avewright/proofwiki-math huggingface dataset:
Theorem: Rational Polynomial is Content Times Primitive Polynomial

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
where:
:$\cont f$ is the content of $\map f X$
:$\map {f^*} X$ is a primitive polynomial.
For a given polynomial $\map f X$, both $\cont f$ and $\map {f^*} X$ are unique.

Proof:
=== Proof of Existence ===
{{:Rational Polynomial is Content Times Primitive Polynomial/Existence}}{{qed|lemma}}
Theorem: Product of Rational Polynomials

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$.
Using Rational Polynomial is Content Times Primitive Polynomial, let these be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g^*} X$
where:
:$\cont f, \cont g$ are the content of $f$ and $g$ respectively
:$f^*, g^*$ are primitive.
Let $\map h X = \map f X \map g X$ be the product of $f$ and $g$.
Then:
:$\map {h^*} X = \map {f^*} X \map {g^*} X$

Proof:
From Rational Polynomial is Content Times Primitive Polynomial:
:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \map {g^*} X$
and this expression is unique.
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \map {g^*} X$ is primitive.
From Content of Rational Polynomial is Multiplicative:
:$\cont f \cont g = \cont f \cont g > 0$.
The result follows.
{{qed}}
Theorem: Equivalence of Definitions of Perfect Set

{{TFAE|def = Perfect Set}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset of $S$.

Proof:
Let $T = \struct {S, \tau}$ be a topological space and let $H \subseteq S$.
Theorem: Complement of Interior equals Closure of Complement

Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.
Let $\map \complement H$ be the complement of $H$ in $T$:
:$\map \complement H = T \setminus H$
Then:
:$\map \complement {H^\circ} = \paren {\map \complement H}^-$
This can alternatively be written:
:$T \setminus H^\circ = \paren {T \setminus H}^-$
which, it can be argued, is easier to follow.

Proof:
Let $\tau$ be the topology on $T$.
Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.
Then:
{{begin-eqn}}
{{eqn | l = T \setminus H^\circ
      | r = T \setminus \bigcup_{K \mathop \in \mathbb K} K
      | c = {{Defof|Interior (Topology)}} of $H$
}}
{{eqn | r = \bigcap_{K \mathop \in \mathbb K} \paren {T \setminus K}
      | c = De Morgan's Laws: Difference with Union
}}
{{end-eqn}}
By the definition of closed set, $K$ is open in $T$ {{iff}} $T \setminus K$ is closed in $T$.
Also, from Set Complement inverts Subsets we have that $T \setminus K \supseteq T \setminus H$.
Now consider the set $\mathbb K'$ defined as:
:$\mathbb K' := \set {K' \subseteq T: \paren {T \setminus H} \subseteq K', K' \text { closed in } T}$
From the above we see that:
:$K \in \mathbb K \iff T \setminus K \in \mathbb K'$
Thus:
{{begin-eqn}}
{{eqn | l = T \setminus H^\circ 
      | r = \bigcap_{K' \mathop \in \mathbb K'} K'
}}
{{eqn | r = \paren {T \setminus H}^-
      | c = {{Defof|Closure (Topology)|index = 2|Closure}} of $T \setminus H$
}}
{{end-eqn}}
{{qed}}
Theorem: Interior of Finite Intersection equals Intersection of Interiors

Let $T$ be a topological space.
Let $n \in \N$.
Let:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
Then:
:$\ds \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ = \bigcap_{i \mathop = 1}^n H_i^\circ$
where $H_i^\circ$ denotes the interior of $H_i$.

Proof:
In the following, $H_i^-$ denotes the closure of the set $H_i$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ
      | r = T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^-
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {\paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i}^-}
      | c = Closure of Finite Union equals Union of Closures
}}
{{eqn | r = T \setminus \paren {\bigcup_{i \mathop = 1}^n T \setminus H_i^\circ}
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n H_i^\circ} }
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \bigcap_{i \mathop = 1}^n H_i^\circ
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}
Theorem: Finite Intersection of Regular Open Sets is Regular Open

