nyuuzyou/wikis huggingface dataset:
Direct Product of C*-Algebras is C*-Algebra

== Theorem ==
Let $\struct {A, \square, \norm {\, \cdot \,}_A}$ and $\struct {B, \diamond, \norm {\, \cdot \,}_B}$ be $\text C^\ast$-algebras.
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 $\tuple {a, b} \in A \times B$.
Let $\norm {\, \cdot \,}_{A \times B}$ be the direct product norm on $A \times B$. 
Then $\struct {A \times B, \ast, \norm {\, \cdot \,}_{A \times B} }$ is a $\text C^\ast$-algebra.
== Proof ==
From Direct Product of Normed Algebras is Normed Algebra with Direct Product Norm, $\struct {A \times B, \norm {\, \cdot \,}_{A \times B} }$ is a normed algebra. 
From Direct Product of Banach Spaces is Banach Space, $\struct {A \times B, \norm {\, \cdot \,}_{A \times B} }$ is a Banach algebra. 
From Direct Product of *-Algebras is *-Algebra, $\struct {A \times B, \ast}$ is a $\ast$-algebra.
It remains to show $(\text C^\ast 5)$, that:
:$\norm {\tuple {a, b} \tuple {a, b}^\ast}_{A \times B} = \norm {\tuple {a, b} }^2_{A \times B}$
for each $\tuple {a, b} \times A \times B$. 
We have:
Category:C*-Algebras
Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup

== Theorem ==
Let $G$ and $H$ be groups.
Let $\map Z G$ denote the center of $G$.
Let $Z$ and $W$ be central subgroups of $G$ and $H$ respectively.
Let:
:$Z \cong W$
where $\cong$ denotes isomorphism.
Let such a group isomorphism be $\theta: Z \to W$.
Let $X$ be the set defined as:
:$X = \set {\tuple {x, \map \theta x^{-1} }: x \in Z}$
Then $X$ is a central subgroup of $G \times H$.
== Proof ==
First note that by Group Homomorphism Preserves Inverses:
:$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1} = \map \phi x^{-1}$
and so there is no amphiboly in the notation used.
It is established that $\tuple {x, \map \theta x^{-1} } \in G \times H$:
:$x \in G$
:$\map \theta x \in H$ and so $\map \theta x^{-1} \in H$.
Let:
:$e_G$ be the identity element of $G$
:$e_H$ be the identity element of $H$
We have that $e_G \in Z$, and so:
Thus $\tuple {e_G, e_H} \in X$ and so $X \ne \O$.
Then we note that:
and so $\tuple {x^{-1}, \map \theta x}$ is the inverse of $\tuple {x, \map \theta x^{-1} }$ in $H$.
Let $x, y \in G$.
Then $x^{-1} \in G$ from .
Hence if:
:$\tuple {x, \map \theta x^{-1} } \in X$
it follows that:
:$\tuple {x^{-1}, \map \theta x} \in X$
Let $x, y^{-1} \in G$.
Then:
:$\tuple {x, \map \theta x^{-1} } \in X$
and:
:$\tuple {y^{-1}, \map \theta y} \in X$
Thus:
So:
:$\tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y} \in X$
and it follows from the One-Step Subgroup Test that $X$ is a subgroup of $G \times H$.
We have that $Z$ is a central subgroup of $G$.
We also have that the image of $\theta$ is $W$.
From Direct Product of Central Subgroups, $Z \times W$ is a central subgroup of $G \times H$.
It follows that $X$ is a central subgroup of $G \times H$.
== Sources ==
* : Chapter $13$: Direct products: Definition $13.9$
Category:Quotient Groups
Category:Group Isomorphisms
Category:Central Subgroups
Direct Product of Central Subgroups

== Theorem ==
Let $G$ and $H$ be groups.
Let $Z$ and $W$ be central subgroups of $G$ and $H$ respectively.
Then $Z \times W$ is a central subgroup of $G \times H$.
== Proof ==
Let $\tuple {z, w} \in Z \times W$.
Let $\tuple {x, y} \in G \times H$.
Then:
$\tuple {x, y}$ is arbitrary.
Thus $\tuple {z, w}$ commutes with all elements of $G \times H$.
Hence $\tuple {z, w}$ is in the center of $G \times H$
That is, $Z \times W$ is a subgroup of the center $G \times H$.
The result follows by definition of central subgroup.
Category:Central Subgroups
Category:Group Direct Products
Direct Product of Group Homomorphisms is Homomorphism

