avewright/proofwiki-math huggingface dataset:
Theorem: Equivalence of Local Uniform Convergence and Compact Convergence

Let $U \subseteq \C$ be an open set of the complex plane.
Let $f_n: U \to \C$ be a sequence of functions which converges pointwise to $f: U \to \C$.
Then $f_n$ converges to $f$ uniformly on all compact subsets of $U$ {{iff}} $f_n$ converges locally uniformly on $U$.

Proof:
=== Sufficient Condition ===
Suppose that $f_n$ converges to $f$ uniformly on all compact subsets of $U$.
Let $z \in U$.
Since $U$ is open, there is an open ball $\map {B_\epsilon} z$, around $z$, of radius $\epsilon$, with the closure of the ball, $\overline{\map {B_\epsilon} z}$, contained in $U$.
Since $\overline {\map {B_\epsilon} z}$ is closed and bounded, it is compact, by the Heine-Borel Theorem. 
{{explain|Currently the link to Heine-Borel Theorem is linked to Real Euclidean Space. However, here it is a compact subset of the complex plane<br/>But the complex plane is a real Euclidean plane so that should work}}
We then have by assumption that $f_n$ converges uniformly on $\overline {\map {B_\epsilon} z}$.
Thus $f_n$ converges locally uniformly on $U$.
{{qed|lemma}}
Theorem: Frattini's Argument

Let $\struct {G, \circ}$ be a group.
Let $K$ be a finite normal subgroup of $G$, and $p$ a prime which divides the order of $K$.
Let $P$ be a Sylow $p$-subgroup of $K$, and $\map {N_G} P$ the normalizer of $P$ in $G$.
Then:
:$G = \map {N_G} P \circ K = K \circ \map {N_G} P$

Proof:
Let $g \in G$.
Since $K$ is normal in $G$, and $P \subset K$, the conjugate $g \circ P \circ g^{-1}$ of $P$ is also a subset of $K$.
From Inner Automorphism is Automorphism, $g \circ P \circ g^{-1}$ is a subgroup of $K$ of the same order as $P$.
Thus $g \circ P \circ g^{-1}$ is also a Sylow $p$-subgroup of $K$.
By Second Sylow Theorem:
:$\exists k \in K: g \circ P \circ g^{-1} = k \circ P \circ k^{-1}$
Then:
:$g^{-1} \circ k \circ P \circ k^{-1} \circ g = P$
and so by definition of normalizer:
:$k^{-1} \circ g \in \map {N_G} P$
We can then write $g$ as
:$g = k \circ \paren {k^{-1} \circ g} \in K \circ \map {N_G} P$
Since $K$ is normal:
:$\map {N_G} P \circ K = K \circ \map {N_G} P$
by Subset Product with Normal Subgroup as Generator.
{{qed}}
{{Namedfor|Giovanni Frattini|cat = Frattini}}
Theorem: Sum of Integrals on Complementary Sets

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A, E \in \Sigma$ with $A \subseteq E$.
Let $f$ be a $\mu$-integrable function on $X$.
Then 
:$\ds \int_E f \rd \mu = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu$

Proof:
Let $\chi_E$ be the characteristic function of $E$.
Because $A$ and $E \setminus A$ are disjoint:
:$A \cap \paren {E \setminus A} = \O$
By Characteristic Function of Union:
:$\chi_E = \chi_A + \chi_{E \mathop \setminus A}$ 
Integrating $f$ over $E$ gives:
{{begin-eqn}}
{{eqn | l = \int_E f \rd \mu
      | r = \int \chi_E \cdot f \rd \mu
      | c = {{Defof|Integral of Measure-Integrable Function over Measurable Set}}
}}
{{eqn | r = \int \paren {\chi_A + \chi_{E \mathop \setminus A} } \cdot f \rd \mu
      | c = by the above argument
}}
{{eqn | r = \int \paren {\chi_A \cdot f + \chi_{E \mathop \setminus A} \cdot f} \rd \mu
      | c = Pointwise Operation on Distributive Structure is Distributive
}}
{{eqn | r = \int \chi_A \cdot f \rd \mu + \int \chi_{E \mathop \setminus A} \cdot f \rd \mu
      | c = Integral of Integrable Function is Additive
}}
{{eqn | r = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu
      | c = {{Defof|Integral of Measure-Integrable Function over Measurable Set}}
}}
{{end-eqn}}
{{qed}}
Theorem: Closure of Pointwise Operation on Algebraic Structure

Let $S$ be a set such that $S \ne \O$.
Let $\struct {T, \circ}$ be an algebraic structure.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $f, g \in T^S$, that is, let $f: S \to T$ and $g: S \to T$ be mappings.
Let $\oplus: T^S \to T^S$ be the pointwise operation on $T^S$ induced by $\circ$.
Then $\oplus$ is closed on $T^S$ {{iff}} $\struct {T, \circ}$ is closed.

