nyuuzyou/wikis huggingface dataset:
Intersection of Subalgebras is Subalgebra

== Theorem ==
Let $K$ be a field.
Let $\struct {A, +, \cdot, \circ}_K$ be an algebra over $K$.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be an $I$-indexed family of subalgebras of $A$.
Let:
:$\ds B = \bigcap_{\alpha \mathop \in I} A_\alpha$
Then $B$ is a subalgebra of $A$.
== Proof ==
From Set of Linear Subspaces is Closed under Intersection, $\struct {B, +, \cdot}_K$ is a vector subspace of $\struct {A, +, \cdot}_K$.
Now let $x, y \in B$.
That is, $x, y \in A_\alpha$ for each $\alpha \in I$.
Since $A_\alpha$ is a subalgebras of $A$ for each $\alpha \in I$, we have:
:$x y \in A_\alpha$ for each $\alpha \in I$.
So $x y \in B$. 
So $B$ is a subalgebra of $A$.
Category:Algebras over Fields
Intersection of Subfields Containing Subset is Smallest

== Theorem ==
Let $\struct {F, +, \circ}$ be a field.
Let $S \subseteq F$ be a subset of $F$.
Let $L$ be the intersection of the set of all subfields of $F$ containing $S$.
Then $L$ is the smallest subfield of $F$ containing $S$. 
== Proof ==
A field is by definition also a division subring.
Thus $L$ is the intersection of the set of all division subrings of $F$ containing $S$.
From Intersection of Division Subrings Containing Subset is Smallest, $L$ is the smallest division subring of $F$ containing $S$.
From Intersection of Subfields is Subfield, $L$ is a subfield of $F$.
As $L$ is the smallest division subring of $F$ containing $S$, and it is a subfield of $F$, there can be no smaller subfield of $F$ containing $S$ as it would then not be a division subring.
So the intersection of the set of all subfields of $F$ containing $S$ is the smallest subfield of $F$ containing $S$. 
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$
Category:Set Intersection
Category:Subfields
Intersection of Subfields is Largest Subfield Contained in all Subfields

== Theorem ==
Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb K$ be a non-empty set of subfields of $F$.
Let $\bigcap \mathbb K$ be the intersection of the elements of $\mathbb K$.
Then $\bigcap \mathbb K$ is the largest subfield of $F$ contained in each element of $\mathbb K$.
== Proof ==
Let $L = \bigcap \mathbb K$.
From Intersection of Subfields is Subfield, $\struct {L, +, \circ}$ is itself a subfield of $\struct {F, +, \circ}$.
A field is by definition also a division subring.
Thus $\struct {L, +, \circ}$ is the largest division subring of $F$ contained in each element of $\mathbb K$.
But as $\struct {F, +, \circ}$ is a field, $\circ$ is commutative.
So by Restriction of Commutative Operation is Commutative, $\struct {L, +, \circ}$ is also a field.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$
Category:Set Intersection
Category:Subfields
Intersection of Subfields is Subfield

== Theorem ==
Let $\struct {F, +, \circ}$ be a field.
Let $\mathbb K$ be a non-empty set of subfields of $F$.
Then the intersection $\ds \bigcap \mathbb K$ of the members of $\mathbb K$ is itself a subfield of $F$.
== Proof ==
Let $\ds L = \bigcap \mathbb K$.
A field is by definition also a division ring.
From Intersection of Division Subrings is Division Subring, $L$ is itself a division subring of $F$.
As $\struct {F, +, \circ}$ is a field, $\circ$ is commutative on $F$.
By Restriction of Commutative Operation is Commutative, it follows that $\circ$ is also commutative on $L$.
Thus $\struct {L, +, \circ}$ is a division ring where $\circ$ is commutative.
Thus, by definition, of $\struct {L, +, \circ}$ is a field.
Hence the result.
== Sources ==
* : Chapter $4$: Fields: $\S 16$. Subfields: Theorem $27$
Category:Set Intersection
Category:Subfields
Intersection of Subgroups is Subgroup

