nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Separable Degree

== Theorem ==
Let $E / F$ be a field extension.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition $1$ is equivalent to Definition $2$ ===
Let $S \subseteq E$ be the separable closure of $F$ in $E$.
Let $\bar F$ be the algebraic closure of $F$.
We need to show that the number of embeddings $E \hookrightarrow \bar F$ fixing $F$ is $\index S F$.
==== Lemma ====
An embedding $\sigma: S \hookrightarrow \bar F$ fixing $F$ uniquely extends to an embedding $\tilde \sigma: E \hookrightarrow \bar F$ fixing $F$.
==== Proof ====
If characteristic of $F$ is $0$, we have $E = S$, nothing to prove.
If characteristic of $F$ is $p > 0$,
since $E / S$ is purely inseparable,
each element of $\alpha \in E \setminus S$ has $\alpha^{p^n} = a$ for some $a \in S$ and $n \in \N_{>0}$.
Then for all embedding $\tilde{\sigma}: E \hookrightarrow \bar F$ that extends $\sigma$,
:$\map {\tilde{\sigma}} \alpha^{p^n} = \map {\tilde{\sigma}} {\alpha^{p^n}} = \map \sigma a$.
But $x^{p^n} = \map \sigma a$ has only one root in $\bar{F}$,
so $\map {\tilde {\sigma}} \alpha$ is uniquely determined by $\map \sigma a$.
Therefore $\sigma$ uniquely extends to $\tilde{\sigma}$.
By the lemma, it suffices to show that the number of embeddings $\sigma: S \hookrightarrow \bar F$ fixing $F$ is $\index S F$.
Since $S / F$ is separable, by Primitive Element Theorem, $S / F$ is generated by a single element $\alpha \in S$
:$S = F \paren \alpha$
Let $P$ be the minimal polynomial of $\alpha$ over $F$. 
Since $P$ is separable, $P$ has $n := \index S F$ distinct roots in $\bar F$.
Let $\alpha_1, \ldots, \alpha_n$ be these roots.
By Simple Algebraic Field Extension consists of Polynomials in Algebraic Number, elements of $F$ are of the form
:$\ds \map f \alpha$ for some polynomial $f \in F \sqbrk x$.
An embedding $\sigma: E \hookrightarrow \bar F$ is uniquely determined by the image of $\alpha$, since 
:$\forall f \in F \sqbrk x: \ds \map \sigma {\map f \alpha} = \map f {\map \sigma \alpha}$
Then we get $n$ distinct embeddings $\sigma_i: S \hookrightarrow \bar F$, $i = 1, \dots, n$ given by
:$\map \sigma \alpha = \alpha_i$
And there can't be other embeddings, since the image of $\alpha$ has to be one of the roots $\alpha_i$ of $P$.
=== Definition $1$ is equivalent to Definition $3$ ===
== Sources ==
* 
Category:Field Extensions
Equivalence of Definitions of Separated Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B \subseteq S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
=== Definition 2 implies Definition 1 ===
Category:Separated Sets
Equivalence of Definitions of Separated Sets/Definition 1 implies Definition 2

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $A, B \subseteq S$ satisfy:
:$A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$ in $T$, and $\O$ denotes the empty set.
Then there exist $U,V \in \tau$ with:
:$A \subset U$ and $U \cap B = \O$
:$B \subset V$ and $V \cap A = \O$
== Proof ==
From Topological Closure is Closed, $B^-$ is closed in $T$.
Let $U = S \setminus B^-$ be the relative complement of $B^-$.
By the definition of a closed set, $U$ is open in $T$.
From Empty Intersection iff Subset of Relative Complement:
:$A \subseteq S \setminus B^- = U$
From Relative Complement of Relative Complement:
:$S \setminus U = B^-$
By the definition of the closure of a subset:
:$B \subseteq B^- = S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
:$U \cap B = \O$
Similarly, let $V = S \setminus A^-$ then $V \in \tau$ with:
:$B \subset V$
and
:$V \cap A = \O$
Category:Separated Sets
Equivalence of Definitions of Separated Sets/Definition 2 implies Definition 1

