nyuuzyou/wikis huggingface dataset:
Maximal Injective Mapping from Ordinals to a Set

== Theorem ==
Let $F$ be a mapping satisfying the following properties:
:The domain of $F$ is $\On$, the class of all ordinals
:For all ordinals $x$, $\map F x = \map G {F \restriction x}$.
:For all ordinals $x$, if $A \setminus \Img x \ne \O$, then $\map G {F \restriction x} \in A \setminus \Img x$ where $\Img x$ is the image of the subset $x$ under $F$.
:$A$ is a set.
Then there exists an ordinal $y$ satisfying the following properties:
:$\forall x \in y: A \setminus \Img x \ne \O$
:$\Img y = A$
:$F \restriction y$ is an injective mapping.
Note that the first third and fourth properties of $F$ are the most important. For any mapping $G$, a mapping $F$ can be constructed satisfying the first two properties using the First Principle of Transfinite Recursion.
== Proof ==
Set $B$ equal to the class of all ordinals $x$ such that $A \setminus \Img x \ne \O$.
Assume $B = \On$. 
Then:
By Condition for Injective Mapping on Ordinals, $A$ is a proper class.
This contradicts the fact that $A$ is a set.
Therefore $B \subsetneq \On$.
Because $B$ is bounded above, $\bigcup B \in \On$.
By Union of Ordinals is Least Upper Bound, the union of ordinals is the least upper bound of $B$.
Setting $\bigcup B = x$:
:$(1): \quad A \setminus \Img x = \O \land \forall y \in x: A \setminus \Img y \ne \O$
The first condition is satisfied.
In addition:
:$(2): \quad A \subseteq \Img x$
Take any $y \in \Img x$.
Then:
This means that:
:$\Img x \subseteq A$
Combining with $(2)$:
:$\Img x = A$
$F$ is a mapping, so $F \restriction x$ is a mapping. 
Take any $y, z \in x$ such that $y$ and $z$ are distinct.
, allow $y \in z$ (justified by Ordinal Membership Trichotomy).
From this, we may conclude that $F$ is injective.
== Also see ==
* Condition for Injective Mapping on Ordinals
* Transfinite Recursion Theorem
* Order Isomorphism between Ordinals and Proper Class
== Sources ==
* : $\S 7.47$
Category:Class of All Ordinals
Category:Class Mappings
Maximal Left and Right Ideal iff Quotient Ring is Division Ring

== Theorem ==
Let $R$ be a ring with unity.
Let $J$ be an ideal of $R$.
Then the following are equivalent:
:$(1): \quad J$ is a maximal left ideal
:$(2): \quad J$ is a maximal right ideal
:$(3): \quad$ the quotient ring $R / J$ is a division ring.
== Proof ==
=== Maximal Left Ideal implies Quotient Ring is Division Ring ===
=== Quotient Ring is Division Ring implies Maximal Left Ideal ===
=== Maximal Right Ideal implies Quotient Ring is Division Ring ===
The proof for Maximal Right Ideal implies Quotient Ring is Division Ring is similar to the proof for Maximal Left Ideal implies Quotient Ring is Division Ring with product orders reversed. It can be found here.
=== Quotient Ring is Division Ring implies Maximal Right Ideal ===
The proof for Quotient Ring is Division Ring implies Maximal Right Ideal is similar to the proof for Quotient Ring is Division Ring implies Maximal Left Ideal with product orders reversed. It can be found here.
Category:Quotient Rings
Category:Maximal Ideals of Rings
Maximal Left and Right Ideal iff Quotient Ring is Division Ring/Maximal Left Ideal implies Quotient Ring is Division Ring