== Theorem ==
Let $\struct {G, \circ}, \struct {H_1, *_1}$ and $\struct {H_2, *_2}$ be groups.
Let $\struct {H_1 \times H_2, *}$ be the group direct product of $H_1$ and $H_2$.
Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be group homomorphisms.
Let $f_1 \times f_2: g \to H_1 \times H_2$ be the direct product of $f_1$ and $f_2$.
Then $f_1 \times f_2$ is a group homomorphism.
== Proof ==
The direct product of $f_1$ and $f_2$ $f_1 \times f_2: g \to H_1 \times H_2$ is defined as:
:$\forall g \in G: \map {\paren {f_1 \times f_2} } g = \tuple {\map {f_1} g, \map {f_2} g}$
From External Direct Product of Groups is Group, the group direct product $H_1 \times H_2$ is a group.
It remains to be shown that $f_1 \times f_2$ fulfils the morphism property.
Let $g, h \in G$.
Then:
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Group Homomorphisms
Category:Group Direct Products
Direct Product of Modules is Module

== Theorem ==
Let $R$ be a ring.
Let $\family {\struct {M_i, +_i, \circ_i} }_{i \mathop \in I}$ be a family of $R$-modules.
Let $\struct {M, +, \circ}$ be their direct product.
Then $\struct {M, +, \circ}$ is a module.
== Proof ==
From External Direct Product of Abelian Groups is Abelian Group it follows that $(M,+)$ is an abelian group.
We need to show that:
$\forall x, y, \in M, \forall \lambda, \mu \in R$:
: $(1): \quad \lambda \circ \paren {x + y} = \paren {\lambda \circ x} + \paren {\lambda \circ y}$
: $(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} + \paren {\mu \circ x}$
: $(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$
Checking the criteria in order:
=== Criterion 1 ===
: $(1): \quad \lambda \circ \paren {x + y} = \paren {\lambda \circ x} + \paren {\lambda \circ y}$
Let $x = \family {x_i}_{i \mathop \in I}, y = \family {y_i}_{i \mathop \in I} \in M$.
So $(1)$ holds.
=== Criterion 2 ===
: $(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} + \paren {\mu \circ x}$
Let $x = \family {x_i}_{i \mathop \in I} \in M$.
So $(2)$ holds.
=== Criterion 3 ===
: $(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$
Let $x = \family {x_i}_{i \mathop \in I} \in M$.
So $(3)$ holds.
Thus all criteria are seen to hold.
The result follows.
== Also see ==
* Direct Product of Unitary Modules is Unitary Module
Category:Module Theory
Category:Direct Products
Direct Product of Normal Subgroups is Normal

== Theorem ==
Let $G$ and $G'$ be groups.
Let:
:$H \lhd G$
:$H' \lhd G'$
where $\lhd$ denotes the relation of being a normal subgroup.
Then:
:$\paren {H \times H'} \lhd \paren {G \times G'}$
where $H \times H'$ denotes the group direct product of $H$ and $H'$
== Proof ==
Let $\tuple {x, x'} \in G \times G'$ and $\tuple {y, y'} \in H \times H'$.
Then:
Hence:
:$\paren {H \times H'} \lhd {G \times G'}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.14 \ \text{(a)}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \theta$
* : Chapter $11$: The Sylow Theorems: Exercise $2$
Category:Group Direct Products
Category:Normal Subgroups
Direct Product of Normed Algebras is Normed Algebra with Direct Product Norm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,}_A}$ and $\struct {B, \norm {\, \cdot \,}_B}$ be normed algebras over $\GF$.
Let $A \times B$ be the direct product of $A$ and $B$.
Let $\norm {\, \cdot \,}_{A \times B}$ be the direct product norm on $A \times B$.
Then $\struct {A \times B, \norm {\, \cdot \,}_{A \times B} }$ is a normed algebra. 
== Proof ==
From Direct Product Norm is Norm, $\struct {A \times B, \norm {\, \cdot \,}_{A \times B} }$ is a normed vector space.
We just need to show that $\norm {\, \cdot \,}_{A \times B}$ is an algebra norm.
Let $\tuple {a, b}, \tuple {a', b'} \in A \times B$. 
We hae:
:$\norm {\tuple {a, b} \tuple {a', b'} }_{A \times B} = \norm {\tuple {a a', b b'} }_{A \times B} = \max \set {\norm {a a'}_A, \norm {b b'}_B}$
Since $\norm {\, \cdot \,}$ is an algebra norm, we have:
:$\norm {a a'}_A \le \norm a_A \norm {a'}_A \le \max \set {\norm a_A, \norm b_B} \max \set {\norm {a'}, \norm {b'}_B}$
and:
:$\norm {b b'}_B \le \norm b_B \norm {b'}_B \le \max \set {\norm a_A, \norm b_B} \max \set {\norm {a'}, \norm {b'}_B}$
Hence we obtain:
:$\max \set {\norm {a a'}_A, \norm {b b'}_B} \le \max \set {\norm a_A, \norm b_B} \max \set {\norm {a'}, \norm {b'}_B}$
that is:
:$\norm {\tuple {a a', b b'} }_{A \times B} \le \norm {\tuple {a, b} }_{A \times B} \norm {\tuple {a', b'} }_{A \times B}$
Category:Normed Algebras
Direct Product of Solvable Groups is Solvable

