avewright/proofwiki-math huggingface dataset:
Theorem: Linear Transformation between Normed Vector Spaces is Bounded iff Bounded as Linear Transformation between Topological Vector Spaces

Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be linear transformations. 
Then $T$ is bounded as a linear transformation between normed vector spaces {{iff}} it is bounded as a linear transformation between topological vector spaces.

Proof:
=== Sufficient Condition ===
Suppose that $T$ is bounded as a linear transformation between topological vector spaces
Then:
:for each von Neumann-bounded subset $E$ of $X$, $T \sqbrk E$ is von Neumann-bounded.
From Characterization of von Neumann-Boundedness in Normed Vector Space, this is equivalent to: 
:if $E \subseteq X$ is such that $\norm x_X < M$ for each $x \in E$ and some $M > 0$, then there exists $M' > 0$ such that $\norm {T x}_Y < M'$ for each $x \in E$.
In particular:
:there exists $M' > 0$ such that $\norm {T x}_Y < M'$ for each $x \in X$ with $\norm x_X < 1$. 
Then for $x \ne \mathbf 0_X$, we have: 
:$\ds \norm {\map T {\frac x {2 \norm x_X} } }_Y < M'$
so that: 
:$\ds \norm {T x}_Y < 2 M' \norm x_X$
for each $x \ne \mathbf 0_X$.
Hence $\norm {T x}_Y \le 2 M' \norm x_X$ for all $x \in X$. 
So $T$ is bounded as a linear transformation between normed vector spaces.
{{qed|lemma}}
Theorem: Homeomorphic Topology of Initial Topology is Initial Topology

Let $\struct {X_\alpha, \tau_\alpha}, \struct {X_\beta, \tau_\beta}$ be topological spaces.
Let $\ds \family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \family {f_i: X_\beta \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.
Let $\tau_\beta$ be the initial topology on $X_\beta$ with respect to $\ds \family {f_i}_{i \mathop \in I}$.
Let $\phi : \struct {X_\alpha, \tau_\alpha} \to \struct {X_\beta, \tau_\beta}$ be a homeomorphism.
Then:
:$\tau_\alpha$ is the initial topology on $X_\alpha$ with respect to $\ds \family {f_i \circ \phi : X_\alpha \to Y_i}_{i \mathop \in I}$

Proof:
Let $\SS_\beta = \set {f_i^{-1} \sqbrk U: i \in I, U \in \tau_i}$.
By definition  of initial topology:
:$\SS_\beta$ is a sub-basis for $\tau_\beta$
From Inverse of Homeomorphism is Homeomorphism:
:$\phi^{-1}$ is a homeomorphism
From Homeomorphic Image of Sub-Basis is Sub-Basis:
:$\SS_\alpha = \set {\phi^{-1} \sqbrk {f_i^{-1} \sqbrk U} : i \in I, U \in \tau_i}$ is a sub-basis for $\tau_\alpha$
From Preimage of Subset under Composite Mapping:
:$\forall i \in I, U \in \tau_i : \phi^{-1} \sqbrk {f_i^{-1} \sqbrk U} = \paren{f_i \circ \phi}^{-1} \sqbrk U$
Hence:
:$\SS_\alpha = \set {\paren {f_i \circ \phi}^{-1} \sqbrk U : i \in I, U \in \tau_i}$ is a sub-basis for $\tau_\alpha$
By definition, $\tau_\alpha$ is the initial topology on $X_\alpha$ with respect to $\ds \family {f_i \circ \phi}_{i \mathop \in I}$
{{qed}}
Category:Homeomorphisms (Topological Spaces)
Category:Initial Topology
50vs0be6j2qdspbmf2ijk05rdpun9bf
Theorem: Net Characterization of von Neumann-Boundedness in Topological Vector Space

Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $E \subseteq X$.
Let $\struct {\Lambda, \preceq}$ be a directed set. 
Let $E$ be von Neumann-bounded.
Then:
:for each net $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda}$ in $E$ and each net $\sequence {\alpha_\lambda}_{\lambda \mathop \in \Lambda}$ in $\GF$ such that $\alpha_\lambda \to 0$, we have $\alpha_\lambda x_\lambda \to {\mathbf 0}_X$.
Conversely, if:
:for each sequence $\sequence {x_n}_{n \in \N}$ in $E$ and each sequence $\sequence {\alpha_n}_{n \in \N}$ in $\GF$ such that $\alpha_n \to 0$ as $n \to \infty$, we have $\alpha_n x_n \to {\mathbf 0}_X$ as $n \to \infty$
then $E$ is von Neumann-bounded.

Proof:
=== Necessary Condition ===
Let $E \subseteq X$ be von Neumann-bounded.
Let $\sequence {x_\lambda}_{\lambda \mathop \in \Lambda}$ be a net in $E$. 
Let $\sequence {\alpha_\lambda}_{\lambda \mathop \in \Lambda}$ be a net in $\GF$ such that $\alpha_\lambda \to 0$. 
Let $U$ be an open neighborhood of ${\mathbf 0}_X$.
From Open Neighborhood of Origin in Topological Vector Space contains Balanced Open Neighborhood, there exists a balanced open neighborhood $V$ of ${\mathbf 0}_X$ such that $V \subseteq X$. 
Since $E$ is von Neumann-bounded, there exists $t > 0$ such that: 
:$E \subseteq t V$ 
so that: 
:$\ds \frac 1 t E \subseteq V$ 
Then we have: 
:$\ds \frac {x_\lambda} t \in V$ 
Since $\alpha_\lambda \to 0$, there exists $\lambda_0 \in \Lambda$ such that: 
:$\ds \cmod {\alpha_\lambda} < \frac 1 t$ for $\lambda \succeq \lambda_0$.
That is: 
:$\ds \cmod {\alpha_n} t = \cmod {\alpha_n t} < 1$ for $\lambda \succeq \lambda_0$.
Since $V$ is balanced, we have: 
:$\alpha_\lambda t V \subseteq V$ 
so that: 
:$\alpha_\lambda x_\lambda \in V$ for $\lambda \succeq \lambda_0$
since $x_\lambda/t \in V$.
So, we have: 
:$\alpha_\lambda x_\lambda \in V$ for $\lambda \succeq \lambda_0$.
So $\alpha_\lambda x_\lambda \to \mathbf 0_X$.
{{qed|lemma}}
Theorem: Primitive of Reciprocal of 1 plus Sine of a x/Corollary

:$\ds \int \frac {\d x} {1 + \sin x} = \map \tan {\frac x 2 - \frac \pi 4} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {1 + \sin a x}
      | r = -\frac 1 a \map \tan {\frac \pi 4 - \frac {a x} 2} + C
      | c = Primitive of $\dfrac 1 {1 + \sin a x}$
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {1 + \sin x}
      | r = -\map \tan {\frac \pi 4 - \frac x 2} + C
      | c = setting $a \gets 1$
}}
{{eqn | r = \map \tan {\frac x 2 - \frac \pi 4} + C
      | c = Tangent Function is Odd
}}
{{end-eqn}}
{{qed}}
Theorem: Newton's Method/Sequence of Approximations Converges Quadratically

Let $\map f x$ be a real function.
Let $\alpha$ be a root of $\map f x$.
Let $\epsilon > 0$ be a positive real number, and $I = \closedint {\alpha - \epsilon} {\alpha + \epsilon}$.
Let $f$ have a continuous second derivative on $I$.
Let $\ds M = \map \sup {\size {\frac {\map {f' '} s} {\map {f'} t} } }$ over all $s, t \in I$.
:For this to be well-defined, it is also necessary that $\map {f'} x$ is non-vanishing on $I$.
Suppose $\epsilon < \dfrac 2 M$.
Then the sequence generated by Newton's Method, starting with any initial guess $x_0 \in I$, converges to $\alpha$ with order $2$.