== Theorem ==
Let $R$ be a ring with unity.
Let $J$ be an ideal of $R$.
If $J$ is a maximal left ideal then the quotient ring $R / J$ is a division ring.
== Proof ==
Since $J \subset R$, it follows from Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a ring with unity.
We now need to prove that every non-zero element of $\struct {R / J, +, \circ}$ has an inverse for $\circ$ in $R / J$.
By Left Inverse for All is Right Inverse it is sufficient to show that $\struct {R / J, +, \circ}$ has a left inverse for every non-zero element.
Let $x + J \in R / J$ not be the zero element of $R / J$.
That is, $x + J \ne J$ and $x \notin J$.
Take $K \subseteq R$ such that $K = \set {j + r \circ x: j \in J, r \in R}$.
That is, $K$ is the subset of $R$ which can be expressed as a sum of an element of $J$ and a left product in $R$ of $x$.
Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$.
So:
:$(1): \quad K \ne \O$
Now let $g, h \in K$.
That is:
: $g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$
Then:
So we have:
:$(2) \quad \forall g, h \in K, g-h \in K$
Now consider $g = \paren {j_1 + r_1 \circ x} \in K, y \in R$.
Then:
Thus:
:$(3) \quad y \circ g \in K$
So Test for Left Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is a left ideal of $R$.
Now:
As $x = 0_R + 1_R \circ x$, it follows that $x \in K$ also.
As $x \notin J$, it follows that $K$ is a left ideal such that $J \subset K \subseteq R$.
As $J$ is a maximal left ideal, it follows that $K = R$.
Thus $1_R \in K$ and thus:
:$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$
So:
:$1_R + \paren {-s \circ x} = j_0 \in J$
Hence:
:$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$
So in the ring with unity $\struct {R / J, +, \circ}$, the left inverse of $x + J$ is $s + J$.
The result follows.
== Sources ==
Category:Maximal Left and Right Ideal iff Quotient Ring is Division Ring
Maximal Left and Right Ideal iff Quotient Ring is Division Ring/Maximal Right Ideal implies Quotient Ring is Division Ring

== Theorem ==
Let $R$ be a ring with unity.
Let $J$ be an ideal of $R$.
If $J$ is a maximal right ideal then the quotient ring $R / J$ is a division ring.
== Proof ==
Since $J \subset R$, it follows from Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a ring with unity.
We now need to prove that every non-zero element of $\struct {R / J, +, \circ}$ has an inverse for $\circ$ in $R / J$.
By Right Inverse for All is Left Inverse it is sufficient to show that $\struct {R / J, +, \circ}$ has a right inverse for every non-zero element.
Let $x + J \in R / J$ not be the zero element of $R / J$.
That is, $x + J \ne J$ and $x \notin J$.
Take $K \subseteq R$ such that $K = \set {j + x \circ r: j \in J, r \in R}$.
That is, $K$ is the subset of $R$ which can be expressed as a sum of an element of $J$ and a right product in $R$ of $x$.
Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + x \circ 0_R = 0_R$.
So:
:$(1): \quad K \ne \O$
Now let $g, h \in K$.
That is:
:$g = j_1 + x \circ r_1, h = j_2 + x \circ r_2$
Then:
So we have:
:$(2) \quad \forall g, h \in K, g - h \in K$
Now consider $g = \paren {j_1 + x \circ r_1} \in K, y \in R$.
Then:
Thus:
: $(3) \quad g \circ y \in K$
So Test for Right Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is a right ideal of $R$.
Now:
Since $x = 0_R + x \circ 1_R$, then $x \in K$ too.
Since $x \notin J$, then $K$ is a right ideal such that $J \subset K \subseteq R$.
Since $J$ is a maximal right ideal, then $K = R$.
Thus $1_R \in K$ and thus:
:$\exists j_0 \in J, s \in R: 1_R = j_0 + x \circ s$
So:
:$1_R + \paren {- x \circ s} = j_0 \in J$
Hence:
:$1_R + J = x \circ s + J = \paren {x + J} \circ \paren {s + J}$
So in the ring with unity $\struct {R / J, +, \circ}$, the right inverse of $x + J$ is $s + J$.
The result follows.
Category:Maximal Left and Right Ideal iff Quotient Ring is Division Ring
Maximal Left and Right Ideal iff Quotient Ring is Division Ring/Quotient Ring is Division Ring implies Maximal Left Ideal

== Theorem ==
Let $R$ be a ring with unity.
Let $J$ be an ideal of $R$.
If the quotient ring $R / J$ is a division ring then $J$ is a maximal left ideal.
== Proof ==
Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subset R$.
Let $x \in K \setminus J$.
As $x \notin J$, then $x + J \ne J$, the zero in $R / J$.
As $R / J$ is a division ring, $x + J \in R / J$ has an inverse, say $s + J$.
That is:
:$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x} + J$
By Left Cosets are Equal iff Product with Inverse in Subgroup then:
:$1_R - s \circ x \in J \subsetneq K$
By the definition of a left ideal then:
:$x \in K$ and $s \in R \implies s \circ x \in K$
:$1_R - s \circ x \in K$ and $s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x } = 1_R \in K$
:$1_R \in K \implies \forall y \in R, y \circ 1_R = y \in K$
Hence $K = R$
The result follows.
== Sources ==
Category:Maximal Left and Right Ideal iff Quotient Ring is Division Ring
Maximal Left and Right Ideal iff Quotient Ring is Division Ring/Quotient Ring is Division Ring implies Maximal Right Ideal