== Theorem ==
Let $G$ and $H$ be groups which are solvable.
Then their (external) direct product $G \times H$ is also solvable.
== Proof ==
By Image of Canonical Injection is Normal Subgroup, $G \times \set {e_H}$ is a normal subgroup of $G \times H$.
Also, by Quotient Group of Direct Products, $\paren {G \times H} / \paren {G \times \set{e_H} }$ is isomorphic to $H$.
The result then follows from Group is Solvable iff Normal Subgroup and Quotient are Solvable.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 75 \delta$
Category:Solvable Groups
Category:Group Direct Products
Direct Product of Sylow p-Subgroups is Sylow p-Subgroup

== Theorem ==
Let $G_1$ and $G_2$ be groups.
Let $H_1$ and $H_2$ be subgroups of $G_1$ and $G_2$ respectively.
Let $H_1$ be a Sylow $p$-subgroup of $G_1$.
Let $H_2$ be a Sylow $p$-subgroup of $G_2$.
Then $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.
== Proof ==
By definition of Sylow $p$-subgroup:
:$\order {H_1} = p^r$, where $p^r$ is the highest power of $p$ which is a divisor of $\order {G_1}$.
:$\order {H_2} = p^s$, where $p^s$ is the highest power of $p$ which is a divisor of $\order {G_2}$.
We have that:
:$\order {H_1 \times H_2} = p^{r + s}$
We also have that $p^{r + s}$ is the highest power of $p$ which is a divisor of $\order {G_1 \times G_2}$.
Hence it follows by definition that $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $2$
Category:Group Direct Products
Category:Sylow p-Subgroups
Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces

== Theorem ==
Let $K$ be a topological field.
Let $I$ be a set. 
Let $\family {X_i}_{i \in I}$ be an $I$-indexed family of topological vector spaces over $K$.
Let:
:$\ds X = \prod_{i \mathop \in I} X_i$
be the direct product of $\family {X_i}_{i \in I}$.
Equip $X$ with the product topology.
Then $X$ is Hausdorff $X_i$ is Hausdorff for each $i \in I$. 
== Proof ==
For each $i \in I$, let $\FF_i$ be the set of open neighborhoods of ${\mathbf 0}_{X_i}$, where ${\mathbf 0}_{X_i}$ is the zero vector of $X_i$.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
=== Lemma ===
=== Sufficient Condition ===
Suppose that $X_i$ is Hausdorff for each $i \in I$.
From Characterization of Hausdorff Topological Vector Space, we have:
:$\bigcap \FF_i = \set { {\mathbf 0}_{X_i} }$
Then we have:
:$\ds \bigcap \FF = \prod_{i \mathop \in I} \set { {\mathbf 0}_{X_i} } = {\mathbf 0}_X$
Then from Characterization of Hausdorff Topological Vector Space, $X$ is Hausdorff.
=== Necessary Condition ===
Suppose that $X$ is Hausdorff.
Then:
:$\bigcap \FF = \set { {\mathbf 0}_X}$.
Hence, we have:
:$\ds \prod_{i \mathop \in I} \bigcap \FF_i = \bigcap \FF = \set { {\mathbf 0}_X} = \prod_{i \mathop \in I} \set { {\mathbf 0}_{X_i} }$
Hence we obtain:
:$\ds \bigcap \FF_i = \set { {\mathbf 0}_{X_i} }$
for each $i \in I$.
Hence from Characterization of Hausdorff Topological Vector Space, $X_i$ is Hausdorff.
Category:Direct Product of Vector Spaces
Category:Topological Vector Spaces
Category:Hausdorff Topological Vector Spaces
Category:Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces
Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces/Lemma

