nyuuzyou/wikis huggingface dataset:
Intersection of Turing Machines

== Theorem ==
Let $T_1, T_2$ be Turing machines.
Let $\Sigma_1, \Sigma_2$ be the input symbols of $T_1, T_2$, respectively.
Let $L_1, L_2$ be the languages accepted by $T_1, T_2$, respectively.
There exists a Turing machine $T$ such that:
* The input symbols of $T$ are $\Sigma_1 \cap \Sigma_2$.
* The language accepted by $T$ is $L_1 \cap L_2$.
* $T$ halts on some input it is in the accepted language, or either $T_1$ or $T_2$ halt on it without accepting.
== Proof ==
Let:
:$T_1 = \tuple {Q_1, \Sigma_1, \Gamma_1, \delta_1, q_1, B, F_1}$
:$T_2 = \tuple {Q_2, \Sigma_2, \Gamma_2, \delta_2, q_2, B, F_2}$
As implied by the notation, the blank symbols of the two machines will be identified with each other, and denoted as $B$.
Construct the $2$-tape Turing machine $T' = \tuple{Q, \Sigma, \Gamma, \delta, q_0, B, F}$ as follows:
:$Q = \set {q_0, q_L} \cup Q_1 \times Q_2$
:$\Sigma = \Sigma_1 \cap \Sigma_2$
:$\Gamma = \Gamma_1 \cup \Gamma_2$
:$F = F_1 \times F_2$
:$\map \delta {q, s, t} = \begin{cases}
\tuple {q_0, s, s, R, R} & : q = q_0 \land s \ne B \\
\tuple {q_L, B, B, L, L} & : q = q_0 \land s = B \\
\tuple {q_L, s, t, L, L} & : q = q_L \land s \ne B \\
\tuple {\tuple {q_1, q_2}, B, B, R, R} & : q = q_L \land s = B \\
\tuple {\tuple {a', b'}, s', t', d_1, d_2} & : q = \tuple {a, b}
\end{cases}$
::where:
:::$\tuple {a', s', d_1} = \map {\delta_1} {a, s}$ if $\map {\delta_1} {a, s}$ is defined and $a \notin F_1$
:::$\tuple {a', s', S} = \tuple {a, s, S}$ if $a \in F_1$
::and likewise for $\delta_2$
The following special case also applies to $\delta$:
:$\map \delta {q, s, t}$ is undefined if $q \in F$
::That is, if $q = \tuple {a, b}$ such that $a \in F_1$ and $b \in F_2$
The states $q_0$ and $q_L$ effectively copy the input string onto tape $2$ without altering tape $1$.
The two machines are represented on the separate tapes in the natural way.
Once $T_1$ or $T_2$ enters an accepting state, emulation stops and it is saved in that state indefinitely.
Once the second machine enters an accepting state as well, the machine $T$ accepts and halts, by the special case.
If either machine halts without accepting, the transition function $\delta$ is undefined for that state, and $T$ halts.
Category:Turing Machines
Intersection of Two Ordinals is Ordinal

== Theorem ==
Let $S$ and $T$ be ordinals.
Then $S \cap T$ is an ordinal.
== Proof ==
Because $S$ and $T$ are ordinals, they are (strictly) well-ordered by the subset relation.
Let $a \in S \cap T$.
Then the initial segments $S_a$ and $T_a$ are such that:
:$S_a = a = T_a$
That is:
:$\set {x \in S: x \subset a} = a = \set {y \in T: y \subset a}$
So:
:$a = \set {z \in S \cap T: z \subset a} = \paren {S \cap T}_a$
Hence it is seen that $\paren {S \cap T}_a$ is an initial segment of both $S$ and $T$.
The result follows from Initial Segment of Ordinal is Ordinal.
== Sources ==
* : $\S 7.9$
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.8$
Category:Ordinals
Category:Set Intersection
Intersection of Unions with Complements is Subset of Union

