nyuuzyou/wikis huggingface dataset:
Matroid with No Circuits Has Single Base

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid with no circuits.
Then:
:$S$ is the only base on $M$.
== Proof ==
From Dependent Subset Contains a Circuit:
:$M$ has no dependent subsets
By definition of dependent subsets:
:Every subset of $S$ is independent
In particular:
:$S \in \mathscr I$
By definition of independent subsets:
:$X \in \mathscr I \implies X \subseteq S$
Hence $S$ is a base on $M$ by definition.
Let $X$ be a base on $M$.
Then $X \subseteq S$.
By definition of a base on $M$:
:$X$ is a maximal independent subset of $M$
As $S \in \mathscr I$, by definition of a maximal set:
:$X = S$
Hence:
:$S$ is the only base on $M$.
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits
Category:Matroid Circuits
Max Operation Equals an Operand

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Then:
:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
where:
:$\max$ denotes the max operation.
== Proof ==
We will prove the result by induction on the number of operands $n$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\exists i \in \closedint 1 1 : x_i = \max \set {x_1}$
By definition of the max operation:
:$\max \set {x_1} = x_1$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$
from which it is to be shown that:
:$\exists i \in \closedint 1 {k + 1} : x_i = \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
=== Induction Step ===
This is the induction step.
By the definition of Max Operation:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }$
By the induction hypothesis:
:$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$
So:
:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {x_i, x_{k + 1} }$
As $\struct {S, \preceq}$ is a totally ordered set, all elements of $S$ are comparable by $\preceq$.
Therefore there are two cases to consider:
==== Case 1: $x_{k + 1} \preceq x_i$ ====
By definition of the max operation:
:$\max \set {x_i, x_{k + 1} } = x_i$
==== Case 2: $x_i \preceq x_{k + 1}$ ====
By definition of the max operation:
:$\max \set {x_i, x_{k + 1} } = x_{k + 1}$
In either case, the result holds.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
Category:Max Operation
Max Operation Preserves Total Ordering

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $a, b, c, d \in S$:
:$a \preceq b, c \preceq d$
Then:
:$\max \set{a, c} \preceq \max \set{b, d}$
where $\max$ denotes the max operation on $\struct {S, \preceq}$.
== Proof ==
From Max Operation Equals an Operand, either:
:$\max \set{a, c} = a$
or
:$\max \set{a, c} = c$
, suppose:
:$\max \set{a, c} = a$
We have:
By transitivity of an ordering:
:$\max \set{a, c} \preceq \max \set{b, d}$
Category:Max Operation
Max Operation Representation on Real Numbers

== Theorem ==
Let $x, y \in \R$.
Then:
:$\max \set{x, y} = \dfrac 1 2 \paren {x + y + \size {x - y} }$
where $\max$ denotes the max operation.
== Proof ==
From the Trichotomy Law for Real Numbers exactly one of the following holds:
:$x < y$ and so $\max \set {x, y} = y$
:$x = y$ and so $\max \set {x, y} = x = y$
:$y < x$ and so $\max \set {x, y} = x$
By the definition of the absolute value function for each case respectively we have:
:$\size {x - y} = y - x$
:$\size {x - y} = 0$
:$\size {x - y} = x - y$
Thus the equation holds by $+$ being commutative and associative as for each case:
:$\dfrac 1 2 \paren {x + y + y - x} = y$
:$\dfrac 1 2 \paren {x + y + 0} = x = y$
:$\dfrac 1 2 \paren {x + y + x - y} = x$
== Also see ==
* Min Operation Representation on Real Numbers
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 18 \ \text {(a)}$
Category:Real Analysis
Category:Max Operation
Max Operation Yields Supremum of Parameters

