nyuuzyou/wikis huggingface dataset:
Finite Group has Composition Series/Proof 1

== Theorem ==
== Proof ==
Let $G$ be a finite group whose identity is $e$.
Either $G$ has a proper non-trivial normal subgroup or it does not.
If not, then:
:$\set e \lhd G$
is the composition series for $G$.
Otherwise, $G$ has one or more proper non-trivial normal subgroup.
Of these, one or more will have a maximum order.
Select one of these and call it $G_1$.
Again, either $G_1$ has a proper non-trivial normal subgroup or it does not.
If not, then:
:$\set e \lhd G_1 \lhd G$
is a composition series for $G$.
By the Jordan-Hölder Theorem, there can be no other composition series which is longer.
As $G_1$ is a proper subgroup of $G$:
:$\order {G_1} < \order G$
where $\order G$ denotes the order of $G$.
Again, if $G_1$ has one or more proper non-trivial normal subgroup, one or more will have a maximum order.
Select one of these and call it $G_2$.
Thus we form a normal series:
:$\set e \lhd G_2 \lhd G_1 \lhd G$
The process can be repeated, and at each stage a normal subgroup is added to the series of a smaller 
order than the previous one.
This process cannot continue infinitely.
Eventually a $G_n$ will be encountered which has no proper non-trivial normal subgroup.
Thus a composition series:
:$\set e \lhd G_n \cdots \lhd G_2 \lhd G_1 \lhd G$
will be the result.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.8$
Category:Finite Group has Composition Series
Finite Group has Composition Series/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a finite group whose identity is $e_G$.
We shall use induction on $|G|$. 
If $G$ is trivial ($|G|=1$), then its composition series is 
:$G=\{e_G\}$. 
Suppose $G$ has a composition series if $|G|<n$, then it suffices to construct a composition series for $G$ with order $n$. 
If $G$ is simple, then its composition series is 
:$G\supsetneq \{e_G\}$. 
Otherwise, $G$ has one or more proper non-trivial normal subgroup. Let $S$ denote the set of all non-trivial normal proper subgroup of $G$. 
Notice that $S$ is non-empty, ordered by inclusion, and $G$ is an upper bound of every chain in $S$, then by Zorn's lemma, $S$ has at least one maximal element, denoted $H$. 
Since $H\in S$, $H$ is a non-trivial normal proper subgroup of $G$, in particular, $|H|<|G|$, so by the induction hypothesis, $H$ has a composition series. 
Now we have 
:$G\supsetneq H\supsetneq H_1\supsetneq \cdots \supsetneq H_n\supsetneq \{e_G\}$. 
In order to show that it is indeed a composition series of $G$, it suffices to check that $G/H$ is simple, which is a direct consequence of the maximal property of $H$, so we are done. 
Category:Finite Group has Composition Series
Finite Group is p-Group iff Order is Power of p

== Theorem ==
Let $p$ be a prime number.
Let $G$ be a finite group.
Then $G$ is a $p$-group the order of $G$ is a power of $p$.
== Proof ==
=== Necessary Condition ===
Let $G$ be a finite group whose order is $p^n$ for some $n \in \Z_{>0}$.
Let $g \in G$.
From Order of Element Divides Order of Finite Group, the order of $g$ is a divisor of $p^n$.
That is, $x$ is a $p$-element by definition.
As $x$ is arbitrary, it follows that all elements of $G$ are $p$-elements.
Thus $G$ is a $p$-group.
=== Sufficient Condition ===
Let $G$ be a finite $p$-group.
By definition, every element of $G$ is a $p$-element.
From Order of Finite $p$-Group is Power of $p$:
:$G$ is a finite group whose order is $p^n$ for some $n \in \Z_{\ge 0}$.
Hence the result.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Definition $11.8$
Category:P-Groups
Category:Finite Groups
Finite Group whose Subsets form Chain is Cyclic P-Group