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $A, B \subseteq S$.
Let $U, V \in \tau$ satisfy:
:$A \subset U$ and $U \cap B = \O$
:$B \subset V$ and $V \cap A = \O$
where $\O$ denotes the empty set.
Then
:$A^- \cap B = A \cap B^- = \O$
where $A^-$ denotes the closure of $A$ in $T$.
== Proof ==
Let $U \in \tau$ be an arbitrary open set of $T$ such that $A \subseteq U$ and $U \cap B = \O$.
From Empty Intersection iff Subset of Relative Complement:
:$B \subseteq S \setminus U$
By the definition of a closed set, the relative complement $S \setminus U$ of $U$ is closed in $T$.
From Set Closure is Smallest Closed Set in Topological Space:
:$B^- \subseteq S \setminus U$
From Empty Intersection iff Subset of Relative Complement:
:$B^- \cap U = \O$
Then
Similarly:
:$A^- \cap B = \O$
Category:Separated Sets
Equivalence of Definitions of Set Equality

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $S = T$ by Definition 1.
Then:
Similarly:
Thus by the Rule of Conjunction:
:$S \subseteq T \land T \subseteq S$
and so $S$ and $T$ are equal by Definition 2.
=== Definition 2 implies Definition 1 ===
Let $S = T$ by Definition 2:
:$S \subseteq T \land T \subseteq S$
First:
Then:
Thus by Biconditional Introduction:
:$\forall x: \paren {x \in S \iff x \in T}$
and so $S$ and $T$ are equal by Definition 1.
== Sources ==
* : Chapter $0$: Subsets and Complements; Union and Intersection
* : $\S 1$: The Axiom of Extension
* : A Note on Set-Theoretic Concepts
* : $\S 1$: Sets and Set Inclusion
* : $\S 1.1$: Theorem $1.1$
* : $\S 1.2$. Subsets
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : $\S 0.2$. Sets
* : $1$ Set Theory: $1$. Sets and Functions: $1.1$: Basic definitions
* : Chapter $1$: The Notation and Terminology of Set Theory: $\S 3$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Theorem $1.1 \ \text{(c)}$
* : Appendix $\text{A}.1$: Sets and Subsets
* : $\S 6.3$: Subsets
* : $\S 1$: Naive Set Theory: $\S 1.1$: What is a Set?: Exercise $1.1.2$
Category:Set Theory
Category:Subsets
Category:Set Equality
Equivalence of Definitions of Set Partition

== Theorem ==
Let $S$ be a set
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\Bbb S$ be a set of subsets $\Bbb S$ of $S$ such that:
:$(1): \quad$ $\Bbb S$ is pairwise disjoint: $\forall S_1, S_2 \in \Bbb S: S_1 \cap S_2 = \O$ when $S_1 \ne S_2$
:$(2): \quad$ The union of $\Bbb S$ forms the whole set $S$: $\ds \bigcup \Bbb S = S$
:$(3): \quad$ None of the elements of $\Bbb S$ is empty: $\forall T \in \Bbb S: T \ne \O$.
By $(3)$ all sets in $\Bbb S$ are non-empty.
 there exists $x \in S$ in more than one set of $\Bbb S$.
That is:
:$\exists S_1, S_2 \in \Bbb S, S_1 \ne S_2: x \in S_1 \land x \in S_2$
Then by definition of set intersection:
:$x \in S_1 \cap S_2$
and so $S_1 \cap S_2 \ne \O$.
This contradicts condition $(1)$.
So no element of $S$ can be in more than one set of $\Bbb S$.
 there exists $x \in S$ in none of the sets of $\Bbb S$.
Then $\ds x \notin \bigcup \Bbb S$ and so $\Bbb S \notin S$.
This contradicts condition $(2)$.
So every element of $S$ is in at least one set of $\Bbb S$.
Thus $x$ lies in exactly one element of $\Bbb S$.
Hence it follows that $\Bbb S$ is a set of non-empty subsets of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
=== Definition 2 implies Definition 1 ===
Let $\Bbb S$ be a set of non-empty subsets of $S$ such that each element of $S$ lies in exactly one element of $\Bbb S$.
$(3)$ is fulfilled automatically by definition.
Consider $\Bbb S$.
From Union of Subsets is Subset: Subset of Power Set:
:$\ds \bigcup \Bbb S \subseteq S$
Let $x \in T$.
We have :
:$\exists S_1 \in \Bbb S: x \in S_1$
That is:
:$\ds x \in \bigcup \Bbb S$
and so:
:$\ds S \subseteq \bigcup \Bbb S$
By definition of set equality:
:$\ds \bigcup \Bbb S = S$
thus demonstrating that $(2)$ holds.
 $\Bbb S$ were not pairwise disjoint, and that:
:$\exists S_1, S_2 \in \Bbb S: S_1 \cap S_2 \ne \O$
Then:
:$\exists x \in S: x \in S_1 \cap S_2$
By definition of set intersection:
:$x \in S_1$ and $x \in S_2$
thus contradicting the supposition that each element of $S$ lies in exactly one element of $\Bbb S$.
So $\Bbb S$ is pairwise disjoint and so $(1)$ holds.
Hence the result.
Category:Set Partitions
Equivalence of Definitions of Sets Separated by Neighborhoods

