nyuuzyou/wikis huggingface dataset:
Condition for Straight Lines in Plane to be Parallel/General Equation

== Theorem ==
Let $L: \alpha_1 x + \alpha_2 y = \beta$ be a straight line in $\R^2$.
Then the straight line $L'$ is parallel to $L$ there is a $\beta' \in \R^2$ such that:
:$L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = \beta'}$
== Proof ==
=== Necessary Condition ===
When $L' = L$, the claim is trivial.
Let $L' \ne L$ be described by the equation:
:$\alpha'_1 x + \alpha'_2 y = \beta'$
, let $\alpha'_1 \ne 0$ (the case $\alpha'_2 \ne 0$ is similar).
Then for $\tuple {x, y} \in L'$ to hold, one needs:
For $L'$ to be parallel to $L$, it is required that then $\tuple {x, y} \notin L$, that is:
It follows that necessarily $\beta - \alpha_1 \frac {\beta'} {\alpha'_1} \ne 0$, or taking $y = 0$ would yield equality.
The only remaining way to obtain the desired inequality for all $y$ is that:
:$\alpha_2 - \alpha_1 \dfrac {\alpha'_2} {\alpha'_1} = 0$
One observes that now $\alpha_1 = 0 \implies \alpha_2 = 0$.
However, as $L: \alpha_1 x + \alpha_2 y = \beta$ is a straight line in $\R^2$, it cannot be that $\alpha_1 = \alpha_2 = 0$.
So $\alpha_1 \ne 0$, and one finds:
:$\alpha'_2 = \dfrac {\alpha'_1} {\alpha_1} \alpha_2$
Hence obtain:
That is, $L'$ is described by an equation of the required form.
=== Sufficient Condition ===
Let $L' \ne L$ be a straight line given by the equation:
:$\alpha_1 x + \alpha_2 y = \beta'$
 we have a point $\mathbf x = \tuple {x, y} \in L \cap L'$.
Then, as $\mathbf x \in L$, it also satisfies:
:$\alpha_1 x + \alpha_2 y = \beta$
It follows that $\beta = \beta'$, so $L = L'$.
This contradiction shows that $L \cap L' = \O$, that is, $L$ and $L'$ are parallel.
The remaining case is when $L' = L$. 
By definition, $L$ is parallel to itself. 
The result follows.
== Also see ==
* Condition for Planes to be Parallel
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
* : Chapter $\text V$: Vector Spaces: $\S 28$: Linear Transformations
Category:Condition for Straight Lines in Plane to be Parallel
Condition for Straight Lines in Plane to be Parallel/Slope Form

== Theorem ==
Let $L_1$ and $L_2$ be straight lines in the Cartesian plane.
Let the slope of $L_1$ and $L_2$ be $\mu_1$ and $\mu_2$ respectively.
Then $L_1$ is parallel to $L_2$ :
:$\mu_1 = \mu_2$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 10$: Formulas from Plane Analytic Geometry: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$: $10.9$
* : $\S 8$: Formulas from Plane Analytic Geometry: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$: $8.9.$
Category:Condition for Straight Lines in Plane to be Parallel
Condition for Straight Lines in Plane to be Parallel/Slope Form/Proof 1

== Theorem ==
== Proof ==
Let $L_1$ and $L_2$ be described by the general equation: 
Then:
:the slope of $L_1$ is $\mu_1 = -\dfrac {l_1} {m_1}$
:the slope of $L_2$ is $\mu_2 = -\dfrac {l_2} {m_2}$.
From Condition for Straight Lines in Plane to be Parallel: General Equation:
:$L_1$ and $L_2$ are parallel $L_2$ is given by the equation:
::$m_1 x + m_2 y = n'$
:for some $n'$.
But then the slope of $L_2$ is $-\dfrac {l_1} {m_1}$.
That is:
:$-\dfrac {l_1} {m_1} = -\dfrac {l_2} {m_2}$
and the result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
Category:Condition for Straight Lines in Plane to be Parallel
Condition for Straight Lines in Plane to be Parallel/Slope Form/Proof 2