== Theorem ==
Let $G$ be a group.
Let $G$ be such that its subgroups form a chain.
Then $G$ is a cyclic $p$-group.
== Proof ==
Suppose $G$ is not a $p$-group.
Then there exist two distinct primes $p_1, p_2$.
By Cauchy's Group Theorem, there exist subgroups $H, K$ such that:
:$\order H = p_1$
:$\order K = p_2$
That is:
:$H = \gen a$
:$k = \gen b$
for some $a, b \in G: a \ne b$, where:
:$\order a = p_1$
:$\order b = p_2$
and so both $H \nsubseteq K$ and $K \nsubseteq H$
Thus, by definition, the subgroups of $G$ do not form a chain.
It follows by the Rule of Transposition that $G$ is a $p$-group.
It remains to be shown that $G$ is cyclic.
Suppose to the contrary that $G$ is not cyclic.
Then it is not the case that $G$ is generated by a single element of $G$.
Thus there exist $a, b \in G, a \ne b$ such that $H = \gen a$ and $K = \gen b$ are subgroups of $G$.
But as $a \in H, b \in K, a \notin K, b \notin H$ it follows that $H \nsubseteq K$ and $K \nsubseteq H$.
It follows by the Rule of Transposition that $G$ is cyclic.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 43 \beta$
Category:Cyclic Groups
Category:P-Groups
Finite Group with 2 Conjugacy Classes has 2 Elements

== Theorem ==
Let $G$ be a finite group.
Let $G$ have exactly $2$ conjugacy classes.
Then $G$ has exactly $2$ elements.
== Proof ==
Let $G$ be of order $n$.
Let $G$ have exactly $2$ conjugacy classes.
Let $x \in G$ such that $x \ne e$.
Let $\conjclass x$ denote the conjugacy class of $x$.
From Identity of Group is in Singleton Conjugacy Class:
$\conjclass e = \set e$
where $\conjclass e$ denotes the conjugacy class of $e$
The other elements of $G$ are in $\conjclass x$.
Thus:
:$\card {\conjclass x} = n - 1$
where $\card {\, \cdot \,}$ denotes the cardinality of a set.
We have that Size of Conjugacy Class is Index of Normalizer.
From Lagrange's Theorem, the index of any subgroup of $G$ is a divisor of $G$.
Thus:
:$\paren {n - 1} \divides n$
where $\divides$ denotes divisibility.
It follows that $n = 2$.
Hence the result by definition of order of group.
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Exercise $5$
Category:Conjugacy Classes
Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product

== Theorem ==
Let $G$ be a finite group whose order is $n$ and whose identity element is $e$.
Let $G$ be such that it has exactly $1$ Sylow $p$-subgroup for each prime divisor of $n$.
Then $G$ is isomorphic to the internal direct product of all its Sylow $p$-subgroups.
== Proof ==
If each of the Sylow $p$-subgroups are unique, they are all normal.
As the order of each one is coprime to each of the others, their intersection is $\set e$.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 59 \theta$
Category:Sylow p-Subgroups
Finite Hausdorff Measure Implies Zero Higher Dimensional Measure

== Theorem ==
Let $n \in \N_{>0}$.
Let $F \subseteq \R^n$ be a subset of the real Euclidean space.
Let $s \in \R_{\ge 0}$.
Let $\map {\HH^s} \cdot$ denote the $s$-dimensional Hausdorff measure.
Then:
:$\map {\HH^s} F < +\infty \implies \forall t \in \R_{>s} : \map {\HH^t} F = 0$
== Proof ==
For each $\delta$-cover $\sequence {U_i}$ of $F$:
Thus:
:$\map {\HH^t_\delta} F \le \delta^{t - s} \map {\HH^s_\delta} F$
Therefore:
Category:Measure Theory
Category:Fractals
Finite Infima Set and Upper Closure is Filter

