nyuuzyou/wikis huggingface dataset:
Intersection of Events is Event

== Theorem ==
Let $\EE$ be an experiment with a probability space $\struct {\Omega, \Sigma, \Pr}$.
The event space $\Sigma$ of $\EE$ has the property that:
:$A, B \in \Sigma \implies A \cap B \in \Sigma$
That is, the intersection of two events is also an event in the event space.
== Proof ==
== Also see ==
* Elementary Properties of Event Space
== Sources ==
* : $1$: Events and probabilities: $1.2$: Outcomes and events: Exercise $1$
Category:Intersections of Events
Category:Event Spaces
Intersection of Exteriors contains Exterior of Union

== Theorem ==
Let $T$ be a topological space.
Let $\mathbb H$ be a set of subsets of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then:
:$\ds \paren {\bigcup_{H \mathop \in \mathbb H} H}^e \subseteq \bigcap_{H \mathop \in \mathbb H} H^e $
where $H^e$ denotes the exterior of $H$.
== Proof ==
In the following, $H^\circ$ denotes the interior of the set $H$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Exteriors
Intersection of Exteriors of Singleton Rationals is Irrationals

== 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:
:$\ds \bigcap_{\alpha \mathop \in \Q} B_\alpha^e = \R \setminus \Q$
where $B_\alpha^e$ denotes the exterior of $B_\alpha$ in $\R$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Rational Numbers: $1$
Category:Rational Number Space
Category:Singletons
Intersection of Family/Examples/Size of y-1 lt n and Size of y+1 gt 1 over n

== Example of Intersection of Family ==
Let $I$ be the indexing set $I = \set {1, 2, 3, \ldots}$
Let $\family {T_n}$ be the indexed family of subsets of the set of real numbers $\R$, defined as:
:$T_n = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}$
Then:
:$\ds \bigcap_{n \mathop \in I} T_n = \openint 0 2$
== Proof ==
From Size of y-1 lt n and Size of y+1 gt 1 over n:
:$T_n = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$
We have that:
:$\paren {\openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n} } \subseteq \paren {\openint {1 - \paren {n + 1} } {-1 - \dfrac 1 {n + 1} } \cup \openint {-1 + \dfrac 1 {n + 1} } {1 + \paren {n + 1} } }$
That is:
:$T_n \subseteq T_{n + 1}$
and so:
:$\ds \bigcap_{n \mathop \in I} T_n = T_1$
Hence:
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text {A}$
Category:Intersections of Families
Intersection of Family is Subset of Intersection of Subset of Family

== Theorem ==
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $J \subseteq I$.
Then:
:$\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in J} A_\alpha$
where $\ds \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}_{\alpha \mathop \in I}$.
== Proof ==
== Also see ==
* Union of Subset of Family is Subset of Union of Family
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $3 \ \text{(a)}$
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $3 \ \text{(a)}$
Category:Subsets
Category:Set Intersection
Category:Indexed Families
Intersection of Finite Set of Submodules is Infimum of Lattice of Submodules

== Theorem ==
Let $R$ be a ring.
Let $\struct {G, +_G}$ be an abelian group.
Let $M = \struct {G, +, \circ}_R$ be an $R$-module.
Let $H_1, H_2, \ldots, H_n$ be submodules of $M$.
Then $H_1 \cap H_2 \cap \cdots \cap H_n$ is the infimum of $\set {H_1, H_2, \ldots, H_n}$ in the complete lattice of all submodules of $M$.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$: Corollary
Category:Submodules
Intersection of Horizontal Sections is Horizontal Section of Intersection