== Theorem ==
== Proof ==
Let $L_1$ and $L_2$ be embedded in a cartesian plane, given by the equations:
Let $\phi_1$ and $\phi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively.
Then by definition of slope of a straight line:
=== Necessary Condition ===
Let $m_1 = m_2$.
Then:
:$\tan \psi_1 = \tan \psi_2$
and so:
:$\psi_1 = \psi_2 + n \pi$
The multiple of $\pi$ makes no difference.
Thus from Equal Corresponding Angles implies Parallel Lines, $L_1$ and $L_2$ are parallel.
=== Sufficient Condition ===
Suppose $L_1 \parallel L_2$.
Then:
Category:Condition for Straight Lines in Plane to be Parallel
Condition for Straight Lines in Plane to be Parallel/Slope Form/Proof 3

== Theorem ==
== Proof ==
Let $\psi$ be the angle between $L_1$ and $L_2$
When $L_1$ and $L_2$ are parallel:
:$\psi = 0$
by definition.
From Angle between Straight Lines in Plane:
:$\psi = \arctan \dfrac {m_1 - m_2} {1 + m_1 m_2}$
The result follows immediately.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The angle between two lines
Category:Condition for Straight Lines in Plane to be Parallel
Condition for Straight Lines in Plane to be Perpendicular/Examples/Arbitrary Example 1

== Examples of Use of Condition for Straight Lines in Plane to be Perpendicular ==
Let $\LL_1$ be the straight line whose equation in general form is given as:
:$3 x - 4 y = 7$
Let $\LL_2$ be the straight line perpendicular to $\LL_1$ which passes through the point $\tuple {1, 2}$.
The equation for $\LL_2$ is:
:$4 x + 3 y = 10$
== Proof ==
From Corollary to Condition for Straight Lines in Plane to be Perpendicular, $\LL_2$ has an equation of the form:
:$(1): \quad 4 x + 3 y = C$
We have that $\tuple {1, 2}$ is on $\LL_2$.
Hence substituting $x = 1$ and $y = 2$ into $(1)$:
Hence the result.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The angle between two lines: Example $\text{(ii)}$
Category:Condition for Straight Lines in Plane to be Perpendicular
Condition for Straight Lines in Plane to be Perpendicular/General Equation

== Theorem ==
Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given in general form:
Then $L_1$ is perpendicular to $L_2$ :
:$l_1 l_2 + m_1 m_2 = 0$
=== Corollary ===
== Proof ==
From the general equation for the straight line:
Hence the slope of $L_1$ and $L_2$ are $-\dfrac {l_1} {m_1}$ and $-\dfrac {l_2} {m_2}$ respectively.
From Condition for Straight Lines in Plane to be Perpendicular: Slope Form we have:
$-\dfrac {l_1} {m_1} = \dfrac {m_2} {l_2}$
from which the result follows.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
Category:Condition for Straight Lines in Plane to be Perpendicular
Condition for Straight Lines in Plane to be Perpendicular/General Equation/Corollary

== Theorem ==
Let $L$ be a straight line in the Cartesian plane.
Let $L$ be described by the general equation for the straight line:
:$l x + m y + n = 0$
Then the straight line $L'$ is perpendicular to $L$ $L'$ can be expressed in the form:
:$m x - l y = k$
== Proof ==
From the general equation for the straight line, $L$ can be expressed as:
:$y = -\dfrac l m x + \dfrac n m$
Hence the slope of $L$ is $-\dfrac l m$.
Let $L'$ be perpendicular to $L$.
From Condition for Straight Lines in Plane to be Perpendicular, the slope of $L'$ is $\dfrac m l$.
Hence $L'$ has the equation:
:$y = \dfrac m l x + r$
for some $r \in \R$.
Hence by multiplying by $l$ and rearranging:
:$m x - l y = -l r$
The result follows by replacing the arbitrary $-l r$ with the equally arbitrary $k$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
Category:Condition for Straight Lines in Plane to be Perpendicular
Condition for Straight Lines in Plane to be Perpendicular/Slope Form

== Theorem ==
Let $L_1$ and $L_2$ be straight lines in the Cartesian plane.
Let the slope of $L_1$ and $L_2$ be $\mu_1$ and $\mu_2$ respectively.
Then $L_1$ is perpendicular to $L_2$ :
:$\mu_1 = -\dfrac 1 {\mu_2}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
This can also be presented in the elegant form:
:$\mu_1 \mu_2 + 1 = 0$
or:
:$\mu_1 \mu_2 = -1$
== Sources ==
* : $\S 10$: Formulas from Plane Analytic Geometry: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$: $10.9$
* : $\S 8$: Formulas from Plane Analytic Geometry: Angle $\psi$ between Two Lines having Slopes $m_1$ and $m_2$: $8.9.$
* 
Category:Condition for Straight Lines in Plane to be Perpendicular
Condition for Straight Lines in Plane to be Perpendicular/Slope Form/Proof 1