Proof:
=== Necessary Condition ===
Let $\struct {T, \circ}$ be closed.
Let $x \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | q = \forall f, g \in T^S
      | l = \map {f \oplus g} x
      | r = \map f x \circ \map g x
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | o = \in
      | r = T
      | c = as $\struct {T, \circ}$ is closed
}}
{{end-eqn}}
So $\oplus$ is closed on $T^S$.
{{qed|lemma}}
Theorem: Pointwise Operation on Distributive Structure is Distributive

Let $S$ be a set.
Let $\struct {T, +, \circ}$ be an algebraic structure with two operations $+$ and $\circ$.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $\struct {T^S, +', \circ'}$ be the structure induced on $T^S$ by $+$ and $\circ$.
Let $\circ$ be distributive over $+$ in $T$.
Then $\circ'$ is distributive over $+'$ in $T$.

Proof:
Let $f, g, h: S \to T$ be elements of $T^S$.
Suppose $S$ is the empty set.
Suppose $T^S$ is the set of all mappings from the empty set, $S$, to $T$.
Suppose $\struct {T^S, +', \circ'}$ is the structure induced on $T^S$ by $+$ and $\circ$.
Suppose $\circ$ is distributive over $+$ in $T$.
{{Finish|Prove that the statement is true when S is the empty set}}
Then $\circ'$ is distributive over $+'$ in $T$, as required.
Suppose $S$ is non-empty.
Let $x \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \circ' \paren {g +' h} } } x
      | r = \map f x \circ \paren {\map g x + \map h x}
      | c = {{Defof|Induced Structure}}
}}
{{eqn | r = \paren {\map f x \circ \map g x} + \paren {\map f x \circ \map h x}
      | c = $\circ$ is distributive over $+$ on $T$
}}
{{eqn | r = \map {\paren {\paren {f \circ' g} +' \paren {f \circ' h} } } x
      | c = {{Defof|Induced Structure}}
}}
{{end-eqn}}
Similarly:
:$\map {\paren {\paren {g +' h} \circ' f} } x = \map {\paren {\paren {g \circ' f} +' \paren {h \circ' f} } } x$
Hence the result.
{{qed}}
Category:Pointwise Operations
Category:Distributive Operations
thqrrhnp9or0cyr4n4hd71xx8penx7k
Theorem: Identity Mapping on Metric Space is Homeomorphism

Let $M = \struct {A, d}$ be a metric space.
The identity mapping $I_A: M \to M$ defined as:
:$\forall x \in A: \map {I_A} x = x$
is a homeomorphism.

Proof:
We have Identity Mapping is Bijection.
We also have Identity Mapping is Continuous.
Hence, by definition, $I_T$ is a homeomorphism.
{{qed}}
Category:Homeomorphisms (Metric Spaces)
Category:Topologically Equivalent Metrics
Category:Identity Mappings
mxwnp6kgls3opibt01kued5od1f5hw8
Theorem: Inverse of Homeomorphism between Metric Spaces is Homeomorphism

Let $M = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: M_1 \to M_2$ be a homeomorphism.
Then $f^{-1}: M_2 \to M_1$ is also a homeomorphism.

Proof:
By definition, a homeomorphism is a bijection such that both $f$ and $f^{-1}$ are continuous.
As $f$ is a bijection then by Bijection iff Inverse is Bijection, so is $f^{-1}$.
So by definition $f^{-1}$ is a bijection such that both $f^{-1}$ and $\paren {f^{-1} }^{-1}$ are continuous.
The result follows from Inverse of Inverse of Bijection.
{{qed}}
Category:Homeomorphisms (Metric Spaces)
q0hz7j17p17rfmibnls055gz1pts37s
Theorem: Composite of Homeomorphisms between Metric Spaces is Homeomorphism

Let $M_1, M_2, M_3$ be metric spaces. 
Let $f: M_1 \to M_2$ and $g: M_2 \to M_3$ be homeomorphisms.
Then $g \circ f: M_1 \to M_3$ is also a homeomorphism.

