nyuuzyou/wikis huggingface dataset:
Finding Center of Circle/Proof 2

== Theorem ==
== Proof ==
:
Draw any chord $AB$ on the circle in question.
Bisect $AB$ at $D$.
Construct $CE$ perpendicular to $AB$ at $D$, where $C$ and $E$ are where this perpendicular meets the circle.
Bisect $CE$ at $F$.
Then $F$ is the center of the circle.
The proof is as follows.
From Perpendicular Bisector of Chord Passes Through Center, $CE$ passes through the center of the circle.
The center must be the point $F$ such that $FE = FC$.
That is, $F$ is the bisector of $CE$.
== Historical Note ==
== Sources ==
* : Book $\text{III}$. Propositions
Category:Finding Center of Circle
Finer Supremum Precedes Supremum

== Theorem ==
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.
Let $X, Y$ be subsets of $S$ such that
:$X$ is finer than $Y$.
Then $\sup X \preceq \sup Y$
where $\sup X$ denotes the supremum of $X$.
== Proof ==
We will prove that
:$\sup Y$ is upper bound for $X$.
Let $x \in X$.
By definition of finer subset:
:$\exists y \in Y: x \preceq y$
By definitions of supremum and upper bound:
:$y \preceq \sup Y$
Thus by definition of transitivity:
:$x \preceq \sup Y$
Hence by definition of supremum:
:$\sup X \preceq \sup Y$
== Sources ==
* 
Category:Complete Lattices
Finished Branch Lemma/Corollary

== Corollary to Finished Branch Lemma ==
Let $\Gamma$ be a finished branch of a propositional tableau $\struct {T, \mathbf H, \Phi}$.
Then $\Phi \sqbrk \Gamma$, the image of $\Gamma$ under $\Phi$, is satisfiable for boolean interpretations.
== Proof ==
Category:Finished Branch Lemma
Finished Propositional Tableau has Finished Branch or is Confutation

== Theorem ==
Let $\struct {T, \mathbf H, \Phi}$ be a finished propositional tableau.
Then one of the following holds:
:$T$ has a finished branch
:$T$ is a tableau confutation.
== Proof ==
Suppose $T$ has no finished branch.
Then since $T$ is finished, every branch of $T$ is contradictory.
Hence $T$ is a tableau confutation.
== Sources ==
* : $\S 1.10$: Completeness
* : $\S 1.11$: Compactness: Corollary $1.11.4$
Category:Propositional Tableaux
Finished Set Lemma

== Lemma ==
Let $\Delta$ be a finished set of WFFs of propositional logic.
Then any model of the set of basic WFFs in $\Delta$ is a model of $\Delta$.
=== Corollary ===
== Proof ==
Let $\MM$ be a model of the set of basic WFFs in $\Delta$.
We need to show that $\MM \models \Delta$.
That is, that $\MM \models \mathbf C$ for each $\mathbf C \in \Delta$.
Now, let $\map R n$ be a propositional function on the set of natural numbers $\N$ such that:
:$\map R n = T$ :
::for every WFF $\mathbf C$, if $\mathbf C \in \Delta$ and $\mathbf C$ has length at most $n$, then $\MM \models \mathbf C$.
It is clear that $\map R 0, \map R 1, \map R 2$ are true because every WFF length 2 or less is basic, and $\MM$ models every basic WFF in $\Delta$.
So, assume $\map R k$ is true for some $k \in \N$.
Suppose $\mathbf C$ has length at most $k + 1$ and belongs to $\Delta$.
By examining each of the cases in the definition of finished set, we see that since $\MM$ models every WFF in $\Delta$ of length at most $k$, then $\MM$ models $\mathbf C$.
Thus $\map R {k + 1}$ is true.
Thus by strong induction, $\map R n$ is true for all $n \in \N$.
Hence the result.
== Sources ==
* : $\S 1.9$: Finished Sets: Lemma $1.9.1$
Category:Finished Set Lemma
Category:Propositional Tableaux
Category:Boolean Interpretations
Finished Set Lemma/Corollary