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\set {E_\alpha : \alpha \mathop \in A}$ be a set of subsets of $X \times Y$. 
Let $y \in Y$. 
Then: 
:$\ds \paren {\bigcap_{\alpha \mathop \in A} E_\alpha}^y = \bigcap_{\alpha \mathop \in A} \paren {E_\alpha}^y$
where:
:$\ds \paren {\bigcap_{\alpha \mathop \in A} E_\alpha}^y$ is the $y$-horizontal section of $\ds \bigcap_{\alpha \mathop \in A} E_\alpha$
:$\paren {E_\alpha}^y$ is the $y$-horizontal section of $E_\alpha$.
== Proof ==
Note that:
:$\ds x \in \bigcap_{\alpha \mathop \in A} \paren {E_\alpha}^y$
: 
:$x \in \paren {E_\alpha}^y$ for all $\alpha \in A$.
From the definition of the $x$-horizontal section, this is equivalent to:
:$\tuple {x, y} \in E_\alpha$ for all $\alpha \in A$. 
This in turn is equivalent to: 
:$\ds \tuple {x, y} \in \bigcap_{\alpha \mathop \in A} E_\alpha$
Again applying the definition of the $x$-horizontal section, this is equivalent to:
:$\ds x \in \paren {\bigcap_{\alpha \mathop \in A} E_\alpha}^y$
So:
:$\ds x \in \bigcap_{\alpha \mathop \in A} \paren {E_\alpha}^y$ $\ds x \in \paren {\bigcap_{\alpha \mathop \in A} E_\alpha}^y$
giving:
:$\ds \paren {\bigcap_{\alpha \mathop \in A} E_\alpha}^y = \bigcap_{\alpha \mathop \in A} \paren {E_\alpha}^y$
Category:Set Intersection
Category:Horizontal Section of Sets
Intersection of Ideal of *-Algebra with its Star is Self-Adjoint Ideal

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $I$ be an ideal of $A$.
Define:
:$I^\ast = \set {x^\ast : x \in I}$
Then $I \cap I^\ast$ is a self-adjoint ideal.
== Proof ==
From Star of Ideal of *-Algebra is Ideal, $I^\ast$ is an ideal.
From Intersection of Algebra Ideals is Ideal, $I \cap I^\ast$ is an ideal.
Further, if $x \in I \cap I^\ast$, then $x \in I$ and there exists $y \in I$ such that $x = y^\ast$.
From $(\text C^\ast 1)$ in the definition of an involution, we have $x^\ast = y^{\ast \ast} = y$. 
So $x^\ast \in I$.
Since $x \in I$ we also clearly have $x^\ast \in I^\ast$.
Hence $x^\ast \in I \cap I^\ast$ for all $x \in I \cap I^\ast$.
Hence $I \cap I^\ast$ is self-adjoint.
Category:*-Algebras
Intersection of Ideals of Ring contains Product

== Theorem ==
Let $R$ be a ring.
Let $I$ be a right ideal of $R$.
Let $J$ be a left ideal of $R$.
Let $I J$ be their product.
Then $I J \subseteq I \cap J$.
== Proof ==
Let $a_1, \ldots, a_n \in I$ and $b_1, \ldots, b_n \in J$ be arbitrary.
Then:
and:
By definition of set intersection:
:$\ds \sum_{k \mathop = 1}^n a_k b_k \in I \cap J$
As $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ were arbitrary, this completes the proof.
Category:Ideal Theory
Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x \in S$.
Then:
:$\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} = x^\ll$
where $\mathit {Ids}$ denotes the set of all ideals in $L$.
== Proof ==
By Supremum of Lower Closure of Element:
:$\map \sup {x^\preceq} = x$
By Lower Closure of Element is Ideal:
:$x^\preceq \in \mathit {Ids}$
Then by definition of reflexivity:
:$x^\preceq \in \set {I \in \mathit {Ids}: x \preceq \sup I}$
We will prove that:
:$\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} \subseteq x^\ll$
Let $z \in \ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I}$
By definition of intersection:
:$\forall I \in \mathit {Ids}: x \preceq \sup I \implies z \in I$
By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
:$z \ll x$
Thus by definition of way below closure:
:$z \in x^\ll$
By definition of set equality, it remains to prove the pposite inclusion.
Let $z \in x^\ll$
By definition of way below closure:
:$z \ll x$
We will prove that
:$\forall Y \in \set {I \in \mathit {Ids}: x \preceq \sup I}: z \in Y$
Let $Y \in \set {I \in \mathit {Ids}: x \preceq \sup I}$
Then
:$Y \in \mathit {Ids}$ and $x \preceq \sup Y$
Thus Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:
:$z \in Y$
Thus by definition of intersection:
:$z \in \ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I}$
== Sources ==
* 
* 
Category:Way Below Relation
Intersection of Image with Subset of Codomain

