nyuuzyou/wikis huggingface dataset:
Is Pi to the power of Euler's Number Rational?

== Open Question ==
It is not known whether $\pi$(pi) to the power of Euler's number $e$:
:$\pi^e$
is rational or irrational.
Its decimal expansion gives its value to be approximately:
:$\pi^e \approx 22 \cdotp 45915 \, 77183 \, 61045 \, 47342 \, 7152 \ldots$
== Progress ==
See Schanuel's Conjecture Implies Transcendence of Pi to the power of Euler's Number.
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.11$
* : $22 \cdotp 459 \, 157 \, 718 \, 361 \, 045 \, 473 \, 427 \, 152 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : $22 \cdotp 45915 \, 77183 \, 61045 \, 47342 \, 7152 \ldots$
Category:Open Questions
Category:Pi
Category:Euler's Number
Is there a Limit to the Multiplicative Persistence of a Number?

== Theorem ==
It has been conjectured that there may be an upper limit to the multiplicative persistence of a natural number.
== Progress ==
It is not known whether there exists a number $n$ such that:
:$\map P n = 12$
where $\map P n$ denotes the multiplicative persistence of $n$, but it is known that it is greater than $10^{200}$.
== Sources ==
* 
* : $10$
* : $10$
Category:Multiplicative Persistence
Category:Open Questions
Is there an Infinite Number of Primes of Form n^2 + 1?

== Open Question ==
Is there an infinite number of prime numbers of the form $n^2 + 1$?
== Also known as ==
This problem is found according to some sources to be referred to as the Fifth Hardy-Littlewood Conjecture, but it is difficult to find hard evidence to back this up.
== Landau's Problems ==
This is the $4$th of Landau's problems.
== Sources ==
* : $\text {II}$: The Series of Primes ($2$): $\S 2.7$. Further Results on formulae for primes: $\text {(iii)}$
* : $1,3727 \ldots$
Category:Open Questions
Category:Prime Numbers
Category:Landau's Problems
Island and Lake Paradox

== Paradox ==
An island is a body of land surrounded entirely by water.
A lake is a body of water surrounded entirely by land.
What if you had a planet whose northern hemisphere was all land, and whose southern hemisphere was all water?
Would you call the northern hemisphere an island, or would you call the southern hemisphere a lake?
Or would you say:
:there is an island in the middle of a lake
:which is in the middle of an island
:which is in the middle of a lake
:which is in the middle of an island
:$\ldots$
?
== Resolution ==
This paradox is an instance of a dialetheia.
Both of the main statements are true, as well as the statement derived from it.
== Sources ==
* : Chapter One: What is a Paradox?
Category:Dialetheia
Island in English Channel Further from France than England

== Problem ==
:There is a certain island, situated between and ,
:and yet, that island is farther from than is.
== Solution ==
At their closest point, and are $34$ kilometres apart.
The , the body of water which lies between and is considerably wider than that through most of its length.
, for example, one of the , is about $42$ kilometres from , but well and truly in the between and .
The is also between and , and considerably further away from than is.
== Sources ==
* : Geographical Paradoxes
* : Rational Amusements for Winter Evenings: $162$
Category:Geography Puzzles
Category:Trick Puzzles
Isolated Point in Metric Space iff Isolated Point in Topological Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $T = \struct {A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Let $x \in H$
Then:
:$x$ is an isolated point of $H$ in $M$ $x$ is an isolated point of $H$ in $T$
== Proof ==
From Open Balls form Local Basis for Point of Metric Space, the set:
:$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0} }$
is a local basis of $x$.
From Local Basis Test for Isolated Point:
:$x$ is an isolated point of $H$ in $T$ $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$
By definition of an isolated point in $M$:
:$x$ is an isolated point of $H$ in $T$ $x$ is an isolated point of $H$ in $M$
Category:Isolated Points
Isolated Point of Closure of Subset is Isolated Point of Subset

