nyuuzyou/wikis huggingface dataset:
Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent

== Theorem ==
Let $\map {y_1} x$ and $\map {y_2} x$ be particular solutions to the homogeneous linear second order ODE:
:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
on a closed interval $\closedint a b$.
Then:
:$y_1$ and $y_2$ are linearly dependent
:
:the Wronskian $\map W {y_1, y_2}$ of $y_1$ and $y_2$ is zero everywhere on $\closedint a b$.
== Proof ==
=== Sufficient Condition ===
Let $y_1$ and $y_2$ are linearly dependent.
Suppose either $y_1$ or $y_2$ is zero everywhere on $\closedint a b$.
Then either ${y_1}'$ or ${y_2}'$ is also zero everywhere on $\closedint a b$.
Thus:
:$y_1 {y_2}' - y_2 {y_1}' = 0$
and so by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.
Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.
, it follows by definition of linearly dependent that:
:$y_2 = C y_1$
for some $C \in \R$.
Thus by Derivative of Constant Multiple:
:${y_2}' = C {y_1}'$
Hence:
Hence by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.
=== Necessary Condition ===
Let $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.
One way for this to happen is for either $y_1$ or $y_2$ to be zero everywhere on $\closedint a b$.
, suppose $y_1$ is zero everywhere on $\closedint a b$.
Then by Real Function is Linearly Dependent with Zero Function, $y_1$ and $y_2$ are linearly dependent.
Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.
As $y_1$ is continuous, there is some closed interval:
:$\closedint c d \subseteq \closedint a b$
on which:
:$\forall x \in \closedint c d: \map {y_1} x \ne 0$
Since $\map W {y_1, y_2} = 0$:
:$\dfrac {y_1 {y_2}' - y_2 {y_1}'} { {y_1}^2} = 0$
on $\closedint c d$.
By Quotient Rule for Derivatives, this is the same as:
:$\paren {\dfrac {y_2} {y_1} }' = 0$
Integrating $x$, this gives:
:$\dfrac {y_2} {y_1} = k$
Explicitly:
:$\forall x \in \closedint c d: \map {y_2} x = k \, \map {y_1} x$
As $\map {y_2} x$ and $k \, \map {y_1} x$ have equal values on $\closedint c d$, their derivatives are equal there as well.
From Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions, it follows that:
:$\forall x \in \closedint a b: \map {y_2} x = k \, \map {y_1} x$.
== Sources ==
* : $\S 3.15$: The General Solution of the Homogeneous Equation: Lemma $2$
Category:Homogeneous LSOODEs
Zero and One are the only Consecutive Perfect Squares/Proof 1

== Theorem ==
== Proof ==
Let $x$ and $h$ be integers such that:
:$x^2 + 1 = \paren {x - h}^2$
Then:
We have that Consecutive Integers are Coprime.
However by the Fundamental Theorem of Arithmetic, both sides must have the same unique prime decomposition.
Therefore $h$ cannot have any prime factors since they cannot be shared by $\paren {h - 1} \paren {h + 1}$.
This leaves $h = -1$, $h = 0$, or $h = 1$ as the only possibilities since they are the only integers with no prime factors.
If $h = -1$ then $h + 1 = 0$, so $2 x h = 0$.
It follows that $x = 0$.
If $h = 1$ then $h - 1 = 0$, so $2 x h = 0$.
It follows that $x = 0$.
If $h = 0$, then $2 x \cdot 0 = \paren {-1} \paren 1$, which is a contradiction.
Therefore the only pairs of consecutive squares are:
:$0^2 = 0$ and $\paren {0 + \paren {-1} }^2 = \paren {-1}^2 = 1$
and:
:$0^2 = 0$ and $\paren {0 + 1}^2 = 1^2 = 1$
Category:Zero and One are the only Consecutive Perfect Squares
Zero and One are the only Consecutive Perfect Squares/Proof 2