Proof:
Suppose that the sequence is produced up to $x_n$.
Suppose also that $x_n \in I$.
{{begin-eqn}}
{{eqn | l = \map f {x_n} + \paren {\alpha - x_n} \map {f'} {x_n} + \frac {\map {f' '} {\zeta_n} } 2 \paren {\alpha - x_n}^2
      | r = \map f \alpha
      | c = Taylor's Theorem, for some $\zeta_n \in \openint {x_n} \alpha$
}}
{{eqn | r = 0
      | c = {{Defof|Root of Function}}
}}
{{eqn | l = \frac {\map f {x_n} } {\map {f'} {x_n} } + \paren {\alpha - x_n} + \frac {\map {f' '} {\zeta_n} } {2\map {f'} {x_n} } \paren {\alpha - x_n}^2
      | r = 0
      | c = $\map {f'} {x_n} \ne 0$
}}
{{eqn | l = \alpha - \paren {x_n - \frac {\map f {x_n} } {\map {f'} {x_n} } }
      | r = -\frac {\map {f' '} {\zeta_n} } {2\map {f'} {x_n} } \paren {\alpha - x_n}^2
}}
{{eqn | l = \alpha - x_{n + 1}
      | r = -\frac {\map {f' '} {\zeta_n} } {2\map {f'} {x_n} } \paren {\alpha - x_n}^2
      | c = Newton's Method
}}
{{eqn | l = \size {\alpha - x_{n + 1} }
      | r = \size {\frac {\map {f''} {\zeta_n} } {2\map {f'} {x_n} } \paren {\alpha - x_n}^2}
}}
{{eqn | o = \le
      | r = \frac M 2 \size {\paren {\alpha - x_n} }^2
      | c = Definition of $M$
}}
{{end-eqn}}
First:
{{begin-eqn}}
{{eqn | l = \size {\alpha - x_{n + 1} }
      | o = \le
      | r = \frac {M \epsilon} 2 \size {\alpha - x_n}
      | c = by assumption of $x_n \in I$
}}
{{eqn | o = <
      | r = \size {\alpha - x_n}
      | c = {{hypothesis}} of $\epsilon < \dfrac 2 M$
}}
{{end-eqn}}
Therefore $x_{n + 1} \in I$.
But as $x_0 \in I$, by Principle of Mathematical Induction, every $x_i \in I$.
Next, apply the Principle of Recursive Definition to define the sequence:
{{begin-eqn}}
{{eqn | l = \epsilon_0
      | r = \size {\alpha - x_0}
}}
{{eqn | l = \epsilon_{n + 1}
      | r = \frac M 2 \epsilon_n^2
}}
{{end-eqn}}
Trivially:
:$\size {\alpha - x_0} \le \epsilon_0$
Assuming $\size {\alpha - x_n} \le \epsilon_n$, it follows that:
{{begin-eqn}}
{{eqn | l = \size {\alpha - x_{n + 1} }
      | o = \le
      | r = \frac M 2 \size {\alpha - x_n}^2
}}
{{eqn | o = \le
      | r = \frac M 2 \epsilon_n^2
}}
{{eqn | r = \epsilon_{n + 1}
}}
{{end-eqn}}
By Principle of Mathematical Induction, every $\size {\alpha - x_n} \le \epsilon_n$.
But:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \frac {\epsilon_{n + 1} } {\epsilon_n^p}
      | r = \lim_{n \mathop \to \infty} \frac M 2
}}
{{eqn | r = \frac M 2
}}
{{end-eqn}}
Therefore, the sequence $\sequence {x_n}$ converges to $\alpha$ with order $2$, by definition.
{{qed}}
Theorem: Primitive of Reciprocal of 1 minus Sine of a x/Corollary 2