Let $T$ be a topological space.
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are regular open in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{- \circ}$
where $H_i^{- \circ}$ denotes the interior of the closure of $H_i$.
Then $\ds \bigcap_{i \mathop = 1}^n H_i$ is regular open in $T$.
That is:
:$\ds \bigcap_{i \mathop = 1}^n H_i = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}
      | r = \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^\circ
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {\paren {T \setminus \bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^\circ
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^-}^\circ
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n H_i^-}^\circ
      | c = Relative Complement of Relative Complement
}}
{{eqn | r = \bigcap_{i \mathop = 1}^n H_i^{- \circ}
      | c = Interior of Finite Intersection equals Intersection of Interiors
}}
{{eqn | r = \bigcap_{i \mathop = 1}^n H_i
      | c = as all $H_i$ are regular open in $T$
}}
{{end-eqn}}
{{qed}}
Theorem: Finite Union of Regular Closed Sets is Regular Closed

Let $T$ be a topological space.
Let $n \in \N$.
Suppose that:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i \subseteq T$
where all the $H_i$ are regular closed in $T$.
That is:
:$\forall i \in \set {1, 2, \dotsc, n}: H_i = H_i^{\circ -}$
where $H_i^{\circ -}$ denotes the closure of the interior of $H_i$
Then $\ds \bigcup_{i \mathop = 1}^n H_i$ is regular closed in $T$.
That is:
:$\ds \bigcup_{i \mathop = 1}^n H_i = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}
      | r = \paren {T \setminus \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^-}^-
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^-
      | c = De Morgan's Laws: Difference with Union
}}
{{eqn | r = \paren {\paren {T \setminus \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^-
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^\circ}^-
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \paren {\bigcup_{i \mathop = 1}^n H_i^\circ}^-
      | c = Relative Complement of Relative Complement
}}
{{eqn | r = \bigcup_{i \mathop = 1}^n H_i^{\circ -}
      | c = Closure of Finite Union equals Union of Closures
}}
{{eqn | r = \bigcup_{i \mathop = 1}^n H_i
      | c = as all $H_i$ are regular closed in $T$
}}
{{end-eqn}}
{{qed}}
Theorem: Boundary is Intersection of Closure with Closure of Complement

Let $T = \struct {S, \tau}$ be a topological space.
Let $X \subseteq S$.
Let $\partial X$ denote the boundary of $X$ in $T$, defined as:
:$\partial X = X^- \setminus X^\circ$
Let $\overline X = S \setminus X$ denote the complement of $X$ in $S$.
Let $X^-$ denote the closure of $X$ in $T$.
Then:
:$\partial X = X^- \cap \paren {\overline X}^-$

Proof:
{{begin-eqn}}
{{eqn | l = \partial X
      | r = X^- \setminus X^\circ
      | c = {{Defof|Boundary (Topology)|index = 1}}: $X^\circ$ is the interior of $X$
}}
{{eqn | r = X^- \cap \overline {\paren {X^\circ} }
      | c = Set Difference as Intersection with Relative Complement
}}
{{eqn | r = X^- \cap \paren {\overline X}^-
      | c = Complement of Interior equals Closure of Complement
}}
{{end-eqn}}
{{qed}}
Theorem: Set is Closed iff it Contains its Boundary

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is closed in $T$ {{iff}}:
:$\partial H \subseteq H$
where $\partial H$ is the boundary of $H$.

Proof:
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the closure of $H$ in $T$.
Hence from Intersection is Subset we have that:
:$\partial H \subseteq H^-$
Then from Closed Set Equals its Closure, $H$ is closed in $T$ {{iff}} $H = H^-$.
Hence the result.
{{qed}}
Theorem: Set is Open iff Disjoint from Boundary

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is open in $T$ {{iff}}:
:$\partial H \cap H = \O$
where $\partial H$ is the boundary of $H$ in $T$.

Proof:
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the closure of $H$.
Hence from Intersection is Subset we have that:
:$\partial H \subseteq \paren {S \setminus H}^-$
But from Closed Set Equals its Closure, $\paren {S \setminus H}^- = S \setminus H$ {{iff}} $S \setminus H$ is closed in $T$.
That is, {{iff}} $H$ is open in $T$.
So $\partial H \subseteq S \setminus H$ {{iff}} $H$ is open in $T$.
From Intersection with Complement is Empty iff Subset it follows that $\partial H \cap H = \O$ {{iff}} $H$ is open in $T$.
{{qed}}
Theorem: Set is Clopen iff Boundary is Empty

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is both closed and open in $T$ {{iff}}:
:$\partial H = \O$
where $\partial H$ is the boundary of $H$.