== Theorem ==
== Proof ==
Suppose that $k, l \in \Z$ are such that their squares are consecutive.
That is:
:$l^2 - k^2 = 1$
Then we can factor the as:
:$l^2 - k^2 = \paren {l + k} \paren {l - k}$
By Invertible Integers under Multiplication, it follows that:
:$l + k = \pm 1 = l - k$
Therefore, it must be that:
:$\paren {l + k} - \paren {l - k} = 0$
That is, $2 k = 0$, from which we conclude $k = 0$.
So if $n$ and $n + 1$ are squares, then necessarily $n = 0$.
The result follows.
Category:Zero and One are the only Consecutive Perfect Squares
Zero and Unity of Subfield

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
Let $\struct {K, +, \times}$ be a subfield of $F$.
=== Zero of Subfield is Zero of Field ===
=== Unity of Subfield is Unity of Field ===
== Sources ==
* 
* 
Category:Subfields
Zero is Accumulation Point of Sequence in Sierpiński Space

== Theorem ==
Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
The sequence in $T$:
:$\sigma = \sequence {0, 1, 0, 1, \ldots}$
has $0$ as an accumulation point.
== Proof ==
By definition, $\alpha$ is an accumulation point of $\sigma$ :
:$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \in U}$ is infinite.
Both $\set 0$ and $\set {0, 1}$ contain $0$, which occurs an infinite number of times in $\sigma$.
Hence, by definition, $0$ is an accumulation point of $\sigma$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$
Category:Accumulation Points
Category:Sierpiński Space
Zero is Identity in Naturally Ordered Semigroup

== Theorem ==
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $0$ be the zero of $\struct {S, \circ, \preceq}$.
Then $0$ is the identity for $\circ$.
That is:
:$\forall n \in S: n \circ 0 = n = 0 \circ n$
== Proof ==
By definition of an ordering:
:$0 \preceq 0$
Thus from :
:$\exists p \in S: 0 \circ p = 0$
By the definition of zero:
:$0 \preceq 0 \circ 0$ and $0 \preceq p$
Thus since $\preceq$ is compatible with $\circ$:
:$0 \circ 0 \preceq 0 \circ p = 0$
Thus:
:$0 \circ 0 \preceq 0$ and $0 \preceq 0 \circ 0$ 
Hence, as $\preceq$ is antisymmetric, it follows that:
:$0 \circ 0 = 0$
Because $\struct {S, \circ, \preceq}$ is a semigroup, $\circ$ is associative. 
So:
:$\forall n \in S: \paren {n \circ 0} \circ 0 = n \circ \paren {0 \circ 0} = n \circ 0$
Thus from :
:$\forall n \in S: n \circ 0 = n$
Similarly:
:$\forall n \in S: 0 \circ \paren {0 \circ n} = \paren {0 \circ 0} \circ n = 0 \circ n$
meaning:
:$\forall n \in S: 0 \circ n = n$
Thus:
:$\forall n \in S: n \circ 0 = n = 0 \circ n$
and so $0$ is the identity for $\circ$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.1$
Category:Naturally Ordered Semigroup
Zero is Limit Point of Integer Reciprocal Space

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then $0$ is the only limit point of $A$ in $\R$.
== Proof ==
There are three cases to consider:
=== Points in $A$ ===
Consider $x \in \R$ such that $x \in A$.
That is, $x = \dfrac 1 n$ for some $n \in \N$.
Let:
:$d = \dfrac 1 n - \dfrac 1 {n + 1}$
Consider the open real interval:
:$I := \openint {\dfrac 1 n - d} {\dfrac 1 n + d} \subseteq \R$
By definition, $I$ is an open set of $\R$.
Thus $I$ is an open set of $\R$ which contains no element of $A$ distinct from $x$.
Thus $x$ is not a limit point of $A$ in $\R$.
=== Non-Zero Points not in $A$ ===
Let $x \in \R$ such that $x \ne 0$ and $x \notin A$.
Let:
:$d := \min \set {\size {x - m}: m \in A}$
that is, the smallest distance from $x$ to an element of $A$.
Consider the open real interval:
:$I := \openint {x - d} {x + d} \subseteq \R$
By definition, $I$ is an open set of $\R$.
Thus $I$ is an open set of $\R$ which contains no element of $A$ (distinct from $x$ or not).
Thus $x$ is not a limit point of $A$ in $\R$.
=== Zero ===
Finally, consider the point $0$.
Let $U$ be an open set of $\R$ which contains $0$.
From Open Sets in Real Number Line, there exists an open interval $I$ of the form:
:$I := \openint {-a} b \subseteq U$
By the Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 b$
and so:
:$\exists n \in \N: \dfrac 1 n < b$
But $\dfrac 1 n \in A$.
Thus an open set $U$ which contains $0$ contains at least one element of $A$ (distinct from $0$).
Thus, by definition, $0$ is a limit point of $A$ in $\R$.
Thus the only limit point of $A$ in $\R$ is $0$.
Category:Integer Reciprocal Space
Category:Examples of Limit Points
Zero is Limit Point of Integer Reciprocal Space Union with Closed Interval

