nyuuzyou/wikis huggingface dataset:
Intersection of Relations Compatible with Operation is Compatible

== Theorem ==
Let $\struct {S, \circ}$ be a closed algebraic structure.
Let $\mathscr F$ be a indexed family of relations on $S$.
Suppose that each element of $\mathscr F$ is compatible with $\circ$.
Let $\QQ = \bigcap \mathscr F$ be the intersection of $\mathscr F$.
Then $\QQ$ is a relation compatible with $\circ$.
== Proof ==
Let $x, y, z \in S$.
Suppose that $x \mathrel \QQ y$.
Then for each $\RR \in \mathscr F$:
:$x \mathrel \RR y$
Then since $\RR$ is a relation compatible with $\circ$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Since this holds for each $\RR \in \mathscr F$:
:$\paren {x \circ z} \mathrel \QQ \paren {y \circ z}$
We have shown that:
:$\forall x, y, z \in S: x \mathrel \QQ y \implies \paren {x \circ z} \mathrel \QQ \paren {y \circ z}$
A similar argument shows that:
:$\forall x, y, z \in S: x \mathrel \QQ y \implies \paren {z \circ x} \mathrel \QQ \paren {z \circ y}$
so $Q$ is a relation compatible with $\circ$.
Category:Compatible Relations
Intersection of Relations is Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\FF$ be a family of relations from $S$ to $T$.
Let $\ds \RR = \bigcap \FF$, the intersection of all the elements of $\FF$.
 
Then $\RR$ is a relation from $S$ to $T$.
== Proof ==
By the definition of a relation from $S$ to $T$, each element of $\FF$ is a subset of $S \times T$.
By Intersection of Subsets is Subset: Set of Sets:
:$\RR \subseteq S \times T$
Therefore, by the definition of a relation from $S$ to $T$, $\RR$ is a relation from $S$ to $T$.
Category:Relation Theory
Intersection of Ring Ideals is Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring
Let $\mathbb L$ be a non-empty set of ideals of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself an ideal of $R$.
== Proof ==
Let $L = \bigcap \mathbb L$.
From Intersection of Subrings is Subring, we have that $L$ is a subring of $R$.
Let $x \in L$ and $y \in R$.
Then:
:$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$
as every element of $\bigcap \mathbb L$, including $T$, is an ideal of $R$.
If $y \in R$, then $x \circ y$ and $y \circ x$ are in every element of $\mathbb L$, and hence in $L$.
So $L$ is an ideal of $R$.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $39$
* : $\S 2.3$: Some properties of subrings and ideals: Lemma $2.13 \ \text{(i)}$
* : Chapter $9$: Rings: Exercise $5$
Category:Set Intersection
Category:Ideal Theory
Intersection of Ring Ideals is Largest Ideal Contained in all Ideals

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring
Let $\mathbb L$ be a non-empty set of ideals of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is the largest ideal of $R$ contained in each member of $\mathbb L$.
== Proof ==
Let $L = \bigcap \mathbb L$.
From Intersection of Ring Ideals is Ideal $L$ is indeed an ideal of $R$.
Let $L = \bigcap \mathbb L$.
From Intersection of Subrings is Largest Subring Contained in all Subrings, we have that $L$ is the largest subring of $R$ contained in each member of $\mathbb L$.
As $L$ is the largest subring of $R$ contained in each member of $\mathbb L$, and it is an ideal of $R$, there can be no larger ideal as it would then not be a subring.
So $L$ is the largest ideal of $R$ contained in each member of $\mathbb L$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$
Category:Set Intersection
Category:Ideal Theory
Intersection of Rings of Sets

== Theorem ==
Let $\RR_k$ be a ring of sets, where $k$ is an element of an arbitrary set of indices.
Then their intersection $\ds \RR = \bigcap_k \RR_k$ is itself a ring of sets.
== Proof ==
Consider the set $\ds \SS = \bigcup_k \RR_k$.
Let $S = \set {X \in Y: Y \in \SS}$.
This contains all the elements of all the sets contained in all the $\RR_k$.
Now consider the power set $\powerset S$ of $S$.
By Power Set is Algebra of Sets and the definition of algebra of sets, we have that $\powerset S$ is a ring of sets.
Thus $\struct {\powerset S, *, \cap}$ is a ring (in the abstract algebraic sense of the term).
From the method of construction of $\powerset S$, it follows that all the $\struct {\RR_k, *, \cap}$ are subrings of $\powerset S$.
The result then follows from Intersection of Subrings is Subring.
Category:Rings of Sets
Intersection of Semilattice Ideals is Ideal