== Corollary to Finished Set Lemma ==
Let $\Delta$ be a finished set of WFFs of propositional logic.
Then $\Delta$ has a model.
== Proof ==
Note that the set of basic WFFs in $\Delta$ has at least one model.
Let $v$ be the boolean interpretation defined as follows:
$\quad \map v p = \begin{cases}
\T & : p \in \Delta \\
\F & : p \notin \Delta
\end{cases}$
Because $\Delta$ is finished, it is not contradictory, and hence $\map v p = \F$ if $\neg p \in \Delta$.
Thus $v$ is a model of the basic WFFs of $\Delta$.
The result follows by the Finished Set Lemma.
== Sources ==
* : $\S 1.9$: Finished Sets: Lemma $1.9.1$
Category:Finished Set Lemma
Category:Boolean Interpretations
Finite-Dimensional Integral Domain over Field is Field

== Theorem ==
Let $k$ be a field.
Let $R$ be a $k$-algebra of finite dimension which is an integral domain.
Then $R$ is a field.
== Proof ==
We will prove this using the Rank - Nullity Theorem applied to $A$ viewing it as a finite dimensional vector space over $K$. Pick a non-zero element $y \in A$ and consider the $K$ - linear transformation $$\begin{eqnarray*} T&\colon&A \longrightarrow A \\
&& x\mapsto yx.\end{eqnarray*} $$
Now the kernel of this linear transformation is trivial because $A$ is an integral domain, and hence by rank-nullity we have that 
$$\begin{eqnarray*} \dim_K A &=& \dim \ker T + \dim \operatorname{im} T \\
&=& 0 + \dim \operatorname{im} T\\
&=& \dim \operatorname{im} T \end{eqnarray*}$$
from which it follows that $T$ is surjective. In particular there exists $z \in A$ such that $zy = 1$, so that since $y$ was arbitrary we have shown that every element in $A$ is invertible, i.e. $A$ is a field.
**Edit:** I should say it is important to include the hypothesis that $A$ is an integral domain otherwise the result is not necessarily true. Consider the $\Bbb{C}$ - algebra 
$$\Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C} $$
as a finite dimensional vector space over $\Bbb{C}$. This is not an integral domain because $(1,0,0) \cdot (0,1,0) = (0,0,0)$. Now we also have 
$$\dim_{\Bbb{C}} \big( \Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C} \big) = 3 < \infty$$
but $\Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C}$ cannot be a field because elements like $(1,0,0)$ don't have an inverse.
If we also drop the hypothesis that $\dim_K A < \infty$ the result is not true as well. For example consider the polynomial ring $\Bbb{C}[T]$ in the indeterminate $T$. This is an integral domain because $\Bbb{C}$ is. Then if we view this as a vector space over $\Bbb{C}$ it is infinite dimensional, but it is already a well known fact that $\Bbb{C}[T]$ is not a field.
== Also see ==
* Finite Integral Domain is Galois Field
* Ring Element is Unit iff Unit in Integral Extension
== Sources ==
Category:Linear Algebra
Category:Integral Domains
Category:Field Extensions
Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Closed