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $T = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $B$ be the uncountable set:
:$B := A \cup \closedint 2 3$
where $\closedint 2 3$ is a closed interval of $\R$.
$2$ and $3$ are to all intents arbitrary, but convenient.
Then $0$ is a limit point of $B$ in $\R$.
== Proof ==
Let $U$ be an open set of $\R$ which contains $0$.
From Open Sets in Real Number Line, there exists an open interval $I$ of the form:
:$I := \openint {-a} b \subseteq U$
By the Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 b$
and so:
:$\exists n \in \N: \dfrac 1 n < b$
But $\dfrac 1 n \in B$.
Thus an open set $U$ which contains $0$ contains at least one element of $B$ (distinct from $0$).
Thus, by definition, $0$ is a limit point of $B$ in $\R$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $1 \ \text{(b)}$
Category:Integer Reciprocal Space
Category:Examples of Limit Points
Zero is Omega-Accumulation Point of Integer Reciprocal Space Union with Closed Interval

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $B$ be the uncountable set:
:$B := A \cup \closedint 2 3$
where $\closedint 2 3$ is a closed interval of $\R$.
$2$ and $3$ are to all intents arbitrary, but convenient.
Then $0$ is an $\omega$-accumulation point of $B$ in $\R$.
== Proof ==
Let $U$ be an open set of $\R$ which contains $0$.
From Open Sets in Real Number Line, there exists an open interval $I$ of the form:
:$I := \openint {-a} b \subseteq U$
By the Axiom of Archimedes:
:$\exists n \in \N: n > \dfrac 1 b$
and so:
:$\exists n \in \N: \dfrac 1 n < b$
Let:
:$M := \set {m \in \N: m \ge n}$
Then:
:$\forall m \in M: 0 < \dfrac 1 m < b$
Thus:
:$\forall m \in \N, m \ge n: \dfrac 1 m \in I \cap B$
Thus an open set $U$ which contains $0$ contains a countably infinite number of elements of $B$ (distinct from $0$).
Thus, by definition, $0$ is an $\omega$-accumulation point of $B$ in $\R$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $1 \ \text{(b)}$
Category:Integer Reciprocal Space
Category:Omega-Accumulation Points
Zero is Smallest Ordinal

== Theorem ==
The natural number $0$ is the smallest ordinal.
== Proof ==
Let $\On$ denote the class of all ordinals.
By Zero is Ordinal, $0$ is an element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$
By Empty Class is Subclass of All Classes:
:$\forall \alpha \in \On: \O \subseteq \alpha$
Hence the result by definition of smallest element.
Category:Ordinals
Category:Zero
Zero is Zero Element for Natural Number Multiplication

