nyuuzyou/wikis huggingface dataset:
Zero of Subfield is Zero of Field

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
The zero of $\struct {K, +, \times}$ is also $0$.
== Proof 1 ==
== Proof 2 ==
Category:Subfields
Category:Zero of Subfield is Zero of Field
Zero of Subfield is Zero of Field/Proof 1

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
== Proof ==
By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both rings.
Thus $\struct {K, +, \times}$ is a subring of $\struct {F, +, \times}$
The result follows from Zero of Subring is Zero of Ring.
Category:Zero of Subfield is Zero of Field
Zero of Subfield is Zero of Field/Proof 2

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
== Proof ==
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K, +}$ and $\struct {F, +}$ are groups such that $K \subseteq F$.
So, by definition, $\struct {K, +}$ is a subgroup of $\struct {F, +}$.
By Identity of Subgroup, the identity of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.
Category:Zero of Subfield is Zero of Field
Zero of Subring is Zero of Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring whose zero is $0$.
Let $\struct {S, +, \times}$ be a subring of $\struct {R, +, \times}$.
The zero of $\struct {S, +, \times}$ is also $0$.
== Proof ==
By definition, $\struct {S, +, \times}$ is a subset of $R$ which is a ring.
By definition of ring, $\struct {S, +}$ and $\struct {R, +}$ are groups such that $S \subseteq R$.
So, by definition, $\struct {S, +}$ is a subgroup of $\struct {R, +}$.
By Identity of Subgroup, the identity of $\struct {S, +}$, which is $0$, is also the identity of $\struct {R, +}$.
Category:Subrings
Zero to the Power of Zero

== Examples ==
In light of some mathematicians' (or, indeed, muggles') reticence to define the zeroth power of zero as $1$, the following are examples of reasons why defining $0^0 = 1$ is a good idea.
=== Empty Product ===
=== Binomial Theorem ===
=== Cardinality of Mappings ===
=== Exponential of Zero ===
=== Derivatives ===
=== As a Limit ===
== Historical Note ==
== Sources ==
* : $0$ Zero
* : $0$ Zero
Category:Zero to the Power of Zero
Category:Powers
Category:Numbers
Category:Zero
Zero to the Power of Zero/As a Limit

== Example of Zero to the Power of Zero ==
Consider the real function:
:$y = x^x$
This function is well defined for $x > 0$.
It is not obvious whether or not the right hand limit:
:$\ds \lim_{x \mathop \to 0^+} y$
exists.
If it does, it would be nice if:
:$\ds \lim_{x \mathop \to 0^+} x^x = 0^0$
Indeed, by Limit of x to the x, we have:
:$\ds \lim_{x \mathop \to 0^+} x^x = 1$
We see that defining $0^0 = 1$ allows $x^x$ to be right-continuous at $x = 0$.
Category:Zero to the Power of Zero
Zero to the Power of Zero/Binomial Theorem

== Example of Zero to the Power of Zero ==
Consider the real polynomial function:
:$y = \paren {x + c}^n$
for $n \in \N, c \in \R$.
By Binomial Theorem for Integral Index, $y$ contains a term of the form:
:$\dbinom n n x^{n - n} c^n$
If we did not define $0^0 = 1$, $y$ would have a discontinuity at $x = 0$.
This would contradict Real Polynomial Function is Continuous.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text{F} \ (14)$
Category:Zero to the Power of Zero
Zero to the Power of Zero/Derivatives

== Example of Zero to the Power of Zero ==
Consider the identity mapping:
:$\map {I_\GF} x = x$
where $\GF \in \set {\R, \C}$.
From Derivative of Identity Function:
:$\dfrac {\d I_\GF} {\d x} = 1$
But $\map {I_\GF} x = x^1$ is also an order one polynomial.
By Power Rule for Derivatives:
:$\dfrac {\d I_\GF} {\d x} = 1 x^0$
As $I_\GF$ is differentiable at $0$, for these theorems to be consistent, we insist that $0^0 = 1$.
Category:Zero to the Power of Zero
Zero to the Power of Zero/Exponential of Zero