== Theorem ==
Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$ be a subset of $T$.
Let $\map \cl H$ denote the closure of $H$.
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
Then $x$ is also an isolated point of $H$.
== Proof 1 ==
== Proof 2 ==
Category:Isolated Points
Category:Set Closures
Category:Isolated Point of Closure of Subset is Isolated Point of Subset
Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 1

== Theorem ==
== Proof ==
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
 that $x$ is not an isolated point of $H$.
Then by the definition of a limit point, it follows that $x$ must be a limit point of $H$.
From Set is Subset of its Topological Closure:
:$H \subseteq \map \cl H$
So by Limit Point of Subset is Limit Point of Set it follows that $x$ is a limit point of $\map \cl H$.
But then $x$ cannot be an isolated point of $\map \cl H$.
This is a contradiction.
Therefore every isolated point of $\map \cl H$ is also an isolated point of $H$.
Category:Isolated Point of Closure of Subset is Isolated Point of Subset
Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 2

== Theorem ==
== Proof ==
Let $x \in \map \cl H$ be isolated in $\map \cl H$.
By definition of isolated point, $x$ is not a limit point of $\map \cl H$.
From Set is Subset of its Topological Closure:
:$H \subseteq \map \cl H$
We have that Limit Point of Subset is Limit Point of Set.
But $x$ is not a limit point of $\map \cl H$ .
So by the Rule of Transposition, $x$ cannot be a limit point of $H$.
As $x$ is not a limit point of $H$, it follows that $x$ must be an isolated point of $H$.
Category:Isolated Point of Closure of Subset is Isolated Point of Subset
Isolated Points in Arens-Fort Space

== Theorem ==
Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Let $q \in S: q \ne \tuple {0, 0}$.
Then $q$ is an isolated point of $T$.
== Proof ==
If $q \ne \tuple {0, 0}$ then from Clopen Points in Arens-Fort Space we have that $\set q$ is both closed and open in $T$.
In particular, $\set q$ is open in $T$.
The result follows from Point in Topological Space is Open iff Isolated.
== Sources ==
* : Part $\text {II}$: Counterexamples: $26$. Arens-Fort Space: $8$
Category:Isolated Points
Category:Arens-Fort Space
Isolated Points in Subsets of Modified Fort Space

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Let $H \subseteq S$ contain more than two points.
Then $H$ contains an isolated point.
== Proof ==
Let $H \subseteq S$ contain more than two points.
From , we can write $S = N \cup \set a \cup \set b$, where $N$ is an infinite set.
Suppose $H \cap N \ne \O$.
Let $x \in H \cap N$.
Then:
:$\set x \subset N$
Therefore:
:$\set x \in \tau_{a, b}$
This shows that $x$ is isolated.
Suppose $H \cap N = \O$.
By Empty Intersection iff Subset of Complement:
:$H \subseteq \relcomp S N = \set {a, b}$
Because $H$ contains at least $2$ elements:
:$H = \set {a, b}$
Because $a \in \relcomp S {\set b}$ and $\relcomp S {\relcomp S {\set b} } = \set b$ is a finite set:
:$\relcomp S {\set b} \in \tau_{a, b}$
Thus:
:$H \cap \relcomp S {\set b} = \set a$
This shows that $a$ is isolated.
Therefore $H$ must contain an isolated point.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $6$
Category:Isolated Points
Category:Modified Fort Space
Isometric Image of Cauchy Sequence is Cauchy Sequence

== Theorem ==
Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces.
Let $f: S_1 \to S_2$ be an isometry.
Let $\sequence {x_n}$ be a Cauchy sequence in $S_1$.
Let $\sequence {y_n} = \sequence {\map f {x_n} }$ be the image of $\sequence {x_n}$ under $f$.
Then $\sequence {y_n}$ is a Cauchy sequence.
== Proof ==
Let $\epsilon \in \R_{>0}$.
By the definition of Cauchy sequence, there is an $N \in \R$ such that:
:$\paren{m > N} \land \paren {n > N} \implies \map {d_1} {x_m, x_n} < \epsilon$
Since $f$ is an isometry, $\map {d_2} {y_m , y_n} = \map {d_1} {x_m, x_n}$ for all $m$ and $n$.
Thus:
:$\paren {m > N} \land \paren {n > N} \implies \map {d_2} {y_m, y_n} < \epsilon$
Hence $\sequence {y_n}$ is a Cauchy sequence.
Category:Isometries (Metric Spaces)
Category:Cauchy Sequences
Isometric Isomorphism is Norm-Preserving