== Theorem ==
The intersection of two subgroups of a group is itself a subgroup of that group:
:$\forall H_1, H_2 \le \struct {G, \circ}: H_1 \cap H_2 \le G$
It also follows that $H_1 \cap H_2 \le H_1$ and $H_1 \cap H_2 \le H_2$.
=== General Result ===
== Proof ==
Let $H = H_1 \cap H_2$ where $H_1, H_2 \le \struct {G, \circ}$.
Then:
As $H \subseteq H_1$ and $H \subseteq H_2$, the other results follow directly.
== Sources ==
* : $\S 5.2$. Subgroups: Example $94$
* : $\S 1.9$: Theorem $17$
* : $\S 36.6$ Subgroups
* : $\S 3.2$: Groups; the axioms
Category:Subgroups
Category:Set Intersection
Intersection of Subgroups is Subgroup/General Result

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\mathbb S$ be a set of subgroups of $\struct {G, \circ}$, where $\mathbb S \ne \O$.
Then the intersection $\ds \bigcap \mathbb S$ of the elements of $\mathbb S$ is itself a subgroup of $G$.
Also, $\ds \bigcap \mathbb S$ is the largest subgroup of $\struct {G, \circ}$ contained in each element of $\mathbb S$.
== Proof ==
Let $\ds H = \bigcap \mathbb S$.
Let $H_k$ be an arbitrary element of $\mathbb S$.
Then:
Now to show that $\struct {H, \circ}$ is the largest such subgroup.
Let $K$ be a subgroup of $\struct {G, \circ}$ such that:
:$\forall S \in \mathbb S: K \subseteq S$
Then by definition $K \subseteq H$.
Let $x, y \in K$.
Then:
:$x \circ y^{-1} \in K \implies x \circ y^{-1} \in H$
Thus any subgroup of all elements of $\mathbb S$ is also a subgroup of $H$ and so no larger than $H$.
Thus $\ds H = \bigcap \mathbb S$ is the largest subgroup of $S$ contained in each element of $\mathbb S$.
== Sources ==
* : $\S 5.2$. Subgroups: Example $94$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.5$
* : $\S 1.9$: Theorem $18$
* : Chapter $\text{II}$: Groups: Exercise $\text J$
* : Chapter $2$: Subgroups and Cosets: $\S 35 \beta$
* : $\S 1.2$
* : $\S 36.7$ Subgroups
* : $\S 3.2$: Groups; the axioms
* : Chapter $4$: Subgroups: Proposition $4.6$
Category:Subgroups
Category:Set Intersection
Intersection of Subgroups of Prime Order

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ and $K$ be subsets of $G$ such that:
:$\order H = \order K = p$
:$H \ne K$
:$p$ is prime.
Then:
: $H \cap K = \set e$
That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity.
== Proof ==
From Intersection of Subgroups is Subgroup:
:$H \cap K \le G$
and:
:$H \cap K \le H$
where $\le$ denotes subgrouphood.
So:
Because $H \ne K$ and $\order H = \order k$, it follows that $H \nsubseteq K$.
So:
== Sources ==
* : $\S 43$. Lagrange's theorem: Worked Example $1$
Category:Prime Groups
Category:Subgroups
Intersection of Submagmas is Largest Submagma

== Theorem ==
Let $\struct {S, \odot}$ be a magma.
Let $\HH$ be a set of submagmas of $\struct {S, \odot}$, where $\HH \ne \O$.
Then the intersection $\bigcap \HH$ of the elements of $\HH$ is the largest submagma of $\struct {S, \odot}$ contained in each element of $\HH$.
== Proof ==
Let $K = \bigcap \HH$.
Let $K_i$ be an arbitrary element of $\HH$.
Then:
That is, $\odot$ is closed in $K$.
Hence by definition $\struct {K, \odot}$ is a submagma of $\struct {S, \odot}$.
Now to show that $\struct {K, \odot}$ is the largest submagma of $\struct {S, \odot}$ contained in each element of $\HH$.
Let $N$ be a submagma of $\struct {S, \odot}$ such that:
:$\forall H \in \HH: N \subseteq H$
Then by definition $N \subseteq K$.
Let $x, y \in N$.
Then:
:$x \odot y \in N \implies x \odot y \in K$
Thus any submagma of all elements of $\HH$ is also a submagma of $K$ and so no larger than $K$.
Thus $K = \bigcap \HH$ is the largest submagma of $H$ contained in each element of $\HH$.
Category:Submagmas
Category:Set Intersection
Intersection of Submodules is 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 $H$ and $K$ be submodules of $M$.
Then $H \cap K$ is also a submodule of $M$.
=== General Result ===
== Proof ==
This is a special case of the General Result with $S = \set {H, K}$.
The proof follows immediately from the proof of the General Result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$
Category:Submodules
Intersection of Submodules is Submodule/General Result