== Theorem ==
Let $R$ be a ring with unity.
Let $J$ be an ideal of $R$.
If the quotient ring $R / J$ is a division ring then $J$ is a maximal right ideal.
== Proof ==
Let $K$ be a right ideal of $R$ such that $J \subsetneq K \subset R$.
Let $x \in K \setminus J$.
As $x \notin J$ then $x + J \ne J$, the zero in $R / J$.
As $R / J$ is a division ring then $x + J \in R / J$ has an inverse, say $s + J$.
That is:
:$1_R + J = \paren {x + J} \circ \paren {s + J} = \paren {x \circ s} + J$
By Left Cosets are Equal iff Product with Inverse in Subgroup then:
:$1_R - x \circ s \in J \subsetneq K$
By the definition of a right ideal:
:$x \in K$ and $s \in R \implies x \circ s \in K$
:$1_R - x \circ s \in K$ and $x \circ s \in K \implies \paren {1_R - x \circ s} + \paren {x \circ s} = 1_R \in K$
:$1_R \in K \implies \forall y \in R, 1_R \circ y = y \in K$
Hence $K = R$
The result follows.
Category:Maximal Left and Right Ideal iff Quotient Ring is Division Ring
Maximal Principles

== Theorem ==
The maximal principles are a collection of theorems which can be considered as forms of Zorn's Lemma.
=== Kuratowski's Lemma ===
=== Tukey's Lemma ===
=== Zorn's Lemma ===
=== Hausdorff's Maximal Principle ===
== Also known as ==
== Also see ==
== Historical Note ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Maximal Principles
Category:Axiom of Choice
Category:Zorn's Lemma
Maximal Principles/Also known as

== Maximal Principles: Also known as ==
The maximal principles are also known as:
* Kuratowski Principles (for )
* Hausdorff Principles (for )
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Maximal Principles
Maximal Principles are Equivalent

== Theorem ==
The Maximal Principles are equivalent:
* Kuratowski's Lemma
* Tukey's Lemma
* Zorn's Lemma 
* Hausdorff's Maximal Principle 
== Proof ==
We have:
* Kuratowski's Lemma implies Tukey's Lemma
* Tukey's Lemma implies Zorn's Lemma
* Zorn's Lemma implies Hausdorff's Maximal Principle
* Hausdorff's Maximal Principle implies Kuratowski's Lemma
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.11$
Category:Maximal Principles
Maximal Principles are Equivalent to Axiom of Choice

== Theorem ==
The Maximal Principles are equivalent to the Axiom of Choice.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Maximal Principles
Category:Axiom of Choice
Maximal Radical implies Primary Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $R$.
Let $\map \Rad {\mathfrak a}$ be the radical of $\mathfrak a$.
Suppose that $\map \Rad {\mathfrak a}$ is a maximal ideal.
Then $\mathfrak a$ is a primary ideal.
== Proof ==
Consider the quotient ring $R / \mathfrak a$.
By and :
:$\Nil {R / \mathfrak a} = \map \Rad {\mathfrak a} / \mathfrak a$
On the other hand, , $\map \Rad {\mathfrak a} / \mathfrak a$ is a maximal ideal of $R / \mathfrak a$.
Thus, in view of :
:$\Nil {R / \mathfrak a}$
is the only maximal ring in $R / \mathfrak a$.
In view of Proper Ideal of Ring is Contained in Maximal Ideal, follows.
Category:Radical of Ideals
Category:Primary Ideals
Category:Prime Ideals of Rings
Maximal Spectrum of Ring is Nonempty

== Theorem ==
Let $A$ be a non-trivial commutative ring with unity.
Then its maximal spectrum is non-empty:
:$\operatorname {Max} \Spec A \ne \O$
== Proof ==
This is a reformulation of Krull's Theorem.
== Also see ==
* Spectrum of Ring is Nonempty
Category:Commutative Algebra
Maximal Subalgebra in Normed Algebra is Closed