Proof:
By definition of homeomorphism, $f$ and $g$ are both bijections.
From Composite of Bijections is Bijection it follows that $g \circ f$ is also a bijection.
Similarly, also by definition of homeomorphism, $f$ and $g$ are both continuous mappings.
From Composite of Continuous Mappings on Metric Spaces is Continuous it follows that $g \circ f$ is also a continuous mapping.
From Inverse of Bijection is Bijection, $f^{-1}: M_2 \to M_1$ and $g^{-1}: M_3 \to M_2$ are also both bijections.
From Composite of Bijections is Bijection it follows that $f^{-1} \circ g^{-1} : M_3 \to M_1$ is also a bijection.
By definition of homeomorphism, $f^{-1}$ and $g^{-1}$ are both continuous mappings.
Again from Composite of Continuous Mappings on Metric Spaces is Continuous it follows that $f^{-1} \circ g^{-1}$ is also a continuous mapping.
Hence the result, from definition of homeomorphism.
{{qed}}
Category:Homeomorphisms (Metric Spaces)
if9e03mfh240offj33289n2yh4bwo26
Theorem: Metrizable Space is Hausdorff

Let $T$ be a metrizable topological space.
Then $T$ is a $T_2$ (Hausdorff) space.

Proof:
By definition, $T$ is homeomorphic to a topological space $\struct {S, \tau_d}$ such that $\tau_d$ is the topology induced by a metric $d$ on $S$.
From Metric Space is Hausdorff, $\struct {S, d}$ is a $T_2$ (Hausdorff) space.
As the open sets of $\struct {S, d}$ are the same as the open sets of $\struct {S, \tau_d}$, it follows that $\struct {S, \tau_d}$ is a $T_2$ (Hausdorff) space.
From $T_2$ Property is Preserved under Homeomorphism it follows that $T$ is also a $T_2$ (Hausdorff) space.
{{qed}}
Theorem: Neighborhood of Point in Metrizable Space contains Closed Neighborhood

Let $T = \struct {S, \tau}$ be a metrizable topological space.
Let $x \in S$ be an arbitrary point of $T$.
Let $N$ be a neighborhood of $x$.
Then $N$ has as a subset a neighborhood $V$ of $x$ such that $V$ is closed.

Proof:
Since $N$ is a neighborhood of $x$, there exists an open set $U \subseteq N$ containing $x$ by definition.
As $\struct {S, \tau}$ is metrizable, the set $U$ is open with respect to some metric space $\struct {S, d}$.
Hence:
:$\exists \epsilon > 0: \map {\BB_\epsilon} x \subseteq U$
where $\map {\BB_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.
Taking $\zeta = \dfrac \epsilon 2$, we have that the closed $\zeta$-ball of $x$ is contained in $\map {\BB_\epsilon} x$, and hence is a subset of $U \subseteq N$. 
This closed ball is a closed neighborhood of $x$ in $\struct {S, \tau}$, and is a subset of $N$.
Hence the result.
{{qed}}
Theorem: Underlying Set of Topological Space is Closed

Let $T = \struct {S, \tau}$ be a topological space.
Then the underlying set $S$ of $T$ is closed in $T$.

Proof:
From the definition of closed set, $U$ is open in $T = \struct {S, \tau}$ {{iff}} $S \setminus U$ is closed in $T$.
From Empty Set is Element of Topology, $\O$ is open in $T$.
From Set Difference with Empty Set is Self:
:$S \setminus \O = S$
Hence $S$ is closed in $T$.
{{qed}}
Theorem: Point in Topological Space has Neighborhood

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then there exists in $T$ at least one neighborhood of $x$.

Proof:
Let $x \in S$.
Then $S$ itself is a neighborhood of $x$.
{{qed}}
Theorem: Point in Topological Space is Element of its Neighborhood

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Then $a \in N$.

Proof:
Trivially follows by definition of neighborhood of $a$.
{{qed}}
Theorem: Superset of Neighborhood in Topological Space is Neighborhood

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Let $N \subseteq N' \subseteq S$.
Then $N'$ is a neighborhood of $x$ in $T$.

