nyuuzyou/wikis huggingface dataset:
Monotone Real Function with Everywhere Dense Image is Continuous/Lemma

== Theorem ==
Let $I$ and $J$ be real intervals.
Let $f: I \to J$ be a monotone real function.
Let $f \sqbrk I$ denote the image of $I$ under $f$.
Let $f \sqbrk I$ be everywhere dense in $J$.
Let $c \in I$.
Then:
:$\ds \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I \subseteq \set {\map f c}$
== Proof ==
From Discontinuity of Monotonic Function is Jump Discontinuity, $\ds \lim_{x \mathop \to c^-} \map f x$ and $\ds \lim_{x \mathop \to c^+} \map f x$ are finite.
Since $f$ is increasing:
:$\ds \lim_{x \mathop \to c^-} \map f x < \lim_{x \mathop \to c^+} \map f x$
Suppose $z \in \ds \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I$.
Then:
:$\ds \exists t \in I : \map f t \in \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}$
=== Case 1 : $t < c$ ===
Suppose that $t < c$.
So we may discard this case.
=== Case 2 : $t = c$ ===
Suppose that $t = c$.
So the proposition holds in this case.
=== Case 3 : $c < t$ ===
Suppose that $t > c$.
So we may discard this case.
So $\map f t = c$, by Proof by Cases.
Thus:
Thus:
:$\ds z \in \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I \implies z \in \set {\map f c}$
Hence the result, by definition of subset.
Category:Real Analysis
Monotonic Sequence Characterization of Left Limit of Function

== Theorem ==
Let $\hointl a b$ be a real interval. 
Let $x \in \hointl a b$. 
Let $f : \hointl a b \to \R$ be a real function.
Let $L \in \R$. 
Then:
:$\ds \lim_{y \mathop \to x^-} \map f y = L$ 
:
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to L$
=== Corollary ===
== Proof ==
=== Necessary Condition ===
Suppose that:
:$\ds \lim_{y \to x^-} \map f y = L$
then:
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, converging to $x$ we have:
::$\map f {x_n} \to L$
from Limit of Function by Convergent Sequences: Corollary.
So in particular: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to L$
=== Sufficient Condition ===
Suppose that: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to L$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence, with $x_n < x$ for each $n$, converging to $x$.
Let $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ be a subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$. 
We aim to show that $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ has a subsequence converging to $L$. 
We will then obtain $\map f {x_n} \to L$ from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
From the Peak Point Lemma, $\sequence {x_{n_j} }_{j \mathop \in \N}$ has a monotonic subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$.
Since $x_n < x$ for each $n$, we must have $x_{n_{j_k} } < x$ for each $k$. 
Further, $x_{n_{j_k} } \to x$ as $k \to \infty$.
So from hypothesis: 
:$\map f {x_{n_{j_k} } } \to L$
So:
:$\sequence {\map f {x_{n_{j_k} } } }_{j \mathop \in \N}$ is a subsequence of $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ converging to $L$.
Since $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ was an arbitrary subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$, we have: 
:$\map f {x_n} \to L$ 
from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have:
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, converging to $x$ we have $\map f {x_n} \to L$.
So, from Limit of Function by Convergent Sequences: Corollary, we have: 
:$\ds \lim_{y \to x^-} \map f y = L$ 
Category:Monotonic Sequence Characterization of Left Limit of Function
Monotonic Sequence Characterization of Left Limit of Function/Corollary

== Corollary ==
Let $\hointl a b$ be a real interval. 
Let $x \in \hointl a b$. 
Let $f : \hointl a b \to \R$ be a real function.
Then $f$ is left-continuous at $x$ :
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to \map f x$
== Proof ==
From the definition of left-continuity, $f$ is left-continuous at $x$ :
:$\ds \lim_{y \mathop \to x^-} \map f y = \map f x$
By Monotonic Sequence Characterization of Left Limit of Function, this is equivalent to:
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n < x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to L$
Category:Monotonic Sequence Characterization of Left Limit of Function
Monotonic Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals

== Theorem ==
Let $\hointr a b$ be a real interval. 
Let $x \in \hointr a b$. 
Let $f : \hointr a b \to \R$ be a real function.
Then $f$ is right-continuous at $x$ :
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to \map f x$
== Proof ==
=== Necessary Condition ===
Suppose $f$ is right-continuous at $x$, then: 
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, converging to $x$ we have:
::$\map f {x_n} \to \map f x$
from Limit of Function by Convergent Sequences: Corollary.
So in particular: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to \map f x$
=== Sufficient Condition ===
Suppose that: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to \map f x$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence, with $x_n > x$ for each $n$, converging to $x$.
Let $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ be a subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$. 
We aim to show that $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ has a subsequence converging to $\map f x$. 
We will then obtain $\map f {x_n} \to \map f x$ from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
From the Peak Point Lemma, $\sequence {x_{n_j} }_{j \mathop \in \N}$ has a monotonic subsequence $\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$.
Since $x_n > x$ for each $n$, we must have $x_{n_{j_k} } > x$ for each $k$. 
Further, $x_{n_{j_k} } \to x$ as $k \to \infty$.
So from hypothesis: 
:$\map f {x_{n_{j_k} } } \to \map f x$
So:
:$\sequence {\map f {x_{n_{j_k} } } }_{j \mathop \in \N}$ is a subsequence of $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ converging to $\map f x$.
Since $\sequence {\map f {x_{n_j} } }_{j \mathop \in \N}$ was an arbitrary subsequence of $\sequence {\map f {x_n} }_{n \mathop \in \N}$, we have: 
:$\map f {x_n} \to \map f x$ 
from Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit.
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have:
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, converging to $x$ we have $\map f {x_n} \to \map f x$.
So, from Limit of Function by Convergent Sequences: Corollary, we have: 
:$f$ is right-continuous at $x$.
Category:Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals
Monotonicity of Real Sequences

== Theorem ==
Let $\mathbb D$ be a subset of $\N$.
Let $\sequence {a_n}: \mathbb D \to \R$ be a real sequence.
Let $\mathbb X$ be a real interval such that $\mathbb D \subseteq \mathbb X$.
Let $f: \mathbb X \to \R, x \mapsto \map f x$ be a differentiable real function.
Suppose that for every $n \in \mathbb D$:
:$\map f n = a_n$
Then:
:If $\forall x \in \mathbb X: D_x \map f x \ge 0$, $\sequence {a_n}$ is increasing
:If $\forall x \in \mathbb X: D_x \map f x > 0$, $\sequence {a_n}$ is strictly increasing
:If $\forall x \in \mathbb X: D_x \map f x \le 0$, $\sequence {a_n}$ is decreasing
:If $\forall x \in \mathbb X: D_x \map f x < 0$, $\sequence {a_n}$ is strictly decreasing
where $D_x$ denotes differentiation $x$.
== Proof ==
Consider the case where $D_x \map f x \ge 0$
Let $n \in \N$ be in the domain of $\sequence {a_n}$.
From Derivative of Monotone Function, the sign of $D_x f$ is indicative of the monotonicity of $f$.
Because Differentiable Function is Continuous and Continuous Real Function is Darboux Integrable, $D_x f$ is integrable.
Hence:
Then:
Hence the result, by the definition of monotone.
The proofs of the other cases are similar.
== Also see ==
* Restriction of Monotone Function is Monotone
Category:Real Sequences
Monte Carlo Method/Repeated Sampling/Examples/Bridge Hands

== Examples of Monte Carlo Method ==
Consider the game of Bridge.
Let us try to estimate the probability that one of the players receives more than $6$ cards of the same suit.
We can obtain a Monte Carlo estimate of this by dealing $1000$ sets of hands, for example, and recording the number of occasions $r$ at which this event occurs.
Hence the probability of this event is given as $p = \dfrac r {1000}$.
If $N$ samples are used, then the probability is $p = \dfrac r N$.
== Sources ==
* 
* 
Category:Examples of Monte Carlo Methods
Monte Carlo Method/Sampling Experiments/Examples/Area of Region