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.
Let $B$ be a subalgebra of $A$ that is maximal with respect to set inclusion.
Then $B$ is closed.
== Proof ==
From Closure of Subalgebra in Normed Algebra is Subalgebra, the closure $B^-$ of $B$ is a subalgebra with $B \subseteq B^-$.
Since $B$ is maximal with respect to set inclusion, we have that $B = B^-$.
From Set is Closed iff Equals Topological Closure, we conclude that $B$ is closed.
Category:Normed Algebras
Maximal implies Completely Irreducible

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $p \in S$ such that
:$\exists k \in S: p$ is maximal in $S \setminus k^\succeq$
Then $p$ is completely irreducible.
== Proof ==
By Join Succeeds Operands:
:$k \preceq p \vee k$ and $p \preceq p \vee k$
By definition of upper closure of element:
:$p \vee k \in k^\succeq$ and $p \vee k \in p^\succeq$
By definition of intersection:
:$p \vee k \in p^\succeq \cap k^\succeq$
By Maximal implies Difference equals Intersection:
:$p^\succeq \setminus \set p = p^\succeq \cap k^\succeq$
By Infimum of Intersection of Upper Closures equals Join Operands
:$\map \inf {p^\succeq \cap k^\succeq} = p \vee k$
By definition:
:$p^\succeq \setminus \set p$ admits a minimum.
Thus by definition:
:$p$ is completely irreducible.
== Sources ==
* 
* 
Category:Meet Irreducible Elements
Maximal implies Difference equals Intersection

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x, y \in S$ such that
:$x$ is maximal in $S \setminus y^\succeq$
Then $x^\succeq \setminus \left\{ {x}\right\} = x^\succeq \cap y^\succeq$
== Proof ==
By definition of maximal:
:$x \in S \setminus y^\succeq$
By definition of difference:
:$x \notin y^\succeq$
By definition of upper closure of element:
:$y \npreceq x$
We will prove that
:$x^\succeq \setminus \left\{ {x}\right\} \subseteq x^\succeq \cap y^\succeq$
Let $a \in x^\succeq \setminus \left\{ {x}\right\}$
By definition of difference:
:$a \in x^\succeq$ and $a \notin \left\{ {x}\right\}$
By definitions of upper closure of element and singleton:
:$x \preceq a$ and $a \ne x$
By definition of strictly precede:
:$x \prec a$
By definition of maximal:
:$a \notin S \setminus y^\succeq$
By definition of difference:
:$a \in y^\succeq$
Thus by definition of intersection:
:$a \in x^\succeq \cap y^\succeq$
We will prove that
:$x^\succeq \cap y^\succeq \subseteq x^\succeq \setminus \left\{ {x}\right\}$
Let $a \in x^\succeq \cap y^\succeq$
By definition of intersection:
:$a \in x^\succeq$ and $a \in y^\succeq$
By definition of upper closure of element:
:$y \preceq a$
Then
:$a \ne x$
By definition of singleton:
:$a \notin \left\{ {x}\right\}$
Thus by definition of difference:
:$a \in x^\succeq \setminus \left\{ {x}\right\}$
Thus by definition of set equality:
:$x^\succeq \setminus \left\{ {x}\right\} = x^\succeq \cap y^\succeq$
== Sources ==
* 
Category:Upper Closures
Maximization Problem for Independence Systems

== Problem ==
Let $\struct{S,\mathscr F}$ be an independence system.
Let $w:S \to \R_{\ge 0}$ be a weight function.
The maximization problem $\struct{\mathscr F, w}$ for $\struct{S,\mathscr F}$ is the computational problem to find a subset $A_0$ of $S$ with the properties:
:$(1)\quad A_0 \in \mathscr F$
:$(2)\quad \map {w^+} {A_0} = \max \set{\map {w^+} A : A \in \mathscr F}$
where $w^+$ is the extended weight function of $w$.
== Greedy Algorithm ==
== Also see ==
* Greedy Algorithm yields Maximal Set where it is shown that the output from the greedy algorithm for the Maximization Problem $\struct{\mathscr F, w}$ is a maximal set of $\mathscr F$.
* Greedy Algorithm may not yield Maximum Weight where it is shown that the maximal set of $\mathscr F$ constructed by the greedy algorithm is not guaranteed to be of maximum weight. 
* Greedy Algorithm guarantees Maximum Weight iff Matroid where it can be seen that the greedy algorithm only guarantees that the chosen maximal set has maximum weight iff $\struct{S, \mathscr F}$ is a matroid.
== Sources ==
* Chapter $19.$ $\S 1.$ The greedy algorithm
* Chapter $13$ Matroids $\S 13.1$ Independence Systems and Matroids
Category:Maximization Problem for Independence Systems
Maximization Problem for Independence Systems/Greedy Algorithm

