nyuuzyou/wikis huggingface dataset:
Zermelo's Well-Ordering Theorem implies Hausdorff's Maximal Principle

== Theorem ==
Let Zermelo's Well-Ordering Theorem hold.
Then Hausdorff's Maximal Principle holds.
== Proof ==
Let $X$ be a non-empty set.
Let $X$ contain at least two elements.
Otherwise, any non-empty ordering on $X$ is trivially a maximal chain.
By Zermelo's Well-Ordering Theorem, $X$ can be well-ordered set.
Fix such a well-ordering.
Let $\le$ be any ordering on $X$.
Let $\map P {a, Y}$ be the predicate:
:$a$ is $\le$-comparable with every $y \in Y$.
Here, $a$ and $Y$ are bound variables.
That is, $\map P {a, Y}$ holds if, for every $y \in Y$, $a \le y$ or $y \le a$.
Define the mapping:
:$\map \rho {f: S_x \to \powerset X} = \begin {cases}
f \sqbrk {S_x} \cup \set x & : \map P {x, S_x} \\
f \sqbrk {S_x} & : \text {otherwise}
\end{cases}$
where $S_x$ is the initial segment in $X$ defined by $x$, and $\map f {\,\cdot\,}$ denotes an image set.
Using the Principle of Recursive Definition for Well-Ordered Sets, we can use $\rho$ to uniquely define:
:$h: X \to \powerset X$: 
:$\map h \alpha = \map \rho {h {\restriction_{S_\alpha} } } = \begin {cases}
h \sqbrk {S_\alpha} \cup \set \alpha & : \map P {\alpha, S_\alpha} \\ h \sqbrk {S_\alpha} & : \text {otherwise} \end{cases}$
Then $\map h \alpha$ is a $\le$-totally ordered set, by virtue of the construction of $h$ in terms of the predicate $P$.
Similarly, $h \sqbrk {S_\alpha}$ is totally ordered, for the same reason.
Then the union:
:$Z = \ds \bigcup_{\alpha \mathop \in X} \map h \alpha$
admits an ordered sum in terms of $\le$ imposed on each $\map h \alpha$.
By Ordered Sum of Tosets is Totally Ordered Set, this is a totally ordered set.
In particular, it is a chain of $\struct {X, \le}$.
We claim this chain is maximal.
To see this, consider $x_0 \notin \ds \bigcup_{\alpha \mathop \in X} \map h \alpha$.
Then $x_0$ is not $\le$-comparable with the elements of $Z$, because $\neg \map P {x, Z}$.
That is, there are no proper supersets of $Z$ that are totally ordered.
Therefore the ordered sum on $Z$ is a maximal chain in $\struct {X, \le}$.
== Also see ==
* Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem
== Sources ==
* $\S 1.11$ Supplementary Exercise $6$
Category:Hausdorff's Maximal Principle
Category:Zermelo's Well-Ordering Theorem
Zermelo's Well-Ordering Theorem is Equivalent to Axiom of Choice

== Theorem ==
Zermelo's Well-Ordering Theorem holds the Axiom of Choice holds.
That is, every set is well-orderable every collection of sets has a choice function.
== Proof ==
=== Necessary Condition ===
Suppose the Axiom of Choice holds.
Then Zermelo's Well-Ordering Theorem holds by Zermelo's Well-Ordering Theorem itself.
That is, every set is well-orderable.
=== Sufficient Condition ===
Let it be supposed that every set is well-orderable.
Then by the converse to Zermelo's Well-Ordering Theorem:
:the Axiom of Choice holds.
== Sources ==
* 
Category:Axiom of Choice
Category:Zermelo's Well-Ordering Theorem
Zero (Category) is Initial Object

== Theorem ==
Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf 0$ be the zero category.
Then $\mathbf 0$ is an initial object of $\mathbf{Cat}$.
== Proof ==
Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category.
By Empty Mapping is Unique, there are unique mappings:
:$F_0: \mathbf 0_0 = \O \to \mathbf C_0$
:$F_1: \mathbf 0_1 = \O \to \mathbf C_1$
making $F: \mathbf 0 \to \mathbf C$ a functor by vacuous truth.
That $F_0$ and $F_1$ are actually mappings follows from $\mathbf C$ being a small category.
Hence the result, by definition of initial object.
== Sources ==
* : $\S 2.2$: Example $2.11$: $2$
Category:Category of Categories
Zero Choose Zero

