nyuuzyou/wikis huggingface dataset:
Intersection of Normal Subgroup with Sylow P-Subgroup

== Theorem ==
Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $N$ be a normal subgroup of $G$.
Then $P \cap N$ is a Sylow $p$-subgroup of $N$.
== Proof ==
Since $N \lhd G$, we see that:
:$\gen {P, N} = P N$
from Subset Product with Normal Subgroup as Generator.
Since $P \cap N \le P$, it follows that:
:$\order {P \cap N} = p^k$
where $k > 0$.
By Order of Subgroup Product:
:$\order {P N} \order {P \cap N} = \order P \order N$
Hence from Lagrange's Theorem:
:$\index N {P \cap N} = \index {P N} P$
By Tower Law for Subgroups:
:$\index G P = \index G {P N} \index {P N} P$
Thus:
:$\index {P N} P \divides \index G P$
where $\divides$ denotes divisibility.
Thus:
:$p \nmid \index {P N} P$
so:
:$p \nmid \index N {P \cap N}$
Thus $P \cap N$ is a Sylow $p$-subgroup of $N$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Proposition $11.14 \ \text{(i)}$
Category:Normal Subgroups
Category:Set Intersection
Category:Sylow p-Subgroups
Intersection of Normal Subgroups is Normal

== Theorem ==
Let $G$ be a group.
Let $I$ be an indexing set.
Let $\family {N_i}_{i \mathop \in I}$ be a non-empty indexed family of normal subgroups of $G$.
Then $\ds \bigcap_{i \mathop \in I} N_i$ is a normal subgroup of $G$.
== Proof ==
Let $\ds N = \bigcap_{i \mathop \in I} N_i$.
From Intersection of Subgroups is Subgroup, $N \le G$.
Suppose $H \in \set {N_i: i \in I}$.
We have that $N \subseteq H$.
Thus from Subgroup is Superset of Conjugate iff Normal:
:$a N a^{-1} \subseteq a H a^{-1} \subseteq H$
Thus $a N a^{-1}$ is a subset of each one of the subgroups in $\set {N_i: i \in I}$, and hence in their intersection $N$.
That is, $a N a^{-1} \subseteq N$.
The result follows by Subgroup is Superset of Conjugate iff Normal.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $7$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.9$
Category:Normal Subgroups
Category:Set Intersection
Intersection of Open Intervals of Positive Reals is Empty

== Theorem ==
Let $\R_{>0}$ be the set of strictly positive real numbers.
For all $x \in \R_{>0}$, let $A_x$ be the open real interval $\openint 0 x$.
Then:
:$\ds \bigcap_{x \mathop \in \R_{>0} } A_x = \O$
== Proof ==
Let $\ds A = \bigcap_{x \mathop \in \R_{>0} } A_x$.
 $A \ne \O$.
Then:
:$\exists y \in \R_{>0}: y \in A$
By definition of open interval:
:$y \notin \openint 0 y = A_y$
and so by definition of intersection of family:
:$y \notin A$
From this contradiction it follows that $A$ has no elements.
That is:
:$\ds \bigcap_{x \mathop \in \R_{>0} } A_x = \O$
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $5$
Category:Real Intervals
Intersection of Open Set with Closure of Set is Subset of Closure of Intersection

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \in \tau$ be an open set of $T$.
Let $K \subseteq S$ be an arbitrary subset of $S$.
Then:
:$H \cap \map \cl K \subseteq \map \cl {H \cap K}$
where $\cl$ denotes set closure.
== Proof ==
Let $x \in H \cap \map \cl K$.
Then:
:$x \in H$
and:
:$x \in \map \cl K$
Let $N_1$ be an arbitrary neighborbood of $x$.
Because $x \in H$, there exists a neighborbood $N_2$ of $x$ entirely within $U$.
Let $N_3 = N_1 \cap N_2$.
By Intersection of Neighborhoods in Topological Space is Neighborhood, $N_3$ is also a neighborbood of $x$.
Because $N_1 \cap N_2 \subseteq N_2$, $N_3$ is also contained entirely within $U$.
Because $x \in \map \cl K$, by definition of closure, there exists $y \in N_3$ which is also in $K$.
From $x \in N_3$ we know that $y \in H$.
Thus we have demonstrated the existence of $y \in N_1$ that is in $H \cap K$.
As $N_1$ is arbitrary, it follows that for every neighborbood of $x$ there exists $y \in N_1$ such that $y \in H$ and $y \in K$.
That is:
:$x \in \map \cl {H \cap K}$
Hence the result, by definition of subset.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 27 \ \text {(iii)}$
Category:Set Closures
Category:Set Intersection
Intersection of Open Sets of Hausdorff Space containing Point is Singleton