== Example of Zero to the Power of Zero ==
By Exponential of Zero:
:$\exp 0 = 1$
From Power Series Expansion for Exponential Function
:$\exp x = \dfrac {x^0} {0!} + \dfrac {x^1} {1!} + \dfrac {x^2} {2!} + \cdots$
For these theorems to be consistent, it is necessary that:
:$\exp 0 = 1 = \dfrac {0^0} {0!} + 0 + 0 + \cdots$
which holds only if $0^0 = 1$.
Category:Zero to the Power of Zero
Zero to the Power of Zero/Historical Note

== Historical Note on Zero to the Power of Zero ==
Whether $0^0 = 0$ or $0^0 = 1$ has been in question since the concept of zero was first raised as a concept.
From one point of view:
:$\forall a \in \R: a^0 = 1$
and so $0^0 = 1$.
From the other point of view:
:$\forall a \in \Z_{\ge 0}: 0^a = 0$
and so $0^0 = 0$.
== Sources ==
* 
* : $0$ Zero
* : $0$ Zero
Category:Historical Notes
Zeroes of Analytic Function are Isolated

== Theorem ==
Let $U \subset \C$ be some open set.
Let $f$ be an analytic function defined on $U$.
Then either $f$ is a constant function, or the set $\set {z \in U: \map f z = 0}$ is totally disconnected.
== Proof ==
Suppose $f$ has no zeroes in $U$.
Then the set described in the theorem is the empty set, and we're done.
So we suppose $\exists z_0 \in U$ such that $\map f {z_0} = 0$.
Since $f$ is analytic, there is a Taylor series for $f$ at $z_0$ which converges for $\cmod {z - z_0} < R$.
Now, since $\map f {z_0} = 0,$ we know $a_0 =0$.
Other $a_j$ may be $0$ as well.
So let $k$ be the least number such that $a_j = 0$ for $0 \le j < k$, and $a_k \ne 0$.
Then we can write the Taylor series for $f$ about $z_0$ as:
:$\displaystyle \sum_{n \mathop = k}^\infty a_n \paren {z - z_0}^n = \paren {z - z_0}^k \sum_{n \mathop = 0}^\infty a_{n + k} \paren {z - z_0}^n$ 
where $a_k \ne 0$ (otherwise, we'd just start at $k + 1$).
Now we define a new function $\map g z$, as the sum on the right hand side, which is clearly analytic in $\cmod {z - z_0} < R$. 
Since it is analytic here, it is also continuous here.
Since $\map g {z_0} = a_k \ne 0, \exists \epsilon >0$ so that $\forall z$ such that $\cmod {z - z_0} < \epsilon, \cmod {\map g z - a_k} < \dfrac {\cmod {a_k} } 2$.
But then $\map g z $ cannot possibly be $0$ in that disk.
Hence the result.
Category:Complex Analysis
Zeroes of Functions of Finite Order

== Theorem ==
Let $\map f z$ be an entire function which satisfies:
:$\map f 0 \ne 0$
:$\cmod {\map f z} \ll \map \exp {\map \alpha {\cmod z} }$
for all $z \in \C$ and some function $\alpha$, where $\ll$ is the order notation.
For $T \ge 1$, let:
:$\map N T = \# \set {\rho \in \C: \map f r = 0, \ \cmod \rho < T}$
where $\#$ denotes the cardinality of a set.
Then:
:$\map N T \ll \map \alpha {2 T}$
== Proof ==
Fix $T \ge 1$ and let $\rho_1, \rho_2, \ldots, \rho_n$ be an enumeration of the zeroes of $f$ with modulus less than $T$, counted with multiplicity.
By Jensen's Formula:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln T - \ln \size {\rho_k} }$
Let $\rho_0 = 1$, $\rho_{n + 1} = T$, $r_k = \size {\rho_k}$.
Then:
and
Moreover, by hypothesis we have that:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta \le \sup_{\theta \mathop \in \closedint 0 {2 \pi} } \ln \size {\map f {T e^{i \theta} } } \ll \map \alpha T$
Putting these facts into Jensen's Formula we have:
:$\map N {\dfrac T 2} \ln 2 + \cmod {\map f 0} \ll \map \alpha T$
which implies:
:$\map N T \ll \map \alpha {2 T}$
Category:Entire Functions
Zeroes of Functions of Finite Order/Corollary