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $A, B \subseteq S$ such that:
:$\exists N_A, N_B \subseteq S: \exists U, V \in \tau: A \subseteq U \subseteq N_A, B \subseteq V \subseteq N_B: N_A \cap N_B = \O$
From Subsets of Disjoint Sets are Disjoint then:
:$U \cap V = \O$
Thus:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V: U \cap V = \O$
=== Definition 2 implies Definition 1 ===
Let $A, B \subseteq S$ such that:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V: U \cap V = \O$
Let $N_A = U$ and $N_B = V$.
From Set is Subset of Itself then:
:$\exists N_A, N_B \subseteq S: \exists U, V \in \tau: A \subseteq U \subseteq N_A, B \subseteq V \subseteq N_B: N_A \cap N_B = \O$
Category:Separation Axioms
Equivalence of Definitions of Sigma-Finite Measure

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 equivalent to Definition 2 ===
This is Measure Space has Exhausting Sequence of Finite Measure iff Cover by Sets of Finite Measure.
=== Definition 3 implies Definition 2 ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a partition of $X$ in $\Sigma$ with:
:$\map \mu {E_n} < \infty$ for each $n \in \N$.
Then: 
:$E_n \cap E_m = \O$ for $n \ne m$
and:
:$\ds X = \bigcup_{n \mathop = 1}^\infty E_n$
with:
:$\map \mu {E_n} < \infty$ for each $n \in \N$.
In particular, $\sequence {E_n}_{n \mathop \in \N}$ is a cover of $X$ in $\Sigma$ with:
:$\map \mu {E_n} < \infty$ for each $n \in \N$.
=== Definition 2 implies Definition 3 ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a cover of $X$ in $\Sigma$ with: 
:$\map \mu {E_n} < \infty$ for each $n \in \N$.
By Countable Union of Measurable Sets as Disjoint Union of Measurable Sets there exists a sequence of measurable sets $\sequence {F_n}_{n \mathop \in \N}$ such that: 
:$F_n \subseteq E_n$ for each $n \in \N$
and:
:$\ds X = \bigcup_{n \mathop = 1}^\infty E_n = \bigcup_{n \mathop = 1}^\infty F_n$
From Measure is Monotone we have: 
:$\map \mu {F_n} < \infty$ for each $n \in \N$.
So $\sequence {F_n}_{n \mathop \in \N}$ is a partition of $X$ in $\Sigma$ with:
:$\map \mu {F_n} < \infty$ for each $n \in \N$.
Category:Sigma-Finite Measures
Equivalence of Definitions of Sigma-Ring

