {\map {\operatorname {fininfs} } A}^\succeq$
By filter in ordered set:
:$F$ is an upper section.
By Set Coarser than Upper Section is Subset:
:$A \subseteq F$
Thus by Finite Infima Set and Upper Closure is Smallest Filter:
:$\paren {\map {\operatorname {fininfs} } A}^\succeq \subseteq F$
Thus by definition of set equality:
:$F = \paren {\map {\operatorname {fininfs} } A}^\succeq$
Hence $A$ is generator set of $F$.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Coarser Topology than Compact Space is Compact

== Theorem ==
Let $S$ be a set.
Let $\tau_1$ and $\tau_2$ be topologies on $S$ such that $\tau_1$ is coarser than $\tau_2$:
:$\tau_1 \subseteq \tau_2$
Let $\struct {S, \tau_2}$ be a compact space.
Then $\struct {S, \tau_1}$ is also compact.
== Proof ==
Let $\struct {S, \tau_2}$ be a compact space as asserted.
Let $I_S: \struct {S, \tau_2} \to \struct {S, \tau_1}$ denote the identity mapping on $S$:
:$\forall x \in S: \map {I_S} x = x$
From Identity Mapping to Coarser Topology is Continuous, $I_S$ is continuous.
We also have the result Identity Mapping is Surjection.
Hence we apply Compactness is Preserved under Continuous Surjection.
The result follows.
== Sources ==
* : $5$: Compact spaces: Exercise $5.10: 4$
Category:Compact Spaces
Code Number for Non-Negative Integer is Primitive Recursive

== Theorem ==
Let $c : \N \to \N$ be defined as:
:$\map c n = m$
where $m$ is the code number for the integer $n : \Z$.
Then $c$ is a primitive recursive function.
== Proof ==
Let $c : \N \to \N$ be defined as:
:$\map c n = \begin{cases}
\map {\operatorname{pred}} {n + n} & n > 0 \\
0 & n = 0
\end{cases}$
That $c$ is primitive recursive follows from:
* Definition by Cases is Primitive Recursive
* Predecessor Function is Primitive Recursive
* Addition is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Equality Relation is Primitive Recursive
Suppose $n = 0$.
Then:
:$m = -2 n = 0$
But:
:$\map c n = 0 = m$
Suppose $n > 0$.
Then:
:$m = 2 n - 1$
But:
:$\map c n = \map {\operatorname{pred}} {n + n} = 2 n - 1$
as $n + n \ge 1$.
Thus, in all cases:
:$\map c n = m$
Category:Primitive Recursive Functions
Code Number for Non-Positive Integer is Primitive Recursive

== Theorem ==
Let $c : \N \to \N$ be defined as:
:$\map c n = m$
where $m$ is the code number for the integer $-n : \Z$.
Then $c$ is a primitive recursive function.
== Proof ==
Let $c : \N \to \N$ be defined as:
:$\map c n = n + n$
which is primitive recursive by:
* Addition is Primitive Recursive
For every $n \in \N$, we have:
:$-n \le 0$
Thus:
:$m = -2 \paren {-n} = 2 n$
Therefore:
:$\map c n = m$
Category:Primitive Recursive Functions
Codomain of Bijection is Domain of Inverse

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Let $f^{-1}: T \to S$ be the inverse of $f$.
Then the domain of $f^{-1}$ equals the codomain of $f$.
== Proof ==
Follows directly from the definition of domain and codomain:
:$\Dom f = S$ and $\Cdm f = T$
:$\Dom {f^{-1} } = T$ and $\Cdm {f^{-1} } = S$
That is:
:$\Dom {f^{-1} } = T = \Cdm f$
== Also see ==
* Domain of Bijection is Codomain of Inverse
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Bijections