== Theorem ==
Let $R, S, T$ be sets.
Then:
:$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$
== Proof ==
Let $x \in \left({R \cup S}\right) \cap \left({\overline R \cup T}\right)$.
Then by definition of set intersection, set union and set complement, we have:
:$\left({x \in R \lor x \in S}\right) \land \left({x \notin R \lor x \in T}\right)$
From Conjunction of Disjunctions with Complements implies Disjunction‎ it follows that:
:$x \in S \lor x \in T$
By definition of set union, it follows that:
:$x \in S \cup T$
By definition of subset it follows that:
:$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$
== Illustration by Venn Diagram ==
A Venn diagram illustrating this result is given below:
:
The red and purple field together mark $R \cup S$.
The blue and purple field together mark $\overline R \cup T$. Note that the outer rectangular area is also included in this, but has been left colorless for clarity.
The purple field marks $\left({R \cup S}\right) \cap \left({\overline R \cup T}\right)$, where the red and blue intersect with each other.
== Sources ==
* : $\S 5$: Complements and Powers
Category:Set Union
Category:Set Intersection
Category:Set Complement
Intersection of Upper Closures is Upper Closure of Join Operands

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $x, y \in S$.
Then $\paren {x \vee y}^\succeq = x^\succeq \cap y^\succeq$
== Proof ==
We will prove that:
:$\paren {x \vee y}^\succeq \subseteq x^\succeq \cap y^\succeq$
Let $a \in \paren {x \vee y}^\succeq$
By definition of upper closure of element:
:$x \vee y \preceq a$
By Join Succeeds Operands:
:$x \preceq x \vee y$ and $y \preceq x \vee y$
By definition of transitivity:
:$x \preceq a$ and $y \preceq a$
By definition of upper closure of element:
:$a \in x^\succeq$ and $a \in y^\succeq$
Thus by definition of intersection:
:$a \in x^\succeq \cap y^\succeq$
We will prove that:
:$x^\succeq \cap y^\succeq \subseteq \paren {x \vee y}^\succeq$
Let $a \in x^\succeq \cap y^\succeq$
Thus bt definition of intersection:
:$a \in x^\succeq$ and $a \in y^\succeq$
By definition of upper closure of element:
:$x \preceq a$ and $y \preceq a$
By definitions of supremum and Upper bound:
:$x \vee y \preceq a$
Thus by definition of upper closure of element:
:$a \in \paren {x \vee y}^\succeq$
Hence by definition of set equality:
:$\paren {x \vee y}^\succeq = x^\succeq \cap y^\succeq$
== Sources ==
* 
Category:Upper Closures
Intersection of Upper Section with Directed Set is Directed Set

== Theorem ==
Let $L = \struct {S, \preceq}$ be an ordered set.
Let $A, B$ be subsets of $S$ such that
:$A \cap B \ne \O$
and
:$A$ is an upper set,
:$B$ is a directed set.
Then $A \cap B$ is a directed set.
== Proof ==
Let $x, y \in A \cap B$.
By definition of intersection:
:$x, y \in A$ and $x, y \in B$
By definition of directed subset:
:$\exists z \in B: x \preceq z \land y \preceq z$
By definition of upper section:
:$z \in A$
Thus by definition of intersection:
:$z \in A \cap B$
Thus:
:$\exists z \in A \cap B: x \preceq z$ and $y \preceq z$
== Sources ==
* 
Category:Upper Sections
Intersection of Vertical Sections is Vertical Section of Intersection