== Theorem ==
Let $P = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.
where
:$\map {\operatorname {fininfs} } X$ denotes the finite infima set of $X$,
:$X^\succeq$ denotes the upper closure of $X$.
== Proof ==
By Finite Infima Set and Upper Closure is Smallest Filter:
:$X \subseteq \map {\operatorname {fininfs} } X^\succeq$
By definition of non-empty set:
:$\map {\operatorname {fininfs} } X^\succeq$ is a non-empty set.
We will prove that
:$\map {\operatorname {fininfs} } X$ is filtered.
Let $x, y \in \map {\operatorname {fininfs} } X$
By definition of finite infima set:
:$\exists A \in \map {\mathit {Fin} } X: x = \inf A \land A$ admits an infimum
and
:$\exists B \in \map {\mathit {Fin} } X: y = \inf B \land B$ admits an infimum
where $\map {\mathit {Fin} } X$ denotes the set of all finite subsets of $X$.
Define $C = A \cup B$.
By Union of Subsets is Subset:
:$C \subseteq X$
By Union of Finite Sets is Finite:
:$C$ is finite.
Then
:$C \in \map {\mathit {Fin} } X$
By Existence of Non-Empty Finite Infima in Meet Semilattice:
:$C \ne \O \implies C$ admits an infimum.
By Union is Empty iff Sets are Empty:
:$C = \O \implies A = \O$
So
:$C$ admits an infimum.
By definition of finite infima set:
:$\inf C \in \map {\operatorname {fininfs} } X$
By Set is Subset of Union:
:$A \subseteq C$ and $B \subseteq C$
Thus by Infimum of Subset:
:$\inf C \preceq x$ and $\inf C \preceq y$
Hence $\map {\operatorname {fininfs} } X$ is filtered.
By Filtered iff Upper Closure Filtered:
:$\map {\operatorname {fininfs} } X^\succeq$ is filtered.
By Upper Closure is Upper Section:
:$\map {\operatorname {fininfs} } X^\succeq$ is upper.
Hence $\map {\operatorname {fininfs} } X^\succeq$ is filter in $P$.
== Sources ==
* 
* 
* 
Category:Order Theory
Finite Infima Set and Upper Closure is Smallest Filter

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $X$ be a non-empty subset of $S$.
Then
:$X \subseteq \map {\operatorname {fininfs} } X^\succeq$ and
:for every a filter $F$ in $L$: $\paren {X \subseteq F \implies \map {\operatorname {fininfs} } X^\succeq \subseteq F}$
where
:$\map {\operatorname {fininfs} } X$ denotes the finite infima set of $X$,
:$X^\succeq$ denotes the upper closure of $X$.
== Proof ==
By Set is Subset of Finite Infima Set:
:$X \subseteq \map {\operatorname {fininfs} } X$
By Upper Closure of Subset is Subset of Upper Closure:
:$X^\succeq \subseteq \map {\operatorname {fininfs} } X^\succeq$
By Set is Subset of Upper Closure:
:$X \subseteq X^\succeq$
Thus by Subset Relation is Transitive:
:$X \subseteq \map {\operatorname {fininfs} } X^\succeq$
Let $F$ be a filter in $L$ such that
:$X \subseteq F$
Let $x \in \map {\operatorname {fininfs} } X^\succeq$
By definition of upper closure of subset:
:$\exists y \in \map {\operatorname{fininfs} } X: y \preceq x$
By definition of finite infima set:
:$\exists A \in \map {\operatorname {Fin} } X: y = \inf A \land A$ admits an infimum
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $S$.
By Subset Relation is Transitive:
:$A \subseteq F$
By Filtered in Meet Semilattice with Finite Infima:
:$A \ne \O \implies \inf A \in F$
By Infimum of Empty Set is Greatest Element and Top in Filter:
:$A = \O \implies \inf A = \top \in F$
where $\top$ denotes the greatest element of $S$.
So
:$y \in F$
Thus by definition of upper section:
:$x \in F$
== Sources ==
* 
Category:Order Theory
Finite Infima Set of Coarser Subset is Coarser than Finite Infima Set