== Theorem ==
== Proof ==
Let $\mu_1 = \tan \phi$.
Then $L_1$ is perpendicular to $L_2$ :
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $5$
Category:Condition for Straight Lines in Plane to be Perpendicular
Condition for Straight Lines in Plane to be Perpendicular/Slope Form/Proof 3

== Theorem ==
== Proof ==
Let $\psi$ be the angle between $L_1$ and $L_2$
From Angle between Straight Lines in Plane:
:$\psi = \arctan \dfrac {m_1 - m_2} {1 + m_1 m_2}$
When $L_1$ and $L_2$ are perpendicular:
:$\psi = \dfrac \pi 2$
by definition.
From Tangent of Right Angle $\tan \dfrac \pi 2$ is undefined.
This happens $1 + m_1 m_2 = 0$. 
The result follows.
== Sources ==
* : $\text {III}$. Analytical Geometry: The Straight Line: The angle between two lines
Category:Condition for Straight Lines in Plane to be Perpendicular
Condition for Subgroup of Monoid to be Normal

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.
Let $\struct {H, \circ}$ be a subgroup of $\struct {S, \circ}$.
Then:
::the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
:and:
::the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
:such that the equivalence relations induced by those partitions are congruence relations for $\circ$
:
:$\struct {H, \circ}$ is a normal subgroup of $\struct {S, \circ}$.
== Proof ==
=== Necessary Condition ===
Let $\struct {H, \circ}$ be a normal subgroup of $\struct {S, \circ}$.
Then by definition:
:$e \in H$
Hence from Condition for Cosets of Subgroup of Monoid to be Partition, the set of left cosets and the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$.
=== Sufficient Condition ===
Let $\struct {H, \circ}$ be a subgroup of $\struct {S, \circ}$ such that:
:the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
and:
:the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
such that the equivalence relations induced by those partitions are congruence relations for $\circ$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.16$
Category:Normal Subgroups
Category:Monoids
Condition for Subset of Group to be Right Transversal

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index in $G$ is $n$:
:$\index G H = n$
Let $S \subseteq G$ be a subset of $G$ of cardinality $n$.
Then $S$ is a right transversal for $H$ in $G$ :
:$\forall x, y \in S: x \ne y \implies x y^{-1} \notin H$
== Proof ==
From , $S$ is a right transversal for $H$ in $G$ every right coset of $H$ contains exactly one element of $S$.
Since there are $n$ right cosets of $H$ and $S$ has cardinality $n$, if $S$ is not a right transversal for $H$ in $G$, at least one right coset of $H$ contain more than one element of $S$.
Thus the contrapositive of the theorem is as follows:
:At least one right coset of $H$ contain more than one element of $S$ :
::$\exists x, y \in S: x \ne y \land x y^{-1} \in H$
This is a consequence of Elements in Same Right Coset iff Product with Inverse in Subgroup.
== Sources ==
* : Chapter $6$: Cosets: Exercise $5$
Category:Transversals (Group Theory)
Condition for Supremum of Subset to equal Supremum of Set

