nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Generalized Pentagonal Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
It is sufficient to verify that the sequence $GP_n$ defined in Definition 2 contains both sequences of pentagonal numbers and second pentagonal numbers, and is in ascending order.
From Closed Form for Pentagonal Numbers:
:$P_n = \dfrac {n \paren {3 n - 1} } 2$
which are the odd terms (and the zeroth term) of $GP_n$.
From definition of second pentagonal numbers:
:$P'_n = \dfrac {n \paren {3 n + 1} } 2$
which are the even terms of $GP_n$.
Finally it is shown that $GP_n$ is increasing.
We have:
:$GP_{2 m - 1} = \dfrac {m \paren {3 m - 1} } 2 < \dfrac {m \paren {3 m + 1} } 2 = GP_{2 m}$
and:
:$GP_{2 m} = \dfrac {m \paren {3 m + 1} } 2 < \dfrac {\paren {m + 1} \paren {3 m + 2} } 2 = GP_{2 m + 1}$
Hence the result.
Category:Generalized Pentagonal Numbers
Equivalence of Definitions of Generated Normal Subgroup

== Theorem ==
Let $G$ be a group.
Let $S \subseteq G$ be a subset.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $H$ be the smallest normal subgroup containing $S$.
Let $\mathbb S$ be the set of normal subgroups containing $S$.
To show the equivalence of the two definitions, we need to show that $\ds H = \bigcap \mathbb S$.
Since $H$ is a normal subgroup containing $S$:
:$H \in \mathbb S$
By Intersection is Subset:
:$\ds \bigcap \mathbb S \subseteq H$
On the other hand, by Intersection of Normal Subgroups is Normal:
:$\ds \bigcap \mathbb S$ is a normal subgroup containing $S$.
Since $H$ be the smallest normal subgroup containing $S$:
:$\ds H \subseteq \bigcap \mathbb S$
By definition of set equality:
:$\ds H = \bigcap \mathbb S$
Hence the result.
Category:Generated Normal Subgroups
Equivalence of Definitions of Generated Subgroup

== Theorem ==
Let $G$ be a group.
Let $S \subset G$ be a subset.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ is equivalent to $(2)$ ===
Let $H$ be the smallest subgroup containing $S$.
Let $\mathbb S$ be the set of subgroups containing $S$.
To show the equivalence of the two definitions, we need to show that $H = \bigcap \mathbb S$.
Since $H$ is a subgroup containing $S$:
:$H \in \mathbb S$
By Intersection is Subset:
:$\bigcap \mathbb S \subseteq H$
On the other hand, by Intersection of Subgroups is Subgroup:
:$\bigcap \mathbb S$ is a subgroup containing $S$.
Since $H$ be the smallest subgroup containing $S$:
:$H \subseteq \bigcap \mathbb S$
By definition of set equality:
:$H = \bigcap \mathbb S$
Hence the result.
=== $(1)$ is equivalent to $(3)$ ===
This is shown in Set of Words Generates Group.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $12$
* : Chapter $\text{II}$: Groups: Exercise $\text{K}$
Category:Generated Subgroups
Equivalence of Definitions of Generated Submodule over Ring with Unity