== Theorem ==
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $A, B$ be subsets of $S$ such that
:$A$ is coarser than $B$.
Then $\map {\operatorname{fininfs} } A$ is coarser than $\map {\operatorname{fininfs} } B$
where $\map {\operatorname{fininfs} } B$ denotes the finite infima set of $B$.
== Proof ==
Let $x \in \map {\operatorname{fininfs} } A$
By definition of finite infima set:
:$\exists Y \in \map {\operatorname {Fin} } A: x = \inf Y$ and $Y$ admits an infimum,
where $\map {\operatorname {Fin} } A$ denotes the set of all finite subsets of $A$.
By definition of coarser subset:
:$\forall y \in Y: \exists z \in B: z \preceq y$
By Axiom of Choice:
:$\exists f: Y \to B: \forall y \in Y: \map f y \preceq y$
We will prove that
:$f \sqbrk Y$ admits an infimum.
In the case when $Y = \O$:
By Image of Empty Set is Empty Set: Corollary 1:
:$f \sqbrk Y = \O$
Thus
:$f \sqbrk Y$ admits an infimum.
In the case when $Y \ne \O$
By definition of non-empty set:
:$\exists y: y \in Y$
By definition of image of set:
:$\map f y \in f \sqbrk Y$
By Image of Mapping from Finite Set is Finite:
:$f \sqbrk Y$ is a finite set.
Thus by Existence of Non-Empty Finite Infima in Meet Semilattice:
:$f \sqbrk Y$ admits an infimum.
We will prove that:
:$\inf f \sqbrk Y$ is lower bound for $Y$.
Let $y \in Y$.
By definition of $f$:
:$\map f y \preceq y$
By definition of image of set:
:$\map f y \in f \sqbrk Y$
By definitions of infimum and lower bound:
:$\inf f \sqbrk Y \preceq \map f y$
Thus by definition of transitivity:
:$\inf f \sqbrk Y \preceq y$
By definition of infimum:
:$\inf f \sqbrk Y \preceq x$
By definition of finite infima set:
:$\inf f \sqbrk Y \in \map {\operatorname{fininfs} } B$
Thus
:$\exists y \in \map {\operatorname{fininfs} } B: y \preceq x$
== Sources ==
* 
* 
Category:Order Theory
Finite Integral Domain cannot be Ordered

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain whose unity is $1_D$ and whose order is finite.
Then $D$ cannot be an ordered integral domain.
== Proof ==
Let $\struct {D, +, \times}$ be an integral domain whose order is $n$.
 there exists a (strict) positivity property $P$ on $\struct {D, +, \times}$.
From Unity of Ordered Integral Domain is Strictly Positive, we have that $\map P {1_D}$.
Let $\struct {D, +}$ be the additive group of $\struct {D, +, \times}$.
From Order of Element Divides Order of Finite Group, the order of $1_D$ is also finite.
So:
:$\exists m \in \N: m \cdot 1_D = 0_D$
where $0_D$ is the zero of $D$.
But then from the General Positivity Rule in Ordered Integral Domain that would mean $\map P {0_D}$ which contradicts the trichotomy law of the positivity property.
Further, note that from Inverse Element is Power of Order Less 1 we have that $-1_D = \paren {m - 1} \cdot 1_D$.
But from the General Positivity Rule in Ordered Integral Domain $\map P {\paren {m - 1} \cdot 1_D}$ and so $\map P {-1_D}$ which is an even more dramatic way of showing that the trichotomy law of the positivity property is contradicted.
Hence an integral domain whose order is finite can not be ordered.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $11$
Category:Ordered Integral Domains
Finite Integral Domain is Galois Field

== Theorem ==
A finite integral domain is a Galois field.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
=== Generalization ===
* Finite-Dimensional Integral Domain over Field is Field
Category:Galois Fields
Category:Integral Domains
Category:Finite Integral Domain is Galois Field
Finite Integral Domain is Galois Field/Proof 1

== Theorem ==
== Proof ==
Let $R$ be a finite integral domain whose unity is $1$ and whose zero is $0$.
Let $a \in R$ such that $a \ne 0$.
We wish to show that $a$ has a product inverse in $R$. So consider the map $f: R \to R$ defined by $f: x \mapsto a x$.
We first show that the kernel of $f$ is trivial.
Consider that:
:$\map \ker f = \set {x \in R: \map f x = 0} = \set {x \in R: a x = 0}$
Since $R$ is an integral domain, it has no proper zero divisors and thus $a x = 0$ means that $a = 0$ or $x = 0$.
Since, by definition, $a \ne 0$, then it must be true that $x = 0$.
Therefore, $\map \ker f = \set 0$ and so $f$ is injective.
Next, the Pigeonhole Principle gives us that an injective mapping from a finite set onto itself is surjective.
Since $R$ is finite, the mapping $f$ is surjective.
Finally, since $f$ is surjective and $1 \in R$, we have:
:$\exists x \in R: \map f x = a x = 1$
So this $x$ is the product inverse of $a$ and the proof is complete.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.4$
* : $\S 1$: Exercise $7$
Category:Finite Integral Domain is Galois Field
Finite Integral Domain is Galois Field/Proof 2