Proof:
From Set is Open iff Disjoint from Boundary we have that:
:$H$ is open in $T$ {{iff}} $\partial H \cap H = \O$
From Set is Closed iff it Contains its Boundary we have that:
:$H$ is closed in $T$ {{iff}} $\partial H \subseteq H$
From Intersection with Subset is Subset:
:$\partial H \subseteq H \iff \partial H \cap H = \partial H $
From the above it follows that $\partial H = \O$ {{iff}} $H$ is closed and $H$ is open.
{{qed}}
Theorem: Boundary of Set is Closed

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\partial H$ denote the boundary of $H$ in $T$.
Then $\partial H$ is closed in $T$.

Proof:
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial H = H^- \cap \paren {S \setminus H}^-$
where $H^-$ is the closure of $H$
From Topological Closure is Closed, both $H^-$ and $\paren {S \setminus H}^-$ are closed in $T$.
From Topology Defined by Closed Sets, the intersection of arbitrarily many (in particular two) closed sets of $T$ is a closed set of $T$.
As $\partial H$ is the intersection of $H^-$ and $\paren {S \setminus H}^-$ the result follows.
{{qed}}
Theorem: Boundary of Boundary is Contained in Boundary

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq T$.
Then:
:$\map \partial {\partial H} \subseteq \partial H$
where $\partial H$ is the boundary of $H$ in $T$.
That is, the boundary of the boundary of $H$ is contained in the boundary of $H$.

Proof:
Let $B = \partial H$.
From Boundary of Set is Closed we have that $B$ is closed in $T$.
Let $B^-$ denote the closure of $B$.
From Boundary is Intersection of Closure with Closure of Complement:
:$\partial B = B^- \cap \paren {S \setminus B}^-$
and so from Intersection is Subset:
:$\partial B \subseteq B^-$
But from Closed Set Equals its Closure:
:$B = B^-$
and so:
:$\partial B \subseteq B$
That is:
:$\map \partial {\partial H} \subseteq \partial H$
{{qed}}
Theorem: Equivalence of Definitions of Exterior

{{TFAE|def = Exterior (Topology)|view = exterior|context = Topology (Mathematical Branch)|contextview = topology}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.

Proof:
Let $H^e$ be defined as:
:$H^e$ is the complement of the closure of $H$ in $T$.
Then:
{{begin-eqn}}
{{eqn | l = H^e
      | r = S \setminus H^-
      | c = 
}}
{{eqn | r = \paren {S \setminus H}^\circ
      | c = Complement of Interior equals Closure of Complement
}}
{{end-eqn}}
Thus:
:$H^e$ is the interior of the complement of $H$ in $T$.
{{qed}}
Theorem: Interior is Subset of Exterior of Exterior

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $H^e$ denote the exterior of $H$ in $T$, and let $H^\circ$ denote the interior of $H$ in $T$.
Then:
:$H^\circ \subseteq \paren {H^e}^e$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {H^e}^e
      | r = \paren {S \setminus H^e}^\circ
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {S \setminus \paren {S \setminus H^-} }^\circ
      | c = Equivalence of Definitions of Exterior
}}
{{eqn | r = \paren {H^-}^\circ
      | c = Relative Complement of Relative Complement
}}
{{eqn | o = \supseteq
      | r = H^\circ
      | c = Interior is Subset of Interior of Closure
}}
{{end-eqn}}
{{qed}}
Theorem: Interior is Subset of Interior of Closure

Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^\circ$ denote the interior of $H$.
Let $H^-$ denote the closure of $H$.
Then:
:$H^\circ \subseteq \left({H^-}\right)^\circ$

Proof:
From Set is Subset of its Topological Closure, we have $H \subseteq H^-$.
The result follows directly from Interior of Subset.
{{qed}}
Category:Set Closures
Category:Set Interiors
hpi3i0ga2cmk8ylx6ricx5jzzr61qz3
Theorem: Exterior of Finite Union equals Intersection of Exteriors

Let $T = \struct {S, \tau}$ be a topological space.
Let $n \in \N$.
Let $\forall i \in \closedint 1 n: H_i \subseteq S$.
Then:
:$\ds \paren {\bigcup_{i \mathop = 1}^n H_i}^e = \bigcap_{i \mathop = 1}^n H_i^e$
where $H_i^e$ denotes the exterior of $H_i$.