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Hausdorff topological vector space over $\GF$.
Let $n \in \N$. 
Let $Y$ be a subspace of $X$ with dimension $n$. 
Then $Y$ is closed.
== Proof ==
Let $f : \GF^n \to Y$ be a vector space isomorphism.
From Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism, $f$ is a homeomorphism. 
Let $B$ be the unit ball in $\GF^n$. 
Let $D$ be the closed unit ball in $\GF^n$. 
Let $S$ be the unit sphere in $\GF^n$ and: 
:$K = f \sqbrk S$ 
In Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism, it was determined that there exists a balanced open neighborhood $V$ of $\mathbf 0_X$ such that: 
:$V \cap K = \O$
Now let $\map \cl Y$ be the closure of $Y$. 
From Set is Closed iff Equals Topological Closure, we just need to show that $Y^- \subseteq Y$. 
From Topological Vector Space as Union of Dilations of Open Neighborhood of Origin, there exists $t > 0$ such that: 
:$p \in t V$
It follows that: 
:$p \in \map \cl {Y \cap t V}$
In Isomorphism from Cartesian Space to Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Homeomorphism, it was determined that: 
:$f^{-1} \sqbrk V \subseteq B$
so that: 
:$Y \cap \paren {t V} \subseteq t f \sqbrk B$ 
by Image of Preimage under Mapping.
From Image of Dilation of Set under Linear Transformation is Dilation of Image, we have: 
:$Y \cap \paren {t V} \subseteq f \sqbrk {t B}$
That is: 
:$Y \cap \paren {t V} \subseteq f \sqbrk {t D}$
Since $D$ is compact, we have: 
:$t D$ is compact
from Dilation of Compact Set in Topological Vector Space is Compact.
From Continuous Image of Compact Space is Compact, $f \sqbrk {t D}$ is compact.
Since $X$ is a Hausdorff topological vector space, $X$ is in particular Hausdorff.
From Compact Subspace of Hausdorff Space is Closed, $f \sqbrk {t D}$ is therefore closed, and we obtain: 
:$\map \cl {Y \cap \paren {t V} } \subseteq f \sqbrk {t D}$ 
So: 
:$p \in f \sqbrk {t D}$
giving $p \in Y$. 
So we have $Y^- = Y$ as desired.
== Sources ==
* : $1.21$: Theorem
Category:Hausdorff Topological Vector Spaces
Finite Abelian Group is Solvable

== Theorem ==
Let $G$ be a finite abelian group.
Then $G$ is solvable.
== Proof ==
Proof by strong induction on the order of $G$:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:All abelian groups of order $n$ and below are solvable.
=== Basis for the Induction ===
$\map P 1$ is true, as the trivial group is trivially solvable.
From Prime Group is Simple‎, both $\map P 2$ and $\map P 3$ hold.
This is the basis for the induction.
=== Induction Hypothesis ===
It remains to be shown that, if $\map P j$ holds, for all $j$ such that $1 \le j \le k$, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ holds.
So this is our induction hypothesis:
:All abelian groups of order $k$ and below are solvable.
Then we need to show:
:All abelian groups of order $k+1$ and below are solvable.
=== Induction Step ===
This is our induction step:
Let $G$ be an abelian group such that:
:$\order G = k + 1$
where $\order G$ denotes the order of $G$.
Then from Positive Integer Greater than 1 has Prime Divisor, there exists some prime $p$ which divides $k + 1$.
From Cauchy's Lemma (Group Theory), $G$ has an element of order $p$.
Since $G$ is abelian, this element generates a subgroup of $G$ which is normal by Subgroup of Abelian Group is Normal.
If $p = n$ and $H = G$ the proof is finished, from Prime Power Group is Solvable.
If $p < n$, then $H$ and $G / H$ (which has order $n / p$) are both solvable from the induction hypothesis.
It follows from Group is Solvable iff Normal Subgroup and Quotient are Solvable‎ that $G$ is solvable.
The result follows by the strong principle of mathematical induction.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75$. Solvable Groups
Category:Abelian Groups
Category:Solvable Groups
Finite Boolean Lattice is a Locale

== Theorem ==
Let $\struct{B, \vee, \wedge, \neg, \preceq}$ be a finite Boolean lattice.
Then:
:$\struct{B, \vee, \wedge, \preceq}$ is a locale.
== Proof ==
By definition of a lattice:
:$B$ admits all finite non-empty suprema and finite non-empty infima.
By definition of a Boolean lattice:
:$B$ has a greatest element and a smallest element.
From Infimum of Empty Set is Greatest Element and Supremum of Empty Set is Smallest Element:
:$B$ admits empty supremum and empty infimum.
Hence
:$B$ admits all supremum and infimum.
By definition of complete lattice:
:$B$ is complete
From Complete Boolean Lattice is a Locale:
:$B$ is a locale.
Category:Boolean Lattices
Category:Locales
Finite Cartesian Product of Non-Empty Sets is Non-Empty