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be normed division rings.
Let $\phi: R \to S$ be a ring isomorphism.
Then $\phi: R \to S$ is an isometric isomorphism $\phi$ satisfies:
:$\forall x \in R: \norm {\map \phi x}_S = \norm x_R $
== Proof ==
Let $d_R$ and $d_S$ be the metric induced by the norms $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively.
=== Necessary Condition ===
Let $\phi: R \to S$ be an isometric isomorphism.
Then for $x \in R$:
The result follows.
=== Sufficient Condition ===
Let $\phi: R \to S$ satisfy:
:$\forall x \in R: \norm {\map \phi x}_S = \norm x_R$
Then for $x, y \in R$:
The result follows.
Category:Normed Division Rings
Category:Isometric Isomorphisms (Normed Division Rings)
Isometric Metric Spaces/Examples/Euclidean Plane is Isometric to Complex Plane

== Example of Isometric Metric Spaces ==
Let $\R^2$ be the real number plane with the Euclidean metric.
Let $\C$ denote the complex plane.
Let $f: \R^2 \to \C$ be the function defined as:
:$\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = x_1 + i x_2$
Then $f$ is an isometry from $\R^2$ to $\C$.
== Proof ==
Let $d: \C^2 \to \R$ denote the mapping defined as:
:$\forall \tuple {z_1, z_2} \in \C: \map d {z_1, z_2} = \cmod {z_1 - z_2}$
From Complex Plane is Metric Space, $\struct {\C, d}$ is a metric space such that $d$ is the same as the Euclidean metric on the real number plane.
Hence from Distance-Preserving Image Isometric to Domain for Metric Spaces, $f$ is an isometry.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Example $2.4.10$
Category:Examples of Isometries
Isometric Metric Spaces are Homeomorphic

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be isometric.
Then $M_1$ and $M_2$ are homeomorphic.
== Proof ==
By the definition of an isometry, there exists a bijection $f: A_1 \to A_2$ such that:
:$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$
From Isometry of Metric Spaces is Homeomorphism, $f$ is a homeomorphism from $M_1$ to $M_2$.
The result follows by definition of homeomorphic metric spaces.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Corollary $7.7$
Category:Isometries (Metric Spaces)
Category:Homeomorphisms (Metric Spaces)
Isometrically Isomorphic Non-Archimedean Division Rings

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be normed division rings.
Let $\phi:R \to S$ be an isometric isomorphism.
Then:
:$\norm {\,\cdot\,}_R$ is a non-archimedean norm $\norm {\,\cdot\,}_S$ is a non-archimedean norm.
== Proof ==
=== Necessary Condition ===
Let $\norm {\,\cdot\,}_R$ be a non-archimedean norm.
Then for all $x,y \in R$:
=== Sufficient Condition ===
Let $\norm {\, \cdot \,}_S$ be a non-archimedean norm.
By Inverse of Isometric Isomorphism, $\phi^{-1}: S \to R$ is an isometric isomorphism.
By the necessary condition, $\norm {\, \cdot \,}_R$ is non-archimedean.
Category:Normed Division Rings
Isometry Invariance of Riemannian Distance Function

== Theorem ==
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be connected Riemannian manifolds with or without boundaries.
Let $\phi : M \to \tilde M$ be an isometry.
Let $x, y \in M$ be points.
Let $\map {d_g} {x, y}$ be the Riemannian distance between $x$ and $y$.
Then:
:$\map {d_{\tilde g} } {\map \phi x, \map \phi y} = \map {d_g} {x, y}$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Riemannian Geometry
Isometry Invariance of Riemannian Length of Admissible Curve