== 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 $S$ be a set of submodules of $M$.
Then the intersection $\ds \bigcap S$ is a submodule of $M$.
== Proof ==
From Intersection of Subgroups is Subgroup:General Result, it follows that $\ds \bigcap S$ is a subgroup of $M$.
As a subgroup is closed for its operation, it follows that for all $x,y \in \ds \bigcap S$, we have $x+y, y+x \in \ds \bigcap S$.
As $M$ is an $R$-module, and the addition $+$ on $\ds \bigcap S$ is the restriction of the addition on $M$, it follows that $x+y = y+x$.
Hence, $\ds \bigcap S$ is an abelian group.
Let $\lambda \in R$, and let $x \in \ds \bigcap S$.
Let $H$ be a submodule of $M$ such that $H \in S$.
By definition of intersection, it follows that $x \in H$.
As a submodule is closed for scalar product, it follows that $\lambda \circ x, x \circ \lambda \in H$.
As this is true for all submodules in the set $S$, it follows that $\lambda \circ x, x \circ \lambda \in \ds \bigcap S$.
Hence, $\ds \bigcap S$ is closed for scalar product.
Next, we show that $\ds \struct {\bigcap S, +, \circ}_R$ satisfy the left module axioms, and the right module axioms.
By definition of $R$-module, each of these axioms are satisfied by $\struct {G, +, \circ}_R$.
As the addition $+$ and the scalar multiplication $\circ$ on $\ds \bigcap S$ are the restrictions of the operations $+, \circ$ on $M = \struct {G, +, \circ}_R$, it follows that the axioms also are satisfied by $\ds \struct {\bigcap S, +, \circ}_R$.
For instance, the can be shown by:
Hence, $\ds \struct {\bigcap S, +, \circ}_R$ is a submodule of $M$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.2$
Category:Module Theory
Intersection of Submonoids with Monoid Identity is Submonoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $I$ be an indexing set.
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of submonoids of $S$.
For each $S_\alpha \in \family {S_\alpha}_{\alpha \mathop \in I}$, let $e_S \in S_\alpha$.
Let $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ denote the intersection of $\family {S_\alpha}$
Then $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is a submonoid of $S$.
== Proof ==
First we show that $\struct {\ds \bigcap_{\alpha \mathop \in I} S_\alpha, \circ}$ is a semigroup:
=== ===
Since $\struct {S, \circ}$ is a monoid whose identity is $e_S$, it follows that $\bigcap_{\alpha \mathop \in I} S_\alpha$ is non-empty.
Therefore:
=== ===
From the above we have that $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$ is closed under $\circ$.
From Restriction of Associative Operation is Associative we have that $\circ$ is associative on $\ds \bigcap_{\alpha \mathop \in I} S_\alpha$.
Hence we have that $\struct {\ds \bigcap_{\alpha \mathop \in I} S_\alpha, \circ}$ is a semigroup.
=== Identity Element ===
We are given that:
:$\forall \alpha \in I: e_S \in S_\alpha$
That is, for all $S_\alpha$, $e_S$ is the identity element of $S_\alpha$
Thus for all $S_\alpha$, $\struct {S_\alpha, \circ}$ is a monoid.
== Sources ==
* : Chapter $1$: Basic Definitions and Results
Category:Submonoids
Intersection of Subrings is Largest Subring Contained in all Subrings