== Theorem ==
== Proof ==
Let $R$ be a finite integral domain with unity $1$ and zero $0$.
Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the zero.
As $R$ is finite, we may enumerate the elements of $R$ as:
:$x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$
Let $R^*$.
Consider the elements:
:$x_k x_1, x_k x_2, \ldots, x_k x_n$ 
which are members of $R$ by closure of the ring product.
Since $R$ is an integral domain, by definition it has no proper zero divisors.
Therefore all of $x_k x_1, x_k x_2, \ldots, x_k x_n$ are elements of $R^*$.
Recall:
:$x_k x_p = x_k x_q \implies x_p = x_q$ 
by the Cancellation Law for Ring Product of Integral Domain.
So the list of products consists of $n$ distinct elements and is therefore equal to the whole $R^*$.
Since $1 \in R^*$, we have that $x_k x_{k^*} = 1$ for some $k^*$.
Therefore, there exists an inverse for every non-zero element, so $R$ is a field.
As the underlying set of $R$ is finite, $R$ is by definition a Galois field.
== Sources ==
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Theorem $25$
* : $\S 3$: Theorem $3.3.4$
Category:Finite Integral Domain is Galois Field
Finite Integral Domain is Galois Field/Proof 4

== Theorem ==
== Proof ==
 $\struct {D, +, \circ}$ is a finite integral domain which is not a field.
From Non-Field Integral Domain has Infinite Number of Ideals, $\struct {D, +, \circ}$ has an infinite number of distinct ideals.
But this contradicts the assertion that $\struct {D, +, \circ}$ is finite.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $9$: Rings: Exercise $7 \ \text {(ii)}$
Category:Finite Integral Domain is Galois Field
Finite Intersection of Absorbing Sets is Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A_1, \ldots, A_n \subseteq X$ be absorbing sets.
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j$ is absorbing.
== Proof ==
Let $x \in X$ and $j \in \set {1, \ldots, n}$.
Since $A_j$ is absorbing, there exists $t_j \in \R_{> 0}$ such that:
:$x \in \alpha A_j$ for $\alpha \in \C$ with $\cmod \alpha \ge t_j$. 
Let $t = \max \set {t_1, \ldots, t_n}$.
Then, we have:
:$\ds x \in \bigcap_{j \mathop = 1}^n \paren {\alpha A_j}$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
From Dilation of Intersection of Subsets of Vector Space, we have:
:$\ds x \in \alpha \bigcap_{j \mathop = 1}^n A_j$ for $\alpha \in \C$ with $\cmod \alpha \ge t$.
So:
:$\ds \bigcap_{j \mathop = 1}^n A_j$ is absorbing.
Category:Absorbing Sets
Finite Intersection of Open Sets of Metric Space is Open

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $U_1, U_2, \ldots, U_n$ be open sets in $M$.
Then $\ds U = \bigcap_{i \mathop = 1}^n U_i$ is open in $M$.
That is, a finite intersection of open subsets is open.
== Proof 1 ==
Let $x \in U$.
By the definition of intersection:
:$\forall i \in \closedint 1 n: x \in U_i$.
Thus, since $U_i$ is open in $M$:
:$\forall i \in \closedint 1 n: \exists \epsilon_i \in \R_{>0}: \map {B_{\epsilon_i} } x \subseteq U_i$
where $\map {B_{\epsilon_i} } x$ is the open $\epsilon_i$-ball of $x$.
Let $\ds \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$.
Then, by Open Ball contains Smaller Open Ball:
:$\forall i \in \closedint 1 n: \map {B_\epsilon} x \subseteq \map {B_{\epsilon_i} } x$
By definition of intersection:
:$\map {B_\epsilon} x \subseteq U$
The result follows.
== Proof 2 ==
 $U$ is not open in $M$.
By definition of open set, there exists a $y \in U$ such that:
:$\ds \forall \epsilon \in \R_{>0}: \map {B_\epsilon} y \setminus \bigcap_{i \mathop = 1}^n U_i \ne \O$.
By De Morgan's law, we have
:$\ds \bigcup_{i \mathop = 1}^n \paren {\map {B_\epsilon} y \setminus U_i} \ne \O$.
Thus, there exists $i \in \closedint 1 n$ such that $\map {B_\epsilon} y \setminus U_i \ne \O$.
As $U_i$ is open and $\epsilon$ is arbitrarily given, by Open Ball of Point Inside Open Ball, $y \notin U_i$. This is a contradiction.
Thus, $U$ is open in $M$.
== Also see ==
* Infinite Intersection of Open Sets of Metric Space may not be Open for how this result breaks down for an infinite number of open sets.
== Sources ==
* : $2.24 c$
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.4$
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Proposition $2.3.17$
Category:Open Sets (Metric Spaces)
Category:Set Intersection
Finite Intersection of Open Sets of Neighborhood Space is Open