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set. 
Let $x, y \in S$.
Then:
:$\max \set {x, y} = \sup \set {x, y}$
where:
:$\max$ denotes the max operation
:$\sup$ denotes the supremum.
=== General Case ===
== Proof ==
As $\struct {S, \preceq}$ be a totally ordered set, all elements of $S$ are $\preceq$-comparable.
Therefore there are two cases to consider:
=== Case 1: $x \preceq y$ ===
In this case:
:$\max \set {x, y} = y = \sup \set {x, y}$
=== Case 2: $y \preceq x$ ===
In this case:
:$\max \set {x, y} = x = \sup \set {x, y}$
In either case, the result holds.
== Warning ==
Note that it is considered abuse of notation to write
:$\max = \sup$
This is because:
:$\max: S \times S \to S$
while:
:$\sup: \powerset S \to S$
where $\powerset S$ is the power set of $S$.
== Also see ==
* Min Operation Yields Infimum of Parameters
Category:Max Operation
Max Operation is Associative

== Theorem ==
The Max operation is associative:
: $\map \max {\map \max {x, y}, z} = \map \max {x, \max \paren{y, z}}$
Thus we are justified in writing $\map \max {x, y, z}$.
== Proof ==
To simplify our notation:
: Let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$
There are the following cases to consider:
:$(1): \quad x \le y \le z$
:$(2): \quad x \le z \le y$
:$(3): \quad y \le x \le z$
:$(4): \quad y \le z \le x$
:$(5): \quad z \le x \le y$
:$(6): \quad z \le y \le x$
Taking each one in turn:
:$(1): \quad$ Let $x \le y \le z$. Then:
:$(2): \quad$ Let $x \le z \le y$. Then:
:$(3): \quad$ Let $y \le x \le z$. Then:
:$(4): \quad$ Let $y \le z \le x$. Then:
:$(5): \quad$ Let $z \le x \le y$. Then:
:$(6): \quad$ Let $z \le y \le x$. Then:
Thus in all cases it can be seen that the result holds.
== Also see ==
* Min Operation is Associative
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.12 \ \text{(a)}$
Category:Max Operation
Category:Examples of Associative Operations
Max Operation is Commutative

== Theorem ==
The Max operation is commutative:
:$\map \max {x, y} = \map \max {y, x}$
== Proof ==
To simplify our notation:
:Let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$
There are three cases to consider:
:$(1): \quad x \le y$
:$(2): \quad y \le x$
:$(3): \quad x = y$
$(1): \quad$ Let $x \le y$.
Then:
$(2): \quad$ Let $y \le x$.
Then:
$(3): \quad$ Let $x = y$.
Then:
Thus $\overline \wedge$, that is $\max$, has been shown to be commutative in all cases.
== Also see ==
* Min Operation is Commutative
Category:Max Operation
Category:Examples of Commutative Operations
Max Operation is Idempotent

== Theorem ==
The max operation operation is idempotent:
:$\map \max {x, x} = x$
== Proof ==
Follows immediately from the definition of max operation:
:$\map \max {a, b} = \begin {cases} b & : a \le b \\ a & : b \le a \end {cases}$
Setting $x = a = b$ returns the result.
== Also see ==
* Min Operation is Idempotent
Category:Max Operation
Category:Examples of Idempotence‎
Max Operation on Continuous Real Functions is Continuous

== Theorem ==
Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.
Let $f$ and $g$ be continuous at a point $a \in \R$.
Let $h: \R \to \R$ be the real function defined as:
:$\map h x := \map \max {\map f x, \map g x}$ 
Then $h$ is continuous at $a$.
== Proof ==
From Max Operation Representation on Real Numbers
:$\max \set {x, y} = \dfrac 1 2 \paren {x + y + \size {x - y} }$
Hence:
:$\max \set {\map f x, \map g x} = \dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$
From Difference Rule for Continuous Real Functions:
:$\map f x - \map g x$ is continuous at $a$.
From Absolute Value of Continuous Real Function is Continuous:
:$\size {\map f x - \map g x}$ is continuous at $a$.
From Sum Rule for Continuous Real Functions:
:$\map f x + \map g x$ is continuous at $a$
and hence:
:$\map f x + \map g x + \size {\map f x - \map g x}$ is continuous at $a$
From Multiple Rule for Continuous Real Functions:
:$\dfrac 1 2 \paren {\map f x + \map g x + \size {\map f x - \map g x} }$ is continuous at $a$.
== Also see ==
* Min Operation on Continuous Real Functions is Continuous
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 18 \ \text {(b)}$
Category:Continuous Real Functions
Category:Max Operation
Max Operation on Natural Numbers forms Monoid