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $x \in S$ be arbitrary.
Then the intersection of all open sets containing $x$ is $\set x$:
:$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$
== Proof ==
Let $x \in S$.
Let $K = \ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} }$, that is, the intersection of all open sets containing $x$.
 there exists $y \in S$ such that $y \in K$ but $y \ne x$.
By definition of Hausdorff space, there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
We have that $y \in V$.
But we also have that $y \in K$.
By Intersection is Subset:
:$K \subseteq U$
and so:
:$y \in U$
So $y \in U$ and $y \in V$ and so $U \cap V \ne \O$.
This directly contradicts our assertion that $U, V \in \tau$ are disjoint.
In turn, this contradicts our supposition that $T$ is a Hausdorff space.
Hence by Proof by Contradiction there can be no such $y$ such that $y \in K$.
The result follows.
== Also see ==
* Space such that Intersection of Open Sets containing Point is Singleton may not be Hausdorff: the converse does not hold
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 4 \ \text {(a)}$
Category:Hausdorff Spaces
Intersection of Orderings is Ordering

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be orderings on $A$.
Then $\RR \cap \SS$ is an ordering on $A$.
== Proof ==
By definition of ordering:
:$\RR$ and $\SS$ are reflexive
:$\RR$ and $\SS$ are transitive
:$\RR$ and $\SS$ are antisymmetric.
We have:
:Intersection of Reflexive Relations is Reflexive
:Intersection of Transitive Relations is Transitive
:Intersection of Antisymmetric Relations is Antisymmetric
and the result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.27 \ \text {(c)}$
Category:Orderings
Category:Set Intersection
Intersection of Ordinals is Ordinal

== Theorem ==
Let $A$ be a non-empty class of ordinals.
Then its intersection $\bigcap A$ is an ordinal.
== Proof ==
Let $\alpha = \bigcap A$.
It will be demonstrated that $\alpha$ is an ordinal according to Definition $2$:
=== Set ===
By Intersection of Non-Empty Class is Set, $\alpha$ is a set.
=== Transitive ===
Let $n \in \alpha$.
Let $m \in n$.
Let $\beta \in A$ be arbitrary.
By the definition of intersection:
:$n \in \beta$
Since $\beta$ is an ordinal, it is transitive.
Thus:
:$m \in \beta$
As $\beta$ is arbitrary, this holds for all $\beta \in A$.
Thus:
:$m \in \alpha$
Thus $\alpha$ is transitive.
=== $\in$-connected ===
Let $x, y \in \alpha$ be arbitrary such that $x \ne y$.
Since $A$ is non-empty, it has an element $\beta$.
By the definition of intersection:
:$x, y \in \beta$
Since $\beta$ is an ordinal, it is $\in$-connected.
Thus $x \in y$ or $y \in x$.
As $x$ and $y$ are arbitrary, this holds for all $x, y \in \alpha$.
That is, $\alpha$ is $\in$-connected.
=== Well-founded ===
Let $\beta$ be a non-empty subset of $\alpha$.
Let $\gamma \in A$ (such exists because $A$ is non-empty).
By Intersection is Largest Subset:
:$\beta \subseteq \gamma$
Since $\gamma$ is an ordinal it is well-founded.
Thus $\beta$ has an element $x$ such that $x \cap \beta = \O$.
As this holds for all such $\beta$, $\alpha$ is well-founded.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Exercise $2.2$
Category:Ordinals
Category:Set Intersection
Intersection of Ordinals is Smallest