== Theorem ==
:$\dbinom 0 0 = 1$
where $\dbinom 0 0$ denotes a binomial coefficient.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $2$
Category:Examples of Binomial Coefficients‎
Category:Zero Choose Zero
Zero Choose Zero/Proof 1

== Theorem ==
== Proof ==
By Zero Choose n:
:$\dbinom 0 n = \delta_{0 n}$
where:
:$\dbinom 0 n$ denotes a binomial coefficient
:$\delta_{0 n}$ denotes the Kronecker delta.
Hence directly:
:$\dbinom 0 0 = 1$
Category:Zero Choose Zero
Zero Choose n

== Theorem ==
:$\dbinom 0 n = \delta_{0 n}$
where:
:$\dbinom 0 n$ denotes a binomial coefficient
:$\delta_{0 n}$ denotes the Kronecker delta.
== Proof ==
By definition of binomial coefficient:
:$\dbinom m n = \begin{cases}\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\&\\0 & : \text { otherwise } \end{cases}$
Thus when $n > 0$:
:$\dbinom 0 n = 0$
and when $n = 0$:
:$\dbinom 0 0 = \dfrac {0!} {0! \paren {0 - 0}!} = 1$
by definition of factorial.
Hence the result by definition of the Kronecker delta.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(48)$
Category:Examples of Binomial Coefficients‎
Zero Complement is Not Empty

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $S^*$ be the zero complement of $S$.
Then $S^*$ is not empty.
== Proof ==
From , we have:
:$\exists m, n \in S: m \ne n$
That is, there are at least two distinct elements in $S$.
Therefore, there must be at least one element in $S^* = S \setminus \set 0$.
So:
:$S^* = S \setminus \set 0 \ne \O$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers
Category:Naturally Ordered Semigroup
Zero Definite Integral of Nowhere Negative Function implies Zero Function

== Theorem ==
Let $\closedint a b \subseteq \R$ be a closed real interval.
Let $h: \closedint a b \to \R$ be a continuous real function such that:
:$\forall x \in \closedint a b: \map h x \ge 0$
Let:
:$\ds \int_a^b \map h x \rd x = 0$
Then:
:$\forall x \in \closedint a b: \map h x = 0$
== Proof ==
 that:
:$\exists c \in \closedint a b: \map h c > 0$
As $h$ is continuous, there exists some closed real interval $\closedint r s \subseteq \closedint a b$ where $r < s$ such that:
:$\exists \epsilon \in \R_{>0}: \forall x \in \closedint r s: \map h x > \dfrac {\map h c} 2$
From Sign of Function Matches Sign of Definite Integral:
:$\ds \int_r^s \map h x \rd x > 0$
This makes a strictly positive contribution to the integral.
$h$ is still continuous on $\closedint a r$ and $\closedint s b$.
So, again from Sign of Function Matches Sign of Definite Integral:
:$\forall x \in \closedint a r: \map f x \ge 0 \implies \ds \int_a^r \map f x \rd x \ge 0$
:$\forall x \in \closedint s b: \map f x \ge 0 \implies \ds \int_s^b \map f x \rd x \ge 0$
Thus as:
:$\ds \int_a^b \map h x \rd x = \int_a^r \map f x \rd x + \int_r^s \map f x \rd x + \int_s^b \map f x \rd x$
it follows that:
:$\ds \int_a^b \map h x \rd x > 0$
But by hypothesis:
:$\ds \int_a^b \map h x \rd x = 0$
Thus by contradiction, there can be no $c \in \closedint a b$ such that $\map h c > 0$.
Hence the result.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Lemma $2.2.10$
Category:Definite Integrals
Zero Derivative implies Constant Complex Function