== Example of Monte Carlo Method ==
Let $A$ be a region of the plane enclosed by a square $S$ of side $a$.
In order to estimate the area of $A$, a Monte Carlo method may be set up as follows.
Select $n$ points at random throughout $S$.
If $r$ of those points are in $A$, then the area of $A$ is approximately $\dfrac {r a^2} n$.
== Sources ==
* 
* 
Category:Examples of Monte Carlo Methods
Montel's Theorem

== Theorem ==
Let $U \subseteq \C$ be an open subset of the complex numbers.
Let $\map \HH U$ be the space of holomorphic mappings on $U$.
Then a family of mappings $\FF \subseteq \map \HH U$ is normal $\FF$ is locally bounded.
== Proof ==
=== Normal implies locally bounded ===
By the Arzelà-Ascoli Theorem, every normal family is locally bounded.
=== Locally bounded implies normal ===
By the Arzelà-Ascoli Theorem, any locally bounded and locally uniformly equicontinuous family is normal. 
Hence it suffices to show that $\FF$ is locally uniformly equicontinuous.
As a Family of Lipschitz Continuous Functions with same Lipschitz Constant is Uniformly Equicontinuous, it suffices to show that every point $z_0 \in U$ has a neighbourhood $N$ such that all functions from $\FF$ have the same Lipschitz constant on $N$.
To this end, take any $z_0 \in U$.
By the assumption that $\FF$ is locally bounded, we may choose $R > 0$ such that for all $z \in B := \map B {x; R}$ (the closed disk of radius $R$ around $x$), we have $\size {\map f z} < M$.
Now take any $r \in \tuple {0, R}$ and let $B' = \map B {x; r}$. 
We then have for any $z, z' \in B'$ and $w \in B$:
:$\size {\paren {w - z} \paren {w - z'} } \ge \paren {R - r}^2$
This allows us to derive:
Therefore, $M \paren {R - r}^{-2}$ is a Lipschitz constant for all functions in $\FF$ on $B'$. 
== Sources ==
* : $\S 7.2$
Category:Complex Analysis
Moore-Osgood Theorem

== Theorem ==
Let $X$ and $Y$ be metric spaces.
Let $S$ be a subspace of $X$.
Let $c$ be a limit point of $S$.
Let $\sequence {f_n}$ be a sequence of mappings $f_n : X \to Y$.
Suppose that:
:$(1): \quad \sequence {f_n}$ is uniformly convergent on $S$
:$(2): \quad \ds \forall n \in \N : \lim_{x \mathop \to c} \map {f_n} x$ exists
Then:
:$\ds \lim_{x \mathop \to c} \lim_{n \mathop \to \infty} \map {f_n} x = \lim_{n \mathop \to \infty} \lim_{x \mathop \to c} \map {f_n} x$
== Proof ==
Category:Real Analysis
Category:Metric Spaces
Moore-Penrose Pseudoinverse/Examples/Arbitrary Example 1

== Example of Moore-Penrose Pseudoinverse ==
Let $\mathbf A$ be the matrix:
:$\mathbf A := \begin {pmatrix} 2/3 & 0 \\ 5/6 & 1/2 \\ 1/3 & 1 \end {pmatrix}$
The Moore-Penrose pseudoinverse of $\mathbf A$ is:
:$\mathbf A^+ := \begin {pmatrix} 5/6 & 2/3 & -1/3 \\ -1/2 & 0 & 1 \end {pmatrix}$
== Proof ==
== Sources ==
* 
Category:Examples of Moore-Penrose Pseudoinverses
More than One Right Zero then No Left Zero