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A \subseteq S$ and $B \subseteq T$.
Then:
:$f \sqbrk {A \cap f^{-1} \sqbrk B} = f \sqbrk A \cap B$
== Proof ==
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $4 \ \text{(d})$
Category:Images
Category:Set Intersection
Intersection of Inductive Set as Subset of Real Numbers is Inductive Set

== Theorem ==
Let $\AA$ be a set of inductive sets defined as subsets of real numbers.
Then their intersection is an inductive set.
== Proof ==
By definition of inductive set:
:$\forall S \in \AA: 1 \in S$
Thus by definition of set intersection:
:$\ds 1 \in \bigcap_{S \mathop \in \AA} S$
Also by definition of inductive set:
:$\forall S \in \AA: x \in S \implies x + 1 \in S$
So:
Hence the result, by definition of inductive set.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $3 \ \text{(a)}$
Category:Inductive Sets
Intersection of Inductive Sets

== Theorem ==
Let $\mathbb S$ be a non-empty indexed family of inductive sets.
Then $\bigcap \mathbb S$ is an inductive set.
== Proof ==
From definition, a set $X$ is an inductive set both the following holds:
:$\O \in X$
:$x \in X \implies x^+ \in X$
For all $S \in \mathbb S$, $S$ is an inductive set.
From definition of an inductive set, $\O \in S$.
By definition of set intersection, $\O \in \bigcap \mathbb S$.
Now suppose $x \in \bigcap \mathbb S$.
By definition of set intersection:
:$\forall S \in \mathbb S: x \in S$
Hence from definition of an inductive set, $x^+ \in S$.
By definition of set intersection, $x^+ \in \bigcap \mathbb S$.
Therefore $\bigcap \mathbb S$ is an inductive set.
== Sources ==
* : $\S 11$: Numbers
Category:Inductive Sets
Category:Set Intersection
Intersection of Injective Image with Relative Complement

== Theorem ==
Let $f: S \to T$ be a mapping.
Then $f$ is an injection :
:$\forall A \subseteq S: f \sqbrk A \cap f \sqbrk {\relcomp S A} = \O$
== Proof ==
From Intersection with Relative Complement is Empty:
:$A \cap \relcomp S A = \O$
From Image of Intersection under Injection:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
 $f$ is an injection.
Hence the result.
== Sources ==
* : Chapter $4$: Mappings: Exercise $8$
Category:Injections
Category:Set Intersection
Category:Relative Complement
Intersection of Integer Ideals is Lowest Common Multiple

== Theorem ==
Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal k$ be the intersection of $\ideal m$ and $\ideal n$.
Then $k = \lcm \set {m, n}$.
== Proof ==
By Intersection of Ring Ideals is Ideal we have that $\ideal k = \ideal m \cap \ideal n$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal k$ are all necessarily principal ideals.
By Subrings of Integers are Sets of Integer Multiples we have that:
:$\ideal m = m \Z, \ideal n = n \Z$
Thus:
:$\ideal k = \set {x \in \Z: n \divides x \land m \divides x}$
The result follows by LCM iff Divides All Common Multiples.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Example $37$
Category:Ideal Theory
Category:Integers
Category:Lowest Common Multiple
Intersection of Interiors contains Interior of Intersection

== Theorem ==
Let $T$ be a topological space.
Let $\mathbb H$ be a set of subsets of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then the interior of the intersection of $\mathbb H$ is a subset of the intersection of the interiors of the elements of $\mathbb H$.
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$
== Proof ==
In the following, $H^-$ denotes the closure of the set $H$.
At this point we note that:
:$(1): \quad \ds \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \supseteq \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$
from Closure of Union contains Union of Closures.
Then we note that:
:$\ds T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} \subseteq T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}$
from $(1)$ and Set Complement inverts Subsets.
Then we continue:
== Also see ==
* Interior of Finite Intersection equals Intersection of Interiors
* Interior of Intersection may not equal Intersection of Interiors
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Interiors
Category:Set Intersection
Intersection of Left Cosets of Subgroups is Left Coset of Intersection