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\set {E_\alpha : \alpha \in A}$ be a set of subsets of $X \times Y$. 
Let $x \in X$. 
Then: 
:$\ds \paren {\bigcap_{\alpha \in A} E_\alpha}_x = \bigcap_{\alpha \in A} \paren {E_\alpha}_x$
where:
:$\ds \paren {\bigcap_{\alpha \in A} E_\alpha}_x$ is the $x$-vertical section of $\ds \bigcap_{\alpha \in A} E_\alpha$
:$\paren {E_\alpha}_x$ is the $x$-vertical section of $E_\alpha$.
== Proof ==
Note that:
:$\ds y \in \bigcap_{\alpha \in A} \paren {E_\alpha}_x$
: 
:$y \in \paren {E_\alpha}_x$ for all $\alpha \in A$.
From the definition of the $x$-vertical section, this is equivalent to:
:$\tuple {x, y} \in E_\alpha$ for all $\alpha \in A$. 
This in turn is equivalent to: 
:$\ds \tuple {x, y} \in \bigcap_{\alpha \in A} E_\alpha$
Again applying the definition of the $x$-vertical section, this is equivalent to:
:$\ds y \in \paren {\bigcap_{\alpha \in A} E_\alpha}_x$
So:
:$\ds y \in \bigcap_{\alpha \in A} \paren {E_\alpha}_x$ $\ds y \in \paren {\bigcap_{\alpha \in A} E_\alpha}_x$
giving:
:$\ds \paren {\bigcap_{\alpha \in A} E_\alpha}_x = \bigcap_{\alpha \in A} \paren {E_\alpha}_x$
Category:Set Intersection
Category:Vertical Section of Sets
Intersection of Weak Lower Closures in Toset

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $a, b \in S$.
Then:
:$a^\preccurlyeq \cap b^\preccurlyeq = \paren {\min \set {a, b} }^\preccurlyeq$
where:
:$a^\preccurlyeq$ denotes the weak lower closure of $a$
:$\min$ denotes the min operation.
== Proof ==
As $\struct {S, \preccurlyeq}$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
, let $b \preccurlyeq a$.
Then it follows by definition of $\min$ that $\min \set {a, b} = b$.
Thus, from Intersection with Subset is Subset, it suffices to show that:
:$b^\preccurlyeq \subseteq a^\preccurlyeq$
By definition of weak lower closure, this comes down to showing that:
:$\forall c \in S: c \preccurlyeq b \implies c \preccurlyeq a$
So let $c \in S$ with $c \preccurlyeq b$.
Recall that $b \preccurlyeq a$.
Now as $\preccurlyeq$ is a total ordering, it is in particular transitive.
Hence $c \preccurlyeq a$.
== Also see ==
* Intersection of Weak Upper Closures in Toset
Category:Total Orderings
Intersection of Weak Upper Closures in Toset

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $a, b \in S$.
Then:
:$a^\succcurlyeq \cap b^\succcurlyeq = \paren {\map \max {a, b} }^\succcurlyeq$
where:
:$a^\succcurlyeq$ denotes weak upper closure of $a$
:$\max$ denotes the max operation.
== Proof ==
As $\struct {S, \preccurlyeq}$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
, let $a \preccurlyeq b$.
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:
:$b^\succcurlyeq \subseteq a^\succcurlyeq$
By definition of weak upper closure, this comes down to showing that:
:$\forall c \in S: b \preccurlyeq c \implies a \preccurlyeq c$
So let $c \in S$ with $b \preccurlyeq c$.
Recall that $a \preccurlyeq b$.
Now as $\preccurlyeq$ is a total ordering, it is in particular transitive.
Hence $a \preccurlyeq c$.
== Also see ==
* Intersection of Weak Lower Closures in Toset
Category:Total Orderings
Intersection of Zero Loci is Zero Locus

== Theorem ==
Let $k$ be a field.
Let $n \in \N_{>0}$.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $\mathbb S \subseteq \powerset A$ be a subset of the power set of $A$.
Then:
:$\ds \bigcap _{S \mathop \in \mathbb S} \map V S = \map V {\bigcup \mathbb S}$
where $\map V \cdot$ denotes the zero locus.
== Proof ==
Category:Algebraic Geometry
Intersection of two Open Sets of Neighborhood Space is Open

== Theorem ==
Let $\struct {S, \NN}$ be a neighborhood space.
Let $U$ and $V$ be open sets of $\struct {S, \NN}$.
Then $U \cap V$ is an open set of $\struct {S, \NN}$.
=== Corollary ===
== Proof ==
Let $U$ and $V$ be open sets of $\struct {S, \NN}$.
Let $x \in S$ such that $x \in U \cap V$.
Then $x \in U$ and $x \in V$, both of which are neighborhoods of $x$ by definition of open set.
By neighborhood space axiom $(\text N 4)$ it follows that $U \cap V$ is a neighborhood of $x$.
As $x$ is arbitrary, it follows that the above is true for all $x \in U \cap V$.
It follows by definition that $U \cap V$ 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
Intersection with Complement