== Corollary to Zeroes of Functions of Finite Order ==
Let $\map f z$ be an entire function which satisfies:
:$\map f 0 \ne 0$
:$\cmod {\map f z} \ll \map \exp {\map \alpha {\cmod z} }$
for all $z \in \C$ and some function $\alpha$, where $\ll$ is the order notation.
Let $f$ have order $1$.
Let $\size {\rho_k}_{k \mathop \ge 1}$ be a non-decreasing enumeration of the zeroes of $f$, counted with multiplicity.
Then for all $\epsilon > 0$, the summation:
:$\ds \sum_{k \mathop \ge 1} \frac 1 {\size {\rho_k}^{1 + \epsilon} }$
converges.
== Proof ==
Let $\epsilon > 0$, $\map N 0 = 0$, so that:
:$\ds \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le \sum_{T \mathop \ge 1} \paren {\map N T - \map N {T - 1} } T^{-1 - \epsilon}$
We have $\map N T \ll 2 T$, so $\map N T - \map N {T - 1}$ is bounded in $T$, say by $C > 0$.
Therefore:
:$\ds \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le C \ \sum_{T \mathop \ge 1} \frac 1 {T^{1 + \epsilon} }$
and the sum on the converges absolutely for $\epsilon > 0$.
Category:Entire Functions
Zeroes of Gamma Function

== Theorem ==
The Gamma function is never equal to $0$.
== Proof ==
Suppose $\exists z$ such that $\map \Gamma z = 0$.
We examine the Euler form of the gamma function, which is defined for $\C \setminus \set {0, -1, -2, \ldots}$.
The Euler form, equated with zero, yields
:$\ds 0 = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1} }$
It is clear that $\dfrac 1 z \ne 0$, so we may divide this out for $z$ in the area of definition.
Now it is clear that as $n \to \infty$, each of the two halves of the term in the product will tend to $1$ for any $z$, and there is no $z$ which yields zero for any $n$ in either of the product terms.
Hence this product will not equal $0$ anywhere.
This leaves only the question of the behavior on $\set {0, -1, -2, \ldots}$, which is discussed at Poles of Gamma Function.
Category:Gamma Function
Zeroes of Infinite Product of Analytic Functions

== Theorem ==
Let $D \subset \C$ be an open connected set.
Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$.
Let $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.
Let $z_0\in D$.
Then:
:$(1): \quad$ $f$ is identically zero some $f_n$ is identically zero 
:$(2): \quad$ $\map {f_n} {z_0} = 0$ for finitely many $n \in \N$
:$(3): \quad$ If $f$ is not identically zero, $\map {\operatorname {mult}_{z_0} } f = \ds \sum_{n \mathop = 1}^\infty \map {\operatorname {mult}_{z_0} } {f_n}$
where $\operatorname {mult}$ denotes multiplicity.
== Proof ==
Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic.
Let $n_0 \in \N$ and $U\subset D$ an open neighborhood of $z_0$ such that $\ds \prod_{n \mathop = n_0}^\infty \map {f_n} z \ne 0$ for $z\in U$.
Let $f$ be identically zero on $U$.
Then $\ds \prod_{n \mathop = 1}^{n_0 - 1} \map {f_n} z$ is identically zero on $U$.
Then some $f_k$ with $k < n_0$ is identically zero on $U$.
By Uniqueness of Analytic Continuation, $f_k$ is identically zero on $D$.
Now let $f$ be not identically zero on $U$.
We have:
== Sources ==
* $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S5$: Weierstrass Factorization Theorem: Theorem $5.9$
Category:Infinite Products
Zeroes of Sine and Cosine/Cosine