Proof:
By definition of neighborhood:
:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
where $U$ is an open set of $T$.
By Subset Relation is Transitive:
:$U \subseteq N'$
The result follows by definition of neighborhood of $x$.
{{qed}}
Theorem: Intersection of Neighborhoods in Topological Space is Neighborhood

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $M, N$ be a neighborhoods of $x$ in $T$.
Then $M \cap N$ is a neighborhood of $x$ in $T$.

Proof:
By definition of neighborhood:
:$\exists U_1 \in \tau: x \in U_1 \subseteq M$
where $U_1$ is an open set of $T$.
:$\exists U_2 \in \tau: x \in U_2 \subseteq N$
where $U_2$ is an open set of $T$.
Thus by Set Intersection Preserves Subsets:
:$U \subseteq M \cap N$
where $U = U_1 \cap U_2$
The result follows by definition of neighborhood of $a$.
{{qed}}
Theorem: Neighborhood in Topological Space has Subset Neighborhood

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $N$ be a neighborhood of $x$ in $T$.
Then there exists a neighborhood $N'$ of $x$ such that:
:$(1): \quad N' \subseteq N$
:$(2): \quad N'$ is a neighborhood of each of its points.

Proof:
By definition of neighborhood:
:$\exists U \in \tau: x \in U \subseteq N \subseteq S$
where $U$ is an open set of $T$.
By Set is Open iff Neighborhood of all its Points, $N' = U$ fulfils the conditions of the statement.
{{qed}}
Theorem: Empty Set is Open in Neighborhood Space

Let $\struct {S, \NN}$ be a neighborhood space.
Then the empty set $\O$ is an open set of $\struct {S, \NN}$.

Proof:
Suppose $\O$ were not an open set of $\struct {S, \NN}$.
Then $\exists x \in \O$ such that $\O$ is not a neighborhood of $\O$.
By definition of empty set, such an $x$ does not exist.
Hence the result.
{{qed}}
Theorem: Whole Space is Open in Neighborhood Space

Let $\struct {S, \NN}$ be a neighborhood space.
Then $S$ itself is an open set of $\struct {S, \NN}$.

Proof:
Let $x \in S$.
Then by neighborhood space axiom $\text N 1$ there exists a neighborhood $N$ of $x$.
As $N \subseteq S$ it follows from neighborhood space axiom $\text N 3$ that $S$ is a neighborhood of $x$.
As this holds for all $x \in S$ it follows that $S$ is an open set of $\struct {S, \NN}$.
{{qed}}
Theorem: Intersection of two Open Sets of Neighborhood Space is Open

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 $\struct {S, \NN}$.

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 $\struct {S, \NN}$.
{{qed}}
Theorem: Union of Open Sets of Neighborhood Space is Open

Let $S$ be a neighborhood space.
Let $I$ be an indexing set.
Let $\family {U_\alpha}_{\alpha \mathop \in I}$ be a family of open sets of $\struct {S, \NN}$ indexed by $I$.
Then their union $\ds \bigcup_{\alpha \mathop \in I} U_i$ is an open set of $\struct {S, \NN}$.

Proof:
Let $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
Then $x \in U_\beta$ for some $\beta \in I$.
By definition of open set, $U_\beta$ is a neighborhood of $x$.
But from Set is Subset of Union:
:$\ds U_\beta \subseteq x \in \bigcup_{\alpha \mathop \in I} U_\alpha$
By neighborhood space axiom $N 3$ it follows that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is a neighborhood of $x$.
As $x$ is arbitrary, it follows that the above is true for all $\ds x \in \bigcup_{\alpha \mathop \in I} U_\alpha$.
It follows by definition that $\ds \bigcup_{\alpha \mathop \in I} U_\alpha$ is an open set of $\struct {S, \NN}$.
{{qed}}
Theorem: Neighborhood Space is Topological Space

Let $\struct {S, \NN}$ be a neighborhood space.
Let $\tau = \set {N: N \in \NN}$ be the set of all open sets of $\struct {S, \NN}$.
Then $\struct {S, \tau}$ forms a topological space.