== Theorem ==
Let $\N$ be the natural numbers.
Then $0$ is a zero element for multiplication:
:$\forall n \in \N: 0 \times n = 0 = n \times 0$
== Proof ==
Proof by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$0 \times n = 0 = n \times 0$
=== Basis for the Induction ===
By definition, we have:
:$0 \times 0 = 0 = 0 \times 0$
Thus $\map P 0$ is seen to be true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis $\map P k$:
:$0 \times k = 0 = k \times 0$
Then we need to show that $\map P {k + 1}$ follows directly from $\map P k$:
:$0 \times \paren {k + 1} = 0 = \paren {k + 1} \times 0$
=== Induction Step ===
This is our induction step:
By definition:
:$\paren {k + 1} \times 0 = 0$
So $\map P k \implies \map P {k + 1}$, and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: 0 \times n = 0 = n \times 0$
== Sources ==
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(g)}$
Category:Natural Numbers
Zero is both Positive and Negative

== Theorem ==
The number $0$ (zero) is the only (real) number which is both:
:a positive (real) number
and
:a negative (real) number.
== Proof ==
Let $x$ be a real number which is both positive and negative.
Thus:
:$x \in \set {x \in \R: x \ge 0}$
and:
:$x \in \set {x \in \R: x \le 0}$
and so:
:$0 \le x \le 0$
from which:
:$x = 0$
== Also defined as ==
In , $0$ is considered to be included in both the set of positive real numbers and the set of negative real numbers.
However, many sources consider $0$ to be neither positive nor negative.
Hence under that convention this result is no longer true.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
Category:Numbers
Zero is not Condensation Point of Integer Reciprocal Space Union with Closed Interval

== Theorem ==
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $B$ be the uncountable set:
:$B := A \cup \closedint 2 3$
where $\closedint 2 3$ is a closed interval of $\R$.
$2$ and $3$ are to all intents arbitrary, but convenient.
Then $0$ is not a condensation point of $B$ in $\R$.
== Proof ==
Let $U$ be an open set of $\R$ which contains $0$.
From Open Sets in Real Number Line, there exists an open interval $I$ of the form:
:$I := \openint {-a} b \subseteq U$
From Zero is Omega-Accumulation Point of Integer Reciprocal Space Union with Closed Interval, there is a countably infinite number of points of $B$ in $U$.
However, when $b < 2$ there is not an uncountable number of points of $B$ in $I$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $1 \ \text{(b)}$
Category:Integer Reciprocal Space
Category:Condensation Points
Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1

== Theorem ==
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:
Then $0$ is not a limit of $\sequence {a_n}$.
== Proof ==
The open interval $\openint {-1} 1$ contains $0$, and also contains all terms of $\sequence {a_n}$ with odd indices greater than $1$.
However, all terms of $\sequence {a_n}$ with even indices are outside $\openint {-1} 1$.
Hence $0$ cannot be a limit of $\sequence {a_n}$.
== Also see ==
* Accumulation Point of Sequence of Reciprocals and Reciprocals + 1 where it is shown that $0$ is an accumulation point of $\sequence {a_n}$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $3$
Category:Sequences
Category:Reciprocals
Zero of Cardinal Product is Zero

== Theorem ==
Let $\mathbf a$ be a cardinal.
Then:
: $\mathbf 0 \mathbf a = \mathbf 0$
where $\mathbf 0 \mathbf a$ denotes the product of the (cardinal) zero and $\mathbf a$.
That is, $\mathbf 0$ is the zero element of the product operation on cardinals.
== Proof ==
Let $\mathbf a = \card A$ for some set $A$.
From the definition of (cardinal) zero, $\mathbf 0$ is the cardinal associated with the empty set $\O$.
We have by definition of product of cardinals that $\mathbf 0 \mathbf a$ is the cardinal associated with $\O \times A$.
But from Cartesian Product is Empty iff Factor is Empty:
:$\O \times A = \O$
Hence the result.
== Sources ==
* : $\S 8$: Theorem $8.4: \ (4)$
Category:Cardinals
Zero of Field is Unique/Proof 2

== Theorem ==
== Proof ==
Let $0_1$ and $0_2$ both be elements of $F$ such that:
:$\forall a \in F: a + 0_1 = a$
:$\forall a \in F: a + 0_2 = a$
Then:
:$0_1 + 0_2 = 0_2$
because $0_1$ is a zero element
:$0_1 + 0_2 = 0_1$
because $0_2$ is a zero element
Hence:
:$0_1 = 0_2$
and the two zero elements are the same.
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $1 \ \text{(i)}$
Category:Zero of Field is Unique
Zero of Integral Domain is Unique