== Theorem ==
Let $A$ be a non-empty set or class of ordinals.
Let $m = \bigcap A$ be the intersection of all the elements of $A$.
Then $m$ is the smallest element of $A$.
== Proof ==
By Intersection of Ordinals is Ordinal, $m$ is an ordinal.
By Intersection is Largest Subset, $m \subseteq a$ for each $a \in A$.
It remains to show that $m \in A$.
Let $m^+ = m \cup \set m$ be the successor of $m$.
By Relation between Two Ordinals:
:for each $a \in A$, either $m^+ \subseteq a$ or $a \in m^+$.
Let $m^+ \subseteq a$ for all $a \in A$.
Then by Intersection is Largest Subset $m^+ \subseteq m$.
This contradicts the fact that Ordinal is not Element of Itself.
Thus there is an $a \in A$ such that $a \in m^+$.
Thus $a \in m$ or $a = m$.
Let $a \in m$.
Then $a \subsetneqq m$.
This contradicts the fact that $m \subseteq a$.
Thus $a = m$.
Thus it follows that $m \in A$.
Category:Ordinals
Intersection of Orthocomplements is Orthocomplement of Closed Linear Span

== Theorem ==
Let $H$ be a Hilbert space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of closed linear subspaces of $H$.
Then:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$
where:
:$\perp$ denotes orthocomplementation
:$\vee$ denotes closed linear span.
=== Corollary ===
Furthermore, the following equality holds:
== Proof ==
By definition of set equality, it suffices to prove the following two inclusions:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp \subseteq \paren {\vee \set {M_i : i \in I} }^\perp$
:$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \ds \bigcap_{i \mathop \in I} M_i^\perp$
=== $\ds \bigcap_{i \mathop \in I} M_i^\perp$ is contained in $\paren {\vee \set {M_i : i \in I} }^\perp$ ===
By Orthocomplement is Closed Linear Subspace and Closed Linear Subspaces Closed under Intersection, both spaces considered are closed linear subspaces of $H$.
By Orthocomplement Reverses Subset, the required containment is equivalent to:
:$\vee \set {M_i : i \in I} \subseteq \ds \paren {\bigcap_{i \mathop \in I} M_i^\perp }^\perp$
For $h \in \ds \bigcap_{i \mathop \in I} M_i^\perp$, by definition one has:
:$h \perp M_i$ for all $i \in I$
That is:
:$M_i \perp \ds \bigcap_{i \mathop \in I} M_i^\perp$
This is equivalent to saying that:
:$M_i \subseteq \ds \paren {\bigcap_{i \mathop \in I} M_i^\perp }^\perp$
Definition $(2)$ of closed linear span now grants the desired subset relation.
=== $\paren {\vee \set {M_i : i \in I} }^\perp$ is contained in $\ds \bigcap_{i \mathop \in I} M_i^\perp$ ===
By definition $(2)$ of closed linear span:
:$\forall i \in I: M_i \subseteq \vee \set {M_i^\perp : i \in I}$
By Orthocomplement Reverses Subset, it follows that:
:$\forall i \ni I: \paren {\vee \set {M_i : i \in I} }^\perp \subseteq M_i^\perp$
Therefore, by definition of set intersection:
:$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \ds \bigcap_{i \mathop \in I} M_i^\perp$
Thus we have established that:
:$h \in \paren {\vee \set {M_i : i \in I} }^\perp \iff h \in \ds \bigcap_{i \mathop \in I} M_i^\perp$
From the definition of set equality, it follows that:
:$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$
== Sources ==
* : $\text{II}.3$ Exercise $3$
Category:Hilbert Spaces
Category:Intersection of Orthocomplements is Orthocomplement of Closed Linear Span
Intersection of Orthocomplements is Orthocomplement of Closed Linear Span/Corollary

== Corollary to Intersection of Orthocomplements is Orthocomplement of Closed Linear Span ==
Let $H$ be a Hilbert space.
Let $\family {M_i}_{i \mathop \in I}$ be an $I$-indexed family of closed linear subspaces of $H$.
Then:
:$\ds \paren {\bigcap_{i \mathop \in I} M_i}^\perp = \vee \set {M_i^\perp : i \in I}$
where:
:$\perp$ denotes orthocomplementation
:$\vee$ denotes closed linear span.
== Proof ==
From Orthocomplement is Closed Linear Subspace, the $M_i^\perp$ are an $I$-indexed family of closed linear subspaces of $H$.
From Intersection of Orthocomplements is Orthocomplement of Closed Linear Span:
:$\ds \bigcap_{i \mathop \in I} \paren {M_i^\perp}^\perp = \paren {\vee \set {M_i^\perp : i \in I} }^\perp$
Taking the orthocomplement of both sides, and using Corollary to Double Orthocomplement is Closed Linear Span yields the result.
== Sources ==
* : $\text{II}.3$ Exercise $3$
Category:Intersection of Orthocomplements is Orthocomplement of Closed Linear Span
Intersection of Plane with Sphere is Circle