== Theorem ==
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be Riemannian manifolds with or without boundary.
Let $\closedint a b$ be a closed real interval.
Let $\gamma : \closedint a b \to M$ with $t \stackrel \gamma \mapsto \map \gamma t$ be an admissible curve.
Let $\phi : M \to \tilde M$ be an isometry.
Let $\map {L_g} \gamma$ be the Riemannian length of $\gamma$ from $t = a$ to $t = b$.
Then:
:$\map {L_g} {\gamma} = \map {L_{\tilde g} } {\phi \circ \gamma}$
where $\circ$ denotes the composition of mappings.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Curves
Category:Riemannian Geometry
Isometry Preserves Congruence

== Theorem ==
Let $\Gamma = \R^n$ denote the real Euclidean space of $n$ dimensions, wher $n = 2$ or $n = 3$.
Let $\phi: \Gamma \to \Gamma$ be an isometry on $\Gamma$.
Let $\FF$ be a geometric figure in $\Gamma$.
The image of $\FF$ under $\phi$ is congruent to $\FF$, either directly or oppositely.
That is, $\phi$ preserves congruence.
== Proof ==
== Sources ==
* 
* 
Category:Isometries (Euclidean Geometry)
Category:Congruence (Geometry)
Isometry Preserves Sequence Convergence

== Theorem ==
Let $M_1 = \struct {S_1, d_1}$ and $M_2 = \struct {S_2, d_2}$ both be metric spaces or pseudometric spaces.
Let $\phi: S_1 \to S_2$ be an isometry.
Let $\sequence {x_n}$ be an infinite sequence in $S_1$.
Suppose that $\sequence {x_n}$ converges to a point $p \in S_1$.
Then $\sequence {\map \phi {x_n}}$ converges to $\map \phi p$.
== Proof 1 ==
== Proof 2 ==
Category:Isometry Preserves Sequence Convergence
Category:Isometries (Metric Spaces)
Category:Sequences
Isometry between Metric Spaces is Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
Then $\phi: M_1 \to M_2$ is a continuous mapping.
=== Corollary ===
== Proof ==
Let $a \in A_1$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \epsilon$.
Then:
So by definition $\phi$ is continuous at $a$.
As $a \in H$ is arbitrary, it follows that $d_H$ is continuous on $H$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Lemma $7.5$
Category:Isometry between Metric Spaces is Continuous
Category:Isometries (Metric Spaces)
Isometry between Metric Spaces is Continuous/Corollary

== Corollary to Isometry between Metric Spaces is Continuous ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
Then its inverse $\phi^{-1}: M_2 \to M_1$ is a continuous mapping.
== Proof ==
From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry.
The result follows from Isometry between Metric Spaces is Continuous.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Lemma $7.5$
Category:Isometry between Metric Spaces is Continuous
Isometry is Homeomorphism of Induced Topologies

== Theorem ==
Let $\struct {S_1, d_1}$ and $\struct {S_2, d_2}$ be metric spaces or pseudometric spaces.
Let $f: S_1 \to S_2$ be an isometry from $\struct {S_1, d_1}$ to $\struct {S_2, d_2}$.
Let $\tau_1$ and $\tau_2$ be the topologies induced on $S_1$ and $S_2$ by the metrics $d_1$ and $d_2$, respectively.
Then $f$ is a homeomorphism from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
== Proof ==
By the definition of an isometry, $f$ is bijective.
By Continuous Mapping is Continuous on Induced Topological Spaces, $f$ is a continuous mapping from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
By Inverse of Isometry of Metric Spaces is Isometry, $f^{-1}$ is an isometry.
By Continuous Mapping is Continuous on Induced Topological Spaces once more, $f^{-1}$ is a continuous mapping from $\struct {S_2, \tau_2}$ to $\struct {S_1, \tau_1}$.
Thus $f$ is a homeomorphism from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$.
Category:Isometries (Metric Spaces)
Category:Homeomorphisms (Topological Spaces)
Isometry of Metric Spaces is Equivalence Relation