== Theorem ==
Let $x \in \R$.
:$\cos x = 0$ $x = \paren {n + \dfrac 1 2} \pi$ for some $n \in \Z$.
== Proof ==
From Sine and Cosine are Periodic on Reals: Corollary:
$\cos x$ is:
:strictly positive on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
and:
:strictly negative on the interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
The result follows directly from Sine and Cosine are Periodic on Reals.
Category:Zeroes of Sine and Cosine
Zeroes of Sine and Cosine/Sine

== Theorem ==
Let $x \in \R$.
:$\sin x = 0$, $x = n \pi$ for some $n \in \Z$.
== Proof ==
From Sine and Cosine are Periodic on Reals: Corollary:
$\sin x$ is:
:strictly positive on the interval $\openint 0 \pi$
and:
:strictly negative on the interval $\openint \pi {2 \pi}$
The result follows directly from Sine and Cosine are Periodic on Reals.
Category:Zeroes of Sine and Cosine
Zeroth Hyperoperation is Successor Function

== Theorem ==
The zeroth hyperoperation is the successor function:
:$H_0 \left({x, y}\right) = y + 1$
== Proof ==
Immediate by definition of the successor function $s: \Z_{\ge 0} \to \Z_{\ge 0}$:
:$\forall y \in \Z_{\ge 0}: s \left({y}\right) = y + 1$
and for the $n$th hyperoperation:
$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases}
y + 1 & : n = 0 \\
x & : n = 1, y = 0 \\
0 & : n = 2, y = 0 \\
1 & : n > 2, y = 0 \\
H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$
setting $n = 0$.
Thus the zeroth hyperoperation degenerates to a mapping with a single operand.
Category:Hyperoperation
Zeroth Power of Real Number equals One

== Theorem ==
Let $a \in \R_{>0}$ be a (strictly) positive real number.
Let $a^x$ be defined as $a$ to the power of $x$.
Then:
:$a^0 = 1$
== Proof ==
== Sources ==
* : $\S 7$: Laws of Exponents: $7.4$
* : $\S 13$: Laws of Exponents: $13.4.$
Category:Powers
Zeta Equivalence to Prime Number Theorem

== Theorem ==
Let $\map \zeta z$ be the Riemann $\zeta$ function.
The Prime Number Theorem is logically equivalent to the statement that the average of the first $N$ coefficients of $\dfrac {\zeta'} {\zeta}$ tend to $-1$ as $N$ goes to infinity.
== Proof ==
The Von Mangoldt Equivalence is equivalent (clearly) to the statement that the average of the coefficients of the function of $z$ defined as:
:$(1): \quad \ds \sum_{n \mathop = 1}^\infty \frac {\map \Lambda n} {n^z}$
tend to $1$.
Let $ \set {p_1, p_2, p_3, \dots}$ be an enumeration of the prime numbers:
:$\set { 2, 3, 5, 7, 11, \dots}$
In the proof of the Von Mangoldt Equivalence, in the sum of von Mangoldt function, the $\map \ln p$ term will appear once for each power of $p$.
So, we expand out $(1)$ as:
This function of $z$ can be recognized as:
Hence the result.
Category:Prime Numbers
Zeta of 2 as Product of Fractions with Prime Numerators

== Theorem ==
where:
:$\zeta$ denotes the Riemann zeta function
:$\ds \prod_p$ denotes the continued product over all prime numbers.
== Proof ==
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta z = \prod_p \frac 1 {1 - p^{-z} }$
where $p$ ranges over the prime numbers.
Thus:
which is the result required.
== Sources ==
* : $1 \cdotp 64493 \, 4066 \ldots$
Category:Riemann Zeta Function at Even Integers
Zhang Qiujian Suanjing/Examples/Example 1