== Theorem ==
Let $\struct {D, +, \times}$ be an integral domain.
Then the zero of $\struct {D, +, \times}$ is unique.
== Proof ==
By definition, an integral domain is a ring.
The result the follows from Ring Zero is Unique.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $1 \ \text{(i)}$
Category:Integral Domains
Zero of Inverse Completion of Integral Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $\struct {K, \circ}$ be the inverse completion of $\struct {D, \circ}$ as defined in Inverse Completion of Integral Domain Exists.
Let $x \in K: x = \dfrac p q$ such that $p = 0_D$.
Then $x$ is equal to the zero of $K$.
That is, any element of $K$ of the form $\dfrac {0_D} q$ acts as the zero of $K$.
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \ominus$ as in the Inverse Completion of Integral Domain Exists.
That is, $\eqclass {\tuple {a, b} } \ominus$ is an equivalence class of elements of $D \times D^*$ under the congruence relation $\ominus$.
$\ominus$ is the congruence relation defined on $D \times D^*$ by:
:$\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
By Inverse Completion of Integral Domain Exists:
:$\dfrac p q \equiv \eqclass {\tuple {p, q} } \ominus$
From Equality of Division Products, two elements $\dfrac a b, \dfrac c d$ of $K$ are equal :
:$a \circ d = b \circ c$
This is consistent with the fact that two elements $\eqclass {\tuple {a, b} } \ominus, \eqclass {\tuple {c, d} } \ominus$ of $K$ are equal $a \circ d = b \circ c$.
Suppose $a = 0_D$ and $a \circ d = b \circ c$.
Hence:
:$\eqclass {\tuple {0_D, b} } \ominus = \eqclass {\tuple {0_D, d} } \ominus$
Thus all elements of $K$ of the form $\eqclass {\tuple {0_D, k} } \ominus$ are equal, for all $k \in D^*$.
To emphasise the irrelevance of the $k$, we will abuse our notation and write:
:$\eqclass {\tuple {0_D, k} } \ominus$
as
:$\eqclass {0_D} \ominus$
Next, by Product of Division Products, we have that $\ds \frac a b \circ \frac c d = \frac {a \circ b} {c \circ d}$.
Again abusing our notation, we will write:
:$\eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {c, d} } \ominus$
to mean:
:$\eqclass {\tuple {a \circ c, b \circ d} } \ominus$
So:
Hence:
:$\eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus$
So $\eqclass {0_D} \ominus$ fulfils the role of a zero for $\tuple {K, \circ}$ as required.
Also we have that:
So $\eqclass {0_D} \ominus$ is idempotent.
It follows that $\eqclass {0_D} \ominus$ can be identified with $0_D$ from the mapping $\psi$ as defined in Construction of Inverse Completion.
Category:Integral Domains
Category:Inverse Completions
Zero of Power Set with Intersection

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Consider the algebraic structure $\struct {\powerset S, \cap}$, where $\cap$ denotes set intersection.
Then the empty set $\O$ serves as the zero element for $\struct {\powerset S, \cap}$.
== Proof ==
From Empty Set is Element of Power Set:
:$\O \in \powerset S$
From Intersection with Empty Set:
:$\forall A \subseteq S: A \cap \O = \O = \O \cap A$
By definition of power set:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$
Thus we see that $\O$ acts as the zero element for $\struct {\powerset S, \cap}$.
== Sources ==
* : Exercise $1.4: \ 10$
* : $\S 4.3$. Units and zeros: Example $75$
Category:Set Intersection
Category:Power Set
Category:Empty Set
Category:Zero Elements
Zero of Power Set with Union