== Theorem ==
Let $M_1$ and $M_2$ be metric spaces.
Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are isometric.
The relation $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $M$ be a topological space.
From Identity Mapping on Metric Space is Isometry, the identity mapping $I_M: M \to M$ is an isometry.
So $M \sim M$, and $\sim$ has been shown to be reflexive.
=== Symmetry ===
Let $M_1$ and $M_2$ be topological spaces such that $M_1 \sim M_2$.
By definition, there exists an isometry $f: M_1 \to M_2$.
From Inverse of Isometry of Metric Spaces is Isometry it follows that $f^{-1}: M_2 \to M_1$ is also an isometry.
So $M_2 \sim M_1$, and $\sim$ has been shown to be symmetric.
=== Transitivity ===
Let $M_1, M_2, M_3$ be metric spaces such that $M_1 \sim M_2$ and $M_2 \sim M_3$.
By definition, there exist isometries $f: M_1 \to M_2$ and $g: M_2 \to M_3$.
From Composite of Isometries is Isometry it follows that $g \circ f: M_1 \to M_3$ is also an isometry.
So $M_1 \sim M_3$, and $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces
Category:Isometries (Metric Spaces)
Category:Examples of Equivalence Relations
Isometry of Metric Spaces is Homeomorphism

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: M_1 \to M_2$ be an isometry.
Then $f$ is a homeomorphism from $M_1$ to $M_2$.
== Proof ==
By the definition of an isometry, $f$ is a bijection $f: A_1 \to A_2$ such that:
:$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$
By Isometry between Metric Spaces is Continuous, $f$ is a continuous mapping from $M_1$ to $M_2$.
By the corollary to Isometry between Metric Spaces is Continuous, $f^{-1}$ is a continuous mapping from $M_2$ to $M_1$.
Thus $f$ is a homeomorphism from $M_1$ to $M_2$.
Hence the result.
Category:Isometries (Metric Spaces)
Category:Homeomorphisms (Metric Spaces)
Isomorphic Ordinals are Equal/Proof 1

== Theorem ==
== Proof ==
Let $S \cong T$.
 that $S \ne T$.
Then from the corollary to Relation between Two Ordinals, either:
:$S$ is an initial segment of $T$
or:
:$T$ is an initial segment of $S$.
But as $S \cong T$, from Well-Ordered Class is not Isomorphic to Initial Segment, neither $S$ nor $T$ can be an initial segment of the other.
From this contradiction it follows that $S = T$.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.10$
Category:Isomorphic Ordinals are Equal
Isomorphic Ordinals are Equal/Proof 2

== Theorem ==
== Proof ==
From Well-Ordered Class is not Isomorphic to Initial Segment, neither $A$ nor $B$ can be an initial segment of the other.
By definition, every element of an ordinal is an initial segment of it.
Hence, neither $A$ nor $B$ can be an element of the other.
By Ordinal Membership is Trichotomy, it follows that $A = B$.
== Sources ==
* : $\S 7.38$
Category:Isomorphic Ordinals are Equal
Isomorphism (Abstract Algebra)/Examples/Addition under Doubling

== Example of Isomorphism ==
Let $\N$ denote the set of natural numbers.
Let $2 \N$ denote the set of even non-negative integers:
:$2 \N := \set {0, 2, 4, 6, \ldots}$
Let $\struct {\N, +}$ and $\struct {2 \N, +}$ be the algebraic structures formed from the above with addition.
Let $f: \N \to 2 \N$ be the mapping defined as:
:$\forall n \in \N: \map f n = 2 n$
Then $f$ is an isomorphism.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Isomorphisms (Abstract Algebra)
Isomorphism (Abstract Algebra)/Examples/Quadratic Integers over 3 with Numbers of Form 2^m 3^n