== Theorem ==
Let $\struct {S, \preceq}$ be a meet semilattice.
Let $I_1, I_2$ be ideals in $\struct {S, \preceq}$.
Then $I_1 \cap I_2$ is ideal in $\struct {S, \preceq}$
=== Set of Sets ===
== Proof ==
=== Directed Subset ===
Let $x, y \in I_1 \cap I_2$.
By definition of intersection:
:$x, y \in I_1$ and $x, y \in I_2$
By definition of directed subset:
:$\exists z_1 \in I_1: x \preceq z_1 \land y \preceq z_1$
and
:$\exists z_2 \in I_2: x \preceq z_2 \land y \preceq z_2$
By Meet Precedes Operands:
:$z_1 \wedge z_2 \preceq z_1$ and $z_1 \wedge z_2 \preceq z_2$
By definition of lower section:
:$z_1 \wedge z_2 \in I_1$ and $z_1 \wedge z_2 \in I_2$
By definition of intersection:
:$z_1 \wedge z_2 \in I_1 \cap I_2$
By definition of meet:
:$z_1 \wedge z_2 = \inf \set {z_1, z_2}$
By definition of infimum:
:$x \preceq z_1 \wedge z_2$ and $y \preceq z_1 \wedge z_2$
Thus by definition:
:$I_1 \cap I_2$ is directed.
=== Lower Section ===
Let $x, y \in S$ such that:
:$x \in I_1 \cap I_2$ and $y \preceq x$
By definition of intersection:
:$x \in I_1$ and $x \in I_2$
By definition of lower section:
:$y \in I_1$ and $y \in I_2$
By definition of intersection:
:$y \in I_1 \cap I_2$
Thus by definition:
:$I_1 \cap I_2$ is a lower section.
=== Non-Empty Set ===
By assumption:
:$I_1 \ne \O$ and $I_2 \ne \O$
By definition of non-empty:
:$\exists x: x \in I_1$ and $\exists y: y \in I_2$
By Meet Precedes Operands:
:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
By definition of lower section:
:$x \wedge y \in I_1$ and $x \wedge y \in I_2$
By definition of intersection:
:$x \wedge y \in I_1 \cap I_2$
Thus by definition:
:$I_1 \cap I_2$ is non-empty.
Thus by definition:
:$I_1 \cap I_2$ is ideal in $\struct {S, \preceq}$.
== Sources ==
* 
Category:Join and Meet Semilattices
Intersection of Semilattice Ideals is Ideal/Set of Sets

== Theorem ==
Let $\struct {S, \preceq}$ be a bounded below join semilattice.
Let $\II$ be a set of ideals in $\struct {S, \preceq}$.
Then $\bigcap \II$ is an ideal in $\struct {S, \preceq}$.
== Proof ==
=== Non-Empty Set ===
By Bottom in Ideal:
:$\forall I \in \II: \bot \in I$
where $\bot$ denotes the smallest element in $S$.
By definition of intersection:
:$\bot \in \bigcap \II$
Hence $\bigcap \II$ is non-empty.
=== Lower Section ===
Let $x \in \bigcap \II$, $y \in S$ such that
:$y \preceq x$
We will prove that:
:$\forall I \in \II: y \in I$
Let $I \in \II$.
By definition of intersection:
:$x \in I$
Thus by definition of lower section:
:$y \in I$
Thus by definition of intersection:
:$y \in \bigcap \II$
=== Directed Subset ===
Let $x, y \in \bigcap \II$.
We will prove that:
:$\forall I \in \II: x \vee y \in I$
Let $I \in \II$.
By definition of intersection:
:$x, y \in I$
By definition of directed:
:$\exists z \in I: x \preceq z \land y \preceq z$
By definition of join:
:$x \vee y = \sup \set {x, y}$
By definition:
:$z$ is an upper bound for $\set {x, y}$.
By definition of supremum:
:$x \vee y \preceq z$
Thus by definition of lower section:
:$x \vee y \in I$
Thus by definition of intersection:
:$x \vee y \in \bigcap \II$
Hence by Directed in Join Semilattice:
:$\bigcap \II$ is directed.
== Sources ==
* 
Category:Set Intersection
Category:Join and Meet Semilattices
Intersection of Set of Submodules containing Subset is Smallest Submodule