:$\ds \int \frac {\d x} {1 - \sin x} = -\map \cot {\frac x 2 - \frac \pi 4} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {1 - \sin a x}
      | r = \map \tan {\frac x 2 + \frac \pi 4} + C
      | c = Primitive of $\dfrac 1 {1 - \sin x}$: Tangent form
}}
{{eqn | r = \dfrac \pi 2 - \map \cot {\frac \pi 4 + \frac x 2} + C
      | c = setting $a \gets 1$
}}
{{eqn | r = 
      | c = 
}}
{{end-eqn}}
{{finish}}
Theorem: Primitive of Reciprocal of 1 minus Sine of a x/Corollary 1

:$\ds \int \frac {\d x} {1 - \sin x} = \map \tan {\frac x 2 + \frac \pi 4} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {1 - \sin a x}
      | r = \frac 1 a \map \tan {\frac \pi 4 + \frac {a x} 2} + C
      | c = Primitive of $\dfrac 1 {1 - \sin a x}$
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {1 - \sin x}
      | r = \map \tan {\frac \pi 4 + \frac x 2} + C
      | c = setting $a \gets 1$
}}
{{eqn | r = \map \tan {\frac x 2 + \frac \pi 4} + C
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Homeomorphic Image of Sub-Basis is Sub-Basis

Let $T_\alpha = \struct{X_\alpha, \tau_\alpha}, T_\beta = \struct{X_\beta, \tau_\beta}$ be topological spaces.
Let $\SS_\alpha \subseteq \tau_\alpha$ be a sub-basis for $\tau_\alpha$.
Let $\phi: T_\alpha \to T_\beta$ be a homeomorphism.
Let $\SS_\beta = \set{\phi \sqbrk S : S \in \SS_\alpha}$.
Then:
:$\SS_\beta$ is a sub-basis for $\tau_\beta$

Proof:
By definition of homeomorphism:
:$\forall U \subseteq X_\alpha : U \in \tau_\alpha \iff \phi \sqbrk U \in \tau_\beta$
By definition of sub-basis:
:$\SS_\alpha \subseteq \tau_\alpha$
Hence:
:$\SS_\beta \subseteq \tau_\beta$
Let $\ds \BB_\alpha = \set {\bigcap \FF: \FF \subseteq \SS_\alpha, \FF \text{ is finite} }$.
Let $\ds \BB_\beta = \set {\bigcap \GG: \GG \subseteq \SS_\beta, \GG \text{ is finite} }$.
Let $B \in \BB_\alpha$.
By definition of $\BB_\alpha$:
:$\exists \FF \subseteq \BB_\alpha : \FF \text{ is finite} : B = \bigcap \set{S : S \in \FF}$
We have:
{{begin-eqn}}
{{eqn | l = \phi \sqbrk B
      | r = \phi \sqbrk {\bigcap \set{S : S \in \FF} }
}}
{{eqn | r = \bigcap \set{\phi \sqbrk S : S \in \FF}
      | c = Image of Intersection under Injection
}}
{{end-eqn}}
Let $\GG = \set{\phi \sqbrk S : S \in \FF}$.
Then:
:$\GG \subseteq \BB_\beta : \GG \text{ is finite} : \phi \sqbrk B = \bigcap \GG$
Hence $\phi \sqbrk B \in \BB_\beta$.
We have shown that:
:$(1): \quad \forall B \in \BB_\alpha : \phi \sqbrk B \in \BB_\beta$
Let $V \in \tau_\beta$.
From Inverse of Homeomorphism is Homeomorphism:
:$\phi^{-1}$ is a homeomorphism
By definition of homeomorphism:
:$\phi^{-1} \sqbrk V \in \tau_\alpha$
By definition of sub-basis:
{{begin-eqn}}
{{eqn | n = 2
      | q = \exists \AA \subseteq \BB_\alpha
      | l = \phi^{-1} \sqbrk V
      | r = \bigcup \AA
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = V
      | r = \phi \sqbrk {\phi^{-1} \sqbrk V}
      | c = Image of Preimage of Subset under Surjection equals Subset
}}
{{eqn | r = \phi \sqbrk {\bigcup \set{B : B \in \AA} }
      | c = From $(2)$ above
}}
{{eqn | r = \bigcup \set{\phi \sqbrk B : B \in \AA}
      | c = From Image of Union under Mapping
}}
{{end-eqn}}
Let $\AA' = \set{\phi \sqbrk B : B \in \BB_\alpha}$
From $(1)$ above:
:$\AA' \subseteq \BB_\beta$
We have:
:$V = \bigcup \AA'$
Hence by definition, $\SS_\beta$ is a sub-basis for $\tau_\beta$
{{qed}}
Category:Homeomorphisms (Topological Spaces)
Category:Topological Bases
n1wsww7xzqpzexjytan2ogaa5xkagng
Theorem: Image of von Neumann-Bounded Set under Equicontinuous Family of Linear Transformations is Contained in von Neumann-Bounded Set

Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be a topological vector space over $\GF$. 
Let $\family {T_\alpha}_{\alpha \in I}$ be an equicontinuous family of linear transformations. 
Let $E \subseteq X$ be von Neumann-bounded. 
Then there exists a von Neumann-bounded set $F \subseteq Y$ such that: 
:$T_\alpha \sqbrk E \subseteq F$ for each $\alpha \in I$.

Proof:
Let:
:$\ds F = \bigcup_{\alpha \mathop \in I} T_\alpha \sqbrk E$
Then $T_\alpha \sqbrk E \subseteq F$ for each $\alpha \in I$.
It is enough to show that $F$ is von Neumann-bounded.
Let $W$ be an open neighborhood of $\mathbf 0_Y$.
Since $\family {T_\alpha}_{\alpha \in I}$ is equicontinuous, there exists an open neighborhood $V$ of $\mathbf 0_X$ such that: 
:$T_\alpha \sqbrk V \subseteq W$ for each $\alpha \in I$. 
Since $E$ is von Neumann-bounded, there exists $s > 0$ such that for all $t > s$ we have $E \subseteq t W$. 
Then for each $t > s$ and $\alpha \in I$, we have:
{{begin-eqn}}
{{eqn	| l = T_\alpha \sqbrk E
		| o = \subseteq
		| r = T_\alpha \sqbrk {t V}
}}
{{eqn	| r = t T_\alpha \sqbrk V
		| c = Image of Dilation of Set under Linear Transformation is Dilation of Image
}}
{{eqn	| o = \subseteq
		| r = t W
}}
{{end-eqn}}
So, from Union of Subsets is Subset:
:$\ds F = \bigcup_{\alpha \in I} T_\alpha \sqbrk E \subseteq t W$
for $t > s$.
So $F$ is von Neumann-bounded.
{{qed}}
Theorem: Integration by Partial Fractions

Let $\map R x = \dfrac {\map P x} {\map Q x}$ be a rational function over $\R$ such that the degree of the polynomial $P$ is strictly smaller than the degree of the polynomial $Q$.
Consider the primitive:
:$\ds \int \map R x \rd x$
Let $\map R x$ be expressible by the partial fractions expansion:
:$\map R x = \ds \sum_{k \mathop = 0}^n \dfrac {\map {p_k} x} {\map {q_k} x}$
where:
:each of the $\map {p_k} x$ are polynomial functions
:the degree of $\map {p_k} x$ is strictly less than the degree of $\map {q_k} x$ for all $k$.
Then:
:$\ds \int \map R x \rd x = \sum_{k \mathop = 0}^n \int \dfrac {\map {p_k} x} {\map {q_k} x} \rd x$
This technique is known as '''Integration by Partial Fractions'''.