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbb L$ be a non-empty set of subrings of $R$.
Then the intersection $\ds \bigcap \mathbb L$ of the members of $\mathbb L$ is the largest subring of $R$ contained in each member of $\mathbb L$.
== Proof ==
Let $\ds L = \bigcap \mathbb L$.
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is the largest subgroup of $\struct {R, +}$ contained in each member of $\mathbb L$.
By Intersection of Subsemigroups, $\struct {L, \circ}$ is the largest subsemigroup of $\struct {R, \circ}$ contained in each member of $\mathbb L$.
Let $S$ be a subring of $R$ such that:
:$\forall K \in \mathbb L: S \subseteq K$
By Intersection is Largest Subset, $S \subseteq L$.
Thus $L$ is the largest subring 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:Subrings
Intersection of Subrings is Subring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\mathbb L$ be a non-empty set of subrings of $R$.
Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself a subring of $R$.
== Proof ==
Let $L = \bigcap \mathbb L$.
By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is a subgroup of $\struct {R, +}$.
By the One-Step Subgroup Test:
:$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$
By Intersection of Subsemigroups, $\struct {L, \circ}$ a subsemigroup of $\struct {R, \circ}$.
So by definition of subsemigroup:
:$\forall x, y \in \struct {L, \circ}: x \circ y \in L$
By the Subring Test it follows that $\struct {L, +, \circ}$ is a subring of $\struct {R, +, \circ}$.
== Sources ==
* : $\S 2.3$: Some properties of subrings and ideals: Lemma $2.13 \ \text{(i)}$
Category:Set Intersection
Category:Subrings
Intersection of Subsemigroups

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ be subsemigroups of $\struct {S, \circ}$.
Then the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$.
=== General Result ===
== Proof ==
Suppose $\struct {S, \circ}$ is a semigroup where $S$ is the empty set. 
Suppose $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ are subsemigroups of $\struct {S, \circ}$. 
Then it follows that $T_1$ and $T_2$ are both empty.
Since $S$, $T_1$ and $T_2$ are empty, by Intersection with Empty Set, it follows that the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is equal to $\struct {S, \circ}$. 
Hence, by Semigroup is Subsemigroup of Itself, it follows that the intersection of $\struct {T_1, \circ}$ and $\struct {T_2, \circ}$ is itself a subsemigroup of that $\struct {S, \circ}$.
Now suppose $\struct {S, \circ}$ is a semigroup, where $S$ is non-empty. 
Let $T = T_1 \cap T_2$ where $T_1, T_2$ are subsemigroups of $\struct {S, \circ}$. Then:
Thus $\struct {T, \circ}$ is closed, and is therefore a semigroup from the Subsemigroup Closure Test.
Category:Set Intersection
Category:Subsemigroups
Category:Intersection of Subsemigroups
Intersection of Subsemigroups/General Result

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\mathbb H$ be a set of subsemigroups of $\struct {S, \circ}$, where $\mathbb H \ne \O$.
Then the intersection $\bigcap \mathbb H$ of the members of $\mathbb H$ is the largest subsemigroup of $\struct {S, \circ}$ contained in each member of $\mathbb H$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.7$
Category:Intersection of Subsemigroups
Intersection of Subsemigroups/General Result/Proof 1

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
== Proof ==
Let $T = \bigcap \mathbb S$.
Then:
So by Subsemigroup Closure Test, $\struct {T, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Now to show that $\struct {T, \circ}$ is the largest such subsemigroup.
Let $x, y \in T$.
Then $\forall K \subseteq T: x \circ y \in K \implies x \circ y \in T$.
Thus $\forall K \in \mathbb S: K \subseteq T$.
Thus $T$ is the largest subsemigroup of $S$ contained in each member of $\mathbb S$.
Category:Intersection of Subsemigroups
Intersection of Subsemigroups/General Result/Proof 2

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
== Proof ==
From Set of Subsemigroups forms Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Hence the result, by definition of infimum.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.12$
Category:Intersection of Subsemigroups
Intersection of Subset with Lower Bounds