== Theorem ==
The intersection of a set and its complement is the empty set:
:$S \cap \map \complement S = \O$
== Proof ==
Substitute $\mathbb U$ for $S$ and $S$ for $T$ in $T \cap \relcomp S T = \O$ from Intersection with Relative Complement is Empty.
== Also see ==
* Union with Complement
Notice the similarity with the Principle of Non-Contradiction.
The complement of a set is similar to the negation of a proposition, and intersection is similar to conjunction.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B v}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(i)}$
* : $\S 2$
* : $\text {(vii)}$
Category:Set Complement
Category:Set Intersection
Category:Empty Set
Intersection with Complement is Empty iff Subset

== Theorem ==
:$S \subseteq T \iff S \cap \map \complement T = \O$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cap T$ denotes the intersection of $S$ and $T$
:$\O$ denotes the empty set
:$\complement$ denotes set complement.
=== Corollary ===
== Proof ==
== Also presented as ==
Some sources present this as an alternative definition of a subset:
:Set $A$ is a subset of set $B$ if set $A$ contains no member that is not also in set $B$
but this is not mainstream.
== Also see ==
* Complement Union with Superset is Universe
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.3 \ \text{(b)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B vi}$
* : Chapter $1$: Sets: Subsets: Definition: $1.7$
Category:Subsets
Category:Set Intersection
Category:Set Complement
Category:Empty Set
Intersection with Empty Set

== Theorem ==
The intersection of any set with the empty set is itself the empty set:
:$S \cap \O = \O$
== Proof ==
== Also see ==
* Union with Empty Set
== Sources ==
* : $\S 4$: Unions and Intersections
* : Exercise $1.1: \ 8 \ \text{(b)}$
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : Chapter $1$: Exercise $1 \ \text {(f)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B ii}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 5 \ \text{(c)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $2 \ \text{(b)}$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
* : $1$: Set Theory and Logic: $\S 1$: Fundamental Concepts
* : $\S 2$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
* : $\text {(iii)}$
* : $\text {(i)}$
Category:Set Intersection
Category:Empty Set
Intersection with Normal Subgroup is Normal

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$.
Then $H \cap N$ is a normal subgroup of $H$.
== Proof ==
By Intersection of Subgroups is Subgroup, $H \cap N$ is a subgroup of $N$.
It remains to be shown that $H \cap N$ is normal in $H$.
Let $g \in H$ and $x \in H \cap N$. 
However, $H \le G$ and $ g \in H$ imply:
:$g \in G$
Because $N \lhd G$:
:$\forall n \in N: \forall g \in G: g n g^{-1} \in N$
Because $H \le G$ and therefore closed: 
:$\forall x \in H \cap N: \forall g \in H: g x g^{-1} \in H$
Therefore, from the definition of normal subgroups:
:$N \lhd G \implies g x g^{-1} \in N$
:$x \in N$
:$g \in G$
Thus:
:$g x g^{-1} \in H \cap N$
Because $H$ is a subgroup, $g x g^{-1} \in H \cap N$, $g \in H$, and the definition of normal subgroups:
:$H \cap N \lhd H$
== Examples ==
== Sources ==
* : Chapter $\text{II}$: Groups: Problem $\text{HH}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \delta$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.8$
Category:Normal Subgroups
Category:Intersection with Normal Subgroup is Normal
Intersection with Normal Subgroup is Normal/Examples/Subset Product of Intersection with Intersection