Proof:
{{begin-eqn}}
{{eqn | l = \int \map R x \rd x
      | r = \int \paren {\sum_{k \mathop = 0}^n \dfrac {\map {p_k} x} {\map {q_k} x} } \rd x
      | c = Definition of $\map R x$: {{hypothesis}}
}}
{{eqn | r = \sum_{k \mathop = 0}^n \int \dfrac {\map {p_k} x} {\map {q_k} x} \rd x
      | c = Linear Combination of Integrals
}}
{{end-eqn}}
{{qed}}
Theorem: Characterization for Topological Evaluation Mapping to be Embedding/Necessary Condition

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping induced by $\family{f_i}_{i \mathop \in I}$.
Let $f$ be an embedding
Then:
:$(1)\quad$ the topology on $X$ is the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separates points

Proof:
=== $(1)$ The Topology on $X$ is the Initial Topology ===
Let $f \sqbrk X$ denote the image of $f$.
Let $\tau_{f \sqbrk X}$ be the subspace topology on $f \sqbrk X$.
By definition of embedding:
:$f \restriction_{X \times f \sqbrk X}$ is a homeomorphism between $X$ and $f \sqbrk X$
From Subspace of Product Space has Initial Topology with respect to Restricted Projections:
:$\tau_{f \sqbrk X}$ is the initial topology on $f \sqbrk X$ with respect to the mappings $\family {\pr_i \restriction_{f \sqbrk X} : f \sqbrk X \to Y_i}_{i \mathop \in I}$
Let $\tau$ be the topology on $X$.
From Homeomorphic Topology of Initial Topology is Initial Topology:
:$\tau$ is the initial topology on $X$ with respect to $\family {\pr_i \restriction_{f \sqbrk X} \circ f \restriction_{X \times f \sqbrk X} : X \to Y_i}_{i \mathop \in I}$
We have:
{{begin-eqn}}
{{eqn | l = \pr_i \restriction_{f \sqbrk X} \circ f \restriction_{X \times f \sqbrk X}
      | r = \pr_i \circ f
      | c = Composition of Mapping with Mapping Restricted to Image
}}
{{eqn | r = f_i
      | c = Composite of Evaluation Mapping and Projection
}}
{{end-eqn}}
Hence:
:$\tau$ is the initial topology on $X$ with respect to $\family {f_i : X \to Y_i}_{i \mathop \in I}$
{{qed|lemma}}
=== $(2)$ The Family $\family {f_i}$ Separates Points ===
By definition of embedding:
:$f$ is a homeomorphism between $X$ and $f \sqbrk X$
By definition of homeomorphism:
:$f$ is an injection
From Evaluation Mapping is Injective iff Mappings Separate Points:
:the family $\family {f_i}$ separates points.
Theorem: Characterization for Topological Evaluation Mapping to be Embedding/Sufficient Condition

Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
Let $f : X \to Y$ be the evaluation mapping  induced by $\family{f_i}_{i \mathop \in I}$.
Let:
:$(1)\quad$ the topology on $X$ be the initial topology with respect to $\family {f_i}_{i \mathop \in I}$
:$(2)\quad$ the family $\family {f_i}$ separates points
Then:
:$f$ is an embedding