== Theorem ==
Let $D \subseteq \C$ be a connected domain of $\C$.
Let $f: D \to \C$ be a complex-differentiable function.
For all $z \in D$, let $\map {f'} z = 0$.
Then $f$ is constant on $D$.
== Proof ==
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined in the Cauchy-Riemann Equations:
:$\map u {x, y} = \map \Re {\map f z}$
:$\map v {x, y} = \map \Im {\map f z}$
By the Cauchy-Riemann Equations:
:$f' = \dfrac {\partial f} {\partial x} = \dfrac {\partial u} {\partial x} + i \dfrac {\partial v} {\partial x}$
:$f' = -i \dfrac {\partial f} {\partial y} = \dfrac {\partial v} {\partial y} - i \dfrac {\partial u} {\partial y}$
As $f' = 0$ by assumption, this implies:
:$0 = \dfrac {\partial u} {\partial x} = \dfrac {\partial u} {\partial y} = \dfrac {\partial v} {\partial x} = \dfrac {\partial v} {\partial y}$
Then Zero Derivative implies Constant Function shows that $\map u {x + t, y} = \map u {x, y}$ for all $t \in \R$.
Similar results apply for the other three partial derivatives.
Let $z, w \in D$.
From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C$ in $D$ with endpoints $z$ and $w$.
The contour $C$ is a concatenation of directed smooth curves that can be parameterized as line segments on one of these two forms:
:$(1): \quad \map \gamma t = z_0 + t r$
:$(2): \quad \map \gamma t = z_0 + i t r$
for some $z_0 \in D$ and $r \in \R$ for all $t \in \closedint 0 1$.
If $z_1 \in D$ lies on the same line segment as $z_0$, it follows that for parameterizations of type $(1)$:
Similarly, for parameterizations of type $(2)$:
As the image of $C$ is connected by these line segments, it follows that for all $z_0$ and $z_1$ in the image of $C$:
:$\map f {z_0} = \map f {z_1}$
In particular:
:$\map f z = \map f w$
so $f$ is constant on $D$.
== Sources ==
* : $\S 1.3$
Category:Complex Analysis
Zero Derivative implies Constant Function

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Suppose that:
:$\forall x \in \openint a b: \map {f'} x = 0$
Then $f$ is constant on $\closedint a b$.
== Proof ==
When $x = a$ then $\map f x = \map f a$ by definition of mapping.
Otherwise, let $x \in \hointl a b$.
We have that:
:$f$ is continuous on the closed interval $\closedint a b$
:$f$ is differentiable on the open interval $\openint a b$
Hence it satisfies the conditions of the Mean Value Theorem on $\closedint a b$.
Hence:
:$\exists \xi \in \openint a x: \map {f'} \xi = \dfrac {\map f x - \map f a} {x - a}$
But by our supposition:
:$\forall x \in \openint a b: \map {f'} x = 0$
which means:
:$\forall x \in \openint a b: \map f x - \map f a = 0$
and hence:
:$\forall x \in \openint a b: \map f x = \map f a$
== Also see ==
* Derivative of Constant: This is the converse. Thus we see that $f$ is a constant function $\forall x: \map {f'} x = 0$.
* Zero Derivative implies Constant Complex Function
== Sources ==
* : $\S 11.7$
Category:Differential Calculus
Category:Constant Mappings
Zero Dimensional Space is T3

== Theorem ==
Let $T = \struct {S, \tau}$ be a zero dimensional topological space.
Then $T$ is a $T_3$ space.
== Proof ==
Let $T = \struct {S, \tau}$ be a zero dimensional space.
Let $F \subseteq S$ be closed in $T$.
Let also $y \notin F$.
Then by definition of closed, $\relcomp S F$ is open in $T$, where $\relcomp S F$ is the complement of $F$ in $S$.
As $T$ is zero dimensional, it has a basis $\BB$ which consists entirely of clopen sets.
As $\BB$ is a basis for $T$, it follows that:
:$\ds \exists \UU \subseteq \BB: \relcomp S F = \bigcup \UU$
that is, $\relcomp S F$ is the union of a subset of elements of $\BB$.
Thus, there is a set $U \in \UU$ such that $y \in U$ and $F \subseteq \relcomp S U$.
But the elements of $\UU$ are clopen sets, so $U$ is itself clopen.
Thus, by definition, $\relcomp S U$ is also clopen.
So we have that $U$ and $\relcomp S U$ are open sets in $T$ such that:
:$\exists W, R \in \tau: F \subseteq W, y \in R: R \cap W = \O$
by setting $R = U$ and $W = \relcomp S U$.
That is, $T$ is a $T_3$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T3 Spaces
Category:Zero Dimensional Spaces
Zero Dimensional Space is not necessarily T0

== Theorem ==
Let $T = \struct {S, \tau}$ be a zero dimensional topological space.
Then $T$ is not necessarily a $T_0$ (Kolmogorov) space.
== Proof ==
Let $T = \struct {S, \tau}$ be a partition space.
From Partition Topology is Zero Dimensional, $T$ is a zero dimensional topological space.
From Partition Topology is not $T_0$, $T$ is not a $T_0$ (Kolmogorov) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T0 Spaces
Category:Zero Dimensional Spaces
Zero Dimensional T0 Space is Totally Separated

