nyuuzyou/wikis huggingface dataset:
Dipper Semigroup is Commutative Semigroup

== Theorem ==
The dipper semigroup is a commutative semigroup.
== Proof ==
Recall the definition of the dipper semigroup:
Let $m, n \in \Z$ be integers such that $m \ge 0, n > 0$.
Let $\N_{< \paren {m \mathop + n} }$ denote the initial segment of the natural numbers:
:$\N_{< \paren {m \mathop + n} } := \set {0, 1, \ldots, m + n - 1}$
Let $+_{m, n}$ be the dipper operation on $\N_{< \paren {m \mathop + n} }$:
:$\forall a, b \in \N_{< \paren {m \mathop + n} }: a +_{m, n} b = \begin{cases} a + b & : a + b < m \\ a + b - k n & : a + b \ge m \end{cases}$
where $k$ is the largest integer satisfying:
:$m + k n \le a + b$
Taking the semigroup axioms in turn:
=== ===
Let $a, b \in \N_{< \paren {m \mathop + n} }$ be arbitrary.
From Dipper Operation is Closed on Initial Segment:
:$a +_{m, n} b \in \N_{< \paren {m \mathop + n} }$
Thus $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ is closed.
=== ===
We have from Dipper Operation is Associative that $+_{m, n}$ is associative.
The semigroup axioms are thus seen to be fulfilled, and so $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ is a semigroup.
Then we note that Integer Addition is Commutative:
:$a + b = b + a$
and so it follows directly that $a +_{m, n} b = b +_{m, n} a$.
Hence $\struct {N_{< \paren {m \mathop + n} }, +_{m, n} }$ is a commutative semigroup.
Category:Dipper Semigroups
Category:Examples of Commutative Semigroups
Dirac's Theorem/Proof 1

== Theorem ==
== Proof ==
Let $P = p_1 p_2 \ldots p_k$ be the longest path in $G$.
If $p_1$ is adjacent to some vertex $v$ not in $P$, then the path $v p_1 p_2 \ldots p_k$ would be longer than $P$, contradicting the choice of $P$.
The same argument can be made for $p_k$.
So both $p_1$ and $p_k$ are adjacent only to vertices in $P$.
Since $\map \deg {p_1} \ge \dfrac n 2$ and $p_1$ cannot be adjacent to itself, $k \ge \dfrac n 2 + 1$.
;Claim
There is some value of $j \in \set {1, 2, \ldots, k}$ such that:
:$p_j$ is adjacent to $p_k$
and:
:$p_{j + 1}$ is adjacent to $p_1$.
 that the claim is not true.
Then since all vertices adjacent to $p_1$ or $p_k$ lie on $P$, there must be at least $\map \deg {p_1}$ vertices on $P$ not adjacent to $p_k$.
Since all the vertices adjacent to $p_k$ and $p_k$ itself also lie on $P$, the path must have at least $\map \deg {p_1} + \map \deg {p_k} + 1 \ge n + 1$ vertices.
But $G$ has only $n$ vertices: a contradiction.
This gives a cycle $C = p_{j + 1} p_{j + 2} \ldots p_k p_j p_{j - 1} \ldots p_2 p_1 p_{j + 1}$.
 $G \setminus C$ is non-empty.
Then since $G$ is connected, there must be a vertex $v \in G \setminus C$ adjacent to some $p_i$.
So the path from $v$ to $p_i$ and then around $C$ to the vertex adjacent to $p_i$ is longer than $P$, contradicting the definition of $P$.
Therefore all vertices in $G$ are contained in $C$, making $C$ a Hamilton cycle.
== Sources ==
* : $\S 3.2$: The Salesman's Problem: An Introduction to Hamiltonian Graphs: Theorem $3.4$
Category:Dirac's Theorem
Dirac Comb is Distribution