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
If $\struct {S, \circ}$ has more than one left zero, then it has no right zero.
Likewise, if $\struct {S, \circ}$ has more than one right zero, then it has no left zero.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure with more than one left zero.
Take any two of these, and call them $z_{L_1}$ and $z_{L_2}$, where $z_{L_1} \ne z_{L_2}$.
Suppose $\struct {S, \circ}$ has a right zero.
Call it $z_R$.
Then, by the behaviour of $z_R$, $z_{L_1}$ and $z_{L_2}$:
:$z_{L_1} = z_{L_1} \circ z_R = z_R$
:$z_{L_2} = z_{L_2} \circ z_R = z_R$
So $z_{L_1} = z_R = z_{L_2}$, which contradicts the supposition that $z_{L_1}$ and $z_{L_2}$ are different.
Therefore, in an algebraic structure with more than one left zero, there can be no right zero.
The same argument can be applied to an algebraic structure with more than one right zero.
Category:Zero Elements
More than one Left Identity then no Right Identity

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
If $\struct {S, \circ}$ has more than one left identity, then it has no right identity.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure with more than one left identity.
Take any two of these, and call them $e_{L_1}$ and $e_{L_2}$, where $e_{L_1} \ne e_{L_2}$.
Suppose $\struct {S, \circ}$ has a right identity.
Call this right identity $e_R$.
Then, by the behaviour of $e_R$, $e_{L_1}$ and $e_{L_2}$:
:$e_{L_1} = e_{L_1} \circ e_R = e_R$
:$e_{L_2} = e_{L_2} \circ e_R = e_R$
So $e_{L_1} = e_R = e_{L_2}$, which contradicts the supposition that $e_{L_1}$ and $e_{L_2}$ are different.
Therefore, in an algebraic structure with more than one left identity, there can be no right identity.
== Also see ==
* More than one Right Identity then no Left Identity
== Sources ==
* : Chapter $5$: Semigroups: Exercise $3 \ \text{(ii)}$
Category:Identity Elements
More than one Right Identity then no Left Identity

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
If $\struct {S, \circ}$ has more than one right identity, then it has no left identity.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure with more than one right identity.
Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$.
Suppose $\struct {S, \circ}$ has a left identity.
Call this left identity $e_L$.
Then, by the behaviour of $e_L$, $e_{R_1}$ and $e_{R_2}$:
:$e_{R_1} = e_{R_1} \circ e_L = e_L$
:$e_{R_2} = e_{R_2} \circ e_L = e_L$
So $e_{R_1} = e_L = e_{R_2}$, which contradicts the supposition that $e_{R_1}$ and $e_{R_2}$ are different.
Therefore, in an algebraic structure with more than one right identity, there can be no left identity.
== Also see ==
* More than one Left Identity then no Right Identity
Category:Identity Elements
Morera's Theorem

== Theorem ==
Let $D$ be a simply connected domain in $\C$.
Let $f: D \to \C$ be a continuous function.
Let $f$ be such that:
:$\ds \int_\gamma \map f z \rd z = 0$
for every simple closed contour $\gamma$ in $D$
Then $f$ is analytic on $D$.
== Proof ==
For a fixed $z_0 \in D$ and $z \in D$ we consider the function:
:$\ds \map F z = \int_\gamma \map f w \rd w$
where $\gamma$ is any (simple) contour starting at $z_0$ and ending at $z$.
By Primitive of Function on Connected Domain, $F$ is a primitive of $f$.
Since $F$ is analytic and $F' = f$, we conclude that $f$ is analytic as well.
== Also see ==
This is the converse of the Cauchy-Goursat Theorem.
== Sources ==
* 
Category:Calculus
Category:Complex Analysis
Morley's Trisector Theorem

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let the internal angles of $\triangle ABC$ be trisected.
Let the points where these angle trisectors first intersect be $D$, $E$ and $F$.
:
Then $\triangle EDF$ is equilateral.
== Dijkstra's Proof ==
== Proof 2 ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
Category:Morley's Trisector Theorem
Category:Triangles
Morley's Trisector Theorem/Dijkstra's Proof