== Theorem ==
Let $\struct {\N, \max}$ denote the algebraic structure formed from the natural numbers $\N$ and the max operation.
Then $\struct {\N, \max}$ is a monoid.
Its identity element is zero.
== Proof ==
By the Well-Ordering Principle, $\N$ is a well-ordered set.
The result follows from Max Operation on Woset is Monoid.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.4$
Category:Natural Numbers
Category:Examples of Monoids
Category:Max Operation
Max Operation on Toset forms Semigroup

== Theorem ==
Let $\struct{S, \preceq}$ be a totally ordered set.
Let $\map \max {x, y}$ denote the max operation on $x, y \in S$.
Then $\struct{S, \max}$ is a semigroup.
== Proof ==
By the definition of the max operation, either:
:$\map \max {x, y}= x$
or
:$\map \max {x, y}= y$
So $\max$ is closed on $S$.
From Max Operation is Associative:
:$\forall x, y, z \in S: \map \max {x, \map \max {y, z}} = \map \max {\map \max {x, y}, z}$
Hence the result, by definition of semigroup.
== Also see ==
* Min Operation on Toset forms Semigroup
Category:Max Operation
Max Operation on Woset is Monoid

== Theorem ==
Let $\struct {S, \preceq}$ be a well-ordered set.
Let $\max \set {x, y}$ denote the max operation on $x, y \in S$.
Then $\struct {S, \max}$ is a monoid.
Its identity element is the smallest element of $S$.
== Proof ==
From Well-Ordering is Total Ordering, we have that $\struct {S, \preceq}$ is a totally ordered set.
Thus, by Max Operation on Toset forms Semigroup, $\struct {S, \max}$ is a semigroup.
By definition, a well-ordered set has a smallest element.
So:
:$\exists m \in S: \forall x \in S: m \preceq x$
where $m$ is that smallest element.
It follows by definition of the max operation that:
:$\forall x \in S: \max \set {m, x} = x$
Thus $m$ is the identity element of $\struct {S, \max}$.
Hence the result, by definition of monoid.
Category:Max Operation
Category:Well-Orderings
Category:Examples of Monoids
Max Semigroup is Commutative

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set. 
Then the semigroup $\struct {S, \max}$ is commutative.
== Proof ==
Let $x, y \in S$. 
From Max Operation is Commutative:
:$\map \max {x, y} = \map \max {y, x}$ 
Hence the result, by definition of commutative semigroup.
== Also see ==
* Min Semigroup is Commutative
	
Category:Max Operation
Category:Examples of Commutative Semigroups
Max Semigroup is Idempotent

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Then the semigroup $\struct {S, \max}$ is an idempotent semigroup.
== Proof ==
The fact that $\struct {S, \max}$ is a semigroup is demonstrated in Max Operation on Toset forms Semigroup.
Then the max operation is idempotent:
:$\forall x \in S: \max \set {x, x} = x$
The result follows by the definition of idempotent semigroup.
== Also see ==
* Min Semigroup is Idempotent
Category:Max Operation
Category:Examples of Idempotent Semigroups‎
Max Semigroup on Toset forms Semilattice

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set. 
Then the max semigroup $\struct {S, \max}$ is a semilattice.
== Proof ==
The Max Semigroup is Commutative and idempotent. 
Hence the result, by definition of a semilattice.
== Also see ==
* Min Semigroup on Toset forms Semilattice
	
Category:Max Operation
Category:Semilattices
Max and Min Operations are Distributive over Each Other