== Theorem ==
Let $G$ be a group.
Let $H, K \le G$ be subgroups of $G$.
Let $a, b \in G$.
Let:
:$a H \cap b K \ne \O$
where $a H$ denotes the left coset of $H$ by $a$.
Then $a H \cap b K$ is a left coset of $H \cap K$.
== Proof ==
Let $x \in a H \cap b K$.
Then:
and similarly:
Hence:
Hence the result by definition of left coset.
== Sources ==
* : Chapter $6$: Cosets: Exercise $3$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $9$
Category:Cosets
Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal

== Theorem ==
Let $\struct {S, \preceq}$ be a meet semilattice.
Let $I$ be an ideal in $\struct {S, \preceq}$.
Let $x \in S$.
Then:
:$\paren {x^\preceq} \cap I = \set {x \wedge i: i \in I}$
where $x^\preceq$ denotes the lower closure of $x$.
== Proof ==
=== First Inclusion ===
Let $z \in \paren {x^\preceq} \cap I$
By definition of intersection:
:$z \in x^\preceq$ and $z \in I$
By definition of lower closure of element:
:$z \preceq x$
By Preceding iff Meet equals Less Operand:
:$x \wedge z = z$
Thus:
:$z \in \set {x \wedge i: i \in I}$
=== Second Inclusion ===
Let $z \in \set {x \wedge i: i \in I}$
Then
:$\exists i \in I: z = x \wedge i$
By Meet Precedes Operands:
:$z \preceq x$ and $z \preceq i$
By definition of ideal in ordered set:
:$I$ is a lower section.
By definitions of lower section and lower closure of element:
:$z \in I$ and $z \in x^\preceq$
Thus by definition of intersection:
:$z \in \struct {x^\preceq} \cap I$
Thus the result by definition of set equality.
== Sources ==
* 
* 
Category:Lower Closures
Intersection of Magmas of Sets is Magma of Sets

== Theorem ==
Let $X$ be a set.
Let $\Phi := \set {\phi_i: i \in I}$ be a collection of partial mappings with codomain $\powerset X$, the power set of $X$.
Let $\SS_j$ be a magma of sets for $\Phi$, for each $j \in J$, for some index set $J$.
Then:
:$\SS := \ds \bigcap_{j \mathop \in J} \SS_j$
is also a magma of sets for $\Phi$.
== Proof ==
For each $i \in I$ and $j \in J$, we have:
:$\map {\phi_i} {\sequence {S_{j, j_i} }_{j_i \mathop \in J_i} } \in \SS_j$
Thus, if for each $j \in J$, one has:
:$(1): \quad \sequence {S_{j_i} }_{j_i \mathop \in J_i} \in \SS_j^{J_i}$
it follows from definition of set intersection that:
:$\map {\phi_i} {\sequence {S_{j_i} }_{j_i \mathop \in J_i} } \in \SS$
The condition $(1)$, for each $j \in J$, comes down to:
:$\sequence {S_{j_i} }_{j_i \mathop \in J_i} \in \SS^{J_i}$
It follows that $\SS$ satisfies the requirement on $\phi_i$ to be a magma of sets, for each $i \in I$.
That is to say, $\SS$ is a magma of sets for $\Phi$.
Category:Magmas of Sets
Intersection of Multiplicative Groups of Complex Roots of Unity

== Theorem ==
Let $\struct {K, \times}$ denote the circle group.
Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let $c = \lcm \set {m, n}$ be the lowest common multiple of $m$ and $n$.
Let $\struct {U_n, \times}$ denote the multiplicative group of complex $n$th roots of unity.
Let $\struct {U_m, \times}$ denote the multiplicative group of complex $m$th roots of unity.
Let $H = U_m \cap U_n$.
Then $H = U_c$.
== Proof ==
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 44 \gamma$
Category:Multiplicative Groups of Complex Roots of Unity
Category:Circle Group
Intersection of Neighborhood of Diagonal with Inverse is Neighborhood