Proof:
From Evaluation Mapping is Injective iff Mappings Separate Points:
:$f$ is an injection
From Injection to Image is Bijection:
:$f \restriction_{X \times f \sqbrk X} \mathop : X \to f \sqbrk X$ is a bijection
From Topological Evaluation Mapping is Continuous:
:$f$ is continuous
From Continuity of Composite of Inclusion on Mapping:
:$f \restriction_{X \times f \sqbrk X}$ is continuous
Let $\SS = \set{ f_i^{-1} \sqbrk V : i \in I, V \subseteq Y_i \text{ is open}}$.
Let $f_i^{-1} \sqbrk V \in \SS$ for some $i \in I, V \subseteq Y_i$ is open.
Let $\pr_i$ denote the projection from $Y$ to $Y_i$.
We have:
{{begin-eqn}}
{{eqn | l = f \sqbrk {f_i^{-1} \sqbrk V}
      | r = f \sqbrk {\paren{\pr_i \circ f}^{-1} \sqbrk V}
      | c = Composite of Evaluation Mapping and Projection
}}
{{eqn | r = f \sqbrk {f^{-1} \sqbrk {\pr_i^{-1} \sqbrk V} }
      | c = Preimage of Subset under Composite Mapping
}}
{{eqn | r = \pr_i^{-1} \sqbrk V \cap f \sqbrk X
      | c = Image of Preimage under Mapping
}}
{{end-eqn}}
By definition of product topology:
:$\pr_i^{-1} \sqbrk V$ is open in $Y$
By definition of subspace:
:$f \sqbrk {f_i^{-1} \sqbrk V} = \pr_i^{-1} \sqbrk V \cap f \sqbrk X$ is open in $f \sqbrk X$
By definition of restriction:
:$f \restriction_{X \times f \sqbrk X} \sqbrk {f_i^{-1} \sqbrk V} = f \sqbrk {f_i^{-1} \sqbrk V}$
We have shown that:
:$\forall U \in \SS : f \restriction_{X \times f \sqbrk X} \sqbrk U \text{ is open in } Y$
By definition of initial topology:
:$\SS$ is a sub-basis for the topology on $X$
From Injection is Open Mapping iff Image of Sub-Basis Set is Open:
:$f \restriction_{X \times f \sqbrk X}$ is an open mapping
By definition, $f \restriction_{X \times f \sqbrk X}$ is a homeomorphism.
By definition, $f$ is an embedding.
Theorem: There exist no 4 Consecutive Triangular Numbers which are all Sphenic Numbers

Let $n \in \N$ be a natural number.
Let $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ be the $n$th, $n + 1$th, $n + 2$th and $n + 3$th triangular numbers respectively.
Then it is not the case that all of $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are sphenic numbers.

Proof:
{{AimForCont}} there exists an $n$ such that $T_n$, $T_{n + 1}$, $T_{n + 2}$, and $T_{n + 3}$ are all sphenic numbers.
Observe from Sequence of Smallest 3 Consecutive Triangular Numbers which are Sphenic that there are no such $n$ for $n < 12$.
Thus from Closed Form for Triangular Numbers:
{{begin-eqn}}
{{eqn | l = T_n
      | r = \dfrac {n \paren {n + 1} } 2
}}
{{eqn | l = T_{n + 1}
      | r = \dfrac {\paren {n + 1} \paren {n + 2} } 2
}}
{{eqn | l = T_{n + 2}
      | r = \dfrac {\paren {n + 2} \paren {n + 3} } 2
}}
{{eqn | l = T_{n + 3}
      | r = \dfrac {\paren {n + 3} \paren {n + 4} } 2
}}
{{end-eqn}}
Recall the definition of sphenic number:
{{:Definition:Sphenic Number}}
Let $S = \set {n, n + 1, n + 2, n + 3, n + 4}$.
Then there exists $x \in S$ such that $4 \divides x$.
If $x = n$ or $x = n + 4$, then one of $n$ or $n + 4$ is divisible by $8$.
Then one of $T_n$ or $T_{n + 3}$ is divisible by $4$, so it cannot be sphenic.
Therefore $x \ne n$ and $x \ne n + 4$.
Hence $x - 1 \in S$ and $x + 1 \in S$.
Note that {{hypothesis}} both $T_{x - 1}$ and $T_x$ are sphenic.
Write $x = 4 m$ for some integer $m$.
Since $x > n \ge 12$, we have $m > 3$.
Note that:
{{begin-eqn}}
{{eqn | l = T_x
      | r = \frac {4 m \paren {4 m + 1} } 2
      | c = Closed Form for Triangular Numbers
}}
{{eqn | r = 2 \times m \times \paren {4 m + 1}
}}
{{end-eqn}}
so as $T_x$ is sphenic, both $m$ and $4 m + 1$ must be prime.
Similarly, since $T_{x - 1}$ is sphenic, $4 m - 1$ must be prime.
One of $4 m - 1, 4 m, 4 m + 1$ is divisible by $3$.
As $4 m + 1 > 4 m - 1 > 3$, $4 m$ must be divisible by $3$.
By Euclid's Lemma, since $3 \perp 4$:
:$3 \divides m$
which contradicts the fact that $m$ is prime.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Number of form 4666...6669 is Divisible by 7