== Theorem ==
Let $S_1, S_2, \ldots, S_n$ be non-empty sets.
Then their cartesian product $S_1 \times S_2 \times \cdots \times S_n$ is non-empty.
== Proof ==
We use mathematical induction.
The base case $n = 2$ is proved in Kuratowski Formalization of Ordered Pair, and the induction step follows directly from the definition of an ordered tuple.
Category:Cartesian Product
Finite Chain is Order-Isomorphic to Finite Ordinal

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a finite chain in $S$.
Then for some finite ordinal $\mathbf n$:
:$\struct {C, {\preceq \restriction_C} }$ is order-isomorphic to $\mathbf n$.
That is:
:$\struct {C, {\preceq \restriction_C} }$ is order-isomorphic to $\N_n$
where $\N_n$ is the initial segment of $\N$ determined by $n$:
:$\N_n = \set {k \in \N: k < n} = \set {0, 1, \ldots, n - 1}$
== Proof ==
By the definition of finite set:
:there exists an $n \in \N$ such that there exists a bijection $f: C \to \N_n$.
This $n$ is unique by Equality of Natural Numbers and Set Equivalence behaves like Equivalence Relation.
Define a mapping $g: \N_n \to C$ recursively as:
:$\map g 0 = \min C$
:$\map g {k + 1} = \map \min {C \setminus \map g {N_k} }$
== Sources ==
* : $\S \text I.3$: Theorem $5$
Category:Total Orderings
Category:Ordinals
Category:Chains (Order Theory)
Finite Character for Sets of Mappings

== Theorem ==
Let $S$ and $T$ be sets.
Let $\FF$ be a set of mappings from subsets of $S$ to $T$.
That is, let $\FF$ be a set of partial mappings from $S$ to $T$.
Then the following are equivalent:
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\FF$ have finite character in the sense of Definition:Finite Character/Mappings.
That is, suppose that for each partial mapping $f \subseteq S \times T$:
:$f \in \FF$ for each finite subset $K$ of the domain of $f$, the restriction of $f$ to $K$ is in $\FF$.
Let $q \subseteq S \times T$.
First suppose that $q \in \FF$.
Let $r$ be a finite subset of $q$.
Then $r$ is a partial mapping from its domain into $T$.
Since $r$ is finite, its domain, $\Dom r$, is a finite subset of the domain of $q$.
Since $\FF$ is of finite character as a set of mappings, $q \restriction \Dom r$ is in $\FF$.
But $q \restriction \Dom r = r$.
Therefore $r \in \FF$.
Suppose instead that for each finite subset $r$ of $q$, $r \in \FF$.
Then for each finite subset $r$ of $q$, $r$ is a partial mapping from $S$ to $T$.
Let $x \in \Dom q$.
Let $y_1, y_2 \in T$.
Suppose that $\tuple {x, y_1}, \tuple {x, y_2} \in q$.
Then $\set {\tuple {x, y_1}, \tuple {x, y_2} }$ is a finite subset of $q$.
Therefore it must be a partial mapping.
Thus $y_1 = y_2$.
As this holds for all such $x, y_1, y_2$, $q$ is a partial mapping.
Let $K$ be a finite subset of $\Dom q$.
Then the restriction of $q$ to $K$ is a finite subset of $q$.
Thus $q \restriction K \in \FF$.
As this holds for all finite subsets $K$ of $\Dom q$, $q \in \FF$.
So we see that $\FF$ has finite character in the sense of Definition:Finite Character.
=== $(2)$ implies $(1)$ ===
Let $\FF$ have finite character in the sense of Definition:Finite Character.
Category:Set Theory
Finite Class is Set