Proof:
In the following, $H_i^\circ$ denotes the interior of the set $H_i$ in $T$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^e
      | r = \paren {S \setminus \bigcup_{i \mathop = 1}^n H_i}^\circ
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {S \setminus H_i} }^\circ
      | c = De Morgan's Laws: Difference with Union
}}
{{eqn | r = \bigcap_{i \mathop = 1}^n \paren {S \setminus H_i}^\circ
      | c = Interior of Finite Intersection equals Intersection of Interiors
}}
{{eqn | r = \bigcap_{i \mathop = 1}^n H_i^e
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{end-eqn}}
{{qed}}
Theorem: Closure of Union contains Union of Closures

Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ denotes the power set of $S$.
Then the union of the closures of the elements of $\mathbb H$ is a subset of the closure of the union of $\mathbb H$:
:$\ds \bigcup_{H \mathop \in \mathbb H} \map \cl H \subseteq \map \cl {\bigcup_{H \mathop \in \mathbb H} H}$

Proof:
Let $\ds K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\ds L = \bigcup_{H \mathop \in \mathbb H} H$.
We have:
:$\forall H \in \mathbb H: H \subseteq L$
so from Topological Closure of Subset is Subset of Topological Closure:
:$\map \cl H \subseteq \map \cl L$
It follows from Union is Smallest Superset: General Result that:
:$K \subseteq \map \cl L$
{{qed}}
Theorem: Set is Subset of its Topological Closure

Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ be the closure of $H$ in $T$.
Then:
:$H \subseteq H^-$

Proof:
From the definition of closure, we have:
:$H^-$ is the union of $H$ and its limit points.
From Subset of Union it follows directly that:
:$H \subseteq H^-$
{{qed}}
Category:Set Closures
4d7jedgi469fn642g55pjs0t2diepw3
Theorem: Set Closure as Intersection of Closed Sets

Let $T$ be a topological space.
Let $H \subseteq T$.
Let the closure of $H$ (in $T$) be defined as:
:$H^- := H \cup H'$
where $H'$ is the derived set of $H$.
Let $\mathbb K$ be defined as:
:$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all closed sets of $T$ which contain $H$.
Then the closure of $H$ (in $T$) can be defined as:
:$\ds H^- := \bigcap \mathbb K$
that is, as the intersection of all the closed sets of $T$ which contain $H$.

Proof:
What needs to be proved here is:
:$\ds H^- = \bigcap \mathbb K$
where:
:$H^- = H \cup H'$
:$H'$ denotes the set of all limit points of $H$
:$\ds \bigcap \mathbb K$ is the intersection of all closed sets of $T$ which contain $H$.
Let $K \in \mathbb K$.
From Topological Closure of Subset is Subset of Topological Closure, we have:
:$H^- \subseteq K^-$
Thus from Closed Set Equals its Closure, we have:
:$K^- = K$
Thus:
:$\ds \forall K \in \bigcap \mathbb K: H^- \subseteq K$
since the choice of $K$ is arbitrary. 
So from Intersection is Largest Subset:
:$\ds H^- \subseteq \bigcap \mathbb K$
{{qed|lemma}}
Conversely, from Topological Closure is Closed, $H^-$ is closed.
From Set is Subset of its Topological Closure, $H \subseteq H^-$.
So $H^-$ is, by definition, a closed set of $T$ which contains $H$.
But we have by its definition that $\ds \bigcap \mathbb K$ is the intersection of '''all''' closed sets in $T$ that contain $H$.
So from Intersection is Subset it follows that:
:$\ds \bigcap \mathbb K \subseteq H^-$
{{qed|lemma}}
Finally, we have that:
:$\ds H^- \subseteq \bigcap \mathbb K$
:$\ds \bigcap \mathbb K \subseteq H^-$
So by definition of set equality:
:$\ds H^- = \bigcap \mathbb K$
which is what we needed to prove.
{{qed}}
Theorem: Set Closure is Smallest Closed Set/Topology

Let $T$ be a topological space.
Let $H \subseteq T$.
Let $H^-$ denote the closure of $H$ in $T$.
Then $H^-$ is the smallest superset of $H$ that is closed in $T$.