== 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 $S \subset M$ be a subset of $M$.
Let $T$ be the set of all submodules of $M$ which contain $S$ as a subset.
Then the intersection $\bigcap T$ is the smallest submodule of $M$ containing $S$.
== Proof ==
By hypothesis, we have:
:$\ds T = \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} }$
From Intersection of Submodules is Submodule:General Result, it follows that $\bigcap T$ is a submodule of $M$.
As $S \subseteq M'$ for all $M' \in T$, it follows that $S \subseteq \bigcap T$.
Let $M' \in T$ be a submodule of $M$ such that $S \subseteq M'$.
From Intersection is Subset:General Result, it follows that $M' \subseteq \bigcap T$.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$
Category:Submodules
Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions

== Theorem ==
Let $S$ be a set of sets.
Let $S$ be such that:
:$\forall x \in S: x$ is closed under chain unions.
Then the intersection $\ds \bigcap S$ of $S$ is also closed under chain unions.
== Proof ==
== Also see ==
* Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Lemma $2.7$
Category:Closure under Chain Unions
Category:Set Intersection
Category:Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions/Proof

== Theorem ==
== Proof ==
First we note that by definition of intersection of $S$:
:$\ds \bigcap S := \set {y: \forall x \in S: y \in x}$
Recall the definition of closed under chain unions:
$S$ is closed under chain unions :
:for every chain $C$ of elements of $S$, $\ds \bigcup C$ is also in $S$.
Let $C_\cap$ be a chain in $\ds \bigcap S$.
Then by :
:$\forall x \in S: C_\cap$ is a chain in $x$.
Hence as $x$ is closed under chain unions for all $x \in S$:
:$\forall x \in S: \ds \bigcup C_\cap$ is in $x$
Hence by :
:$\ds \bigcup C_\cap$ is in $S$
Hence the result by definition of closed under chain unions.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Exercise $2.1$
Category:Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping

== Theorem ==
Let $S$ be a set of sets.
Let $g$ be a mapping such that:
:for every $x \in S$, $x$ is closed under $g$.
Then the intersection $\bigcap S$ of $S$ is also closed under $g$.
== Proof ==
== Also see ==
* Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Lemma $2.7$
Category:Closedness under Mappings
Category:Set Intersection
Category:Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping/Proof

== Theorem ==
== Proof ==
The domain of $g$ is not made clear, but the assumption is that:
:$\forall x \in S: \forall y \in x: y \in \Dom g$
First we note that by definition of intersection of $S$:
:$\bigcap S := \set {y: \forall x \in S: y \in x}$
Then:
Hence the result by definition of closed under $g$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Exercise $2.1$
Category:Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
Intersection of Sets of Integer Multiples

== Theorem ==
Let $m, n \in \Z$ such that $m n \ne 0$.
Let $m \Z$ denote the set of integer multiples of $m$.
Then:
:$m \Z \cap n \Z = \lcm \set {m, n} \Z$
where $\lcm$ denotes lowest common multiple.
== Proof ==
Let $x \in m \Z \cap n \Z$.
Then by definition of set intersection:
:$m \divides x$ and $n \divides x$
So from LCM Divides Common Multiple:
:$\lcm \set {m, n} \divides x$
and so $x \in \lcm \set {m, n} \Z$
That is:
:$m \Z \cap n \Z \subseteq \lcm \set {m, n} \Z$
Now suppose $x \in \lcm \set {m, n} \Z$.
Then $\lcm \set {m, n} \divides x$.
Thus by definition of lowest common multiple:
:$m \divides x$ 
and:
:$n \divides x$
and so:
:$x \in m \Z \land x \in n \Z$
That is:
:$x \in m \Z \cap n \Z$
and so:
:$\lcm \set {m, n} \Z \subseteq m \Z \cap n \Z$
The result follows by definition of set equality.
== Examples ==
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $10$
Category:Sets of Integer Multiples
Category:Lowest Common Multiple
Category:Intersection of Sets of Integer Multiples
Intersection of Sets of Integer Multiples/Examples