Let $x$ be a natural number in the form:
:$\sqbrk {4 \underbrace {666 \cdots 6}_n 9}_{10}$ 
Then $x$ is divisible by $7$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \sqbrk {4 \underbrace {666 \cdots 6}_n 9}_{10}
      | r = 4 \times 10^{n + 1} + 6 \times 10^n + \cdots + 6 \times 10^1 + 9
      | c = 
}}
{{eqn | r = 4 \times 10^{n + 1} + \dfrac {6 \times \paren {10^n - 1} } {\paren {10 - 1} } \times 10 + 9
      | c = Sum of Geometric Sequence
}}
{{eqn | r = 4 \times 10^{n + 1} \times \dfrac 3 3 + \dfrac {6 \times \paren {10^n - 1} } 9 \times 10 + 9 \times \dfrac 3 3 
      | c = multiplying by 1 to obtain Common Denominator
}}
{{eqn | r = \dfrac {12 \times 10^{n + 1} } 3 + \dfrac {2 \times \paren {10^{n + 1} - 10} } 3 + \dfrac {27} 3
      | c = 
}}
{{eqn | r = \dfrac {14 \times 10^{n + 1} + 7} 3
      | c = 
}}
{{end-eqn}}
The sum of the digits in $14 \times 10^{n + 1} + 7$ is in fact $12 = 1 + 4 + 7$ which is a multiple of $3$.
Hence, by Divisibility by $3$, $\dfrac {14 \times 10^{n + 1} + 7} 3$ is indeed an integer.
Then:
{{begin-eqn}}
{{eqn | l = (14 \times 10^{n + 1} + 7) \div 3
      | o = \equiv 
      | r = (0 \times 3^{n + 1} + 0) \div 3
      | rr = \pmod 7
      | c = Congruence of Powers
}}
{{eqn | o = \equiv 
      | r = (0 + 0) \div 3
      | rr = \pmod 7
      | c = Congruence of Product
}}
{{eqn | o = \equiv 
      | r = 0 \div 3
      | rr = \pmod 7
      | c = Congruence of Quotient
}}
{{eqn | o = \equiv 
      | r = 0
      | rr = \pmod 7
}}
{{end-eqn}}
{{qed}}
Category:Divisibility by 7
bnoyf6n78whcl3d9oejtdsb4kle4rkd
Theorem: Number of form 28000...0007 is Divisible by 7

Let $x$ be a natural number in the form:
:$\sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}$ 
Then $x$ is divisible by $7$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}
      | r = 2 \times 10^{n + 2} + 8 \times 10^{n + 1} + 0 \times 10^n + \cdots + 0 \times 10^1 + 7
      | c = 
}}
{{eqn | r = 28 \times 10^{n + 1} + 7
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 28 \times 10^{n + 1} + 7
      | o = \equiv 
      | r = 0 \times 3^{n + 1} + 0
      | rr= \pmod 7
      | c = Fermat's Little Theorem and Congruence of Powers
}}
{{eqn | o = \equiv 
      | r = 0 + 0
      | rr= \pmod 7
      | c = Congruence of Product
}}
{{eqn | o = \equiv 
      | r = 0
      | rr= \pmod 7
}}
{{end-eqn}}
{{qed}}