== Theorem ==
Let $\phi \in \map \DD \R$ be a test function.
Suppose $\map {\operatorname {III} } 0$ is a Dirac comb such that:
:$\ds \map {\map {\operatorname {III} } 0} \phi := \sum_{n \mathop \in \Z} \map \phi n$
Then $\map {\operatorname {III} } 0$ is a distribution.
== Proof ==
By definition of test function, $\phi$ is supported on a compact subset of $\R$.
Hence:
:$\exists N \in \N : \forall x \in \R \setminus \closedint {-N} N : \map \phi x = 0$
Therefore:
:$\ds \sum_{n \mathop \in \Z} \map \phi n = \sum_{n \mathop = - N}^N \map \phi n$
This is a finite sequence of real numbers.
Thus, the sequence converges, and $\map {\operatorname {III} } 0$ is a mapping such that $\map {\operatorname {III} } 0 : \map \DD \R \to \R$.
=== Linearity ===
Follows from Summation is Linear.
=== Convergence in $\map \DD \R$ ===
Let $\sequence {\phi_n}$ be a convergent sequence in $\map \DD \R$ with the limit $\mathbf 0$.
By definition of convergence, all $\sequence {\phi_n}$ are supported on a compact subset of $\R$, say, $\closedint {-K} K$ with $K \in \N$.
By definition of convergence, $\sequence {\phi_n}$ converges to $\mathbf 0$ uniformly.
Hence:
:$\ds \forall k \in \R : \size k < K : \lim_{n \mathop \to \infty} \map {\phi_n} k = 0$
Therefore:
By definition, the Dirac comb is a distribution.
== Also see ==
* Closed Real Interval is Compact
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Convergence
Category:Examples of Distributions
Dirac Delta Distribution is Tempered Distribution

== Theorem ==
Let $\delta$ be the Dirac delta distribution.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then:
:$\delta \in \map {\SS'} \R$
== Proof ==
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Consider the mapping $\phi \stackrel \delta \longrightarrow \map \phi 0 : \map \SS \R \to \C$.
We have that the Schwartz space is a vector space.
Let $\phi$ be a linear superposition of Schwartz test functions.
Then $\map \phi 0$ is also a linear supperposition evaluated at $x = 0$.
Hence, the mapping $\phi \stackrel \delta \longrightarrow \map \phi 0 : \map \SS \R \to \C$ is linear.
Suppose, $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
We have that:
:$\ds \forall x \in \R : \forall l, m \in \N : 0 \le \size {x^l \map {\phi_n^{\paren m} } x} \le \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x}$
Specifically, for $x = 0$, $l = 0$, and $m = 0$ it holds that:
:$\ds 0 \le \size {\map {\phi_n} 0} \le \sup_{x \mathop \in \R} \size {\map {\phi_n} x}$
Taking the limit $n \to \infty$ and applying the squeeze theorem yields:
:$\ds \lim_{n \mathop \to \infty} \map {\phi_n} 0 = 0$.
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Dirac Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Then $\delta_x$ is a measure.
== Proof ==
Let us verify in turn that $\delta_x$ satisfies the axioms for a measure.
=== Axiom $(1)$ ===
By definition of the Dirac measure, $\map {\delta_x} E \ge 0$ for all $E \in \Sigma$.
=== Axiom $(2)$ ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets.
It follows that if for some $m \in \N$, $x \in E_m$, it must be that $n \ne m$ implies $x \notin E_n$.
Now suppose $x \in E_m$ for some $m \in \N$.
Then by definition of set union, $x \in \ds \bigcup_{n \mathop \in \N} E_n$.
Thus:
because $\map {\delta_x} {E_n} = 0$ $n \ne m$, and $1$ otherwise.
Finally, if $x \notin E_n$ for all $n \in \N$, then by definition of set union:
:$x \notin \ds \bigcup_{n \mathop \in \N} E_n$
so that:
Hence, from Proof by Cases:
:$\ds \sum_{n \mathop \in \N} \map {\delta_x} {E_n} = \map {\delta_x} {\bigcup_{n \mathop \in \N} E_n}$
=== Axiom $(3)$ ===
By definition of the Dirac measure, $\map {\delta_x} x = 1$.
Hence there is an $E \in \Sigma$ such that $\map {\delta_x} E$ is finite.
Thus, $\delta_x$, satisfying all the axioms, is a measure.
Category:Measures
Category:Dirac Measure
Dirac Measure is Probability Measure