== Example of Problem from by ==
:A man, who had stolen a horse, rode away on its back.
:When he had gone $37$ miles, the owner discovered the theft and pursued the thief for $145$ miles;
:then he returned, believing himself unable to overtake him.
:When he turned back, the thief was riding $23$ miles ahead of him;
:if he had continued the pursuit without coming back, in how many further miles would he have overtaken him?
== Solution ==
:$238 \dfrac {3} {14}$ miles.
== Proof ==
The pursuer gains $37 - 23 = 14$ miles in $145$ miles.
So he will catch up $37$ miles in $145 \times \dfrac {37} {14}$ miles.
Thus the extra distance needed to travel is:
:$145 \times \dfrac {37} {14} - 145 = 238 \dfrac {3} {14}$ miles.
== Sources ==
* 
* 
* : Sun Tsu Suan Ching: $73$
Category:Zhang Qiujian Suanjing
Zorn's Lemma

== Theorem ==
=== Formulation 1 ===
=== Formulation 2 ===
== Zorn's Lemma and Axiom of Choice ==
=== Axiom of Choice implies Zorn's Lemma ===
=== Zorn's Lemma implies Axiom of Choice ===
== Warning ==
== Proof Outline ==
== Also see ==
== Historical Note ==
== Sources ==
* 
Category:Set Theory
Category:Equivalents of Axiom of Choice
Category:Zorn's Lemma
Zorn's Lemma/Formulation 1

== Theorem ==
Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set such that every non-empty chain in $S$ has an upper bound in $S$.
Then $S$ has at least one maximal element.
=== Axiom of Choice implies Zorn's Lemma ===
=== Zorn's Lemma implies Axiom of Choice ===
== Warning ==
== Proof Outline ==
== Also see ==
== Historical Note ==
== Sources ==
* 
* : $\S 16$: Zorn's Lemma
* : Introduction to Set Theory: $3$. The Axiom of Choice and Its Equivalents
* 
* 
Category:Zorn's Lemma
Zorn's Lemma/Formulation 2

== Theorem ==
Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set.
Let $T \subseteq \powerset S$ be the set of subsets of $S$ that are totally ordered by $\preceq$.
Then every element of $T$ is a subset of a maximal element of $T$ under the subset relation.
That is, $T$ is of type $M$.
=== Axiom of Choice implies Zorn's Lemma ===
=== Zorn's Lemma implies Axiom of Choice ===
== Warning ==
== Proof Outline ==
== Also see ==
== Historical Note ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
* 
Category:Zorn's Lemma
Zorn's Lemma/Historical Note

== Historical Note on Zorn's Lemma ==
Zorn's Lemma was published by in $1935$.
However, it had previously been reported by in $1922$, in a slightly different (and simpler) form.
Hence this result is sometimes known as the Kuratowski-Zorn Lemma.
 himself fully acknowledged in his own publication of this result.
He was apparently somewhat embarrassed to have had it named after him, rather than Kuratowski.
 himself had not attached any particular importance to it, considering it a straightforward corollary of Hausdorff's Maximal Principle.
Category:Historical Notes
Zorn's Lemma/Warning

== Theorem ==
== Warning ==
The statement of Zorn's Lemma supposes the existence of an upper bound in $S$ for any (non-empty) chain $A$.
It does not guarantee the existence of an upper bound for $A$ in $A$ itself.
The conclusion is that:
:$\forall a \in S: a \le x \implies a = x$
Category:Zorn's Lemma
Zorn's Lemma Implies Zermelo's Well-Ordering Theorem