== Examples of Use of Intersection of Sets of Integer Multiples ==
=== Example: $2 \Z \cap 7 \Z$ ===
=== Example: $\paren {3 \Z \cap 6 \Z} \cup 18 \Z$ ===
=== Example: $6 \Z \cap 15 \Z$ ===
Category:Examples of Use of Intersection of Sets of Integer Multiples
Intersection of Sets of Integer Multiples/Examples/(3 Z cap 6 Z) cup 18 Z

== Examples of Use of Intersection of Sets of Integer Multiples ==
Let $m \Z$ denote the set of integer multiples of $m$.
Then:
:$\paren {3 \Z \cap 6 \Z} \cup 18 \Z = 6 \Z$
== Proof ==
From Intersection of Sets of Integer Multiples:
:$3 \Z \cap 6 \Z = \lcm \set {3, 6} \Z = 6 \Z$
Then we have that:
:$18 \Z \subseteq 6 \Z$
and so from Union with Superset is Superset:
:$6 \Z \cup 18 \Z = 6 \Z$
Hence the result.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $2$
Category:Examples of Use of Intersection of Sets of Integer Multiples
Intersection of Sets of Integer Multiples/Examples/2 Z cap 7 Z

== Examples of Use of Intersection of Sets of Integer Multiples ==
Let $m \Z$ denote the set of integer multiples of $m$.
Then:
:$2 \Z \cap 7 \Z = 14 \Z$
== Proof ==
From Intersection of Sets of Integer Multiples
:$2 \Z \cap 7 \Z = \lcm \set {2, 7} \Z$
We have that:
:$2 \perp 7$
where $\perp$ denotes coprimality.
Then from LCM equals Product iff Coprime:
:$\lcm \set {2, 7} = 2 \times 7 = 14$
Hence the result.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $2$
Category:Examples of Use of Intersection of Sets of Integer Multiples
Intersection of Sets of Integer Multiples/Examples/6 Z cap 15 Z

== Examples of Use of Intersection of Sets of Integer Multiples ==
Let $m \Z$ denote the set of integer multiples of $m$.
Then:
:$6 \Z \cap 15 \Z = 30 \Z$
== Proof ==
From Intersection of Sets of Integer Multiples
:$6 \Z \cap 15 \Z = \lcm \set {6, 15} \Z$
From Lowest Common Multiple of $6$ and $15$:
:$\lcm \set {6, 15} = 30$
Hence the result.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $2$
Category:Examples of Use of Intersection of Sets of Integer Multiples
Intersection of Sigma-Algebras

== Theorem ==
Let $X$ be a set.
Let $\family {\Sigma_i}_{i \mathop \in I}$ be a indexed set of $\sigma$-algebras on $X$.
Then $\Sigma := \ds \bigcap_{i \mathop \in I} \Sigma_i$ is also a $\sigma$-algebra on $X$.
Here, $\ds \bigcap$ denotes set intersection.
== Proof ==
Verify the axioms for a $\sigma$-algebra in turn:
=== Axiom $(1)$ ===
As all the $\Sigma_i$ are $\sigma$-algebras, $X \in \Sigma_i$ for all $i \in I$.
Hence $X \in \Sigma$ by definition of intersection.
=== Axiom $(2)$ ===
Let $E \in \Sigma$. 
Then for all $i \in I$, $E \in \Sigma_i$, by definition of intersection.
As all the $\Sigma_i$ are $\sigma$-algebras, also, for all $i \in I$:
:$X \setminus E \in \Sigma_i$
Hence $X \setminus E \in \Sigma$ by definition of intersection.
=== Axiom $(3)$ ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\Sigma$.
Then it is also a sequence in every $\Sigma_i$.
Hence, for all $i \in I$, as the $\Sigma_i$ are all $\sigma$-algebras:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma_i$
By definition of intersection, it follows that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
== Sources ==
* : $3.4 \ \text{(i)}$
* : $1.1$: Algebras and Sigma-Algebras: Proposition $1.1.2$
Category:Sigma-Algebras
Intersection of Singleton