== Theorem ==
The Max and Min operations are distributive over each other:
:$\map \max {x, \map \min {y, z} } = \map \min {\map \max {x, y}, \map \max {x, z} }$
:$\map \max {\map \min {x, y}, z} = \map \min {\map \max {x, z}, \map \max {y, z} }$
:$\map \min {x, \map \max {y, z} } = \map \max {\map \min {x, y}, \map \min {x, z} }$
:$\map \min {\map \max {x, y}, z} = \map \max {\map \min {x, z}, \map \min {y, z} }$
== Proof ==
To simplify our notation, let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$, and let $\map \min {x, y}$ be (temporarily) denoted $x \underline \vee y$.
Note that, once we have proved:
:$x \overline \wedge \paren {y \underline \vee z} = \paren {x \overline \wedge y} \underline \vee \paren {x \overline \wedge z}$
:$x \underline \vee \paren {y \overline \wedge z} = \paren {x \underline \vee y} \overline \wedge \paren {x \underline \vee z}$
then the other results follow immediately from Max Operation is Commutative and Min Operation is Commutative.
There are the following cases to consider:
:$(1): \quad x \le y \le z$
:$(2): \quad x \le z \le y$
:$(3): \quad y \le x \le z$
:$(4): \quad y \le z \le x$
:$(5): \quad z \le x \le y$
:$(6): \quad z \le y \le x$
$(1): \quad $ Let $x \le y \le z$.
Then:
$(2): \quad $ Let $x \le z \le y$.
Then:
$(3): \quad $ Let $y \le x \le z$.
Then:
$(4): \quad $ Let $y \le z \le x$.
Then:
$(5): \quad $ Let $z \le x \le y$.
Then:
$(6): \quad $ Let $z \le y \le x$.
Then:
Thus in all cases it can be seen that the result holds.
Category:Max and Min Operations
Category:Examples of Distributive Operations
Max and Min Operations on Real Numbers are Isomorphic

== Theorem ==
Let $\R$ denote the set of real numbers.
Let $\vee$ and $\wedge$ denote the max operation and min operation respectively.
Let $\struct {\R, \vee}$ and $\struct {\R, \wedge}$ denote the algebraic structures formed from the above.
Then $\struct {\R, \vee}$ and $\struct {\R, \wedge}$ are isomorphic.
== Proof ==
First we note that from:
:Min Operation on Toset forms Semigroup
and:
:Max Operation on Toset forms Semigroup
both $\struct {\R, \vee}$ and $\struct {\R, \wedge}$ are semigroups.
Let $\phi: \R \to \R$ defined as:
:$\forall x \in \R: \map \phi x = -x$
We have that:
:$-x = -y \iff x = y$
which demonstrates that $\phi$ is a bijection.
Then we have:
which demonstrates that $\phi$ is a (semigroup) homomorphism.
The result follows by definition of (semigroup) isomorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.6 \ \text {(a)}$
Category:Max and Min Operations
Category:Examples of Semigroup Isomorphisms
Max and Min of Function on Closed Real Interval

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ reaches a maximum and a minimum on $\closedint a b$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Real Intervals
Category:Max and Min of Function on Closed Real Interval
Max and Min of Function on Closed Real Interval/Proof 1

== Theorem ==
== Proof ==
From Image of Closed Real Interval is Bounded, we have that $f$ is bounded on $\closedint a b$.
Let $d$ be the supremum of $f$ on $\closedint a b$.
Consider a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to d$ as $n \to \infty$.
From the corollary to Limit of Sequence to Zero Distance Point, this can always be found.
Now $\closedint a b$ is a closed interval
So from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which converges to some $\xi \in \closedint a b$.
Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.
So $\map f \xi = d$ and thus the supremum $d$ is indeed a maximum.
A similar argument shows that the infimum is a minimum.
== Sources ==
* : $\S 9.12$
Category:Max and Min of Function on Closed Real Interval
Max is Half of Sum Plus Absolute Difference

== Theorem ==
For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
:$\max \set {a, b} = \frac 1 2 \paren {a + b + \size {a - b} }$
== Proof ==
== Also see ==
* Min is Half of Sum Less Absolute Difference
Category:Max Operation
Max of Subfamily of Operands Less or Equal to Max