Proof:
Each of the open set axioms is examined in turn:
Theorem: Subset in Neighborhood Space is Neighborhood iff it contains Open Set

Let $\struct {S, \NN}$ be a neighborhood space.
Let $x \in S$ be a point of $S$.
Let $N \subseteq S$ be a subset of $S$.
Then $N$ is a neighborhood of $x$ {{iff}} there exists an open set $U$ of $\struct {S, \NN}$ such that $x \in U \subseteq N$.

Proof:
=== Necessary Condition ===
Let $N$ be a neighborhood of $x$.
Then by neighborhood space axiom $\text N 5$, $N$ contains a neighborhood $U$ of $x$ such that $U$ is a neighborhood of each of its points.
By neighborhood space axiom $\text N 2$, $x \in U$.
{{qed|lemma}}
Theorem: Induced Neighborhood Space is Neighborhood Space

Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\struct {S, \tau}$.
Then $\struct {S, \NN}$ is a neighborhood space.

Proof:
Let $x \in S$.
Let $\NN_x$ be the neighborhood filter of $x$.
From Basic Properties of Neighborhood in Topological Space, $\NN_x$ fulfils the neighborhood space axioms.
Hence the result.
{{qed}}
Category:Neighborhood Spaces
az6sirkh6uxvz9p958oypwr5nuh6on5
Theorem: Topological Space Induced by Neighborhood Space is Topological Space

Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\struct {S, \NN}$.
Then $\struct {S, \tau}$ is a topological space.

Proof:
From Neighborhood Space is Topological Space, $\struct {S, \NN}$ is a topological space.
Consequently, the open sets of $\struct {S, \tau}$ are exactly the open sets of $\struct {S, \NN}$.
{{qed}}
Category:Neighborhood Spaces
lfo4lhob0eucnlcfixr1sexh9omba2s
Theorem: Topological Space induced by Neighborhood Space induced by Topological Space

Let $S$ be a set.
Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\struct {S, \tau'}$ be the topological space induced by $\NN$ on $S$.
Then $\tau = \tau'$.

Proof:
Let $U \in \tau$ be an open set of $\struct {S, \tau}$.
By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points.
By definition, $U$ is an open set of $\struct {S, \NN}$.
Thus by definition of neighborhood space induced by $\tau$ on $S$:
:$U \in \NN$
Then, by definition of the topological space induced by $\NN$ on $S$:
:$U \in \tau'$.
Thus $\tau \subseteq \tau'$.
Now let $U \in \tau'$ be an open set of $\struct {S, \tau'}$.
Then, by definition, in the neighborhood space $\struct {S, \NN}$, $U$ is an open set of $\struct {S, \NN}$.
That is, $U$ is a neighborhood in $\struct {S, \NN}$ of each of its points.
But the neighborhoods of $\struct {S, \NN}$ are the neighborhoods of $\struct {S, \tau}$.
Thus by Set is Open iff Neighborhood of all its Points, $U$ is an open set of $\struct {S, \tau}$.
That is:
:$U \in \tau$.
It follows by definition of set equality that $\tau = \tau'$.
{{qed}}
Theorem: Correspondence between Neighborhood Space and Topological Space

Let $S$ be a set.
Let $\struct {S, \tau}$ be a topological space.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\phi: \struct {S, \tau} \to \struct {S, \NN}$ be the mapping defined as:
:$\forall x \in S: \map \phi x = x$
:$\forall U \in \tau: \phi \sqbrk U = U \in \NN$
Let $\struct {T, \NN'}$ be a neighborhood space.
Let $\struct {T, \tau'}$ be the topological space induced by $\NN$ on $S$.
Let $\psi: \struct {T, \NN'} \to \struct {T, \tau'}$ be the mapping defined as:
:$\forall y \in T: \map \psi y = y$
:$\forall V \in \NN': \psi \sqbrk V = V \in \tau'$
Then:
:$(1): \quad \phi^{-1} \sqbrk {\phi \sqbrk {\struct {S, \tau} } } = \struct {S, \tau}$
:$(2): \quad \psi^{-1} \sqbrk {\psi \sqbrk {\struct {T, \NN'} } } = \struct {T, \NN'}$

Proof:
From the construction of:
:the neighborhood space induced by $\tau$ on $S$
:the topological space induced by $\NN$ on $S$
the mappings $\phi$ and $\psi$ are well-defined mappings.
From Topological Space induced by Neighborhood Space induced by Topological Space, $\phi$ is a bijection.
From Neighborhood Space induced by Topological Space induced by Neighborhood Space, $\psi$ is a bijection.
Hence the result.
{{qed}}
Theorem: Neighborhood Space induced by Topological Space induced by Neighborhood Space

Let $\struct {S, \NN}$ be a neighborhood space.
Let $\struct {S, \tau}$ be the topological space induced by $\NN$ on $S$.
Let $\struct {S, \NN'}$ be the neighborhood space induced by $\tau$ on $S$.
Then $\NN = \NN'$.