== Example of Use of Intersection with Normal Subgroup is Normal ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H_1, H_2$ be subgroups of $G$.
Let:
:$N_1 \lhd H_1$
:$N_2 \lhd H_2$
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$\paren {H_1 \cap N_2} \paren {H_2 \cap N_1} \lhd \paren {H_1 \cap H_2}$
== Proof ==
Consider arbitrary $x_h \in H_1 \cap H_2$.
By definition of set intersection, it follows that:
:$x_h \in H_1 \land x_h \in H_2$
We aim to show that:
:$x_h \paren {H_1 \cap N_2} \paren {H_2 \cap N_1} x_h^{-1} = \paren {H_1 \cap N_2} \paren {H_2 \cap N_1}$
thus demonstrating:
:$\paren {H_1 \cap N_2} \paren {H_2 \cap N_1} \lhd \paren {H_1 \cap H_2}$ 
by the Normal Subgroup Test.
First note that from Intersection with Normal Subgroup is Normal:
:$H_1 \cap N_2 \lhd H_1$
:$H_2 \cap N_1 \lhd H_2$
We have:
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \nu$
Category:Intersection with Normal Subgroup is Normal
Intersection with Normal Subgroup is Normal/Examples/Subset Product of Normal Subgroup with Intersection

== Example of Use of Intersection with Normal Subgroup is Normal ==
Let $\struct G$ be a group whose identity is $e$.
Let $H_1, H_2$ be subgroups of $G$.
Let:
:$N_1 \lhd H_1$
:$N_2 \lhd H_2$
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$
== Proof ==
Consider arbitrary $x_n \in N_1, x_h \in H_1 \cap H_2$.
Thus:
:$x_n x_h \in N_1 \paren {H_1 \cap H_2}$
Note that as $x_h \in H_1 \cap H_2$ it follows that $x_h \in H_1$.
We aim to show that:
:$x_n x_h N_1 \paren {H_1 \cap N_2} \paren {x_n x_h}^{-1} \subseteq N_1 \paren {H_1 \cap H_2}$
thus demonstrating $N_1 \paren {H_1 \cap N_2} \lhd N_1 \paren {H_1 \cap H_2}$ by the Normal Subgroup Test.
We have:
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \nu$
Category:Intersection with Normal Subgroup is Normal
Intersection with Relative Complement is Empty

== Theorem ==
The intersection of a set and its relative complement is the empty set:
:$T \cap \relcomp S T = \O$
== Proof ==
== Also see ==
* Union with Relative Complement
== Sources ==
* : $\S 5$: Complements and Powers
* : Chapter $1$: Exercise $1 \ \text {(b)}$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.2$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(h)}$
Category:Relative Complement
Category:Set Intersection
Intersection with Set Difference is Set Difference with Intersection

== Theorem ==
Let $R, S, T$ be sets.
Then:
: $\left({R \setminus S}\right) \cap T = \left({R \cap T}\right) \setminus S$
where:
: $R \setminus S$ denotes set difference
: $R \cap T$ denotes set intersection.
== Proof 1 ==
== Proof 2 ==
Category:Set Difference
Category:Set Intersection
Category:Intersection with Set Difference is Set Difference with Intersection
Intersection with Subclass is Subclass

== Theorem ==
Let $A$ and $B$ be classes.
Then:
:$A \subseteq B \iff A \cap B = A$
where:
:$A \subseteq B$ denotes that $A$ is a subclass of $B$
:$A \cap B$ denotes the intersection of $A$ and $B$.
== Proof ==
Let $A \cap B = A$.
Then by the definition of class equality:
:$A \subseteq A \cap B$
Thus:
 
Now let $A \subseteq B$.
 
We also have:
 
So as we have:
 
it follows from the definition of class equality that:
:$A \cap B = A$
== Also see ==
* Union with Superclass is Superclass
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(e)}$
Category:Subclasses
Category:Class Intersection
Intersection with Subgroup Product of Superset