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T_*$ be the set of all lower bounds of $T$ in $S$.
Then $T_* \cap T \ne \O$ :
:$T$ has a smallest element $m$
and
:$T_* \cap T$ is a singleton such that $T_* \cap T = \set m$
== Proof ==
Suppose $T_* \cap T = \O$, where $\O$ denotes the empty set.
That means $T$ contains none of its lower bounds, if indeed it has any.
From Smallest Element is Lower Bound, if $T$ had a smallest element, it would be a lower bound contained in $T$.
It follows that $T$ can have no smallest element.
Otherwise $T_* \cap T \ne \O$.
That means $T$ contains at least one of its lower bounds.
Suppose $\exists a, b \in T_* \cap T$.
From Intersection is Subset it follows that $a, b \in T$.
Then:
:$\forall y \in T: a \preceq y$
:$\forall y \in T: b \preceq y$
Thus both $a$ and $b$ fulfil the criteria for being a smallest element of $T$.
From Smallest Element is Unique it follows that $a = b$ and so $T_* \cap T$ is a singleton containing the smallest element of $T$.
== Also see ==
* Intersection of Subset with Upper Bounds
== Sources ==
* : $\S 14$: Order
Category:Order Theory
Intersection of Subset with Upper Bounds

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T^*$ be the set of all upper bounds of $T$ in $S$.
Then $T^* \cap T \ne \O$ :
:$T$ has a greatest element $M$
and
:$T^* \cap T$ is a singleton such that $T^* \cap T = \set M$
== Proof ==
Suppose $T^* \cap T = \O$, where $\O$ denotes the empty set.
That means $T$ contains none of its upper bounds, if indeed it has any.
From Greatest Element is Upper Bound, if $T$ had a greatest element, it would be an upper bound contained in $T$.
It follows that $T$ can have no greatest element.
Otherwise $T^* \cap T \ne \O$.
That means $T$ contains at least one of its upper bounds.
Suppose $\exists a, b \in T^* \cap T$.
From Intersection is Subset it follows that $a, b \in T$.
Then:
:$\forall y \in T: y \preceq a$
:$\forall y \in T: y \preceq b$
Thus both $a$ and $b$ fulfill the criteria for being a greatest element of $T$.
From Greatest Element is Unique it follows that $a = b$ and so $T^* \cap T$ is a singleton containing the greatest element of $T$.
== Also see ==
* Intersection of Subset with Lower Bounds
== Sources ==
* : $\S 14$: Order
Category:Order Theory
Intersection of Subsets is Subset/Set of Sets

== Theorem ==
Let $T$ be a set.
Let $\mathbb S$ be a non-empty set of sets.
Suppose that for each $S \in \mathbb S$:
:$S \subseteq T$
Then:
:$\bigcap \mathbb S \subseteq T$
== Proof ==
Let $x \in \bigcap \mathbb S$.
Then by the definition of intersection:
:$\forall S \in \mathbb S: x \in S$.
Since $\mathbb S$ is non-empty by the premise, it has some element $S$.
Then $x \in S$.
Since $S \in \mathbb S$, the premise shows that $S \subseteq T$.
By the definition of subset, $x \in T$.
Since this holds for each $x \in \bigcap \mathbb S$:
:$\bigcap \mathbb S \subseteq T$
Category:Set Intersection
Category:Subsets
Intersection of Sylow p-Subgroup with Subgroup not necessarily Sylow p-Subgroup

== Theorem ==
Let $G$ be a group.
Let $P$ be a Sylow $p$-subgroup of $G$.
Let $H$ be a subgroup of $G$.
Then $P \cap H$ is not necessarily a Sylow $p$-subgroup of $H$.
== Proof ==
We note that from Intersection of Subgroups is Subgroup that $P \cap H$ is a subgroup of $G$ and also of $H$.
Let $G$ be the dihedral group $D_3$, given by its group presentation:
By definition of Sylow $p$-subgroup, $\gen a$ is a Sylow $3$-subgroup of $G$.
However, $\gen b$ is also a subgroup of $G$, of order $2$.
But:
:$\gen b \cap \gen a = e$
and $e$ is not a Sylow $3$-subgroup of $G$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $3$
Category:Sylow p-Subgroups
Category:Set Intersection
Intersection of Topologies is Topology