== Algorithm ==
Let $\struct{S,\mathscr F}$ be an independence system.
Let $w : S \to \R_{\ge 0}$ be a weight function.
The purpose of this algorithm is to produce a maximal set of $\mathscr F$ with maximum weight for the maximization problem $\struct{\mathscr F, w}$.
The greedy algorithm for the maximization problem $\struct{\mathscr F, w}$ proceeds as follows:
:Step 1: Start with $A_0 = \O$.
:Step 2: Choose $x \in S \setminus A_0$ such that:
::a): $A_0 \cup \set x \in \mathscr F$
::b): $\map w x = \max \set{\map w y : y \in S \setminus A_0 \land A_0 \cup \set y \in \mathscr F}$.
:Step 3: If no such $x$ exists, stop. Otherwise, add $x$ to $A_0$.
:Step 4: Go to Step 2.
The above constitutes an algorithm, for the following reasons:
=== Finiteness ===
For each iteration through the algorithm, step 3 is executed, which increases the number of elements in $A_0$ by $1$ if it does not stop.
As $S$ is a finite set the algorithm will terminate after a finite number of iterations.
=== Definiteness ===
:Step 1: Trivially definite.
:Step 2: As the elements of $S$ can be arranged in order of weight, there is a definite element (or set of elements) with maximum weight. It is straightforward to select an element $x$ which extends $A_0$ to another set in $\mathscr F$.
:Step 3: Trivially definite.
:Step 4: Trivially definite.
=== Inputs ===
The input to this algorithm is the maximization problem $\struct{\mathscr F, w}$.
=== Outputs ===
The output to this algorithm is a maximal set of $\mathscr F$.
=== Effective ===
Each step of the algorithm is basic enough to be done exactly and in a finite length of time.
== Also see ==
* Greedy Algorithm yields Maximal Set where it is shown that the output from the greedy algorithm for the maximization problem $\struct{\mathscr F, w}$ is a maximal set of $\mathscr F$.
* Greedy Algorithm may not yield Maximum Weight where it is shown that the maximal set of $\mathscr F$ constructed by the greedy algorithm is not guaranteed to be of maximum weight. 
* Greedy Algorithm guarantees Maximum Weight iff Matroid where it can be seen that the greedy algorithm only guarantees that the chosen maximal set has maximum weight iff $\struct{S, \mathscr F}$ is a matroid.
* Kruskal's Algorithm
* Prim's Algorithm
== Sources ==
* Chapter $19.$ $\S 1.$ The greedy algorithm
Category:Maximization Problem for Independence Systems
Category:Greedy Algorithms
Maximum Abscissa for Loop of Folium of Descartes

== Theorem ==
Consider the folium of Descartes defined in parametric form as:
:$\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$
:
The point on the loop at which the $x$ value is at a maximum occurs when $t = \sqrt [3] {\dfrac 1 2}$, corresponding to the point $P$ defined as:
:$P = \tuple {2^{2/3} a, 2^{1/3} a}$
== Proof ==
We calculate the derivative of $x$ $t$:
Thus $x$ is stationary when:
From Behaviour of Parametric Equations for Folium of Descartes according to Parameter, it is clear from the geometry that $x$ is a local maximum for this value of $t$.
Then we have:
and:
Hence the result.
Category:Folium of Descartes
Maximum Area of Isosceles Triangle