== Theorem ==
Let $X, Y, Z$ be subgroups of a group $\struct {G, \circ}$.
Let $Y \subseteq X$.
Then:
:$X \cap \paren {Y \circ Z} = Y \circ \paren {X \cap Z}$
where $Y \circ Z$ denotes subset product.
== Proof ==
By definition of set equality, it suffices to prove:
:$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
and:
:$Y \circ \paren {X \cap Z} \subseteq X \cap \paren {Y \circ Z}$
=== $X \cap \paren {Y \circ Z}$ is contained in $Y \circ \paren {X \cap Z}$ ===
Let $s \in X \cap \paren {Y \circ Z}$.
Then:
So we have:
Thus by definition of set intersection:
: $z \in X \cap Z$
So:
:$s = y \circ z \in Y \circ \paren {X \cap Z}$
By definition of subset:
:$X \cap \paren {Y \circ Z} \subseteq Y \circ \paren {X \cap Z}$
=== $Y \circ \paren {X \cap Z}$ is contained in $X \cap \paren {Y \circ Z}$ ===
Then:
We have established that:
:$x \in Y \circ \paren {X \cap Z} \iff x \in X \cap \paren {Y \circ Z}$
From the definition of set equality:
:$Y \circ \paren {X \cap Z} = X \cap \paren {Y \circ Z}$
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 35 \theta$
Category:Subset Products
Category:Subgroups
Intersection with Subset is Subset

== Theorem ==
:$S \subseteq T \iff S \cap T = S$
where:
:$S \subseteq T$ denotes that $S$ is a subset of $T$
:$S \cap T$ denotes the intersection of $S$ and $T$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Union with Superset is Superset
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
* : $\S 4$: Unions and Intersections
* : Exercise $1.1: \ 9$
* : $\S 2.1$: Sets: Exercise $\text{B} \ 1 \ \text{(a), (b)}$
* : Chapter $1$: Exercise $1 \ \text {(d)}$
* : Chapter $1$. Sets: Exercise $2$
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.3 \ \text{(c)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B iv}$
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 5 \ \text{(d)}$
* : $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(c)}$
* : Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(b)}$
* : $\S 7.3 \ \text {(ii)}$: Unions and Intersections
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(vii)}$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Intersection with Subset is Subset
Category:Subsets
Category:Set Intersection
Intersection with Subset is Subset/Proof 1

== Theorem ==
== Proof ==
Let $S \cap T = S$.
Then by the definition of set equality, $S \subseteq S \cap T$.
Thus:
 
Now let $S \subseteq T$.
From Intersection is Subset we have $S \supseteq S \cap T$.
We also have:
 
So as we have:
 
it follows from the definition of set equality that:
:$S \cap T = S$
So we have:
 
and so:
:$S \subseteq T \iff S \cap T = S$
from the definition of equivalence.
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(f)}$
* : Appendix $\text{A}.2$: Theorem $\text{A}.11$
Category:Intersection with Subset is Subset
Intersection with Universal Set

== Theorem ==
The intersection of a set with the universal set is the set itself:
:$\mathbb U \cap S = S$
== Proof ==
== Also see ==
* Union with Universal Set
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.1$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B ii}$
* : $\S 2$
* : $\text {(iv)}$
Category:Universal Set
Category:Set Intersection
Intersections of Line joining Conjugate Points with Circle form Harmonic Range

== Theorem ==
Let $\CC$ be a circle.
Let $P$ and $Q$ be conjugate points with respect to $\CC$.
Let $A$ and $B$ be the points of intersection of $PQ$ with $\CC$.
Then $A$ and $B$ are harmonic conjugates with respect to $P$ and $Q$.
== Proof ==
By definition of conjugate points, the polar of $P$ passes through $Q$.
Hence by Harmonic Property of Pole and Polar wrt Circle, $A$ and $B$ are harmonic conjugates with respect to $P$ and $Q$.
== Sources ==
* : Chapter $\text {III}$. The Circle: $8$. Reciprocal property of pole and polar
Category:Conjugate Points
Interval Defined by Absolute Value

== Theorem ==
Let $\xi, \delta \in \R$ be real numbers.
Let $\delta > 0$.
Then:
=== Open Interval Defined by Absolute Value ===
=== Closed Interval Defined by Absolute Value ===
== Also see ==
* Complement of Interval Defined by Absolute Value
Category:Real Intervals
Category:Absolute Value Function