== Theorem ==
Let $T = \struct {S, \tau}$ be a zero dimensional topological space which is also a $T_0$ (Kolmogorov) space.
Then $T$ is totally separated.
== Proof ==
Let $T = \struct {S, \tau}$ be a zero dimensional space which is also a $T_0$ (Kolmogorov) space.
As $T$ is zero dimensional, there exists a basis $\BB$ whose sets are all clopen.
Let $x, y \in S$.
As $T$ is a $T_0$ space:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
, suppose that $\exists U \in \tau: x \in U, y \notin U$.
Then:
:$\exists V \in \BB: x \in V$ and $V \subseteq U$
by definition of basis.
The set $V$ is clopen by the definition of $\BB$.
But then $x \in V$ which is open and $y \in S \setminus V$ which is also open.
$\set {V \mid S \setminus V}$ is a partition and hence $T$ is totally separated.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T0 Spaces
Category:Zero Dimensional Spaces
Category:Totally Separated Spaces
Zero Divides Zero

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
:$0 \divides n \implies n = 0$
That is, zero is the only integer divisible by zero.
== Proof ==
== Sources==
* : Chapter $3$: The Integers: $\S 10$. Divisibility
* : Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(ii)}$
Category:Divisors
Zero Divisor Product is Zero Divisor

== Theorem ==
The ring product of a zero divisor with any ring element is a zero divisor.
== Proof ==
Let $\struct {R, +, \circ}$ be a ring.
Let $x \divides 0_R$ in $R$.
Then:
So $z \circ x \divides 0_R$ in $R$.
The same thing happens if we form the product $\paren {x \circ y} \circ z$.
Category:Zero Divisors
Zero Divisor of Ring/Examples/Order 2 Square Matrices/Example 1

== Examples of Zero Divisors of Rings ==
Let $R$ be the ring square matrices of order $2$ over a field with unity $1$ and zero $0$.
Let:
Then:
:$\mathbf A \mathbf B = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix} = \mathbf B \mathbf A$
Thus both $\mathbf A$ and $\mathbf B$ are zero divisors of $R$.
== Sources ==
* 
* 
Category:Examples of Zero Divisors
Zero Divisor of Ring/Examples/Order 2 Square Matrices/Example 2

== Examples of Zero Divisors of Rings ==
Let $R$ be the ring square matrices of order $2$ over the real numbers.
Then:
:$\begin {bmatrix} 0 & 1 \\ 0 & 0 \end {bmatrix} \begin {bmatrix} 1 & 0 \\ 0 & 0 \end {bmatrix} = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$
demonstrating that $\begin {bmatrix} 0 & 1 \\ 0 & 0 \end {bmatrix}$ and $\begin {bmatrix} 1 & 0 \\ 0 & 0 \end {bmatrix}$ are zero divisors of $R$.
== Sources ==
* 
Category:Examples of Zero Divisors
Zero Divisor of Ring/Examples/Order 2 Square Matrices/Example 3

== Examples of Zero Divisors of Rings ==
Let $R$ be the ring square matrices of order $2$ over the real numbers.
Then:
:$\begin {bmatrix} 0 & 0 \\ 1 & 1 \end {bmatrix} \begin {bmatrix} 0 & 1 \\ 0 & -1 \end {bmatrix} = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$
demonstrating that $\begin {bmatrix} 0 & 0 \\ 1 & 1 \end {bmatrix}$ and $\begin {bmatrix} 0 & 1 \\ 0 & -1 \end {bmatrix}$ are zero divisors of $R$.
== Sources ==
* 
Category:Examples of Zero Divisors
Zero Divisor of Ring/Examples/Order 2 Square Matrices/Example 4