== Theorem ==
Let $R$ be a ring with unity..
Let $\struct {M, +, \circ}_R$ be a unitary $R$-module.
Let $S\subset M$ be a subset.
=== Definition for $R$-module ===
=== Definition for unitary $R$-module ===
== Proof ==
Let:
:$\ds H_1 := \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} }$
be the intersection defined in first definition.
From Intersection of Submodules is Submodule:General Result, it follows that $H_1$ is a submodule of $M$.
Let:
:$\ds H_2 := \set { \sum_{i \mathop = 1}^n \lambda_i \circ x_i: n \in \N_{\ge 1}, \lambda_i \in R, x_i \in S}$
be the set defined in second definition.
To show equivalence of the definitions, we prove that $H_1 = H_2$.
=== $H_2$ is a subset of $H_1$ ===
Let $\ds \sum_{i \mathop = 1}^n \lambda_i \circ x_i \in H_2$, where $\lambda_i \in R, x_i \in S$ for all $i \in \set {1, \ldots, n}$.
Let $M'$ be a submodule of $M$ such that $S \subseteq M'$.
It follows that $x_1 ,\ldots, x_n \in M'$.
As an $R$-module is an $R$-algebraic structure , it follows that $M'$ is closed for addition $+$, and closed for scalar product $\circ$.
As $M'$ is closed for scalar product, it follows that $\lambda_i \circ x_i \in M'$ for all $i \in \set {1, \ldots, n}$.
As $M'$ is closed for addition , it follows that $\ds \sum_{i \mathop = 1}^n \lambda_i \circ x_i \in M'$.
Hence, $H_2 \subseteq M'$.
By definition of intersection, it follows that $H_2 \subseteq H_1$.
=== $H_1$ is a subset of $H_2$ ===
Let $x \in S$.
Then $1_R \circ x = x$, where $1_R$ is the unity of the ring $R$.
As $1_R \circ x$ is a linear combination, it follows that $x \in H_2$.
Hence, $S \subseteq H_2$.
Let:
:$\ds \sum_{i \mathop = 1}^n \lambda_i \circ x_i, \sum_{i \mathop = 1}^m \mu_i \circ y_i \in H_2$ 
where $m, n \in \N_{\ge 1}$, $\lambda_1, \ldots, \lambda_n, \mu_1, \ldots, \mu_m \in R$, and $x_1, \ldots, x_n, y_1, \ldots, y_m \in S$.
From definition of linear combination, it follows that the sum:
:$\ds \sum_{i \mathop = 1}^n \lambda_i \circ x_i + \sum_{i \mathop = 1}^m \mu_i \circ y_i $
is a linear combination of the set $\set {x_1, \ldots, x_n, y_1, \ldots, y_m}$.
It follows that $H_2$ is closed for addition.
Let $\lambda \in R$.
From , it follows that:
:$\ds \lambda \circ \paren{\sum_{i \mathop = 1}^n \lambda_i \circ x_i} = \sum_{i \mathop = 1}^n \paren{ \lambda \circ \lambda_i } \circ x_i $
which is a linear combination of elements of $S$.
It follows that $H_2$ is closed for scalar product.
From the Submodule Test, it follows that $H_2$ is a submodule of $M$.
It follows that $H_2$ is one of the submodules in the intersection that defines:
:$\ds H_1 := \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} }$
From Intersection is Subset:Family of Sets, it follows that $H_1 \subseteq H_2$.
Category:Generators of Modules
Equivalence of Definitions of Generator of Module

== Theorem ==
Let $R$ be a ring.
Let $M$ be an $R$-module.
Let $S \subseteq M$ be a subset of $M$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
By definition of generated submodule, it follows that:
:$\ds M := \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} }$
Suppose that $M'$ is a proper submodule of $M$ such that $S \subseteq M'$.
It follows that there exists $x \in M \setminus M'$.
Hence, $x \notin M'$.
By definition of intersection, it follows that:
:$\ds x \notin \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} } = M$
This is a contradiction.
Hence, the condition of Definition 2 is satisfied.
=== Definition 2 implies Definition 1 ===
Let $M'$ be a submodule of $M$ that contains $S$.
By hypothesis, it follows that $M'$ is not a proper submodule of $M$.
It follows that $M' = M$.
Then:
:$\ds \bigcap \set { M' \subseteq M : S \subseteq M', \textrm {$M'$ is a submodule of $M$} } = \bigcap \set { M } = M$
By definition of generated submodule, it follows that $M$ is generated by $S$.
Category:Generators of Modules
Equivalence of Definitions of Generator of Unitary Module

== Theorem ==
Let $R$ be a ring with unity.
Let $M$ be a unitary $R$-module.
Let $S \subseteq M$ be a subset.
=== Definition by linear combinations ===
$S$ is a generator of $M$ every element of $M$ is a linear combination of elements of $S$.
=== Definition by intersection of submodules ===
That is, $M$ is equal to the intersection of all submodules of $M$ containing $S$.
=== Definition by proper submodules ===
== Proof ==
=== Definition by linear combinations implies Definition by intersection of submodules ===
=== Definition by intersection of submodules implies Definition by linear combinations ===
By definition of submodule, it follows that $M$ can be considered as a submodule of itself.
From Equivalence of Definitions of Generated Submodule over Ring with Unity, it follows that:
:the set of all linear combinations of elements of $S$
is equal to:
:the intersection of all submodules of $M$ containing $S$.
Hence, the two definitions of generator of unitary module by linear combinations and by intersection of submodules are equivalent.
=== Definition by intersection of submodules implies Definition by proper submodules ===
=== Definition by proper submodules implies Definition by intersection of submodules ===
By definition of module, it follows that $M$ is an $R$-module.
The equivalence of the definitions now follows from Equivalence of Definitions of Generator of Module.
Category:Generators of Modules
Equivalence of Definitions of Generic Point of Topological Space