== Example of Isomorphism ==
Let $\Z \sqbrk {\sqrt 3}$ denote the set of quadratic integers over $3$:
:$\Z \sqbrk {\sqrt 3} = \set {a + b \sqrt 3: a, b \in \Z}$
Let $S$ be the set defined as:
:$S := \set {2^m 3^n: m, n \in \Z}$
Let $\struct {\Z \sqbrk {\sqrt 3}, +}$ and $\struct {S, \times}$ be the algebraic structures formed from the above with addition and multiplication respectively.
Then $\struct {\Z \sqbrk {\sqrt 3}, +}$ and $\struct {S, \times}$ are isomorphic.
== Proof ==
Let us define the mapping $\phi: \Z \sqbrk {\sqrt 3} \to S$ as:
:$\forall a + b \sqrt 3 \in \Z \sqbrk {\sqrt 3}: \map \phi {a + b \sqrt 3} = 2^a 3^b$
Let $a_1 + b_1 \sqrt 3$ and $a_2 + b_2 \sqrt 3$ be arbitrary elements of $\Z \sqbrk {\sqrt 3}$.
Then we have:
This demonstrates that $\phi$ is a homomorphism.
Now we have:
demonstrating that $\phi$ is injective.
Then:
demonstrating that $\phi$ is surjective.
The result follows by definition of isomorphic.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.7 \ \text {(c)}$
Category:Examples of Isomorphisms (Abstract Algebra)
Isomorphism (Category Theory) is Epic

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be an isomorphism.
Then $f: C \twoheadrightarrow D$ is epic.
== Proof ==
Since $f$ is an isomorphism, it is a fortiori a split epimorphism.
The result follows from Split Epimorphism is Epic.
== Sources ==
* : $\S 2.1$: Proposition $2.6$
Category:Category Theory
Isomorphism (Category Theory) is Monic

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be an isomorphism.
Then $f: C \rightarrowtail D$ is monic.
== Proof ==
Since $f$ is an isomorphism, it is a fortiori a split monomorphism.
The result follows from Split Monomorphism is Monic.
== Sources ==
* : $\S 2.1$: Proposition $2.6$
Category:Category Theory
Isomorphism Class of Total Orderings

== Theorem ==
Let $S$ be a finite set with $n$ elements.
There is exactly one isomorphism class containing the total orderings on $S$.
That is, every total ordering on $S$ is (order) isomorphic to every other total ordering.
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.5$
Category:Total Orderings
Isomorphism Classes for Order 2 Simple Graphs

== Theorem ==
There are $2$ equivalence classes for simple graphs of order $2$ under graph isomorphism:
:the edgeless graph of order $2$
and
:the complete graph of order $2$.
== Proof ==
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs
Category:Graph Isomorphisms
Isomorphism Classes for Order 4 Size 3 Simple Graphs

== Theorem ==
There are $3$ equivalence classes for simple graphs of order $4$ and size $3$ under isomorphism:
:
== Proof ==
The fact that the $3$ graphs given are not isomorphic follows from Vertex Condition for Isomorphic Graphs.
The vertices have degrees as follows:
:Graph $1$: $2, 2, 1, 1$
:Graph $2$: $3, 1, 1, 1$
:Graph $3$: $2, 2, 2, 0$
The fact that there are no more isomorphism classes of such graphs can be proved constructively.
Let the $4$ vertices be named $A, B, C$ and $D$.
Lemma: There must be intersections among the edges.
Proof: If there were no intersection at all, it requires at least $3 \times 2 = 6$ vertices.
, let $2$ of the edges be $AB$ and $AC$.
To place the last edge, there are $\dbinom 4 2 = 6$ potential choices:
:$AB$: this makes the graph not simple
:$AC$: this makes the graph not simple
:$AD$: this is isomorphic to graph $2$
:$BC$: this is isomorphic to graph $3$
:$BD$: this is isomorphic to graph $1$
:$CD$: this is isomorphic to graph $1$.
Hence, by Proof by Cases, these $3$ are the only isomorphism classes.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs
Category:Graph Isomorphisms
Isomorphism Preserves Associativity

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is associative $*$ is associative.
== Proof 1 ==
== Proof 2 ==
Category:Isomorphism Preserves Associativity
Category:Isomorphisms (Abstract Algebra)