== Theorem ==
Let $A$ be a finite class.
Then $A$ is a set.
== Proof ==
Let it be assumed that all classes are subclasses of a basic universe $V$.
The proof proceeds by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:If $A$ is a finite class with $n$ elements, then $A$ is a set.
The gives that the empty class $\O$ is a set.
From Empty Set is Unique, $\O$ is the only set (and hence class) with $0$ elements.
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
Let $A$ be a singleton class.
Thus by definition it has $1$ element, which we will call $x$.
 $x$ is a class containing $1$ or more elements, one of which we may call $y$.
We have that $A$ is a subclass of $V$.
As $V$ is a basic universe, the holds.
Hence:
:$y \in x \land x \in A \implies y \in A$
But then $x \in A$ and $y \in A$ where $x \ne y$.
This contradicts our assertion that $A$ is a singleton class.
Hence:
:$x = \O$
It has been established that $\O$ is a set.
Hence $A$ is a class whose $1$ element is a set.
From Singleton Class of Set is Set, it follows that $A$ is a set.
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:If $A$ is a finite class with $k$ elements, then $A$ is a set.
from which it is to be shown that:
:If $A$ is a finite class with $k + 1$ elements, then $A$ is a set.
=== Induction Step ===
This is the induction step:
Let $A$ have $k + 1$ elements.
Let each of those elements be assigned a label:
:$a_1, a_2, \ldots, a_{k + 1}$
according to a bijection $\phi: A \to k$:
:$\map \phi {a_k} = k - 1$
This bijection $\phi$ is guaranteed to exist by definition of finite class.
Consider $A$ as the union of the classes:
:$\set {a_1, a_2, \ldots, a_k} \cup \set {a_{k + 1} }$
By the basis for the induction:
:$\set {a_{k + 1} }$ is a set.
By the induction hypothesis:
:$\set {a_1, a_2, \ldots, a_k}$ is a set.
By the :
:$\set {a_1, a_2, \ldots, a_k} \cup \set {a_{k + 1} }$ is a set.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore, for all $\forall n \in \N$:
:If $A$ is a finite class with $n$ elements, then $A$ is a set.
That is, all finite classes are sets.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets: Exercise $6.1 \ (1)$
Category:Finite Classes
Category:Finite Sets
Finite Complement Space is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a connected space.
== Proof ==
We have that a Finite Complement Space is Irreducible.
The result follows from Irreducible Space is Connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $9$
Category:Connected Spaces
Category:Finite Complement Topology
Finite Complement Space is Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is irreducible.
== Proof ==
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
By definition, $S$ is infinite.
By definition, $\relcomp S {U_1}$ and $\relcomp S {U_2}$ are finite.
From Complement of Finite Subset of Infinite Set is Infinite, it follows that $U_1$ and $U_2$ are both infinite.
From Infinite Subset of Finite Complement Space Intersects Open Sets:
:$U_1 \cap U_2\ne \O$
That is, $U_1$ and $U_2$ intersect each other.
As $U_1$ and $U_2$ are arbitrary, the result follows by definition of irreducible space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $6$
Category:Irreducible Spaces
Category:Finite Complement Topology
Finite Complement Space is Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a locally connected space.
== Proof ==
We have that a Finite Complement Space is Irreducible.
The result follows from Irreducible Space is Locally Connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $9$
Category:Locally Connected Spaces
Category:Finite Complement Topology
Finite Complement Space is T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on any set $S$ which contains at least two points.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
Let $x, y \in S$.
Then:
:$S \setminus \set x$ is an open set of $T$ containing $y$ but not $x$.
:$S \setminus \set y$ is an open set of $T$ containing $x$ but not $y$.
Hence the result from definition of $T_1$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $5$
Category:T1 Spaces
Category:Finite Complement Topology
Finite Complement Space is not T2

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is not a $T_2$ (Hausdorff) space.
== Proof ==
We have:
:Finite Complement Space is Irreducible
:Irreducible Hausdorff Space is Singleton
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 1$
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $6$
Category:Hausdorff Spaces
Category:Finite Complement Topology
Finite Complement Space is not T3, T4 or T5