== Theorem ==
Consider the set of sets $A$ such that $A$ consists of just one set $x$:
:$A = \set x$
Then the intersection of $A$ is $x$:
:$\bigcap A = x$
== Proof ==
Let $A = \set x$.
Then from the definition:
:$\bigcap \set x = \set {y: \forall z \in \set x: y \in z}$
from which it follows directly:
:$\bigcap \set x = \set {y: y \in x}$
as $x$ is the only set in $\set x$.
That is:
:$\bigcap A = x$
== Also see ==
* Union of Singleton
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Unions and Intersections
* : $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.2: \ \text{(i)}$
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.5. \ \text {(b)}$
Category:Set Intersection
Category:Singletons
Intersection of Straight Line in Homogeneous Cartesian Coordinates with Axes

== Theorem ==
Let $\LL$ be a straight line embedded in a cartesian plane $\CC$.
Let $\LL$ be given in homogeneous Cartesian coordinates by the equation:
:$l X + m Y + n Z = 0$
such that $l$ and $m$ are not both zero.
Then $\LL$ intersects:
:the $x$-axis $Y = 0$ at the point $\tuple {-n, 0, l}$
:the $y$-axis $X = 0$ at the point $\tuple {0, -n, m}$
When $l = 0$, $\LL$ is parallel to the $x$-axis with its point at infinity at $\tuple {-n, 0, 0}$
When $m = 0$, $\LL$ is parallel to the $y$-axis with its point at infinity at $\tuple {0, -n, 0}$.
== Proof ==
The intersection of $\LL$ with the $x$-axis is the point $\tuple {X, Y, Z}$ satisfied by:
which is satisfied by setting $X = -n$ and $Z = l$, while $Y = 0$.
The intersection of $\LL$ with the $y$-axis is the point $\tuple {X, Y, Z}$ satisfied by:
which is satisfied by setting $Y = -n$ and $Z = m$, while $X = 0$.
When $l = 0$ we have:
When $m = 0$ we have:
The result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Homogeneous Cartesian Coordinates
Intersection of Straight Lines in General Form

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, given by the equations:
The point of intersection of $\LL_1$ and $\LL_2$ has coordinates given by:
:$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
This point exists and is unique $l_1 m_2 \ne l_2 m_1$.
=== Determinant Form ===
This result can also be expressed in the form:
== Proof ==
First note that by the parallel postulate $\LL_1$ and $\LL_2$ have a unique point of intersection they are not parallel.
From Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ are parallel $l_1 m_2 = l_2 m_1$.
Let the equations for $\LL_1$ and $\LL_2$ be given.
Let $\tuple {x, y}$ be the point on both $\LL_1$ and $\LL_2$.
We have:
Similarly:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $8$. Intersection of two lines
Category:Straight Lines
Intersection of Straight Lines in General Form/Determinant Form

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, given by the equations:
The point of intersection of $\LL_1$ and $\LL_2$ has coordinates given by:
:$\dfrac x {\begin {vmatrix} m_1 & n_1 \\ m_2 & n_2 \end {vmatrix} } = \dfrac y {\begin {vmatrix} n_1 & l_1 \\ n_2 & l_2 \end {vmatrix} } = \dfrac 1 {\begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} }$
where $\begin {vmatrix} \cdot \end {vmatrix}$ denotes a determinant.
This point exists and is unique $\begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} \ne 0$.
== Proof ==
From Intersection of Straight Lines in General Form, the point of intersection of $\LL_1$ and $\LL_2$ has coordinates given by:
:$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
which exists and is unique $l_1 m_2 \ne l_2 m_1$.
The result follows by Determinant of Order 2.
Category:Straight Lines
Intersection of Straight Lines in Homogeneous Cartesian Coordinate Form