== Examples of Zero Divisors of Rings ==
Let $R$ be the ring of square matrices of order $2$ over the real numbers.
Then:
:$\begin {bmatrix} 1 & 1 \\ 1 & 1 \end {bmatrix} \begin {bmatrix} -1 & 1 \\ 1 & -1 \end {bmatrix} = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$
demonstrating that $\begin {bmatrix} 1 & 1 \\ 1 & 1 \end {bmatrix}$ and $\begin {bmatrix} -1 & 1 \\ 1 & -1 \end {bmatrix}$ are zero divisors of $R$.
== Sources ==
* 
Category:Examples of Zero Divisors
Zero Element Generates Null Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
For $r \in R$, let $\ideal r$ denote the ideal generated by $r$.
Then $\ideal {0_R}$ is the null ideal.
== Proof ==
By definition:
:$\ideal {0_R} = \set {r \circ 0_R: r \in R}$
but for each $r \in R$ we have by Ring Product with Zero that $r \circ 0_R = 0_R$ for all $r \in R$.
Therefore $\ideal {0_R}$ is the null ideal.
Category:Ideal Theory
Zero Element is Unique

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure that has a zero element $z \in S$.
Then $z$ is unique.
== Proof ==
Suppose $z_1$ and $z_2$ are both zeroes of $\struct {S, \circ}$.
Then by the definition of zero element:
:$z_2 \circ z_1 = z_1$ by dint of $z_1$ being a zero
:$z_2 \circ z_1 = z_2$ by dint of $z_2$ being a zero.
So $z_1 = z_2 \circ z_1 = z_2$.
So $z_1 = z_2$ and there is only one zero after all.
== Sources ==
* : $\S 4.3$. Units and zeros
* : $\S 3.1$: Monoids: Exercise $(6)$
Category:Zero Elements
Zero Element of Multiplication on Numbers

== Theorem ==
On all the number systems:
* natural numbers $\N$
* integers $\Z$
* rational numbers $\Q$
* real numbers $\R$
* complex numbers $\C$
the zero element of multiplication is zero ($0$).
== Proof ==
This is demonstrated by showing that:
:$n \times 0 = 0 = 0 \times n$
for all $n$ in all standard number systems.
== Also see ==
* Identity Element of Addition on Numbers
* Identity Element of Multiplication on Numbers
== Sources ==
* : $\S 4.3$. Units and zeros: Example $74$
Category:Numbers
Zero Element of Union and Intersection

== Source Work ==
:
: Introduction to Set Theory
:: $1$. Elementary Operations on Sets
This mistake can be seen in the 2009 Dover edition: ISBN 0-486-47222-1
== Mistake ==
:The operations $A \cup B$ and $A \cap B$ are meaningful for every pair of sets $A, B$ ... The empty set $\O$ plays the role of the zero element.
This is technically incorrect.
From Zero of Power Set with Intersection, $\O$ does indeed play the role of the zero element on the algebraic structure $\struct {\powerset S, \cap}$.
However, from Zero of Power Set with Union, the role of the zero element on the algebraic structure $\struct {\powerset S, \cup}$ is actually played by $S$.
In defence, this book is not about abstract algebra but about topology.
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
Category:Mistakes/Point Set Topology
Zero Equals One

== Paradox ==
== Resolution ==
This is a falsidical paradox arising from incorrect reasoning about the nature of square roots.
=== Explanation ===
While this line is correct:
:$\paren {4 - \dfrac 9 2}^2 = \paren {5 - \dfrac 9 2}^2$
note that:
:$4 - \dfrac 9 2 = -\dfrac 1 2$
but:
:$5 - \dfrac 9 2 = \dfrac 1 2$
and the paradox is resolved.
Category:Mistakes
Category:Falsidical Paradoxes
Zero Function Vanishes in Neighborhood of Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Define a function $f : X \to \C$ by:
:$\map f x = 0$
for each $x \in X$. 
Then $f$ vanishes in a neighborhood of infinity.
== Proof ==
From Empty Set is Compact Space:
:$\O \subseteq X$ is compact.
We then have: 
:$\map f x = 0$ for all $x \in X \setminus \O$
Hence $f$ vanishes in a neighborhood of infinity.
Category:Complex-Valued Functions Vanishing in Neighborhood of Infinity
Zero Locus of Larger Set is Smaller

== Theorem ==
Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials in $n$ variables over $k$.
Let $I, J \subseteq A$ be subsets, and $\map V I$ and $\map V J$ their zero loci.
Let $I \subseteq J$.
Then $\map V I \supseteq \map V J$.
== Proof ==
Category:Algebraic Geometry
Zero Locus of Set is Zero Locus of Generated Ideal