== Theorem ==
The intersection of a plane with a sphere is a circle.
== Proof ==
Let $S$ be a sphere of radius $R$ whose center is located for convenience at the origin.
Let $P$ be a plane which intersects $S$ but is not a tangent plane to $S$.
It is to be shown that $S \cap P$ is a circle.
Let $S$ and $P$ be embedded in a (real) cartesian space of $3$ dimensions.
Let this space be rotated until $P$ is parallel to the plane $z = 0$.
Thus from the Equation of Plane we have that $P$ can be described as:
:$z = c$
where $\size c < R$ (or $P$ would not intersect $S$).
Let $A = \tuple {x, y, z}$ be an arbitrary point on $S \cap P$.
We have that:
The result follows from Equation of Circle.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $2$. The spherical triangle.
Category:Spheres
Intersection of Plane with Sphere is Great Circle iff Passing through Center

== Theorem ==
Let $S$ be a sphere.
Let $P$ be a plane which intersects $S$ but is not tangent to $S$.
Let $C$ be the circle formed by the intersection between $P$ and $S$.
Then $C$ is a great circle of $S$ $P$ passes through the center of $S$.
Otherwise $C$ is a small circle of $S$.
== Proof ==
From Intersection of Plane with Sphere is Circle it is confirmed that $C$ is indeed a circle.
The result follows directly.
 
== Sources ==
* 
* 
Category:Small Circles
Category:Radii of Circles
Intersection of Positive Set and Negative Set is Null Set

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $P$ be a $\mu$-positive set. 
Let $N$ be a $\mu$-negative set.
Then: 
:$P \cap N$ is a $\mu$-null set.
== Proof ==
Note that, from Sigma-Algebra Closed under Countable Intersection:
:$P \cap N \in \Sigma$
We aim to show that:
:for each $E \in \Sigma$ with $E \subseteq P \cap N$ we have $\map \mu E = 0$.
Note first that from Intersection is Subset, we have: 
:$P \cap N \subseteq P$
so that: 
:$E \subseteq P$
So, since $P$ is $\mu$-positive, we have: 
:$\map \mu E \ge 0$
We also have that, from Intersection is Subset: 
:$P \cap N \subseteq N$
so that: 
:$E \subseteq N$
Since $N$ is $\mu$-negative, we have: 
:$\map \mu E \le 0$
Since: 
:$\map \mu E \ge 0$ and $\map \mu E \le 0$
we have: 
:$\map \mu E = 0$
So:
:for each $E \in \Sigma$ with $E \subseteq P \cap N$ we have $\map \mu E = 0$.
So:
:$P \cap N$ is a $\mu$-null set.
Category:Positive Sets
Category:Negative Sets
Intersection of Power Sets

== Theorem ==
The intersection of the power sets of two sets $S$ and $T$ is equal to the power set of their intersection:
:$\powerset S \cap \powerset T = \powerset {S \cap T}$
== Proof ==
== Also see ==
* Union of Power Sets
== Sources ==
* : $\S 5$: Complements and Powers
* : $\S 2$. Sets of sets: Exercise $5 \ \text{(a)}$
* : Chapter $1$: Sets and Logic: Exercise $7 \ \text{(ii)}$
Category:Power Set
Category:Set Intersection
Intersection of Primitive Recursive Sets