== Theorem ==
Consider two line segments $A$ and $B$ of equal length $a$ which are required to be the legs of an isosceles triangle $T$.
Then the area of $T$ is greatest when the apex of $T$ is a right angle.
The area of $T$ in this situation is equal to $\dfrac {a^2} 2$.
== Proof ==
:
Let $\triangle OAB$ be the isosceles triangle $T$ formed by the legs $OA$ and $OB$.
Thus the apex of $T$ is at $O$.
Let $\theta$ be the angle $\angle AOB$.
We see that by keeping $OA$ fixed, $B$ can range over the semicircle $AOB$.
Thus $\theta$ can range from $0$ to $180 \degrees$, that is, $2$ right angles.
From Area of Triangle in Terms of Two Sides and Angle, the area $\AA$ of $T$ is:
:$\AA = \dfrac 1 2 a^2 \sin \theta$
This is a maximum when $\sin \theta = 1$, that is, when $\theta$ is a right angle.
The result follows.
== Sources ==
* : The Area Enclosed Against The Seashore: $33$
Category:Isosceles Triangles
Maximum Cardinality of Separable Hausdorff Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space which is separable.
Then $S$ can have a cardinality no greater than $2^{2^{\aleph_0} }$.
== Proof ==
Let $D$ be an everywhere dense subset of $S$ which is countable, as is guaranteed as $T$ is separable.
Consider the mapping $\Phi: S \to 2^{\powerset D}$ defined as:
:$\forall x \in S: \map {\map \Phi x} A = 1 \iff A = D \cap U_x$ for some neighborhood $U_x$ of $x$
It is seen that if $T$ is a Hausdorff space, then $\Phi$ is an injection.
It follows that:
:$\card S \le \card {2^{\powerset D} } = 2^{2^{\aleph_0} }$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Compactness Properties and the $T_i$ Axioms
Category:Hausdorff Spaces
Category:Separable Spaces
Maximum Degree of Vertex in Simple Graph

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph.
Let $\card V$ denote the order of $G$.
Then no vertex of $G$ has a degree higher than $\card V - 1$.
== Proof ==
By definition, a simple graph has no loops or multiple edges.
So a vertex can be incident to only as many edges that will join it to all the other vertices once each.
There are $\card V - 1$ other vertices.
Hence the result.
== Examples ==
Category:Vertices of Graphs
Category:Degrees of Vertices
Category:Maximum Degree of Vertex in Simple Graph
Maximum Degree of Vertex in Simple Graph/Examples/Impossible Order 5 Graph

== Examples of Degrees of Vertices ==
There exists no simple graph whose vertices have degrees $2, 3, 4, 4, 5$.
== Proof ==
 there exists a simple graph $G = \struct {V, E}$ with $5$ vertices whose degrees are $2, 3, 4, 4, 5$.
One of the vertices of $G$ has degree $5$.
This contradicts Maximum Degree of Vertex in Simple Graph, which states that no vertex of $G$ can have a degree higher than $4$.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $3$
Category:Degrees of Vertices
Maximum Function in terms of Absolute Value

== Theorem ==
Let $x$ and $y$ be real numbers.
Then:
:$\ds \max \set {x, y} = \frac 1 2 \paren {\paren {x + y} + \size {x - y} }$
== Proof ==
We aim to show that: 
:$\ds \frac 1 2 \paren {\paren {x + y} + \size {x - y} } = \begin{cases}y & x \le y \\ x & x > y\end{cases}$
Let $x \le y$. 
Then:
:$x - y \le 0$
Then, from the definition of the absolute value, we have: 
:$\size {x - y} = y - x$ 
So: 
Let $x > y$.
Then: 
:$x - y > 0$
Then, from the definition of the absolute value, we have: 
:$\size {x - y} = x - y$
So:
So:
:$\ds \frac 1 2 \paren {\paren {x + y} + \size {x - y} } = \begin{cases}y & x \le y \\ x & x > y\end{cases}$
as required.
Category:Absolute Value Function
Maximum Function is Primitive Recursive

== Theorem ==
The maximum function $\max: \N^2 \to \N$, defined as:
:$\max \left({n, m}\right) = \begin{cases}
m: & n \le m \\
n: & m \le n
\end{cases}$
is primitive recursive‎.
== Proof ==
We see that:
:$\max \left({n, m}\right) = \left({n \ \dot - \ m}\right) + m$
because:
:$(1):\quad n > m \implies \left({n \ \dot - \ m}\right) + m = n - m + m = n$
:$(2):\quad n < m \implies \left({n \ \dot - \ m}\right) + m = 0 + m = m$
:$(3):\quad n = m \implies \left({n \ \dot - \ m}\right) + m = 0 + m = m = n$
Hence we see that $\max$ is obtained by substitution from the primitive recursive function $n \ \dot - \ m$.
Hence the result.
Category:Primitive Recursive Functions
Maximum Likelihood Estimator for Expectation of Normal Distribution is Unbiased

== Theorem ==
Let $X$ be a continuous random variable with a normal distribution.
Let $\mu$ be the expectation of $X$.
The maximum likelihood estimator for $\mu$ $X$ is unbiased, and is the sample mean $\overline x$ of $X$.
== Proof ==
== Sources ==
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Category:Maximum Likelihood Estimators