== Theorem ==
Let $\struct {X, \AA}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Then $\delta_x$ is a probability measure.
== Proof ==
By Dirac Measure is Measure, $\delta_x$ is a measure.
Also, $\map {\delta_x} X = 1$ because $x \in X$.
Hence $\delta_x$ is a probability measure.
Category:Probability Measures
Direct Image Mapping is Bijection iff Mapping is Bijection

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping of $\RR$.
Then $\RR \subseteq S \times T$ is a bijection $\RR^\to: \powerset S \to \powerset T$ is a bijection.
== Proof ==
Category:Bijections
Category:Direct Image Mappings
Direct Image Mapping of Domain is Image Set of Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $\powerset S$ and $\powerset T$ be their power sets.
Let $f \subseteq S \times T$ be a mapping from $S$ to $T$.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$:
:$\forall X \in \powerset S: \map {f^\to} X = \begin {cases} \set {t \in T: \exists s \in X: \map f s = t} & : X \ne \O \\ \O & : X = \O \end {cases}$
Then:
:$\map {f^\to} S = \Img f$
where $\Img f$ is the image set of $f$.
== Proof ==
Category:Direct Image Mappings
Direct Image Mapping of Domain is Image Set of Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\powerset S$ and $\powerset T$ be their power sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping of $\RR$:
:$\forall X \in \powerset S: \map {\RR^\to} X = \begin {cases} \set {t \in T: \exists s \in X: \tuple {x, t} \in \RR} & : X \ne \O \\ \O & : X = \O \end {cases}$
Then:
:$\map {\RR^\to} {\Dom \RR} = \Img \RR$
where:
:$\Dom \RR$ is the domain of $\RR$
:$\Img \RR$ is the image set of $\RR$.
== Proof ==
Category:Direct Image Mappings
Direct Image Mapping of Left-Total Relation is Empty iff Argument is Empty

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR: S \to T$ be a left-total relation on $S \times T$.
Let $\RR^\to$ be the direct image mapping of $\RR$:
:$\RR^\to: \powerset S \to \powerset T: \map {\RR^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \RR}$
Then:
:$\map {\RR^\to} X = \O \iff X = \O$
== Proof ==
=== Sufficient Condition ===
Let $\map {\RR^\to} X = \O$.
By definition of direct image mapping:
:$\set {t \in T: \exists s \in X: \tuple {s, t} \in \RR} = \O$
That is:
:$\neg \exists s \in X: \tuple {s, t} \in \RR$
But as $\RR$ is a left-total relation:
:$\forall s \in X: \exists t \in T: \tuple {s, t} \in \RR$
Thus:
:$\neg \exists s \in X$
and so:
:$X = \O$
=== Sufficient Condition ===
Let $X = \O$.
Then by definition of direct image mapping:
:$\map {\RR^\to} X = \O$
Category:Direct Image Mappings
Direct Image Mapping of Mapping is Empty iff Argument is Empty