== Theorem ==
Let $A, B \subseteq \N$ be subsets of the set of natural numbers $\N$.
Let $A$ and $B$ both be primitive recursive.
Then $A \cap B$, the intersection of $A$ and $B$, is primitive recursive.
== Proof ==
$A$ and $B$ are primitive recursive, therefore so are their [Definition:Characteristic Function of Set|characteristic functions]] $\chi_A$ and $\chi_B$.
Let $n \in \N$ be a natural number.
From Characteristic Function of Intersection: Variant 1:
:$\chi_{A \cap B} \left({n}\right) = \chi_A \left({n}\right) \times \chi_B \left({n}\right)$
So:
: $\chi_{A \cap B} \left({n}\right) = \operatorname{mult} \left({\chi_A \left({n}\right), \chi_B \left({n}\right)}\right)$
Thus $A \cap B$ is defined by substitution from the primitive recursive functions $\operatorname{mult}$, $\chi_A$ and $\chi_B$.
Hence the result.
Category:Primitive Recursive Functions
Category:Set Intersection
Intersection of Quotient Groups

== Theorem ==
Let $N \lhd G$ be a normal subgroup of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {A \cap B} = \map \alpha A \cap \map \alpha B$
== Proof ==
From the proof of the Correspondence Theorem:
:$\map \alpha H \subseteq G / N$
Then:
== Sources ==
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.16 \ \text{(i)}$
Category:Quotient Groups
Intersection of Real Intervals is Real Interval

== Theorem ==
Let $I_1$ and $I_2$ be real intervals.
Then $I_1 \cap I_2$ is also a real interval.
== Proof ==
Let $x, y \in I_1 \cap I_2$. 
From the definition of a real interval, it suffices to show that:
:for each $z \in \R$ with $x \le z \le y$ we have $z \in I_1 \cap I_2$.
Let $z$ be a real number with:
:$x \le z \le y$
Since $x, y \in I_1$, we have: 
:$z \in I_1$
from the definition of a real interval.
Similarly, since $x, y \in I_2$, we have:
:$z \in I_2$
from the definition of a real interval.
So:
:$z \in I_1 \cap I_2$
as required.
== Sources ==
* : $\S 2.1$: Sets: Exercise $\text{B} \ 7$
Category:Real Intervals
Intersection of Recursively Enumerable Sets

== Theorem ==
Let $S, T \subseteq \N$ be recursively enumerable.
Then $S \cap T$ is recursively enumerable.
== Proof ==
By Set is Recursively Enumerable iff Domain of Recursive Function, there exist recursive $f, g : \N \to \N$ such that:
:$S = \Dom f$
:$T = \Dom g$
Define:
:$\map h x = \map f x + \map g x$
By:
* Addition is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
it follows that $h$ is recursive.
Let $x \in S \cap T$.
Then $x \in \Dom f$ and $x \in \Dom g$.
Thus, there exist $y, z \in \N$ such that:
:$y = \map f x$
:$z = \map g x$
Therefore:
:$\map h x = y + z$
Hence, $x \in \Dom h$.
Then, by definition of subset, $S \cap T \subseteq \Dom h$.
Let $x \in \Dom h$.
By definition of substitution, there must exist $y, z \in \N$ such that:
:$y = \map f x$
:$z = \map g x$
Therefore, $x \in \Dom f$ and $x \in \Dom g$.
It follows that $x \in S \cap T$.
Therefore, by definition of subset, $\Dom h \subseteq S \cap T$.
By definition of set equality, $S \cap T = \Dom h$.
Thus, by Set is Recursively Enumerable iff Domain of Recursive Function, $S \cap T$ is recursively enumerable.
Category:Recursion Theory
Intersection of Reflexive Relations is Reflexive

== Theorem ==
The intersection of two reflexive relations is also a reflexive relation.
== Proof ==
Let $\RR_1$ and $\RR_2$ be reflexive relations on a set $S$.
From Relation Contains Diagonal Relation iff Reflexive, we have that:
: $\Delta_S \subseteq \RR_1$
: $\Delta_S \subseteq \RR_2$
Hence from Intersection is Largest Subset:
: $\Delta_S \subseteq \RR_1 \cap \RR_2$
Hence the result, from Relation Contains Diagonal Relation iff Reflexive.
Category:Reflexive Relations
Category:Set Intersection
Intersection of Regular Closed Sets is not necessarily Regular Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ and $V$ be regular closed sets of $T$.
Then $U \cap V$ is not also necessarily a regular closed set of $T$.
== Proof ==
Proof by Counterexample:
By Closed Real Interval is Regular Closed, the closed real intervals:
:$\closedint 0 {\dfrac 1 2}, \closedint {\dfrac 1 2} 1$
are both regular closed sets of $\R$.
Consider $A$, the intersection of the two half-unit closed intervals:
:$A := \closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2} = \closedint {\dfrac 1 2} {\dfrac 1 2}$
From Interior of Closed Real Interval is Open Real Interval:
:$A^\circ = \openint {\dfrac 1 2} {\dfrac 1 2} = \O$
From Closure of Empty Set is Empty Set:
:$A^{\circ -} = \O \ne A$
Thus $A$ is not a regular closed set of $\R$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $6$
Category:Regular Closed Sets
Category:Closed Sets
Category:Set Intersection
Intersection of Relation Segments of Approximating Relations equals Way Below Closure