== Theorem ==
Let $S$ be a set and let $\powerset S$ be its power set.
Consider the algebraic structure $\struct {\powerset S, \cup}$, where $\cup$ denotes set union.
Then $S$ serves as the zero element for $\struct {\powerset S, \cup}$.
== Proof ==
We note that by Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$ from the definition of the power set.
From Union with Superset is Superset‎, we have:
:$A \subseteq S \iff A \cup S = S = S \cup A$.
By definition of power set:
:$A \subseteq S \iff A \in \powerset S$
So:
:$\forall A \in \powerset S: A \cup S = S = S \cup A$
Thus we see that $S$ acts as the zero.
== Sources ==
* : Exercise $1.4: \ 10$
* : $\S 4.3$. Units and zeros: Example $75$
Category:Set Union
Category:Power Set
Category:Zero Elements
Zero of Ring of Bounded Continuous Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\map {C^*} {S, \R}, +, *}$ be the ring of bounded continuous real-valued functions from $S$.
Then:
:the zero of $\struct{\map {C^*} {S, \R}, +, *}$ is the constant mapping $0_{\R^S} : S \to \R$ defined by:
::$\forall s \in S : \map {0_{\R^S}} s = 0$
== Proof ==
Let $\struct{\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
From Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a subring of $\struct{\map C {S, \R}, +, *}$
From Zero of Ring of Continuous Real-Valued Functions:
:the zero of $\struct{\map C {S, \R}, +, *}$ is the constant mapping $0_{\R^S} : S \to R$ defined by:
::$\forall s \in S : \map {0_{\R^S}} s = 0$
From Zero of Subring is Zero of Ring:
:$0_{\R^S}$ is the zero of $\struct{\map {C^*} {S, \R}, +, *}$
== Also see ==
* Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions
* Additive Inverse in Ring of Bounded Continuous Real-Valued Functions
* Unity of Ring of Bounded Continuous Real-Valued Functions
* Ring of Bounded Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Bounded Continuous Real-Valued Functions
Zero of Ring of Continuous Mappings

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring with zero $0_R$.
Let $\struct{\map C {S, R}, +, *}$ be the ring of continuous mappings from $S$ to $R$.
Then:
:the zero of $\struct{\map C {S, R}, +, *}$ is the constant mapping $0_{R^S} : S \to R$ defined by:
::$\forall s \in S : \map {0_{R^S}} s = 0_R$
== Proof ==
Let $\struct {R^S, +, *}$ be the ring of mappings from $S$ to $R$.
From Ring of Continuous Mappings is Subring of All Mappings:
:$\struct{\map C {S, R}, +, *}$ is a subring of $\struct {R^S, +, *}$
From Induced Structure Identity:
:the zero of $\struct {R^S, +, *}$ is the constant mapping $0_{R^S} : S \to R$ defined by:
::$\forall s \in S : \map {0_{R^S}} s = 0_R$
From Zero of Subring is Zero of Ring:
:$0_{R^S} \in \map C {S, R}$
== Also see ==
* Ring of Continuous Mappings is Subring of All Mappings
* Additive Inverse in Ring of Continuous Mappings
* Unity of Ring of Continuous Mappings
* Commutativity of Ring of Continuous Mappings
Category:Rings of Continuous Mappings
Zero of Ring of Continuous Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
Then:
:the zero of $\struct{\map C {S, \R}, +, *}$ is the constant mapping $0_{\R^S} : S \to \R$ defined by:
::$\forall s \in S : \map {0_{\R^S}} s = 0$
== Proof ==
By definition of ring of continuous real-valued functions:
:$\struct {\map C {S, \R}, +, *}$ is the ring of continuous mappings from $S$ to $\R$.
From Zero of Ring of Continuous Mappings:
:the zero of $\struct{\map C {S, \R}, +, *}$ is the constant mapping $0_{\R^S} : S \to \R$ defined by:
::$\forall s \in S : \map {0_{\R^S}} s = 0$
== Also see ==
* Ring of Continuous Real-Valued Functions is Ring
* Unity of Ring of Continuous Real-Valued Functions
* Ring of Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.3$
Category:Rings of Continuous Real-Valued Functions