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $f^\to$ be the direct image mapping of $f$:
:$f^\to: \powerset S \to \powerset T: \map {f^\to} X = \set {t \in T: \exists s \in X: \map f s = t}$
Then:
:$\map {f^\to} X = \O \iff X = \O$
== Proof ==
By definition, a mapping is a left-total relation.
The result then follows from Direct Image Mapping of Left-Total Relation is Empty iff Argument is Empty.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections
Category:Direct Image Mappings
Direct Image Mapping of Mapping is Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping on $S \times T$.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$:
:$\forall X \in \powerset S: \map {f^\to} X = \set {t \in T: \exists s \in X: \map f s = t}$
Then $f^\to$ is indeed a mapping.
== Proof ==
$f$, being a mapping, is also a relation.
Hence Direct Image Mapping of Relation is Mapping can be applied directly.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections
Category:Direct Image Mappings
Direct Image Mapping of Relation is Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping of $\RR$:
:$\forall X \in \powerset S: \map {\RR^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \RR}$
Then $\RR^\to$ is indeed a mapping.
== Proof ==
Take the general relation $\RR \subseteq S \times T$.
Let $X \subseteq S$, that is $X \in \powerset S$.
Suppose $X = \O$.
Then from Image of Empty Set is Empty Set:
:$\map {\RR^\to} X = \O \subseteq T$
Suppose $X = S$.
Then from Image is Subset of Codomain: Corollary 1:
:$\map {\RR^\to} X = \Img \RR \subseteq T$
Finally, suppose $\O \subset X \subset S$.
From Image is Subset of Codomain, again:
:$\map {\RR^\to} X \subseteq T$
Now, from the definition of the power set, we have that:
:$Y \subseteq T \iff Y \in \powerset T$
We defined $\RR^\to \subseteq \powerset S \times \powerset T$ such that:
:$\RR^\to: \powerset S \to \powerset T: \map {\RR^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \RR}$
By definition, there is only one $\map {\RR^\to} X$ for any given $X$, and so $\RR^\to$ is many-to-one.
We have shown that:
:$\forall X \subseteq S: \map {\RR^\to} X \in \powerset T$
So:
:$\forall X \in \powerset S: \exists_1 Y \in \powerset T: \map {\RR^\to} X = Y$
So:
:$\RR^\to$ is defined for all $X \in \powerset S$
and therefore $\RR^\to$ is a mapping.
== Also see ==
* Inverse Image Mapping of Relation is Mapping
Category:Direct Image Mappings
Direct Image Mapping of Surjection is Surjection

== Theorem ==
Let $f: S \to T$ be a surjection.
Then the direct image mapping of $f$:
:$f^\to: \powerset S \to \powerset T$
is a surjection.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Mapping is Surjection iff Direct Image Mapping is Surjection
Category:Surjections
Category:Direct Image Mappings
Category:Power Set
Category:Direct Image Mapping of Surjection is Surjection
Direct Image Mapping of Surjection is Surjection/Proof 1

== Theorem ==
== Proof ==
Let $f: S \to T$ be a surjection.
Then:
:$\forall y \in T: \exists x \in S: f \paren x = y$
From the Quotient Theorem for Surjections, there is one and only one bijection $r: S / \RR_f \to T$ such that $r \circ q_{\RR_f} = f$.
Each element of $S / \RR_f$ is a subset of $S$ and therefore an element of $\powerset S$.
Thus as $r: S / \RR_f \to T$ is a bijection, and hence a fortiori also an injection:
:$\forall X_1, X_2 \in \powerset S: \map r {X_1} = \map r {X_2} \implies X_1 = X_2$
Because $f$ is a surjection, every $y \in T$ is mapped to by exactly one element of the partition of $S$ defined by $\RR_f$.
Let $T = \set {y_1, y_2, \ldots}$.
Let the partition defined by $\RR_f$ be $\bigcup \paren {X_1, X_2, \ldots}$ where $\map r {X_n} = y_n$.
Let $Y_r \in \powerset T$, such that $Y_r = \set {y_{r_1}, y_{r_2}, \ldots}$.
Then:
:$\map {f^\to} {X_r} = Y_r$
where $\ds X_r = \bigcup \paren {X_{r_1}, X_{r_2}, \ldots}$.
As $\set {X_1, X_2, \ldots}$ is a partition of $S$, $\forall Y_r \in \powerset T: X_r$ is unique.
Thus $f^\to: \powerset S \to \powerset T$ is a surjection.
Category:Direct Image Mapping of Surjection is Surjection
Direct Image Mapping of Surjection is Surjection/Proof 2

== Theorem ==
== Proof ==
Let $f: S \to T$ be a surjection.
By definition, $f^\to$ is defined by sending subsets of $S$ to their image under $f$.
That is:
:$\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$
To prove that $f^\to$ is a surjection, we need to show that every subset of $T$ is the image under $f^\to$ of some subset of $S$.
Let $Y \subseteq T$.
Since $f$ is a surjection, by definition we have that:
:$\forall t \in T: \exists s \in S: \map f s = t$
Hence (possibly requiring the ), we can select for each $y \in Y$ some $s_y \in S$ such that $\map f {s_y} = y$.
Define $X_Y$ to be:
:$X_Y := \set {s_y: y \in Y}$
Then:
Thus $f^\to: \powerset S \to \powerset T$ is a surjection.
Category:Direct Image Mapping of Surjection is Surjection
Direct Image Mapping of Surjection is Surjection/Proof 3