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\text {SR}$ be a ring of sets which is closed under countable unions.
We have:
which are exactly $\text {SR} 1$ and $\text {SR} 2$.
Then as $\text {SR}$ is closed under countable unions:
:$\ds A_1, A_2, \ldots \in \text {SR} \implies \bigcup_{n \mathop = 1}^\infty A_n \in \text {SR}$
and so $\text {SR} 3$ is fulfilled.
=== Definition 2 implies Definition 1 ===
Let $\text {SR}$ be a system of sets such that:
As noted above, $\text {SR} 1$ and $\text {SR} 2$ are exactly $\text {RS} 1_2$ and $\text {RS} 2_2$.
Let $A, B \in \text {SR}$.
Let $A_1 = A, A_2 = B$ and $A_n = \O$ for all $n = 3, 4, \ldots$
Then:
:$\ds \forall A_n \in \text {SR}: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n = A \cup B \in \text {SR}$
Thus criterion $(\text {RS} 3_2)$ is fulfilled.
So $\text {SR}$ is a ring of sets which is closed under countable unions.
=== Definition 2 implies Definition 3 ===
$\text {SR} 1$ is exactly $\text {SR} 1'$ and $\text {SR} 2$ is exactly $\text {SR} 2'$.
$\text {SR} 3$ allows arbitrary countable unions, and therefore a fortiori implies $\text {SR} 3'$.
=== Definition 3 implies Definition 2 ===
As noted above, $\text {SR} 1$ and $\text {SR} 2$ are exactly $\text {SR} 1'$ and $\text {SR} 2'$.
Let $A_1, A_2, \ldots \in \Sigma$.
We have to show that $\ds \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$.
Let $B_n = A_n \setminus \ds\bigcup_{i \mathop = 1}^{n-1}A_i$ for all $n\in\N$.
Because:
:$B_n = \paren {\cdots \paren {\paren {A_n \setminus A_{n - 1} } \setminus A_{n - 2} } \setminus \cdots \setminus A_1}$
and $\text {SR} 2'$, we have:
:$\forall n \in \N: B_n \in \Sigma$
As $B_n \subseteq A_n$ for all $n \in \N$, it follows that:
:$\ds \bigcup_{n \mathop = 1}^\infty B_n \subseteq \bigcup_{n \mathop = 1}^\infty A_n$
Let:
:$x \in \ds\bigcup_{n \mathop = 1}^\infty A_n$
and:
:$k = \min \set {n \in \N: x \in A_n}$
Then:
:$x \in B_k$
Therefore:
:$x \in \ds \bigcup_{n \mathop = 1}^\infty B_n$
Hence:
:$\ds \bigcup_{n \mathop = 1}^\infty A_n \subseteq \bigcup_{n \mathop = 1}^\infty B_n$
Together with
:$\ds \bigcup_{n \mathop = 1}^\infty B_n \subseteq \bigcup_{n \mathop = 1}^\infty A_n$
it follows that:
:$\ds \bigcup_{n \mathop = 1}^\infty B_n = \bigcup_{n \mathop = 1}^\infty A_n$
Let $i, j \in \N, i < j$.
Then:
:$B_i \cap B_j \subseteq A_i \cap B_j = A_i \cap \paren {A_j \setminus \ds \bigcup_{k \mathop = 1}^{j - 1} A_k} = \O$
So $B_1, B_2, \ldots$ is pairwise disjoint.
Hence $\ds\bigcup_{n \mathop = 1}^\infty A_n = \ds \bigsqcup_{n \mathop = 1}^\infty B_n \in \Sigma$, as $B_n \in \Sigma$ for all $n \in \N$ and $\text {SR} 3'$.
== Sources ==
* : $\S 1.1$: Rings and Algebras: Theorem $1.1.1$
Category:Sigma-Rings
Equivalence of Definitions of Signed Stirling Numbers of the First Kind

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
in the sense that the coefficients of the powers in the summand are uniquely defined by the given recurrence relation.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:the coefficients of the powers in the expression $\ds x^{\underline n} = \sum_k \map s {n, k} x^k$ are uniquely defined by $\map s {n, k} := \map s {n - 1, k - 1} - \paren {n - 1} \map s {n - 1, k}$
where $\map s {n, k} = \delta_{n k}$ where $k = 0$ or $n = 0$.
First the case where $n = 0$ is attended to.
From Signed Stirling Number of the First Kind of 0 we have:
:$\map s {0, k} = \delta_{0 k}$
Hence the result holds for $n = 0$.
=== Basis for the Induction ===
$\map P 1$ is the case:
We have:
Then:
Thus, in the expression:
:$\ds x^{\underline k} = \sum_k \map s {1, k} x^1$
we have:
:$\map s {1, k} = 1$
and for all $k \in \Z$ where $k \ne 1$:
:$\map s {1, k} = 0$
That is:
:$\map s {1, k} = \delta_{1 k}$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:The coefficients in the expression $\ds x^{\underline r} = \sum_k \map s {r, k} x^k$ are uniquely defined by $\map s {r, k} = \map s {r - 1, k - 1} + \paren {r - 1} \map s {r - 1, k}$
from which it is to be shown that:
:The coefficients in the expression $\ds x^{\underline {r + 1} } = \sum_k \map s {r + 1, k} x^k$ are uniquely defined by $\map s {r + 1, k} = \map s {r, k - 1} + r \map s {r, k}$
=== Induction Step ===
This is the induction step:
Thus the coefficients of the falling factorial powers are defined by the recurrence relation:
:$\map s {r + 1, k} = \map s {r, k - 1} + r \map s {r,k}$
as required.
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:for all $n \in \Z_{\ge 0}$, the coefficients of the powers in the expression $\ds x^{\underline n} = \sum_k \map s {n, k} x^k$ are uniquely defined by:
::$\map s {n, k} = \map s {n - 1, k - 1} + \paren {n - 1} \map s {n - 1, k}$
:where $\map s {n, k} = \delta_{n k}$ when $k = 0$ or $n = 0$.
== Also see ==
* Equivalence of Definitions of Unsigned Stirling Numbers of the First Kind
* Equivalence of Definitions of Stirling Numbers of the Second Kind
Category:Stirling Numbers
Equivalence of Definitions of Simplex