== Theorem ==
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$.
Let $\LL_1$ and $\LL_2$ be given in homogeneous Cartesian coordinates by the equations:
The point of intersection of $\LL_1$ and $\LL_2$ is unique and has homogeneous Cartesian coordinates given by:
:$\tuple {X, Y, Z} = \tuple {m_1 n_2 - m_2 n_1, n_1 l_2 - n_2 l_1, l_1 m_2 - l_2 m_1}$
or any multiple of these.
This can also be expressed as:
:$\tuple {X, Y, Z} = \tuple {\begin {vmatrix} m_1 & n_1 \\ m_2 & n_2 \end {vmatrix}, \begin {vmatrix} n_1 & l_1 \\ n_2 & l_2 \end {vmatrix} , \begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} }$
== Proof ==
First note that by the parallel postulate $\LL_1$ and $\LL_2$ have a unique point of intersection they are not parallel.
So, first let it be the case that $\LL_1$ and $\LL_2$ are not parallel.
Let the equations for $\LL_1$ and $\LL_2$ be given.
Let $P = \tuple {X, Y, Z}$ be the point on both $\LL_1$ and $\LL_2$ expressed in homogeneous Cartesian coordinates.
By definition:
where $P = \tuple {x, y}$ described in conventional Cartesian coordinates.
From Intersection of Straight Lines in General Form:
:$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
Hence:
:$\dfrac X Z \dfrac 1 {m_1 n_2 - m_2 n_1} = \dfrac Y Z \dfrac 1 {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$
and so multiplying by $Z$:
:$\dfrac X {m_1 n_2 - m_2 n_1} = \dfrac Y {n_1 l_2 - n_2 l_1} = \dfrac Z {l_1 m_2 - l_2 m_1}$
Hence we have:
for arbitrary $Z \in \R$ such that $Z \ne 0$.
So, let $Z = l_1 m_2 - l_2 m_1$.
Thus we have:
From Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ are parallel $l_1 m_2 = l_2 m_1$.
So, let $l_1 m_2 = l_2 m_1$.
Hence:
:$Z = 0$
and it is seen that the point of intersection is now the point at infinity.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Straight Lines
Category:Homogeneous Cartesian Coordinates
Intersection of Strict Lower Closures in Toset

== Theorem ==
Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $a,b \in S$.
Then:
:$a^\prec \cap b^\prec = \left({\min \left({a, b}\right)}\right)^\prec$
where:
: $a^\prec$ denotes strict lower closure of $a$
: $\min$ denotes the min operation.
== Proof ==
As $\left({S, \preceq}\right)$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$.
Since both sides are seen to be invariant upon interchanging $a$ and $b$, let WLOG $b \preceq a$.
Then it follows by definition of $\min$ that $\min \left({a, b}\right) = b$.
Thus, from Intersection with Subset is Subset, it suffices to show that $b^\prec \subseteq a^\prec$.
By the definition of strict lower closure, this comes down to showing that:
:$\forall c \in S: c \prec b \implies c \prec a$
So let $c \in S$ with $c \prec b$, and recall that $b \preceq a$.
By Strictly Precedes is Strict Ordering, $c \prec a$.
== Also see ==
* Intersection of Strict Upper Closures in Toset
Category:Total Orderings
Intersection of Strict Upper Closures in Toset

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $a, b \in S$.
Then:
:$a^\succ \cap b^\succ = \paren {\map \max {a, b} }^\succ$
where:
:$a^\succ$ denotes strict upper closure of $a$
:$\max$ denotes the max operation.
== Proof ==
As $\struct {S, \preceq}$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
, then, let $a \preceq b$.
Then it follows by definition of $\max$ that:
:$\map \max {a, b} = b$
Thus, from Intersection with Subset is Subset, it suffices to show that $b^\succ \subseteq a^\succ$.
By definition of strict upper closure, this comes down to showing that:
:$\forall c \in S: b \prec c \implies a \prec c$
So let $c \in S$ with $b \prec c$, and recall that $a \preceq b$.
By Strictly Precedes is Strict Ordering, $a \prec c$.
== Also see ==
* Intersection of Strict Lower Closures in Toset
Category:Total Orderings