== Theorem ==
Zorn's Lemma implies Zermelo's Well-Ordering Theorem.
== Proof ==
Let $X$ be a set.
If $X = \O$ the theorem holds vacuously.
Assume $X$ is not empty.
Let $\WW$ be the collection of pairs $\tuple {W, \preceq}$ such that:
:$W \subseteq X$
:$\preceq$ well-orders $W$
Next, define the partial ordering $\preccurlyeq$ on $\WW$ by $\tuple {W, \preceq} \preccurlyeq \tuple {W', \preceq'}$ :
:$W \subseteq W'$
:$\preceq$ is the restriction of $\preceq'$ to $W$
:For all $w \in W$ and $w' \in W' \setminus W$: $w \preceq' w'$
To apply Zorn's Lemma, we need to show that every chain in $\WW$ has an upper bound.
Let $\CC \subseteq \WW$ be such a chain.
Then we claim $\bigcup \CC \in \WW$ and $\bigcup \CC$ is an upper bound for $\CC$, where we define $\bigcup \CC$ as:
:$\ds \bigcup \CC := \tuple { \bigcup_{ \tuple{W, \preceq} \in \CC } W, \bigcup_{ \tuple{W, \preceq} \in \CC } \preceq }$
First to show $\bigcup \CC \in \WW$.
By Union of Subsets is Subset: Set of Sets:
:$\ds \bigcup_{\tuple{W, \preceq} \mathop \in \CC} W \subseteq X$
Next, let $S \subseteq \bigcup_{\tuple {W, \preceq} \mathop \in \CC} W$ be non-empty.
Fix $s \in S$.
Then for some $\tuple {W, \preceq} \in \bigcup \CC$, we have $s \in W$.
Let $w$ be the smallest element of $S \cap W$ with respect to $\preceq$.
Now let $s' \in S$ be arbitrary.
If $s' \in W$, then $w \preceq s'$ and therefore $w \preceq_{\bigcup \CC} s'$.
If $s' \notin W$, then $s' \in W'$ for some $\tuple {W', \preceq'} \in \bigcup \CC$.
But since $\CC$ is a chain, we have:
:$\tuple {W, \preceq} \preccurlyeq \tuple {W', \preceq'}$
and therefore, $w \preceq' s'$ since $w \in W$ and $s' \in W' \setminus W$.
Then, by definition, $w \preceq_{\bigcup \CC} s'$.
Thus $w$ is the smallest element of $S$ and we conclude $\bigcup \CC$ is well-ordered.
Therefore $\bigcup \CC \in \WW$.
Now to show $\bigcup \CC$ is the sought upper bound of $\CC$.
We observe, for all $\tuple {W, \preceq} \in \CC$:
:$W \subseteq \bigcup_{\tuple {W, \preceq} \mathop \in \CC} W$ by Set is Subset of Union
:$\preceq$ is the restriction of $\preceq_{\bigcup \CC}$ to $W$
:For all $w \in W$ and $w' \in \bigcup \CC \setminus W$, $w \preceq_{\bigcup \CC} w'$
and conclude that:
:$\ds \tuple {W, \preceq} \preccurlyeq \bigcup \CC$
That is, $\ds \bigcup \CC$ is an upper bound of $\CC$.
Thus the hypotheses of Zorn's Lemma hold and we can conclude that $\WW$ has a maximal element.
Let $\tuple {E, \preceq}$ be the maximal element of $\WW$.
Suppose that $E \ne X$.
Then there exists $x_0 \in X \setminus E$. 
Define an ordering $\preceq'$ on $E \cup \set {x_0}$ as follows:
:$y \preceq' x$ $x = x_0$ or $x, y \in E$ and $y \preceq x$.
This is a well-order on $E \cup \set {x_0}$ with:
:$\struct {E, \preceq} \preccurlyeq \struct {E \cup \set {x_0}, \preceq'}$
This contradicts that $\tuple {E, \preceq}$ is the maximal element of $\WW$. 
So $E = X$. 
Hence $X$ is well-orderable.
== Sources ==
* : $(\text P.3)$
Category:Axiom of Choice
Category:Well-Orderings
Category:Zermelo's Well-Ordering Theorem
Zorn's Lemma Implies Zermelo's Well-Ordering Theorem/Mistake