== Theorem ==
== Proof ==
Let $f^\gets$ be the inverse image mapping of $f$.
Let $Y \in \powerset T$.
Let $X = \map {f^\gets} Y$.
By Subset equals Image of Preimage iff Mapping is Surjection:
:$\map {f^\to} X = Y$
As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is surjective.
Category:Direct Image Mapping of Surjection is Surjection
Direct Image of Intersection with Inverse Image

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let:
:$f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$
:$f^\gets: \powerset T \to \powerset S$ denote the inverse image mapping of $f$
where $\powerset S$ denotes the power set of $S$.
Then:
:$\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$
== Proof ==
Let $A \in \powerset S, B \in \powerset T$ be arbitrary.
Then:
Let $x \in \map {f^\to} A \cap B$.
Thus we have:
:$\map {f^\to} {A \cap \map {f^\gets} B} \subseteq \map {f^\to} A \cap B$
and:
:$\map {f^\to} A \cap B \subseteq \map {f^\to} {A \cap \map {f^\gets} B}$
and so:
:$\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$
by definition of set equality.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $5$
Category:Direct Image Mappings
Category:Inverse Image Mappings
Category:Set Intersection
Direct Image of Inverse Image of Direct Image equals Direct Image Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Let:
:$f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$
:$f^\gets: \powerset T \to \powerset S$ denote the inverse image mapping of $f$
where $\powerset S$ denotes the power set of $S$.
Then:
:$f^\to \circ f^\gets \circ f^\to = f^\to$
where $\circ$ denotes composition of mappings.
== Proof ==
Thus we have:
:$\forall A \in \powerset S: \map {f^\to} A \subseteq \map {\paren {f^\to \circ f^\gets \circ f^\to} } A \subseteq \map {f^\to} A$
and the result follows.
== Also see ==
* Inverse Image of Direct Image of Inverse Image equals Inverse Image Mapping
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $4$
Category:Direct Image Mappings
Category:Inverse Image Mappings
Direct Iteration/Examples/x = cos x

== Example of Direct Iteration ==
The following is a graph of the direct iteration $x_{n + 1} = \map \cos {x_n}$ for $n = 0, 1, 2, 3, 4$ where $x_0$ is an arbitrary point such that $1 < x_0 < \dfrac \pi 2$:
:
== Sources ==
* 
* 
Category:Examples of Direct Iteration
Direct Product Norm is Norm

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}}$ and $\struct {Y, \norm {\, \cdot \,}}$ be normed vector spaces.
Let $V = X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.
Let $\norm {\tuple {x, y} }$ be the direct product norm.
Then $\norm {\tuple {x, y} }$ is a norm on $V$.
== Proof ==
=== Positive Definiteness ===
Let $\tuple {x , y} \in V$.
Then:
Suppose $\norm {\tuple {x, y} } = 0$.
Then:
Hence, $\norm x = 0$
=== Positive Homogeneity ===
=== Triangle Inequality ===
Together:
:$\map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2}} \le \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }$
Then:
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Examples of Norms
Category:Direct Product of Vector Spaces
Category:Normed Vector Spaces
Direct Product iff Nontrivial Idempotent