== Lemma ==
Let $S$ be a real set.
Let $T$ be a subset of $S$.
Let $S$ and $T$ admit suprema.
Then:
:$\sup S = \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
== Proof ==
=== Necessary Condition ===
Let $\sup S = \sup T$.
The aim is to establish that $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.
We have:
=== Sufficient Condition ===
Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.
The aim is to establish that $\sup S = \sup T$.
We have:
Category:Real Analysis
Condition for Trivial Relation to be Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR = S \times T$ be the trivial relation in $S$ to $T$.
Then $\RR$ is a mapping either:
:$(2): \card S = 0$
or:
:$(1): \card T = 1$
where $\card {\, \cdot \,}$ denotes cardinality.
== Proof ==
By definition, the trivial relation in $S$ to $T$ is the set:
:$\RR = \set {\tuple {s, t}: s \in S, t \in T}$
Proof by Cases:
:$(1)$: Let $\card S = 0$.
That is, $S = \O$.
By definition of the trivial relation, in this case $\RR = \O$.
This is the empty mapping.
From Empty Mapping is Mapping, this is a mapping.
:$(2)$: Let $\card S \ne 0$.
That is, $S \ne \O$.
::$(2) \, \text (a)$: Let $\card T = 0$.
That is, $T = \O$.
From Relation to Empty Set is Mapping iff Domain is Empty, $\RR$ cannot be a mapping for $S \ne \O$.
::$(2) \, \text (b)$: Let $\card T = 1$.
So, let $T = \set t$.
Then the trivial relation in $S$ to $T$ is:
:$\RR = \set {\tuple {s, t}: s \in S}$
We have that each element of $S$ is associated with exactly one element of $T$.
Hence $\RR$ is a mapping by definition.
::$(2) \, \text (c)$: Let $\card T > 1$.
Then:
:$\exists t_1, t_2 \in T: t_1 \ne t_2$
We have that $S \ne \O$, so $\exists s \in S$.
Thus by definition of trivial relation:
:$\tuple {s, t_1} \in \RR$
and:
:$\tuple {s, t_2} \in \RR$
but $t_1 \ne t_2$.
Thus $\RR$ is not a mapping.
Hence the result.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Functions
Category:Trivial Relation
Condition for Uniqueness of Increasing Mappings between Tosets

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets.
Let $f: S \to T$ and $g: S \to T$ be increasing mappings from $S$ to $T$.
Let $H \subseteq S$ be a subset of $S$.
Let $f$ and $g$ agree on $H$.
Let $K = f \sqbrk H$ be the image set of $H$ under $f$.
Let the intersection of $K$ with every set of the form:
:$\set {y \in T: u < y < v: u, v \in T, u < v}$
be non-empty.
Then $f = g$.
== Proof ==
By hypothesis, let the intersection of $K$ with every set of the form:
:$\set {y \in T: u \prec y \prec v: u, v \in T, u \prec v}$
be non-empty.
 $f \ne g$.
Then:
:$\exists x \in S: \map f x \ne \map g x$
, suppose that $\map f x < \map g x$.
Let:
:$a \in H$ such that $a \preceq x$
:$b \in H$ such that $b \succeq x$
We have that $f$ and $g$ are increasing.
Thus:
:$\map f a \preccurlyeq \map f x$
:$\map g b \succcurlyeq \map g x$
Thus no element of $H$ maps to the set:
:$\set {y \in T: \map f x \prec y \prec \map g x}$
That is, the intersection of $K$ with this set, which is of the form:
:$\set {y \in T: u \prec y \prec v: u, v \in T, u \prec v}$
is empty.
This contradicts our hypothesis.
Thus, by Proof by Contradiction:
:$f = g$
== Sources ==
* : Chapter $0$: Orderings: Theorem $11$
Category:Total Orderings
Condition for Valid Time Indication

== Theorem ==
Consider an analogue clock $C$ with an hour hand $H$ and a minute hand $M$.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock.
Then $C$ displays a valid time indication :
:$12 \phi \mod 360 = \theta$
=== Corollary ===
== Proof ==
Let $T$ be a time of day specified in hours $h$ and minutes $m$, where:
:$1 \le h \le 12$ is an integer
:$0 \le m < 60$ is a real number
whether a.m. or p.m. is immaterial.
From Speed of Minute Hand, $M$ travels $6 \degrees$ per minute.
So at time $m$ minutes after the hour, $\theta = 6 m$.
From Speed of Hour Hand, $H$ travels $\dfrac 1 2 \degrees$ per minute. 
The hour marks are at $30 \degrees$ intervals.
So at time $m$ minutes after the hour, $\phi = 30 h + \dfrac m 2 \degrees$ past hour $h$.
That is:
:$\phi = 30 h + \dfrac 1 2 \dfrac \theta 6$
or:
:$12 \phi = 360 h + \theta$
where $h$ is an integer.
Thus:
:$12 \phi \mod 360 = \theta$
Category:Clocks
Condition for Valid Time Indication/Corollary