== Source Work ==
:
:Prologue
::$2$. Orderings: $(\text P.3)$
== Mistake ==
:Every nonempty set $X$ can be well-ordered.
:Consider the collection $\WW$ of well orderings of subsets of $X$.
:Such well orderings may be regarded as subsets of $X \times X$, so $\WW$ is partially ordered by inclusion.
:It is easy to verify that the hypotheses of Zorn's lemma are satisfied, so $\WW$ has a maximal element.
== Correction ==
It is insufficient for the well-orderings to be ordered by inclusion $\subseteq$.
We additionally need some kind of stability of the initial segment through a chain, for example as established in the main proof.
Proceeding with just $\subseteq$ as partial ordering of $\WW$, consider the sequence of well-ordered sets:
:$W_n = \set {-i: 0 < i \le n} \subseteq \Z$
endowed with the standard ordering.
Then $\sequence {W_n}_{n \mathop \in \N}$ is a chain in $\WW$, but there is no upper bound, for:
:$\ds \bigcup_{n \mathop \in \N} W_n = \set {-i: i \in \N_{>0} } = \Z_{<0}$
has no smallest element with respect to $<$.
Hence the hypotheses of Zorn's Lemma are not satisfied.
== Sources ==
* : $(\text P.3)$
Category:Mistakes/Real Analysis: Modern Techniques and their Applications
Zorn's Lemma implies Axiom of Choice

== Theorem ==
Let Zorn's Lemma be accepted.
Then the Axiom of Choice holds.
== Proof ==
Let $X$ be a set.
Let $\FF$ be the set of partial choice functions defined as:
:$f \in \FF \iff \begin{cases}
\Dom f \subseteq \powerset X & \ \\
\Img f \subseteq X & \ \\
\forall A \in \Dom f: \map f A \in A & \ \end{cases}$
Let $\preceq$ be the relation defined on $\FF$ as:
:$\forall f_1, f_2 \in \FF: f_1 \preceq f_2 \iff f_2$ is an extension of $f_1$.
Straightforwardly, $\preceq$ is a partial ordering on $\FF$.
We can also see that the Empty Mapping is an element of $\FF$.
Let $C \subseteq \FF$ be a non-empty chain in $\FF$.
Let $U$ be the union of all domains of mappings in $C$.
Furthermore, let $f$ be the union of all graphs of mappings in $C$.
For each $x \in U$, all mappings $g \in C$ with $x \in \Dom g$ have the same value at $x$.
Thus there is a unique $y \in X$ such that $\tuple {x, y} \in f$.
Hence $f: U \rightarrow X$ is a mapping.
By construction, we also have $\map f x \in x$ for all $x \in \Dom f = U$.
That is, every non-empty chain in $\FF$ has an upper bound.
Suppose Zorn's Lemma holds.
Then there exists a maximal element of $\FF$.
We then show by contraposition that if $g$ is such a maximal element, then:
:$\Dom g = \powerset X \setminus \O$
In that case, we will have constructed a choice function $\powerset X \setminus \set \O \rightarrow X$.
Suppose that $\Dom g \ne \powerset X \setminus \O$.
Then there is an $A \in \paren {\powerset X \setminus \O} \setminus \Dom g$.
Let $x \in A$.
We can then define the mapping $\hat g: \set A \cup \Dom g$ by defining:
:$\forall S \in \Dom g: \forall \map {\hat g} A = x: \map {\hat g} S = \map g S$
That way, we clearly have $\hat g \ne g$ and $\hat g \preceq g$.
Thus $g$ is not maximal in $\FF$.
== Sources ==
* : $\S 16$: Zorn's Lemma: Exercise
Category:Zorn's Lemma
Category:Axiom of Choice
Zorn's Lemma implies Hausdorff's Maximal Principle

== Theorem ==
Let Zorn's Lemma be accepted as true.
Then Hausdorff's Maximal Principle holds.
== Proof ==
Recall Zorn's Lemma:
Recall Hausdorff's Maximal Principle:
It is seen directly that Hausdorff's Maximal Principle is a special case of Zorn's Lemma where the ordering is the subset relation.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles: Proposition $5.9$
Category:Zorn's Lemma
Category:Hausdorff's Maximal Principle