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Then $R$ is the direct product of two non-trivial rings $R$ contains an idempotent element not equal to $0_R$ or $1_R$.
== Proof ==
=== Necessary Condition ===
Suppose $R_1$ and $R_2$ are non-trivial rings, such that $R = R_1 \times R_2$.
Then:
Thus $\tuple {1_{R_1}, 0_{R_2} }$ is an idempotent element of $R$ not equal to $0_R$ or $1_R$.
=== Sufficient Condition ===
Suppose there exists an idempotent element $e \in R$ not equal to $0_R$ or $1_R$.
That is:
:$0_R, 1_R \ne e$
:$e^2 = e$
Note that:
:$\paren {1 - e}^2 = 1 - e$
Let $R_1 = \ideal e$, the ideal generated by $e$.
Let $R_2 = \ideal {1 - e}$, the ideal generated by $1 - e$.
By Ring by Idempotent $\struct {R_1, +, \circ}$ and $\struct {R_2, +, \circ}$ are commutative rings with unity.
Define $R_1 \times R_2$ to be the direct product ring of $\struct {R_1, +, \circ}$ and $\struct {R_2, +, \circ}$.
We define a ring homomorphism $\phi : R \to R_1 \times R_2$ by:
:$\map \phi a := \tuple {e a, \paren {1 - e} a}$
By Ring Homomorphism by Idempotent $\phi$ is a ring homomorphism.
Let $a \in \map \ker \phi$.
Then:
:$e a = \paren {1 - e} a = 0$.
It follows that:
:$a = e a + \paren {1 - e} a = 0_R$
So:
:$\map \ker \phi = 0$
Thus $\phi$ is injective.
Let $\struct {e c, \paren {1 - e} d} \in R_1 \times R_2$ for arbitrary $c, d \in R$.
Then:
Thus $\phi$ is surjective.
Hence $\phi$ is a ring isomorphism.
Category:Commutative Algebra
Direct Product of *-Algebras is *-Algebra

== Theorem ==
Let $\tuple {A, \square}$ and $\tuple {B, \diamond}$ be $\ast$-algebras over $\C$. 
Let $A \times B$ be the direct product of $A$ and $B$.
Define $\ast : A \times B \to A \times B$ by:
:$\tuple {a, b}^\ast = \tuple {a^\square, b^\diamond}$
for each $\tuple {a, b} \in A \times B$. 
Then $\tuple {A \times B, \ast}$ is a $\ast$-algebra.
== Proof ==
We show that $\ast$ is a involution. 
=== Proof of $(\text C^\ast 1)$ ===
Let $\tuple {a, b} \in A \times B$.
We have:
=== Proof of $(\text C^\ast 2)$ ===
Let $\tuple {a, b}, \tuple {a', b'} \in A \times B$.
We have:
=== Proof of $(\text C^\ast 3)$ ===
Let $\tuple {a, b}, \tuple {a', b'} \in A \times B$. 
Then:
=== Proof of $(\text C^\ast 4)$ ===
Let $\lambda \in \C$ and $\tuple {a, b} \in A \times B$.
We have:
Category:*-Algebras
Direct Product of Algebras is Algebra

== Theorem ==
Let $K$ be a field.
Let $A$ and $B$ be algebras over $K$. 
Let $\tuple {A \times B, +_{A \times B}, \cdot_{A \times B} }$ be the direct product of $A$ and $B$ as vector spaces.
Define $\cdot_{A \times B} : A \times B \to A \times B$ by:
:$\tuple {a, b} \tuple {a', b'} = \tuple {a a', b b'}$
for each $\tuple {a, b}, \tuple {a', b'} \in A \times B$. 
Then $\tuple {A \times B, +_{A \times B}, \cdot_{A \times B}, \circ_{A \times B} }$ is an algebra.
== Proof ==
We show that $\circ_{A \times B}$ is bilinear.
Let $\lambda \in K$ and $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {a_3, b_3} \in A \times B$ .
We have:
We also have:
So $\circ_{A \times B}$ is bilinear and $\tuple {A \times B, +_{A \times B}, \cdot_{A \times B}, \circ_{A \times B} }$ is an algebra.
Category:Algebras over Fields
Direct Product of Banach Spaces is Banach Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.
Let $V = X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.
Let $\norm {\, \cdot \,}_{X \times Y}$ be the direct product norm.
Suppose $X$ and $Y$ are Banach spaces.
Then $V$ is a Banach space.
== Proof ==
Let $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ be a Cauchy sequence in $V$:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n > N: \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }_{X \times Y} < \epsilon$
We have that:
Hence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy sequence in $X$.
By assumption, $X$ is a Banach space.
Therefore, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$.
By analogous arguments, $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in Y$.
So, by Convergence in Direct Product Norm, we have: 
:$\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ converges to $\tuple {x, y}$.
By definition, $V$ is a Banach space.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Banach Spaces
Category:Direct Product of Vector Spaces