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is not a $T_3$ space, $T_4$ space or $T_5$ space.
== Proof ==
We have that a Finite Complement Space is a $T_1$ space.
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
We then have that a Finite Complement Space is not $T_2$.
From Regular Space is $T_2$ Space, $T$ is not a regular space.
By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space.
But $T$ is a $T_0$ space and not a regular space.
So it follows that $T$ cannot be a $T_3$ space.
Next we have that a Normal Space is a $T_3$ Space.
But as $T$ is not a $T_3$ space, $T$ cannot be a normal space.
By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space.
But $T$ is a $T_1$ space and not a normal space.
So it follows that $T$ cannot be a $T_4$ space.
Finally we have that a $T_5$ Space is a $T_4$ Space.
But as $T$ is not a $T_4$ space, $T$ cannot be a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $6$
Category:T3 Spaces
Category:T4 Spaces
Category:T5 Spaces
Category:Finite Complement Topology
Finite Complement Topology is Minimal T1 Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement space.
Let $\tau'$ be a topology on $S$ such that $T' = \struct {S, \tau'}$ is a $T_1$ (Fréchet) space.
Then $\tau$ is comparable with $\tau'$ such that $\tau$ is coarser than $\tau'$.
That is, of all the topologies on $S$ fulfilling the $T_1$ separation axiom, the finite complement space is the smallest.
Thus the finite complement topology is known as the minimal $T_1$ topology on any given set.
== Proof ==
Let $T = \struct {S, \tau}$ be the finite complement space on $S$.
Let $U \in \tau$ be any open set of $T$.
Let $H = \relcomp S U$ be the complement of $U$ relative to $S$.
By definition of finite complement topology, $H \subseteq S$ is a finite subset of $S$.
Let $T' = \struct {S, \tau'}$ be any arbitrary $T_1$ (Fréchet) space on $S$.
From Equivalence of Definitions of $T_1$ Space:
:$\forall x \in S: \set x$ is closed in $T'$.
We can write $H$ as:
:$\ds H = \bigcup_{x \mathop \in H} \set x$
As $H$ is finite, it follows that $H$ is a finite union of closed sets of $T'$.
By Topology Defined by Closed Sets, $H$ is therefore closed in $T'$.
By Relative Complement of Relative Complement we have that $U = \relcomp S H$.
So, by definition of closed set, we have that $U$ is open in $T'$.
So we have shown that for any arbitrary $U \in \tau$ it follows that $U \in \tau'$.
So $\tau \subseteq \tau'$ and so by definition $\tau$ is coarser than $\tau'$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $8$
Category:T1 Spaces
Category:Finite Complement Topology
Finite Complement Topology is Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is a separable space.
== Proof ==
Let $H$ be a countably infinite subset of $S$.
From Closure of Infinite Subset of Finite Complement Space, the closure of $H$ is $S$.
So by definition $H$ is everywhere dense in $T$.
Hence the result by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $1$
Category:Separable Spaces
Category:Finite Complement Topology
Finite Complement Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement space.
Then $\tau$ is a topology on $T$.
== Proof ==
By definition, we have that $\O \in \tau$.
We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite.
Let $A, B \in \tau$.
Let $H = A \cap B$.
Then:
But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both finite.
Hence their union is also finite.
Thus $\relcomp S H$ is finite.
So $H = A \cap B \in \tau$ as its complement is finite.
Let $\UU \subseteq \tau$.
From De Morgan's laws: Complement of Union:
:$\ds \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$
But as:
:$\forall U \in \UU: \relcomp S U \in \tau$
each of the $\relcomp S U$ is finite.
Hence so is their intersection.
So $\ds \relcomp S {\bigcup \UU}$ is finite.
So by definition:
:$\ds \bigcup \UU \in \tau$
$\tau$ obeys all of the open set axioms, so it is a topology on $T$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Example $6$
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Example $3.1.7$
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology
Category:Finite Complement Topology