== Theorem ==
== Dijkstra's Proof ==
Choose angles $\alpha$, $\beta$ and $\gamma$ all greater than $0$ such that $\alpha + \beta + \gamma = 60 \degrees$.
Draw an equilateral triangle $\triangle XYZ$.
Construct $\triangle AXY$ and $\triangle BXZ$ with the angles as indicated:
:
We have that:
Because $\angle AXB = 180 \degrees - \paren {\alpha + \beta}$ , it follows that:
:if $\angle BAX = \alpha + x$ then $\angle ABX = \beta - x$
Using the Sine Rule $3$ times, in $\angle AXB$, $\angle AXY$ and $\angle BXZ$, we have:
In the range in which these angles lie, the of the above is a strictly increasing function of $x$.
Thus we conclude that $x = 0$.
The result follows.
== Historical Note ==
== Sources ==
* 
Category:Morley's Trisector Theorem
Morley's Trisector Theorem/Dijkstra's Proof/Historical Note

== Historical Note on Dijkstra's Proof of Morley's Trisector Theorem ==
This proof was published by on $30$ December $1975$ in an open letter to .
As he put it:
:the other day I encountered your delightful booklet "". On account of Chapter 8, I concluded that you might be interested in the following proof of Morley's Theorem "The adjacent pairs of the trisectors of a triangle always meet at the vertices of an equilateral triangle." ... I found this proof in the early sixties, but am afraid that I did not publish it.
== Sources ==
* 
Category:Historical Notes
Morphism Class Equivalence is Equivalence

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $C$ be an object of $\mathbf C$.
Let $\map {\mathbf {Sub}_{\mathbf C} } C$ be the category of subobjects of $C$.
Let $\sim$ denote morphism class equivalence on the morphisms of $\map {\mathbf {Sub}_{\mathbf C} } C$.
Then $\sim$ is an equivalence.
== Proof ==
Let us check the three conditions for an equivalence in turn.
=== Reflexivity ===
To show that $\sim$ is reflexive, note that for $f: m \to m'$:
:$f \sim f$
holds $m \sim m$ and $m' \sim m'$.
By Equivalence of Subobjects is Equivalence, these conditions are satisfied.
=== Symmetry ===
It is to be shown that for $f: m \to n$ and $f': m' \to n'$:
:$f \sim f'$ implies $f' \sim f$
By definition of $\sim$, this is equivalent to:
:$m \sim m' \land n \sim n'$ implies $m' \sim m \land n' \sim n$
Again by Equivalence of Subobjects is Equivalence, this is satisfied.
=== Transitivity ===
Given $f: m \to n$, $f': m' \to n'$ and $f' ': m' ' \to n' '$, it is to be shown that:
:$f \sim f' \land f' \sim f' '$ implies $f \sim f' '$
By definition of $\sim$, this is equivalent to:
:$m \sim m' \land n \sim n' \land m' \sim m' ' \land n' \sim n' '$ implies $m \sim m' ' \land n \sim n' '$
which also follows from Equivalence of Subobjects is Equivalence.
Thence $\sim$ is proven an equivalence.
Category:Subobjects
Morphism Property Preserves Cancellability

== Theorem ==
Let:
:$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$
be a mapping from one algebraic structure:
:$\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$
to another:
:$\struct {T, *_1, *_2, \ldots, *_n}$
Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$.
Then if an element $a \in S$ is either left cancellable or right cancellable under $\circ_k$, then $\map \phi a$ is correspondingly left cancellable or right cancellable under $*_k$.
Thus, the morphism property is seen to preserve cancellability.
== Proof ==
First let $S$ be the empty set.
It follows from the definition of an Definition:Morphism Property that $\circ_1, \circ_2, \ldots, \circ_n$ are all empty maps.
It also follows from the definition of an Definition:Morphism Property that $*_1, *_2, \ldots, *_n$ are all empty maps.
By Image of Empty Set is Empty Set, $T$ is also the empty set.
Hence, the morphism property is seen to preserve cancellability when $S$ is the empty set.
Now let $S \ne \O$.
We need to demonstrate the following properties:
$(1): \quad$ If $a \in S$ has the property that:
:$\forall x, y \in S: x \circ_k a = y \circ_k a \implies x = y$
then:
:$\forall x, y \in S: \map \phi x *_k \map \phi a = \map \phi y *_k \map \phi a \implies \map \phi x = \map \phi y$
Thus:
and thus left cancellability is demonstrated.
$(2): \quad$ If $a \in S$ has the property that:
:$\forall x, y \in S: a \circ_k x = a \circ_k y \implies x = y$
then:
:$\forall x, y \in S: \map \phi a *_k \map \phi x = \map \phi a *_k \map \phi y \implies \map \phi x = \map \phi y$
Thus:
and thus right cancellability is demonstrated.
Category:Morphism Property
Morphism Property Preserves Closure