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below meet-continuous lattice.
Let $\map {\operatorname {App} } L$ be the set of all auxiliary approximating relations on $S$.
Let $x \in S$.
Then
:$\ds \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L} = x^\ll$
== Proof ==
By Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element:
:$\ds \bigcap \set {I \in \operatorname {Ids}: x \preceq \sup I} = x^\ll$
where $\operatorname {Ids}$ denotes the set of all ideals in $L$.
For all $I \in \operatorname {Ids}$ define a mapping $m_I: S \to \operatorname {Ids}$:
:$\forall x \in S: x \preceq \sup I \implies \map {m_I} x = \set {x \wedge i: i \in I}$
and
:$\forall x \in S: x \npreceq \sup I \implies \map {m_I} x = x^\preceq$
By Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element:
:$\ds \forall x \in S: \bigcap \set {\map {m_I} x: I \in \operatorname {Ids} } = x^\ll$
We will prove that
:$\set {\map {m_I} x: I \in \operatorname {Ids} } \subseteq \set {x^{\RR}: \RR \in \map {\operatorname {App} } L}$
Let $a \in \set {\map {m_I} x: I \in \operatorname {Ids} }$
Then
:$\exists I \in \operatorname {Ids}: a = \map {m_I} x$
By Down Mapping is Generated by Approximating Relation:
:$\exists \RR \in \map {\operatorname {App} } L: \forall s \in S: \map {m_I} s = s^\RR$
Then
:$a = x^\RR$
Thus
:$a \in \set {x^\RR: \RR \in \map {\operatorname {App} } L}$
By Intersection of Family is Subset of Intersection of Subset of Family:
:$\ds \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L} \subseteq x^\ll$
We will prove that
:$\set {x^\RR: \RR R \in \map {\operatorname {App} } L} \subseteq \set {I \in \operatorname {Ids}: x \preceq \sup I}$
Let $a \in \set {x^\RR: \RR \in \map {\operatorname {App} } L}$
Then
:$\exists \RR \in \map {\operatorname {App} } L: a = x^\RR$
By definition of approximating relation:
:$x = \sup a$
By Relation Segment of Auxiliary Relation is Ideal:
:$a \in \operatorname {Ids}$
Thus by definition of reflexivity:
:$a \in \set {I \in \operatorname {Ids}: x \preceq \sup I}$
By Intersection of Family is Subset of Intersection of Subset of Family:
:$\ds x^\ll \subseteq \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L}$
Hence the result by definition of set equality.
== Sources ==
* 
* 
Category:Approximating Relations
Category:Meet-Continuous Lattices
Intersection of Relation with Inverse is Symmetric Relation

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Then $\RR \cap \RR^{-1}$, the intersection of $\RR$ with its inverse, is symmetric.
== Proof ==
Let $\tuple {x, y} \in \RR \cap \RR^{-1}$
By definition of intersection:
:$\tuple {x, y} \in \RR$
:$\tuple {x, y} \in \RR^{-1}$
By definition of inverse relation:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR^{-1}$
:$\tuple {x, y} \in \RR^{-1} \implies \tuple {y, x} \in \paren {\RR^{-1} }^{-1}$
By Inverse of Inverse Relation the second statement may be rewritten:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR^{-1}$
:$\tuple {x, y} \in \RR^{-1} \implies \tuple {y, x} \in \RR$
Then by definition of intersection:
:$\tuple {y, x} \in \RR \cap \RR^{-1}$
Hence $\RR \cap \RR^{-1}$ is symmetric.
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $6$
Category:Set Intersection
Category:Inverse Relations
Category:Symmetric Relations