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $x \in S$ be an element of $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\set x^-$ denote the closure of $\set x$ in $T$.
Let $\map \UU x$ denote the system of open neighborhoods of $x$.
We have:
The result follows.
Category:Generic Points
Equivalence of Definitions of Golden Mean

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 is equivalent to Definition 3 ===
Let $AB : AC = AC : BC$.
Let $\dfrac {AB} {AC} = \dfrac {AC} {BC} = \phi$.
Then:
=== Definition 2 equivalent to Definition 3 ===
Of these two roots, only $\dfrac {1 + \sqrt 5} 2$ is positive.
Category:Golden Mean
Equivalence of Definitions of Greatest Common Divisor

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\gcd \set{a, b}$ be the greatest common divisor of integers $a$ and $b$ by definition 1.
Then by definition:
:$\gcd \set{a, b}$ is the largest $d \in \Z_{>0}$ such that $d \divides a$ and $d \divides b$.
From Common Divisor Divides GCD:
:$c \divides a \land c \divides b \implies c \divides d$
Thus $\gcd \set{a, b}$ is the greatest common divisor of integers $a$ and $b$ by definition 2.
=== $(2)$ implies $(1)$ ===
Let $\gcd \set{a, b}$ be the greatest common divisor of integers $a$ and $b$ by definition 2.
Then by definition $\gcd \set{a, b}$ is the (strictly) positive integer $d \in \Z_{>0}$ such that:
: $(1): \quad d \divides a \land d \divides b$
: $(2): \quad c \divides a \land c \divides b \implies c \divides d$
From $d \divides a \land d \divides b$, we see that $d$ is a common divisor of $a$ and $b$.
From $c \divides a \land c \divides b$, we see that $c$ is also a common divisor of $a$ and $b$.
Also, we have that $c \divides d$.
From Absolute Value of Integer is not less than Divisors, we see that (in the domain of $\Z_{>0}$): 
:$c \divides d \implies c \le d$
Thus, whatever $c$ may be, it is no larger than $d$.
Therefore, $d$ must be the greatest of all the common divisors of $a$ and $b$.
Thus $\gcd \set{a, b}$ is the greatest common divisor of integers $a$ and $b$ by definition 1.
== Sources ==
* : Chapter $1$: Properties of the Natural Numbers: $\S 23 \alpha$
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 6$
Category:Greatest Common Divisor
Equivalence of Definitions of Hadamard Matrix

== Theorem ==
=== Definition 1 ===
A Hadamard matrix $H$ is a square matrix such that:
=== Definition 2 ===
A Hadamard matrix $H$ is a square matrix such that:
== Proof ==
Let $H$ be a Hadamard matrix whose arbitrary element in the $j$th row and $k$th column is given by $\sqbrk a_{j k}$.
By either definition of Hadamard matrix:
:$a_{j k} = \pm 1$
and so:
:${a_{j k} }^2 = 1$
By definition of transpose, column $j$ of $H_\intercal$ is ${r_j}^\intercal$, where $r_j$ is row $j$ of $H$.
=== $(1)$ implies $(2)$ ===
Let $H = \sqbrk a_n$ be a Hadamard matrix by definition $1$.
We have that the rows of $H$ are pairwise orthogonal.
That is:
:$r_j \cdot {r_k}^\intercal = 0$
when $j \ne k$.
Let $r_j$ be an arbitrary row of $H$:
:$r_j = \begin {bmatrix} a_{j 1} & a_{j 2} & \cdots & a_{j n} \end {bmatrix}$
Then by definition of matrix product:
:$h_{j j} = \ds \sum_{k \mathop = 1}^n {a_{j k} }^2$
where by definition of Hadamard matrix (either one) each of $a_{j k}$ is either $1$ or $-1$.
Hence:
:$\forall k \in \set {1, 2, \ldots, n}: {a_{j k} }^2 = 1$
Then:
Similarly:
Thus $H H^\intercal$ is of the form:
:$h_{j k} = n \delta_{j k}$
where $\delta_{j k}$ is the Kronecker delta.
So by definition of identity matrix:
:$H H^\intercal = n \mathbf I_n$
Thus $H$ is a Hadamard matrix by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $H = \sqbrk a_n$ be a Hadamard matrix by definition $2$.
We have that:
:$H H^\intercal = n \mathbf I_n$
From that last line:
:$j \ne k \implies r_j \cdot {r_k}^\intercal = 0$
That is, the rows of $H$ are pairwise orthogonal.
Thus $H$ is a Hadamard matrix by definition $1$.
== Sources ==
* 
* 
Category:Hadamard Matrices
Equivalence of Definitions of Hausdorff Topological Vector Space