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $R$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct{X \times X, \tau_{X \times X}}$.
Let $R^{-1}$ denote the inverse relation of $R$ where $R$ is viewed as a relation on $X \times X$. 
Then:
:$R \cap R^{-1}$ is a neighborhood of $\Delta_X$ in $\struct{X \times X, \tau_{X \times X}}$.
== Proof ==
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$\forall \tuple{x, x} \in \Delta_X : R$ is a neighborhood of $\tuple{x, x}$
From Inverse of Neighborhood of Diagonal Point is Neighborhood:
:$\forall \tuple{x, x} \in \Delta_X : R^{-1}$ is a neighborhood of $\tuple{x, x}$
From Intersection of Neighborhoods in Topological Space is Neighborhood
:$\forall \tuple{x, x} \in \Delta_X : R \cap R^{-1}$ is a neighborhood of $\tuple{x, x}$
From Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$R \cap R^{-1}$ is a neighborhood of $\Delta_X$ in $\struct{X \times X, \tau_{X \times X}}$
Category:Neighborhoods
Category:Diagonal Relation
Category:Relations
Intersection of Neighborhoods in Metric Space is Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$ be a point in $M$.
Let $N, N'$ be neighborhoods of $a$ in $M$.
Then $N \cap N'$ is a neighborhood of $a$ in $M$.
== Proof ==
By definition of neighborhood:
:$\exists \epsilon_1 \in \R_{>0}: \map {B_{\epsilon_1} } a \subseteq N$
where $\map {B_{\epsilon_1} } a$ is the open $\epsilon_1$-ball of $a$ in $M$.
:$\exists \epsilon_2 \in \R_{>0}: \map {B_{\epsilon_2} } a \subseteq N'$
where $\map {B_{\epsilon_2} } a$ is the open $\epsilon_2$-ball of $a$ in $M$.
Thus by definition of intersection:
:$\map {B_\epsilon} a \subseteq N \cap N'$
where $\epsilon = \min \set {\epsilon_1, \epsilon_2}$
The result follows by definition of neighborhood of $a$.
== Also see ==
* Intersection of Neighborhoods in Topological Space is Neighborhood
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Theorem $4.8: \ N \, 4$
Category:Neighborhoods
Intersection of Neighborhoods in Topological Space is Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $M, N$ be a neighborhoods of $x$ in $T$.
Then $M \cap N$ is a neighborhood of $x$ in $T$.
That is:
:$\forall x \in S: \forall M, N \in \NN_x: M \cap N \in N_x$
where $\NN_x$ is the neighborhood filter of $x$.
== Proof ==
By definition of neighborhood:
:$\exists U_1 \in \tau: x \in U_1 \subseteq M$
where $U_1$ is an open set of $T$.
:$\exists U_2 \in \tau: x \in U_2 \subseteq N$
where $U_2$ is an open set of $T$.
Thus by Set Intersection Preserves Subsets:
:$U \subseteq M \cap N$
where $U = U_1 \cap U_2$
The result follows by definition of neighborhood of $a$.
== Also see ==
* Intersection of Neighborhoods in Metric Space is Neighborhood
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.1: \ N 4$
Category:Neighborhoods
Intersection of Nested Closed Subsets of Compact Space is Non-Empty