== Theorem ==
Consider an analogue clock $C$ with an hour hand $H$ and a minute hand $M$.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock.
Let $\rho \degrees$ be the angle made by the hour hand with respect to the hour just past.
Then $C$ displays a valid time indication :
:$\rho = \dfrac \theta {12}$
== Proof ==
Follows directly.
Category:Clocks
Condition for Variance Ratio to have Snedecor's F-Distribution

== Theorem ==
Let $V$ be the variance ratio obtained as the result of an analysis of variance exercise.
Suppose that the errors are:
:independently normally distributed
:with zero expectation
:with equal variances.
Under the null hypothesis of no difference between treatment means, $V$ has Snedecor's F-distribution.
High values of variance ratio indicate rejection of the null hypothesis.
== Proof ==
== Sources ==
* 
* 
Category:Variance Ratio
Category:Snedecor's F-Distribution
Condition for Vector Field to satisfy Laplace's Equation

== Theorem ==
Let $\mathbf V$ be a vector field over a region of space $R$.
Then:
:$\mathbf V$ is both solenoidal and conservative
:
:$\mathbf V$ is the gradient of a scalar field $F$ over $R$ which satisfies Laplace's equation:
::$\nabla^2 F \equiv 0$
== Proof ==
=== Sufficient Condition ===
Let $\mathbf V$ be both solenoidal and conservative.
Then from Vector Field is Expressible as Gradient of Scalar Field iff Conservative:
:$\mathbf V = \grad F$
for some scalar field $F$ over $R$.
Because $\mathbf V$ is solenoidal, we have:
:$\operatorname {div} \mathbf V = 0$
that is:
:$\operatorname {div} \grad F = 0$
Hence by Laplacian on Scalar Field is Divergence of Gradient:
:$\nabla^2 F = 0$
where $\nabla^2$ is the Laplacian on $F$.
This holds throughout $R$, and so the equality is an equivalence:
:$\nabla^2 F \equiv 0$
=== Necessary Condition ===
Let $\mathbf V$ be the gradient of a scalar field $F$ over $R$ which satisfies Laplace's equation:
:$\nabla^2 F \equiv 0$
Thus $F$ is such that:
:$\mathbf V = \grad F$
and from Vector Field is Expressible as Gradient of Scalar Field iff Conservative it follows that $\mathbf V$ is conservative.
Then by Laplacian on Scalar Field is Divergence of Gradient:
:$\operatorname {div} \grad F = 0$
That is:
:$\operatorname {div} \mathbf V = 0$
and so by definition $\mathbf V$ is solenoidal.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(i)}$
Category:Solenoidal Vector Fields
Category:Conservative Vector Fields
Category:Laplace's Equation
Condition for Vector Field to satisfy Poisson's Equation

== Theorem ==
Let $\mathbf V$ be a vector field over a region of space $R$.
Then:
:$\mathbf V$ is conservative but not solenoidal
:
:$\mathbf V$ is the gradient of a scalar field $F$ over $R$ which satisfies Poisson's equation over $R$:
::$\nabla^2 F = \phi$
:where $\phi$ is a function which is not identically zero.
== Proof ==
=== Sufficient Condition ===
Let $\mathbf V$ be conservative but not solenoidal.
From Vector Field is Expressible as Gradient of Scalar Field iff Conservative:
:$\mathbf V = \grad F$
for some scalar field $F$ over $R$.
Because $\mathbf V$ is not solenoidal, we have:
:$\exists \mathbf v \in R: \operatorname {div} \mathbf v \ne 0$
that is:
:$\operatorname {div} \grad F \ne 0$
for at least some $\mathbf v \in R$.
Hence by Laplacian on Scalar Field is Divergence of Gradient:
:$\nabla^2 F = \phi$
where:
:$\phi$ is not identically zero
:$\nabla^2$ is the Laplacian on $F$.
Hence $F$ satisfies Poisson's equation.
=== Necessary Condition ===
Let $\mathbf V$ be the gradient of a scalar field $F$ over $R$ which satisfies Poisson's equation:
:$\nabla^2 F = \phi$
where $\phi$ is not identically zero.
Thus $F$ is such that:
:$\mathbf V = \grad F$
and from Vector Field is Expressible as Gradient of Scalar Field iff Conservative it follows that $\mathbf V$ is conservative.
Then by Laplacian on Scalar Field is Divergence of Gradient:
:$\exists \mathbf v \in R: \operatorname {div} \grad F \ne 0$
That is:
:$\exists \mathbf v \in R: \operatorname {div} \mathbf V \ne 0$
and so by definition $\mathbf V$ is specifically not solenoidal.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $7$. The Classification of Vector Fields: $\text {(ii)}$
Category:Solenoidal Vector Fields
Category:Conservative Vector Fields
Category:Poisson's Differential Equation
Condition for Vectors to have Same Syndrome