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$. 
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
From Topological Vector Space is Hausdorff iff T1, $\struct {X, \tau}$ is Hausdorff it is $T_1$ space.
From Definition 2 of a $T_1$ space, it follows that $\struct {X, \tau}$ is Hausdorff :
:for each $x \in X$, $\set x$ is closed in $\struct {X, \tau}$.
Category:Hausdorff Topological Vector Spaces
Equivalence of Definitions of Heptagonal Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $H_n$ be a heptagonal number by definition 1.
Let $n = 0$.
By definition:
:$H_0 = 0$
By vacuous summation:
:$\ds H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4} = 0$
By definition of summation:
and so:
Thus $H_n$ is a heptagonal number by definition 2.
=== Definition 2 implies Definition 1 ===
Let $H_n$ be a heptagonal number by definition 2.
Then:
Then:
:$\ds H_0 = \sum_{i \mathop = 1}^0 \paren {5 i - 4}$
is a vacuous summation and so:
:$H_0 = 0$
Thus $H_n$ is a heptagonal number by definition 1.
=== Definition 1 equivalent to Definition 3 ===
We have by definition that $H_0 = 0 = \map P {7, 0}$.
Then:
Thus $\map P {7, n}$ and $H_n$ are generated by the same recurrence relation.
Category:Heptagonal Numbers
Equivalence of Definitions of Hereditarily Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $T = \struct {S, \tau}$ be hereditarily compact by definition 1.
Let $\family {U_i}_{i \mathop \in I}$ be an indexed family of open sets of $T$.
We have that:
:$\ds \bigcup_{i \mathop \in I} U_i \subset S$
By hypothesis, $\ds \bigcup_{i \mathop \in I} U_i$ is compact when considered as a subspace of $T$.
Furthermore, $\ds U_i = U_i \cap \bigcup_{i \mathop \in I} U_i$ is open in $\ds \bigcup_{i \mathop \in I} U_i$, by definition of the subspace topology.
Therefore $\family {U_i}_{i \mathop \in I}$ is an open cover for $\ds \bigcup_{i \mathop \in I} U_i$.
Since $\ds \bigcup_{i \mathop \in I} U_i$ is compact, there exists a finite subcover of $\family {U_i}_{i \mathop \in I}$ for $\ds \bigcup_{i \mathop \in I} U_i$, say $\family {U_j}_{j \mathop \in J}$.
Then by definition of open cover:
:$\ds \bigcup_{j \mathop \in J} U_j = \bigcup_{i \mathop \in I} U_i$
Thus $T = \struct {S, \tau}$ is hereditarily compact by definition 2.
=== Definition 2 implies Definition 1 ===
Let $T = \struct {S, \tau}$ be hereditarily compact by definition 2.
Let $Y \subset S$ be a subspace of $T$.
Let $\family {V_i}_{i \mathop \in I}$ be an open cover for $Y$:
:$\ds \bigcup_{i \mathop \in I} V_i = Y$
Then by definition of the subspace topology:
:$V_i = U_i \cap Y$
for a certain $V_i \in \tau$
But then $\family {U_i}_{i \mathop \in I}$ is an indexed family of open sets of $T$.
By hypothesis, there exists a finite subset $J \subset I$ such that:
:$\ds \bigcup_{j \mathop \in J} U_i = \bigcup_{i \mathop \in I} U_i$
But then:
Thus $\family {V_j}_{j \mathop \in J}$ is a finite subcover of $\family {V_i}_{i \mathop \in I}$ for $Y$. 
Thus $Y$ is a compact subspace of $T$.
Thus $T = \struct {S, \tau}$ is hereditarily compact by definition 1.
Category:Hereditarily Compact Spaces
Equivalence of Definitions of Heronian Triangle