Proof:
Define:
:$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$
That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$.
The claim is that $H^-$ is the smallest set of $\mathbb K$.
From Set is Subset of its Topological Closure:
:$H \subseteq H^-$
From Topological Closure is Closed, $H^-$ is closed in $T$.
Thus $H^- \in \mathbb K$.
Let $K \in \mathbb K$.
From Set Closure as Intersection of Closed Sets:
:$\ds H^- = \bigcap \mathbb K$
Therefore, from Intersection is Subset: General Result:
:$H^- \subseteq K$
Thus by definition $H^-$ is the smallest set of $\mathbb K$.
{{qed}}
Theorem: Equivalence of Definitions of Interior (Topology)

{{TFAE|def = Interior (Topology)|view = interior|context = Topology (Mathematical Branch)|contextview = topology}}
Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$.
=== Definition 1 ===
{{:Definition:Interior (Topology)/Definition 1}}
=== Definition 2 ===
{{:Definition:Interior (Topology)/Definition 2}}
=== Definition 3 ===
{{:Definition:Interior (Topology)/Definition 3}}

Proof:
=== Definition $1$ is equivalent to Definition $2$ ===
Let $\mathbb K$ be defined as:
:$\mathbb K := \set {K \in \tau: K \subseteq H}$
That is, let $\mathbb K$ be the set of all open sets of $T$ contained in $H$.
Then from definition 1 of the interior of $H$, we have:
:$\ds H^\circ = \bigcup_{K \mathop \in \mathbb K} K$
That is, $H^\circ$ is the union of all the open sets of $T$ contained in $H$.
Let $K \subseteq T$ such that $K$ is open in $T$ and $K \subseteq H$.
That is, let $K \in \mathbb K$.
Then from Subset of Union it follows directly that $K \subseteq H^\circ$.
So any open set in $T$ contained in $H$ is a subset of $H^\circ$, and so $H^\circ$ is the largest open set of $T$ contained in $H$.
That is, $H^\circ$ is also the interior of $H$ by definition 2.
Hence both definitions are equivalent.
{{qed|lemma}}
Theorem: Intersection of Interiors contains Interior of Intersection

Let $T$ be a topological space.
Let $\mathbb H$ be a set of subsets of $T$.
That is, let $\mathbb H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then the interior of the intersection of $\mathbb H$ is a subset of the intersection of the interiors of the elements of $\mathbb H$.
:$\ds \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ \subseteq \bigcap_{H \mathop \in \mathbb H} H^\circ$

Proof:
In the following, $H^-$ denotes the closure of the set $H$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ
      | r = T \setminus \paren {T \setminus \bigcap_{H \mathop \in \mathbb H} H}^-
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}
      | c = De Morgan's Laws: Difference with Intersection
}}
{{end-eqn}}
At this point we note that:
:$(1): \quad \ds \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \supseteq \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$
from Closure of Union contains Union of Closures.
Then we note that:
:$\ds T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} \subseteq T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}$
from $(1)$ and Set Complement inverts Subsets.
Then we continue:
{{begin-eqn}}
{{eqn | l = T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}
      | r = T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} T \setminus H^\circ}
      | c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = T \setminus \paren {T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^\circ} }
      | c = De Morgan's Laws: Difference with Intersection
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb H} H^\circ
      | c = Relative Complement of Relative Complement
}}
{{end-eqn}}
{{qed}}
Theorem: Intersection of Exteriors contains Exterior of Union

Let $T = \struct {S, \tau}$ be a topological space.
Let $\mathbb H$ be a set of subsets of $S$.
That is, let $\mathbb H \subseteq \powerset S$ where $\powerset S$ is the power set of $S$.
Then:
:$\ds \paren {\bigcup_{H \mathop \in \mathbb H} H}^e \subseteq \bigcap_{H \mathop \in \mathbb H} H^e$
where $H^e$ denotes the exterior of $H$ in $T$.

Proof:
In the following, $H^\circ$ denotes the interior of the set $H$ in $T$.
{{begin-eqn}}
{{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e
      | r = \paren {S \setminus \bigcup_{H \mathop \in \mathbb H} H}^\circ
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{eqn | r = \paren {\bigcap_{H \mathop \in \mathbb H} \paren {S \setminus H} }^\circ
      | c = De Morgan's Laws: Difference with Union
}}
{{eqn | o = \subseteq
      | r = \bigcap_{H \mathop \in \mathbb H} \paren {S \setminus H}^\circ
      | c = Intersection of Interiors contains Interior of Intersection
}}
{{eqn | r = \bigcap_{H \mathop \in \mathbb H} H^e
      | c = {{Defof|Exterior (Topology)|Exterior}}
}}
{{end-eqn}}
{{qed}}