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
Let $\set{k_1, k_2, \dotsc, k_m} \subseteq \set{1, 2, \dotsc, n}$
Then:
:$\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} \preceq \max \set {x_1, x_2, \dotsc, x_n}$
where:
:$\max$ denotes the max operation
== Proof ==
From Max Operation Yields Supremum of Operands:
:$\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} = \sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} }$
and
:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
Since $\set {k_1, k_2, \dotsc, k_m} \subseteq \set {1, 2, \dotsc, n}$ then:
:$\set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \subseteq \set {x_1, x_2, \dotsc, x_n}$
From Supremum of Subset:
:$\sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \preceq \sup \set {x_1, x_2, \dotsc, x_n}$
Category:Max Operation
Maximal Algebraic Extension is Subfield

== Theorem ==
Let $L / K$ be a field extension.
Let $K^a$ be the maximal algebraic extension of $K$ contained in $L$.
Then $K^a$ is a subfield of $L$.
== Proof ==
Let $\alpha, \beta \in K^a$.
By Field Adjoined Algebraic Elements is Algebraic, $\map K {\alpha, \beta} / K$ is algebraic.
By definition, $\map K {\alpha, \beta}$ is a field.
Therefore $\alpha \beta$, $\alpha^{-1}$ and $\alpha - \beta$ all lie in $\map K {\alpha, \beta}$.
Hence all are algebraic over $K$.
Also:
$K \subseteq K^a$
so:
:$K^a \ne \O$
By the Subfield Test:
:$K^a \le L$ is a subfield of $L$.
Category:Field Extensions
Maximal Annihilator of Module is Associated Prime

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $M$ be a module over $A$.
Let $\mathbf p$ be a maximal element of the set:
:$\set { \map {\operatorname {Ann}_A} x : x \in M , x \ne 0 }$
with respect to the subset relation.
Then $\mathfrak p$ is an associated prime of $M$.
== Proof ==
Category:Commutative Algebra
Maximal Commutative Subalgebra of Unital Algebra is Unital

== Theorem ==
Let $K$ be a field.
Let $A$ be a unital algebra over $K$.
Let $B$ be a commutative subalgebra of $A$ that is maximal with respect to set inclusion.
Then $B$ is unital.
== Proof ==
We show that $B + K {\mathbf 1}_A$ is a subalgebra of $A$.
First we show that $B + K {\mathbf 1}_A$ is a vector subspace of $A$.
Let $x + t {\mathbf 1}_A, y + s {\mathbf 1}_A \in B + K {\mathbf 1}_A$ for $x, y \in B$ and $s, t \in K$.
Let $\lambda \in K$.
We have:
:$\paren {x + t {\mathbf 1}_A} + \lambda \paren {y + s {\mathbf 1}_A} = {x + \lambda y} + \paren {t + \lambda s} {\mathbf 1}_A$
Since $B$ is a subalgebra of $A$, it is in particular a vector subspace of $A$.
Then $x + \lambda y \in B$. 
So we have that $\paren {x + t {\mathbf 1}_A} + \lambda \paren {y + s {\mathbf 1}_A} \in B + K {\mathbf 1}_A$.
Hence $B + K {\mathbf 1}_A$ is a vector subspace of $A$.
We now show that $B + K {\mathbf 1}_A$ is closed under multiplication.
Let $x + t {\mathbf 1}_A, y + s {\mathbf 1}_A \in B + K {\mathbf 1}_A$ for $x, y \in B$ and $s, t \in K$.
We have:
:$\paren {x + t {\mathbf 1}_A} \paren {y + s {\mathbf 1}_A} = \paren {x y + s x + t y} + t s {\mathbf 1}_A$
Since $B$ is a subalgebra of $A$, we have $x y \in B$.
Since $B$ is a vector subspace of $A$, we have that $x y + s x + t y \in B$.
So, we have:
:$\paren {x + t {\mathbf 1}_A} \paren {y + s {\mathbf 1}_A} \in B + K {\mathbf 1}_A$
We conclude that $B + K {\mathbf 1}_A$ is a subalgebra of $A$.
Now note that we have:
:$B = B + 0_K {\mathbf 1}_A \subseteq B + K {\mathbf 1}_A$
Since $B$ is maximal with respect to set inclusion, we have:
:$B = B + K {\mathbf 1}_A$
Since ${\mathbf 0}_A \in B$, we have that ${\mathbf 1}_A \in B$.
Category:Algebras over Fields
Category:Unital Algebras
Category:Commutative Algebras
Category:Subalgebras
Maximal Element/Examples/Finite Subsets of Natural Numbers