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $S_1$ denote the set obtained by definition 1.
Let $S_2$ denote the set obtained by definition 2.
Let $x \in S_2$ for some $x \in \R^n$.
Then, by definition 2, a sequence of real numbers $\sequence {\theta_i}_{i \mathop = 0}^n$ exists such that:
:$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \closedint 0 1$
:$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
:$x = \ds \sum_{i \mathop = 0}^n \theta_i \alpha_i$
We have that $\sequence {\theta_i}_{i \mathop = 0}^n$ also satisfies the three conditions:
:$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
:$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$
:$x = \ds \sum_{i \mathop = 0}^n \theta_i \alpha_i$
Hence $x \in S_1$ by definition 1.
Then by definition of subset:
:$S_2 \subseteq S_1$
Let $x \in S_1$ for some $x \in \R^n$
Then, by definition 1, a sequence of real numbers $\sequence {\theta_i}_{i \mathop = 0}^n$ exists such that:
:$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$
:$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
:$x = \ds \sum_{i \mathop = 0}^n \theta_i \alpha_i$
:
:$\exists j \in \set {0, 1, 2, \ldots, n}: \theta_j > 1$
Then:
:$\ds \sum_{i \mathop = 0}^n \theta_i \ge \theta_j > 1$
which is a contradiction.
So we see that:
:$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \le 1$
Because:
:$\ds \sum_{i \mathop = 0}^n \theta_i = 1$
and:
:$\forall i \in \set {0, 1, 2, \ldots, n}: \theta_i \in \hointr 0 \to$
it follows by definition 2 that:
:$x \in S_2$
Then by definition of subset:
:$S_1 \subseteq S_2$
Thus we have:
:$S_1 \subseteq S_2$
and:
:$S_2 \subseteq S_1$
and by definition of set equality:
:$S_1 = S_2$
Category:Simplices
Equivalence of Definitions of Sine and Cosine

== Theorem ==
In the following, $\theta$ understood to take values in $\hointr 0 {2 \pi}$.
=== ===
==== Definition from Circle ====
==== Analytic Definition ====
=== ===
==== Definition from Circle ====
==== Analytic Definition ====
== Proof ==
Consider the following vector-valued function $\mathbf f : \R \to {\closedint {-1} 1}^2$:
:$\map {\mathbf f} t = \tuple {\cos t, \sin t}$
where $\cos t$ and $\sin t$ are defined analytically.
Then, for any $t$ the distance to the origin is:
Therefore, $\map {\mathbf f} t$ always lies on the unit circle.
For arbitrary $\theta \in \hointr 0 {2 \pi}$, the arc length on $\closedint 0 \theta$ is:
Thus, by the definition of radians, the angle made by $\map {\mathbf f} \theta$ with the $x$-axis is $\theta$.
A straight line segment from $\tuple {0, \sin \theta}$ to $\tuple {\cos \theta, \sin \theta}$ is perpendicular to the $y$-axis.
Its length is $\size {\cos \theta}$, and it is on the side of the $y$-axis corresponding to the sign of $\cos \theta$.
But that is the unit circle definition for cosine.
Similarly, a straight line segment from $\tuple {\cos \theta, 0}$ to $\tuple {\cos \theta, \sin \theta}$ is perpendicular to the $x$-axis, with length $\size {\sin \theta}$ and appropriate direction.
Again, that is the unit circle definition for sine.
Category:Sine Function
Category:Cosine Function
Category:Definition Equivalences
Equivalence of Definitions of Sine of Angle