== Theorem ==
Let $\family {\tau_i}_{i \mathop \in I}$ be an arbitrary indexed family of topologies on a set $S$.
Then $\tau := \ds \bigcap_{i \mathop \in I} \tau_i$ is also a topology on $S$.
== Proof ==
Each of the open set axioms are examined in turn:
=== ===
Let $\family {U_j}_{j \mathop \in J}$ be an arbitrary indexed family, such that:
:$\forall j \in J: U_j \in \tau$
Thus we have by definition of set intersection that:
:$\forall i \in I: \forall j \in J: U_j \in \tau_i$
Since $\tau_i$ is a topology for every $i \in I$, by definition we have:
:$\ds \forall i \in I: \bigcup_{j \mathop \in J} {U_j} \in \tau_i$
Therefore we have:
:$\ds \bigcup_{j \mathop \in J} {U_j} \in \bigcap_{i \mathop \in I} \tau_i = \tau$
=== ===
Let $U_1, U_2 \in \tau$.
Then by definition of set intersection:
:$\forall i \in I: U_1, U_2 \in \tau_i$
Since $\tau_i$ is a topology for each $i \in I$, we obtain that:
:$\forall i \in I: U_1 \cap U_2 \in \tau_i$
Therefore we have:
:$\ds U_1 \cap U_2 \in \bigcap_{i \mathop \in I} \tau_i = \tau$
=== ===
By the definition of a topology:
:$\forall i \in I: S \in \tau_i$
Thus by definition of set intersection we have that:
:$\ds S \in \bigcap_{i \mathop \in I} \tau_i = \tau$
Thus, by definition, $\tau = \ds \bigcap_{i \mathop \in I} \tau_i$ is a topology.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 4$
* : $\S 2$: Topological Spaces: Topologies
Category:Topology
Intersection of Transitive Relations is Transitive

== Theorem ==
The intersection of two transitive relations is also a transitive relation.
=== General Result ===
== Proof ==
Let $\RR_1$ and $\RR_2$ be transitive relations on an arbitrary set $S$.
Let $\tuple {s_1, s_2} \in \RR_1 \cap \RR_2$ and $\tuple {s_2, s_3} \in \RR_1 \cap \RR_2$.
Then by definition of intersection:
:$\tuple {s_1, s_2} \in \RR_1$ and $\tuple {s_1, s_2} \in \RR_2$
:$\tuple {s_2, s_3} \in \RR_1$ and $\tuple {s_2, s_3} \in \RR_2$
Then as $\RR_1$ and $\RR_2$ are both transitive:
:$\tuple {s_1, s_3} \in \RR_1$ and $\tuple {s_1, s_3} \in \RR_2$
and by definition of intersection:
:$\tuple {s_1, s_3} \in \RR_1 \cap \RR_2$
hence $\RR_1 \cap \RR_2$ is transitive.
Category:Intersection of Transitive Relations is Transitive
Category:Transitive Relations
Category:Set Intersection
Intersection of Transitive Relations is Transitive/General Result

== Theorem ==
Let $\family {\RR_i: i \mathop \in I}$ be an $I$-indexed set of transitive relations on a set $S$.
Then their intersection $\ds \bigcap_{i \mathop \in I} \RR_i$ is also a transitive relation on $S$.
== Proof ==
Let $\ds \tuple {x, y} \in \bigcap_{i \mathop \in I} \RR_i$ and $\ds \tuple {y, z} \in \bigcap_{i \mathop \in I} \RR_i$ be transitive relations on an arbitrary set $S$.
Then by definition of intersection:
:$\tuple {x, y} \in \RR_i$ for all $i \in I$
:$\tuple {y, z} \in \RR_i$ for all $i \in I$
Since each $\RR_i$ is transitive:
:$\tuple {x, z} \in \RR_i$ for all $i \in I$
so, by definition of intersection:
:$\ds \tuple {x, z} \in \bigcap_{i \mathop \in I} \RR_i$
Hence $\ds \bigcap_{i \mathop \in I} \RR_i$ is transitive.
Category:Intersection of Transitive Relations is Transitive