== Theorem ==
Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomial functions in $n$ variables over $k$.
Let $T \subseteq A$ be a subset, and $\map V T$ the zero locus of $T$.
Let $J = \ideal T$ be the ideal generated by $T$.
Then:
:$\map V T = \map V J$
== Proof ==
Let $x \in \map V T$, so $\map f x = 0$ for all $f \in T$.
By definition, $J$ is the set of linear combinations of elements of $T$ over $k$.
So any $g \in J$ is of the form:
:$g = k_1 t_1 + \cdots + k_r t_r$
with $k_i \in k$ and $t_i \in T$.
Therefore:
Therefore:
:$x \in \map V J$
Conversely, if $x \in \map V J$, then $\map f x = 0$ for all $f \in J$.
But $T \subseteq J$, so in particular $\map f x = 0$ for all $f \in T$.
So:
:$x \in \map V t$
Category:Algebraic Geometry
Zero Locus of Vanishing Ideal is Closure

== Theorem ==
Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $\mathbb A^n_k$ be the standard affine space over $k$.
Let $\mathbb A^n_k$ be equipped by Zariski topology.
Let $S \subseteq \mathbb A^n_k$.
Then:
:$\map V {\map I S} = S^-$
where:
:$\map I \cdot$ denotes the vanishing ideal
:$\map V \cdot$ denotes the zero locus
:$S^-$ is the closure of $S$
== Proof ==
Category:Algebraic Geometry
Zero Matrix is Identity for Hadamard Product

== Theorem ==
Let $\struct {S, \cdot}$ be a monoid whose identity is $e$.
Let $\map {\MM_S} {m, n}$ be an $m \times n$ matrix space over $S$.
Let $\mathbf e = \sqbrk e_{m n}$ be the zero matrix of $\map {\MM_S} {m, n}$.
Then $\mathbf e$ is the identity element for Hadamard product.
== Proof ==
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_S} {m, n}$.
Then:
Similarly:
== Also see ==
* Zero Matrix is Identity for Matrix Entrywise Addition
Category:Hadamard Product
Category:Zero Matrix
Zero Matrix is Identity for Matrix Entrywise Addition

== Theorem ==
Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.
Let $\mathbf 0 = \sqbrk 0_{m n}$ be the zero matrix of $\map \MM {m, n}$.
Then $\mathbf 0$ is the identity element for matrix entrywise addition.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Zero Matrix is Identity for Hadamard Product
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $3$
Category:Matrix Entrywise Addition
Category:Zero Matrix
Category:Zero Matrix is Identity for Matrix Entrywise Addition
Zero Matrix is Identity for Matrix Entrywise Addition/Proof 1

== Theorem ==
== Proof ==
From:
:Integers form Ring
:Rational Numbers form Ring
:Real Numbers form Ring
:Complex Numbers form Ring
the standard number systems $\Z$, $\Q$, $\R$ and $\C$ are rings whose zero is the number $0$ (zero).
Hence we can apply Zero Matrix is Identity for Matrix Entrywise Addition over Ring.
The above cannot be applied to the natural numbers $\N$, as they do not form a ring.
However, from Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\struct {\N, +}$ is a commutative monoid whose identity is $0$ (zero).
By definition, matrix entrywise addition is the Hadamard product with respect to addition of numbers.
The result follows from Zero Matrix is Identity for Hadamard Product.
Category:Zero Matrix is Identity for Matrix Entrywise Addition
Zero Matrix is Identity for Matrix Entrywise Addition/Proof 2

== Theorem ==
== Proof ==
Let $\mathbf A = \sqbrk a_{m n} \in \map \MM {m, n}$.
Then:
Similarly:
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.10$
Category:Zero Matrix is Identity for Matrix Entrywise Addition
Zero Matrix is Identity for Matrix Entrywise Addition over Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.
Let $\mathbf 0_R = \sqbrk {0_R}_{m n}$ be the zero matrix of $\map {\MM_R} {m, n}$.
Then $\mathbf 0_R$ is the identity element for matrix entrywise addition.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Zero Matrix is Identity for Hadamard Product
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Matrix Entrywise Addition
Category:Zero Matrix
Category:Zero Matrix is Identity for Matrix Entrywise Addition
Zero Matrix is Identity for Matrix Entrywise Addition over Ring/Proof 2

== Theorem ==
== Proof ==
By definition, matrix entrywise addition is the Hadamard product with respect to ring addition.
We have from that the identity element of ring addition is the ring zero $0_R$.
The result then follows directly from Zero Matrix is Identity for Hadamard Product.
Category:Zero Matrix is Identity for Matrix Entrywise Addition