== Theorem ==
Let $C$ be a linear $\tuple {n, k}$-code whose master code is $\map V {n, p}$
Let $G$ be a (standard) generator matrix for $C$.
Let $P$ be a standard parity check matrix for $C$.
Let $u, v \in \map V {n, p}$.
Then $u$ and $v$ have the same syndrome they are in the same coset of $C$.
== Proof ==
Let $u, v \in \map V {n, p}$.
Let $\map S u$ denote the syndrome of $u$.
Then:
Hence the result from Elements in Same Coset iff Product with Inverse in Subgroup.
== Sources ==
* : Chapter $6$: Error-correcting codes: Corollary $6.21$
Category:Linear Codes
Condition for Woset to be Isomorphic to Ordinal

== Theorem ==
Let $\struct {S, \preceq}$ be a woset.
Let $\struct {S, \preceq}$ be such that $\forall a \in S$, the initial segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.
Then $\struct {S, \preceq}$ itself is order isomorphic to an ordinal.
== Proof ==
For each $a \in S$, let $g_a: S_a \to \map Z a$ be an order isomorphism from $S_a$ to an ordinal $\map Z a$.
By Isomorphic Ordinals are Equal‎ and Order Isomorphism between Wosets is Unique, both $\map Z a$ and $g_a$ are unique.
So this defines a mapping $Z$ on $S$.
Let the image of $Z$ be $W$:
:$W = \set {\map Z a: a \in S}$
Now we define $f: S \to W$ as:
:$\map f a = \map Z a$
Now we show that $x, y \in S, x \prec y \implies \map Z x \subset \map Z y$.
So, let $x, y \in S$ such that $x \prec y$.
Then:
:$(1) \quad g_x: S_x \cong \map Z x$
Also, because:
we have:
:$(2): \quad \paren {g_y \restriction S_x}: S_x \to \paren {\map Z y}_{\map {g_y} x}$ is an order isomorphism
So by Isomorphic Ordinals are Equal, we have:
:$(3): \quad \map Z x = \paren {\map Z y}_{\map {g_y} x}$
So, in particular:
:$\map Z x \subset \map Z y$
By this result, $f: S \to W$ is a bijection.
Also by this result, $f: S \to \struct {W, \subseteq}$ is an order isomorphism.
This means that $W$ is well-ordered by $\subseteq$.
Now we show that $W$ is an ordinal.
Let $y \in S$.
Since $\map Z y$ is an ordinal, we have:
:$x \prec y \implies \paren {\map Z y}_{\map {g_y} x} = \map {g_y} x$
So by $(3)$ above, we have:
:$(4) \quad x \prec y \implies \map Z x = \map {g_y} x$
Hence:
As $y$ is an arbitrary element of $S$, it follows that $\map Z y$ is an arbitrary element of $W$.
So $W$ is an ordinal, as we wanted to prove.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.11$
Category:Ordinals
Condition for Woset to be Isomorphic to Ordinal/Mistake

== Source Work ==
:
$\S 1$: Naive Set Theory:
:$\S 1.7$: Well-Orderings and Ordinals:
::Theorem $1.7.11$
== Mistake ==
:Also, since:
we have
:$(2): \quad \paren {g_y X_x}: X_x \cong \paren {\map Z y}_{\map {g_y} x}$.
Now, $\map Z y$ is an ordinal, so by Theorem $1.7.6$, $\paren {\map Z y}_{\map {g_y} x}$ is an ordinal.
...
== Correction ==
Statement $(2)$ has a misprint.
It should read:
:$(2): \quad \paren {g_y \restriction X_x}: X_x \cong \paren {\map Z y}_{\map {g_y} x}$
thereby providing the crucial information that the mapping under consideration is the restriction of $g_y$ to $X_x$.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.11$
Category:Mistakes/The Joy of Sets