Proof:
Let $x \in S$.
Let $\NN_x$ be the set of all neighborhoods of $x$.
Let $N \in \NN_x$ be a neighborhood of $x$.
From Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is the superset of some open set $U$ in $\struct {S, \NN}$.
By Neighborhood Space is Topological Space we have that $U$ is an open set of $\struct {S, \tau}$.
Thus, by definition, $N$ is a neighborhood of $x$ in the context of the topological space $\struct {S, \tau}$.
Thus by definition of the neighborhood space induced by $\tau$ on $S$, it follows that:
:$N \in \NN'_x$
where $\NN'_x$ is the set of all neighborhoods of $x$ in $\struct {S, \NN'}$.
Thus $\NN_x \subseteq \NN'_x$.
{{qed|lemma}}
Now suppose that $N \in \NN'_x$.
Then, by definition of neighborhood, $N$ is the superset of some open set $U$ in $\struct {S, \tau}$ such that $x \in U$.
Since $U \in \tau$ it follows that $U$ is an open setin $\struct {S, \NN}$.
So by Subset in Neighborhood Space is Neighborhood iff it contains Open Set, $N$ is a neighborhood of $x$ in $\struct {S, \NN}$.
That is:
:$N \in \NN_x$
Thus it follows that $\NN_x \subseteq \NN'_x$
{{qed|lemma}}
It follows by definition of set equality that $\NN_x = \NN'_x$.
As $x$ is arbitrary, it applies to all $x \in S$.
Hence the result.
{{qed}}
Theorem: Binomial Coefficient with Zero/Integer Coefficients

:$\forall n \in \N: \dbinom n 0 = 1$
where $\dbinom n 0$ denotes a binomial coefficient.

Proof:
From the definition:
{{begin-eqn}}
{{eqn | l = \binom n 0
      | r = \frac {n!} {0! \ n!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n!} {1 \cdot n!}
      | c = {{Defof|Factorial}} of $0$
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Form of Geometric Sequence of Integers in Lowest Terms

Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting of positive integers only.
Let $r$ be the common ratio of $G_n$.
Let the elements of $G_n$ be the smallest positive integers such that $G_n$ has common ratio $r$.
Then the $j$th term of $G_n$ is given by:
:$a_j = p^{n - j} q^j$
where $r = \dfrac q p$.
That is:
:$G_n = \tuple {p^n, p^{n - 1} q, p^{n - 2} q^2, \ldots, p q^{n - 1}, q^n}$

Proof:
From Form of Geometric Sequence of Integers the $j$th term of $G_n$ is given by:
:$a_j = k p^{n - j} q^j$
where the common ratio is $\dfrac q p$.
Thus:
:$a_0 = k p^n$
:$a_n = k q^n$
From Geometric Sequence in Lowest Terms has Coprime Extremes it follows that $k = 1$.
Hence the result.
{{qed}}
Category:Geometric Sequences of Integers
cwfob7tfrj3rzjzscijwag34jsinvz6
Theorem: Form of Geometric Sequence of Integers with Coprime Extremes

Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting of positive integers only.
Let $a_1$ and $a_n$ be coprime.
Then the $j$th term of $Q_n$ is given by:
:$a_j = q^j p^{n - j}$

Proof:
Let $r$ be the common ratio of $Q_n$.
Let the elements of $Q_n$ be the smallest positive integers such that $Q_n$ has common ratio $r$.
From Geometric Sequence with Coprime Extremes is in Lowest Terms, the elements of $Q_n$ are the smallest positive integers such that $Q_n$ has common ratio $r$.
From Form of Geometric Sequence of Integers in Lowest Terms the $j$th term of $P$ is given by:
:$a_j = q^j p^{n - j}$
where $r = \dfrac p q$.
Hence the result.
{{qed}}
Category:Geometric Sequences of Integers
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