== Theorem ==
Let $\theta$ be an angle.
=== Definition from Triangle ===
=== Definition from Circle ===
=== Definition from Real Numbers ===
== Proof ==
=== Definition from Triangle implies Definition from Circle ===
Let $\sin \theta$ be defined as $\dfrac {\text{Opposite}} {\text{Hypotenuse}}$ in a right triangle.
Consider the triangle $\triangle OAP$.
By construction, $\angle OAP$ is a right angle.
Thus:
That is:
: $\sin \theta = AP$
=== Definition from Circle implies Definition from Triangle ===
Let $\sin \theta$ be defined as the length of $AP$ in the triangle $\triangle OAP$.
Compare $\triangle OAP$ with $\triangle ABC$ in the diagram above.
We have that:
:$\angle CAB = \angle POA = \theta$
:$\angle ABC = \angle OAP$ which is a right angle
Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAP$ and $\triangle ABC$ are similar.
By definition of similarity:
That is:
:$\dfrac {\text{Opposite} } {\text{Hypotenuse} } = \sin \theta$
=== Definition from Circle equivalent to Definition from Real Numbers ===
This is demonstrated in Arcsine as Integral.
Category:Sine Function
Equivalence of Definitions of Singular Matrix

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\mathbf A$ be an element of the ring of square matrices $\struct {\map {\MM_R} n, +, \times}$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Follows directly from Matrix is Nonsingular iff Determinant has Multiplicative Inverse.
Category:Singular Matrices
Equivalence of Definitions of Small Circle

== Theorem ==
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
Let $S$ be a sphere.
=== Definition $(1)$ implies Definition $(2)$ ===
Let $C$ be a small circle on $S$ by definition $1$.
Thus $C$ is a small circle on $S$ by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $C$ be a small circle on $S$ by definition $2$.
Thus $C$ is a small circle on $S$ by definition $1$.
Category:Small Circles
Equivalence of Definitions of Sober Space

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let each closed irreducible subspace of $T$ have a unique generic point.
Let $U \ne S$ be a meet-irreducible open set.
Let $F = S \setminus U$.
From Meet-Irreducible Open Set iff Complement is Closed Irreducible Subspace:
:$F$ is closed irreducible subspace
We have :
:$\exists ! x \in S : F = \set x^-$
From Relative Complement inverts Subsets of Relative Complement:
:$\exists ! x \in S : U = S \setminus \set x^-$
The result follows.
=== Definition 2 implies Definition 1 ===
Let for every meet-irreducible open set $V \ne S$ there exists a unique $y \in S$ such that:
:$V = S \setminus \set y^-$
Let $F$ closed irreducible subspace of $T$.
Let $U = S \setminus F$.
From Meet-Irreducible Open Set iff Complement is Closed Irreducible Subspace:
:$U$ is a meet-irreducible open set
We have :
:$\exists ! x \in S : U = S \setminus \set x^-$
From Relative Complement inverts Subsets of Relative Complement:
:$\exists ! x \in S : F = \set x^-$
The result follows.
Category:Sober Spaces
Equivalence of Definitions of Square Function

== Theorem ==
Let $\F$ denote one of the standard classes of numbers: $\N$, $\Z$, $\Q$, $\R$, $\C$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By definition of $n$th power (for positive $n$):
:$x^n = \begin {cases} 1 & : n = 0 \\ x \times x^{n - 1} & : n > 0 \end {cases}$
Thus:
Hence the result.
Category:Square Function
Equivalence of Definitions of Square Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 equivalent to Definition 3 ===
By the Odd Number Theorem:
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
=== Definition 2 equivalent to Definition 3 ===
By the Corollary to the Odd Number Theorem:
:$S_n = \ds \sum_{j \mathop = 1}^{n - 1} + 2 n - 1$
and so by Definition 2:
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1} = S_{n-1} + 2 n - 1$
=== Definition 2 equivalent to Definition 4 ===
We have by definition that $S_n = 0 = \map P {4, n}$.
Then:
Thus $\map P {4, n}$ and $S_n$ are generated by the same recurrence relation.
Category:Square Numbers