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\triangle ABC$ be a triangle whose sides $a$, $b$ and $c$ are rational in length.
Let $\AA$ be the area of $\triangle ABC$.
Let $h_a$, $h_b$ and $h_c$ be the altitudes from $A$, $B$ and $C$ respectively.
From Area of Triangle in Terms of Side and Altitude, we have:
:$\AA = \dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$
We are given that $a$, $b$ and $c$ are rational.
It follows from Rational Multiplication is Closed and Rational Division is Closed that $\AA$ is rational $h_a$, $h_b$ and $h_c$ are rational.
Hence the result.
== Sources ==
* : Bachet
Category:Heronian Triangles
Equivalence of Definitions of Hexagonal Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $H_n$ be a hexagonal number by definition 1.
Let $n = 0$.
By definition:
:$H_0 = 0$
By vacuous summation:
:$\ds H_0 = \sum_{i \mathop = 1}^0 \paren {4 \paren {i - 1} + 1} = 0$
By definition of summation:
and so:
Thus $H_n$ is a hexagonal number by definition 2.
=== Definition 2 implies Definition 1 ===
Let $H_n$ be a hexagonal number by definition 2.
Then:
Then:
:$\ds H_0 = \sum_{i \mathop = 1}^0 \paren {4 \paren {i - 1} + 1}$
is a vacuous summation and so:
:$H_0 = 0$
Thus $H_n$ is a hexagonal number by definition 1.
=== Definition 1 equivalent to Definition 3 ===
We have by definition that $H_0 = 0 = \map P {6, 0}$.
Then:
Thus $\map P {6, n}$ and $H_n$ are generated by the same recurrence relation.
Category:Hexagonal Numbers
Equivalence of Definitions of Hilbert Space

== Theorem ==
Let $H$ be a vector space over $\mathbb F \in \set {\R, \C}$.
=== Definition 1===
=== Definition 2===
== Proof ==
=== Definition 1 implies Definition 2 ===
By definition of complete metric space, every Cauchy sequence in $H$ is convergent.
By definition of Banach space, it follows that $\struct { H, \norm {\,\cdot\,}_H }$ is a Banach space.
=== Definition 2 implies Definition 1 ===
By definition of inner product space, it follows that $\struct { H, \innerprod \cdot \cdot_H }$ is an inner product space.
Denote the inner product norm as $\norm {\,\cdot\,}_*$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {H, \norm {\,\cdot\,}_H}$.
By definition of Banach space, it follows that $\sequence {x_n}_{n \mathop \in \N}$ is a convergent sequence in $\struct {H, \norm {\,\cdot\,}_H}$.
By Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide, it follows that $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {H, \norm {\,\cdot\,}_*}$.
By Convergent Sequences in Vector Spaces with Equivalent Norms Coincide, it follows that $\sequence {x_n}_{n \mathop \in \N}$ is a convergent sequence in $\struct {H, \norm {\,\cdot\,}_*}$.
Let $d: H \times H \to \R_{\ge 0}$ be the metric induced by the inner product norm $\norm {\,\cdot\,}_*$.
By definition of complete metric space, it follows that $\struct {H, d}$ is a complete metric space. 
Category:Hilbert Spaces
Equivalence of Definitions of Homeomorphism between Metric Spaces