== Corollary to Intersection of two Open Sets of Neighborhood Space is Open ==
Let $\struct {S, \NN}$ be a neighborhood space.
Let $n \in \N_{>0}$ be a natural number.
Let $\ds \bigcap_{i \mathop = 1}^n U_i$ be a finite intersection of open sets of $\struct {S, \NN}$.
Then $\ds \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\struct {S, \NN}$.
== Proof ==
Proof by induction:
Let $U_1, U_2, \ldots$ be open sets of $\struct {S, \NN}$.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\struct {S, \NN}$.
$\map P 1$ is true, as this just says:
:$U_1$ is an open set of $\struct {S, \NN}$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$U_1 \cap U_2$ is an open set of $\struct {S, \NN}$
which has been proved above.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\struct {S, \NN}$.
Then we need to show:
:$\ds \bigcap_{i \mathop = 1}^{k + 1} U_i$ is an open set of $\struct {S, \NN}$.
=== Induction Step ===
This is our induction step:
We have that:
:$\ds \bigcap_{i \mathop = 1}^{k + 1} U_i = \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$
From the induction hypothesis:
:$\ds \bigcap_{i \mathop = 1}^k U_i$ is an open set of $\struct {S, \NN}$.
From the basis for the induction:
:$\ds \bigcap_{i \mathop = 1}^k U_i \cap U_{k + 1}$ is an open set of $\struct {S, \NN}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{>0}: \bigcap_{i \mathop = 1}^n U_i$ is an open set of $\struct {S, \NN}$.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Lemma $3.6$
Category:Neighborhood Spaces
Finite Intersection of Open Sets of Normed Vector Space is Open

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $U_1, U_2, \ldots, U_n$ be open in $M$.
Then $\ds \bigcap_{i \mathop = 1}^n U_i$ is open in $M$.
That is, a finite intersection of open subsets is open.
== Proof ==
Let $\ds x \in \bigcap_{i \mathop = 1}^n U_i$.
For each $i \in \closedint 1 n$, we have $x \in U_i$.
Thus:
:$\exists \epsilon_i > 0: \map {B_{\epsilon_i}} x \subseteq U_i$
where $\map {B_{\epsilon_i}} x$ is the open $\epsilon_i$-ball of $x$.
Let $\ds \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$.
So:
:$\epsilon > 0$.
Let $y \in \map {B_\epsilon} x$.
Then $\norm {x - y} < \epsilon$
Hence:
:$\forall i \in \closedint 1 n : \norm {x - y} < \epsilon_i$
In other words:
:$\map {B_\epsilon} x \subseteq \map {B_{\epsilon_i}} x \subseteq U_i$
for all $i \in \closedint 1 n$.
So:
:$\ds \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$
The result follows.
== Also see ==
* Finite Intersection of Open Sets is Open
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Set Intersection
Category:Open Sets (Normed Vector Spaces)
Finite Intersection of Regular Open Sets is Regular Open

== Theorem ==
Let $T$ be a topological space.
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are regular open in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$
where $H_i^{- \circ}$ denotes the interior of the closure of $H_i$.
Then $\ds \bigcap_{i \mathop = 1}^n H_i$ is regular open in $T$.
That is:
:$\ds \bigcap_{i \mathop = 1}^n H_i = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}$
== Proof ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Regular Open Sets
Finite Irreducible Space is Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite irreducible topological space.
Then $T$ is path-connected.
== Proof ==
By Power Set of Finite Set is Finite, the power set $\powerset S$ is finite.
By Subset of Finite Set is Finite, $\tau \subseteq \powerset S$ is finite.
The result follows from Irreducible Space with Finitely Many Open Sets is Path-Connected.
== Also see ==
* Irreducible Space is not necessarily Path-Connected
Category:Irreducible Spaces
Category:Path-Connected Spaces
Category:Finite Topological Spaces