== Theorem ==
Let $\struct {T, \tau}$ be a compact topological space.
Let $\sequence {V_n}$ be a sequence of non-empty closed subsets of $T$ with:
:$V_{i + 1} \subseteq V_i$
for each $i$.
Then:
:$\ds \bigcap_{n \mathop = 1}^\infty V_n \ne \O$
== Proof ==
From Closed Subspace of Compact Space is Compact:
:$V_n$ is compact for each $n$. 
 that:
:$\ds \bigcap_{n \mathop = 1}^\infty V_n = \O$
Then:
Since each $V_n$ is closed in $T$, from Closed Set in Topological Subspace: Corollary we have:
:$V_n$ is closed in $V_1$. 
So:
:$V_1 \setminus V_n$ is open for each $n$.
So:
:$\set {V_1 \setminus V_n : n \in \N}$
is a open cover of $V_1$.
Since $V_1$ is compact, there exists a finite subcover:
:$\set {V_1 \setminus V_{n_1}, V_1 \setminus V_{n_2}, \cdots, V_1 \setminus V_{n_j} }$
with:
:$n_1 < n_2 < \cdots < n_j$
so that:
:$\ds \bigcup_{i \mathop = 1}^j \paren {V_1 \setminus V_{n_i} } = V_1$
We then have, by De Morgan's Laws: Difference with Intersection:
:$\ds V_1 \setminus \paren {\bigcap_{i \mathop = 1}^j V_{n_i} } = V_1$
By construction we have:
:$\ds \bigcap_{i \mathop = 1}^j V_{n_i} \subseteq V_{n_1} \subseteq V_1$
so:
:$\ds \bigcap_{i \mathop = 1}^j V_{n_i} = \O$
Since each $V_{n_i}$ is non-empty, for every $x \in V_{n_j}$, there exists some $1 \le k < j$ such that: 
:$x \notin V_{n_k}$
But this is impossible since $V_{n_j} \subseteq V_{n_k}$, and so we have reached a contradiction.
So, we must have: 
:$\ds \bigcap_{n \mathop = 1}^\infty V_n \ne \O$
Category:Compact Spaces
Category:Closed Sets
Intersection of Non-Empty Class is Set/Corollary

== Theorem ==
Let $x$ be a non-empty set.
Let $\bigcap x$ denote the intersection of $x$.
Then $\bigcap x$ is a set.
== Proof ==
It is assumed that $x$ is an element of a basic universe.
Hence from the , every set is a class.
Hence Intersection of Non-Empty Class is Set applies directly.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Theorem $5.1 \ (1)$
Category:Intersection of Non-Empty Class is Set
Intersection of Non-Empty Class is Set/Proof 1

== Theorem ==
== Proof ==
Let $V$ denote the basic universe such that $A \subseteq V$.
We are given that $A$ is non-empty.
Then $\exists x \in A$ where $x$ is a set.
By definition of intersection of class, every element of $\bigcap A$ is an element of all elements of $A$.
Thus:
:$\bigcap A \subseteq x$
We are given that $A$ is a subclass of the basic universe $V$.
Thus $x \in V$ by definition of basic universe.
By the , $V$ is a swelled class.
By definition of swelled class, every subclass of a set $x \in V$ is a set.
It follows $\bigcap A$ is a set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Theorem $5.1 \ (1)$
Category:Intersection of Non-Empty Class is Set
Intersection of Non-Empty Class is Set/Proof 2

== Theorem ==
== Proof ==
Since $A$ is a non-empty class, there exists $S \in A$.
Since $S$ is an element of a class, it is not a proper class, and is thus a set.
By definition of class intersection:
:$x \in \bigcap A \implies x \in S$
By the subclass definition:
:$\bigcap A \subseteq S$
By Subclass of Set is Set, $\bigcap A$ is a set.
== Sources ==
* : Chapter $1$: Separation Schema
Category:Intersection of Non-Empty Class is Set
Intersection of Normal Subgroup with Center in p-Group

== Theorem ==
Let $p$ be a prime number
Let $G$ be a $p$-group.
Let $N$ be a non-trivial normal subgroup of $G$.
Let $\map Z G$ denote the center of $G$.
Then:
:$N \cap \map Z G$ is a non-trivial normal subgroup of $G$.
== Proof ==
First we note that:
:Center of Group is Normal Subgroup
and from Intersection of Normal Subgroups is Normal:
:$N \cap \map Z G$ is normal in $G$.
Suppose $G$ is abelian.
By definition:
:$\map Z G = G$
Then:
:$N \cap \map Z G = N$
which is non-trivial.
From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$.
So, suppose $G$ is non-abelian.
From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial.
From Union of Conjugacy Classes is Normal:
:$N = \ds \bigcup_{x \mathop \in N} \conjclass x$
Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $N$ is partitioned.
From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element.
From Number of Conjugates is Number of Cosets of Centralizer:
:$\order {\conjclass {x_j} } \divides \order G$
for all $j \in \set {1, 2, \ldots, m}$.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $21$
Category:Normal Subgroups
Category:P-Groups