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a bijection.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
In order to prove the assertion it is sufficient to prove that the conditions for homeomorphism in definitions $2$ to $4$ are necessary and sufficient conditions for $f$ and $f^{-1}$ to be continuous on $M_1$ and $M_2$ respectively.
=== Definition 1 implies Definition 2 ===
Let $f$ and $f^{-1}$ both be continuous by definition $2$.
Let $U \subseteq A_1$ be open in $M_1$.
As $f^{-1}$ is continuous, $\paren {f^{-1} }^{-1} \sqbrk U = f \sqbrk U$ is open in $M_2$.
That is, $f \sqbrk U = f \sqbrk U$ is open in $M_2$.
Let $f \sqbrk U \subseteq A_2$ be open in $M_2$.
Then as $f$ is continuous, $f^{-1} \sqbrk {f \sqbrk U} = U$ is open in $M_1$.
Thus:
:for all $U \subseteq A_1$, $U$ is an open set of $M_1$ $f \sqbrk U$ is an open set of $M_2$.
=== Definition 2 implies Definition 4 ===
Let definition $2$ hold.
Let $a \in A_1$.
Let $N \subseteq A_1$.
Then:
:$N$ is a neighborhood of $a$
:
:$N$ contains an open set $U$ containing $a$
:
:$f \sqbrk N$ contains an open set $U' = f \sqbrk U$ containing $\map f a$
:
:$f \sqbrk N$ is a neighborhood $\map f a$.
=== Definition 4 implies Definition 1 ===
Let definition $4$ hold.
Let $a \in A_1$.
Let $U \subseteq A_2$ be a neighborhood of $\map f a$.
Then $f \sqbrk {f^{-1} \sqbrk U}$ is a neighborhood $\map f a$.
Hence $f^{-1} \sqbrk U$ is a neighborhood of $a$.
Thus $f$ is continuous.
Similarly, let $b \in A_2$.
Let $V \subseteq A_1$ be a neighborhood of $\map {f^{-1} } b$.
Then $f^{-1} \sqbrk {f \sqbrk V}$ is a neighborhood $\map {f^{-1} } b$.
Hence $f \sqbrk V$ is a neighborhood of $\map f {\map {f^{-1} } b} = b$.
Thus $f^{-1}$ is continuous.
=== Definition 2 iff Definition 3 ===
This is demonstrated in Continuity of Mapping between Metric Spaces by Closed Sets.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Theorem $7.10$
Category:Homeomorphism (Metric Spaces)
Equivalence of Definitions of Homeomorphisms between Topological Spaces

== Theorem ==
Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.
Let $f: T_\alpha \to T_\beta$ be a bijection.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 iff Definition 2 ===
Let $f$ and $f^{-1}$ both be continuous.
As $f$ is continuous, then by definition:
:$V \in \tau_\beta \implies f^{-1} \sqbrk V \in \tau_\alpha$
and as $f^{-1}$ is continuous, then by definition:
:$U \in \tau_\alpha \implies \paren {f^{-1} }^{-1} \sqbrk U = f \sqbrk U \in \tau_\beta$
That is:
:$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$
Conversely, suppose that:
:$\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$
By definition:
:$U \in \tau_\alpha \implies f \sqbrk U \in \tau_\beta$ $f^{-1}$ is continuous.
Also by definition:
:$f \sqbrk U \in \tau_\beta \implies U = f^{-1} \sqbrk {f \sqbrk U} \in \tau_\alpha$ $f$ is continuous.
That is, both $f$ and $f^{-1}$ are continuous.
=== Definition 1 iff Definition 3 ===
Let $f$ and $f^{-1}$ both be continuous.
Then by Bijection is Open iff Inverse is Continuous, $f$ is open.
Thus $f$ is both open and continuous.
Conversely, let $f$ be both open and continuous.
Then also by Bijection is Open iff Inverse is Continuous, $f^{-1}$ is continuous.
Thus both $f$ and $f^{-1}$ are continuous.
=== Definition 3 iff Definition 4 ===
Let $f$ be both open and continuous.
Then by Bijection is Open iff Closed it follows that $f$ is both closed and continuous.
Conversely, let $f$ be both both closed and continuous.
Then also by Bijection is Open iff Closed it follows that $f$ is both open and continuous.
Category:Homeomorphisms (Topological Spaces)
Equivalence of Definitions of Hyperbola

== Theorem ==
=== Equidistance Property ===
=== Focus-Directrix Property ===
== Proof ==
Let $K$ be an hyperbola aligned in a cartesian plane in reduced form.
Thus its foci are at $\tuple {\mathop \pm c, 0}$.
Let:
:the transverse axis of $K$ have length $2 a$
:the conjugate axis of $K$ have length $2 b$.
From Equation of Hyperbola in Reduced Form, the equation of $K$ is:
:$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$
This has been derived from the equidistance property of $K$.
From Focus of Hyperbola from Transverse and Conjugate Axis:
:$c^2 - a^2 = b^2$
and so $K$ can be expressed as:
:$(1): \quad \dfrac {x^2} {a^2} - \dfrac {y^2} {c^2 - a^2} = 1$
It remains to be shown that $K$ possesses the focus-directrix property.
:
Let $D$ be the straight line whose equation is $x = \dfrac {a^2} c$.
It will be shown that the locus of points $P = \tuple {x, y}$ such that:
:the distance from $P$ to $F_1$ is $\dfrac c a$ of the distance from $P$ to $D$
is precisely equation $(1)$.
We have that:
:$P F_2 = \dfrac c a d$
where:
:$d = x - \dfrac {a^2} c$
Thus:
Category:Hyperbolas