== Examples of Minimal Elements ==
Let $\FF$ denote the set of finite subsets of the natural numbers $\N$.
Consider the ordered set $\struct {\FF, \subseteq}$.
There are no maximal elements of $\struct {\FF, \subseteq}$.
== Proof ==
 $A \in \FF$ is a maximal element of $\struct {\FF, \subseteq}$.
Then $A$ consists of a finite number of natural numbers.
Let $M = \map \max A$ denote the maximum of the elements of $A$.
Now consider the set:
:$A' = A \cup \set {M + 1}$
We have that $A'$ is a finite subset of $\N$.
Hence $A' \in \FF$.
But also from Set is Subset of Union:
:$A \subseteq A'$
while because $M + 1 \notin A$:
:$A' \ne A$
Hence $A$ cannot be a maximal element of $\struct {\FF, \subseteq}$.
The result follows from Proof by Contradiction.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(a)}$
Category:Examples of Maximal Elements
Maximal Element/Examples/Finite Subsets of Natural Numbers less Empty Set

== Examples of Minimal Elements ==
Let $\FF$ denote the set of finite subsets of the natural numbers $\N$.
Let $\GG$ denote the set $\FF \setminus \O$, that is, $\FF$ with the empty set excluded.
Consider the ordered set $\struct {\GG, \subseteq}$.
There are no maximal elements of $\struct {\GG, \subseteq}$.
== Proof ==
 $A \in \GG$ is a maximal element of $\struct {\GG, \subseteq}$.
Then $A$ consists of a finite number of natural numbers.
Let $M = \map \max A$ denote the maximum of the elements of $A$.
Now consider the set:
:$A' = A \cup \set {M + 1}$
We have that $A'$ is a finite subset of $\N$.
Hence $A' \in \GG$.
But also from Set is Subset of Union:
:$A \subseteq A'$
while because $M + 1 \notin A$:
:$A' \ne A$
Hence $A$ cannot be a maximal element of $\struct {\GG, \subseteq}$.
The result follows from Proof by Contradiction.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(a)}$
Category:Examples of Maximal Elements
Maximal Element in Toset is Unique and Greatest

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $M$ be a maximal element of $\struct {S, \preceq}$.
Then:
:$(1): \quad M$ is the greatest element of $\struct {S, \preceq}$.
:$(2): \quad M$ is the only maximal element of $\struct {S, \preceq}$.
== Proof ==
By definition of maximal element:
:$\forall y \in S: M \preceq y \implies M = y$
As $\struct {S, \preceq}$ is a totally ordered set, by definition $\preceq$ is a connected.
That is:
:$\forall x, y \in S: y \preceq x \lor x \preceq y$
It follows that:
:$\forall y \in S: M = y \lor y \preceq M$
But as $M = y \implies y \preceq M$ by definition of $\preceq$, it follows that:
:$\forall y \in S: y \preceq M$
which is precisely the definition of greatest element.
Hence $(1)$ holds.
Suppose $M_1$ and $M_2$ are both maximal elements of $S$.
By $(1)$ it follows that both are greatest elements.
It follows from Greatest Element is Unique that $M_1 = M_2$.
That is, $(2)$ holds.
== Also see ==
* Minimal Element in Toset is Unique and Smallest
* Maximal Element need not be Greatest Element
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(b)}$
Category:Total Orderings
Category:Maximal Elements
Category:Greatest Elements
Maximal Element is Dual to Minimal Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$, and $a \in T$.
The following are dual statements:
:$a$ is a maximal element of $T$
:$a$ is a minimal element of $T$
== Proof ==
By definition, $a$ is a maximal element of $T$ :
:$\forall t \in T: a \preceq t$ implies $a = t$
The dual of this statement is:
:$\forall t \in T: t \preceq a$ implies $a = t$
by Dual Pairs (Order Theory).
By definition, this means $a$ is a minimal element of $T$.
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
Category:Maximal Elements
Category:Minimal Elements
Category:Dual Pairs (Order Theory)