== Theorem ==
Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping from one algebraic structure $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ to another $\struct {T, *_1, *_2, \ldots, *_n}$.
Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$.
Then the following properties hold:
:If $S' \subseteq S$ is closed under $\circ_k$, then $\phi \sqbrk {S'}$ is closed under $*_k$
:If $T' \subseteq T$ is closed under $*_k$, then $\phi^{-1} \sqbrk {T'}$ is closed under $\circ_k$
where $\phi \sqbrk {S'}$ denotes the image of $S'$.
== Proof ==
Suppose that $\circ_k$ has the morphism property under $\phi$.
Suppose that $S' \subseteq S$ is closed under $\circ_k$.
Thus, for non-empty $S'$:
:$s_1, s_2 \in S' \implies s_1 \circ_k s_2 \in S'$
Similarly, suppose that $T' \subseteq T$ is closed under $*_k$.
Thus, non-empty $T'$:
:$t_1, t_2 \in T' \implies t_1 *_k t_2 \in T'$
First we prove that $\phi \sqbrk {S'}$ is closed under $*_k$:
Then we prove that $\phi^{-1} \sqbrk {T'}$ is closed under $\circ_k$:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.1$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.2$
Category:Morphism Property
Morphism from Integers to Group

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $g \in G$.
Let $\phi: \Z \to G$ be the mapping defined as:
:$\forall n \in \Z: \map \phi n = g^n$.
Then:
:If $g$ has infinite order, then $\phi$ is a group isomorphism from $\struct {\Z, +}$ to $\gen g$.
:If $g$ has finite order such that $\order g = m$, then $\phi$ is a group epimorphism from $\struct {\Z, +}$ to $\gen g$ whose kernel is the principal ideal $\paren m$.
:Thus $\gen g$ is isomorphic to $\struct {\Z, +}$, and $m$ is the smallest (strictly) positive integer such that $g^m = e$.
== Proof ==
By Epimorphism from Integers to Cyclic Group, $\phi$ is an epimorphism from $\struct {\Z, +}$ onto $\gen g$.
By Kernel of Group Homomorphism is Subgroup, the kernel $K$ of $G$ is a subgroup of $\struct {\Z, +}$.
Therefore by Subgroup of Integers is Ideal and Principal Ideals of Integers, $\exists m \in \N_{>0}: K = \paren m$.
Thus $\gen g \cong \struct {\Z, +}$.
By Quotient Epimorphism from Integers by Principal Ideal:
:$\forall m \in \N_{>0}: \order {\Z_m} = m$
So, if $\gen g$ is finite, and if $\gen g \cong \struct {\Z, +}$, then $m = \order g$.
Furthermore, $m$ is the smallest (strictly) positive integer such that $g^m = e$, since $m$ is the smallest (strictly) positive integer in $\paren m$ from Principal Ideals of Integers.
If $\gen g$ is infinite, then $m = 0$ and so $\phi$ is a (group) isomorphism from $\struct {\Z, +}$ onto $\gen g$.
== Also see ==
* Morphism from Ring with Unity to Module
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.5$
Category:Group Isomorphisms
Category:Group Epimorphisms
Category:Integers
Category:Ideal Theory