== Lemma ==
Let $K$ be a topological field.
Let $I$ be a set. 
Let $\family {X_i}_{i \in I}$ be an $I$-indexed family of topological vector spaces over $K$.
For each $i \in I$, let $\FF_i$ be the set of open neighborhoods of ${\mathbf 0}_{X_i}$, where ${\mathbf 0}_{X_i}$ is the zero vector of $X_i$.
Let $\FF$ be the set of open neighborhoods of ${\mathbf 0}_X$.
Then:
:$\ds \prod_{i \mathop \in I} \bigcap \FF_i = \bigcap \FF$
== Proof ==
We first show:
:$\ds \prod_{i \mathop \in I} \bigcap \FF_i \subseteq \bigcap \FF$
Let:
:$\ds \family {x_i}_{i \mathop \in I} \in \prod_{i \mathop \in I} \bigcap \FF_i$
Let $U \in \FF$. 
From Product Space Local Basis Induced from Factor Spaces Local Bases, there exists $i_1, \ldots, i_n \in I$ and open neighborhoods $U_{i_j}$ of ${\mathbf 0}_{X_{i_j} }$ in $X_{i_j}$ such that:
:$\ds \prod_{i \mathop \in I} U_i \subseteq U$
where we define $U_i = X$ for $i \in I \setminus \set {i_1, \ldots, i_n}$.
For each $j \in \set {1, \ldots, n}$, we have:
:$\ds x_{i_j} \in \bigcap \FF_{i_j} \subseteq U_{i_j}$
Hence, we have:
:$\ds \family {x_i}_{i \mathop \in I} \in \prod_{i \mathop \in I} U_i$
So we obtain that $\family {x_i}_{i \mathop \in I} \in U$.
Since $U \in \FF$ was arbitrary, we have that $\family {x_i}_{i \mathop \in I} \in \bigcap \FF$.
So we obtain:
:$\ds \prod_{i \mathop \in I} \bigcap \FF_i \subseteq \bigcap \FF$
Conversely, let $\family {x_i}_{i \mathop \in I} \in \bigcap \FF$.
Let $i \in I$.
We want to show that $x_i \in \bigcap \FF_i$.
Let $U_i$ be an open neighborhood of ${\mathbf 0}_{X_i}$.
For $j \in I \setminus \set i$, let $U_j = X$. 
Then:
:$\ds U = \prod_{i \mathop \in I} U_i$ is an open neighborhood of ${\mathbf 0}_X$ in $X$. 
So we have $U \in \FF$. 
That is, $\family {x_i}_{i \mathop \in I} \in U$.
Hence $x_i \in U_i$.
Category:Direct Product of Topological Vector Spaces is Hausdorff iff Hausdorff Factor Spaces
Direct Product of Unique Sylow p-Subgroups is Unique Sylow p-Subgroup

== Theorem ==
Let $G_1$ and $G_2$ be groups.
Let $H_1$ and $H_2$ be subgroups of $G_1$ and $G_2$ respectively.
Let $G_1$ be such that $H_1$ is the unique Sylow $p$-subgroup of $G_1$.
Let $G_2$ be such that $H_2$ is the unique Sylow $p$-subgroup of $G_2$.
Then $H_1 \times H_2$ is the unique Sylow $p$-subgroup of $G_1 \times G_2$.
== Proof ==
From Direct Product of Sylow p-Subgroups is Sylow p-Subgroup, $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.
By Sylow $p$-Subgroup is Unique iff Normal, each of $H_1$ and $H_2$ are normal in $G_1$ and $G_2$ respectively.
By Direct Product of Normal Subgroups is Normal, $H_1 \times H_2$ is normal in $G_1 \times G_2$.
Again by Sylow $p$-Subgroup is Unique iff Normal, $H_1 \times H_2$ is the unique Sylow $p$-subgroup of $G_1 \times G_2$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Exercise $2$
Category:Group Direct Products
Category:Sylow p-Subgroups
Direct Product of Unitary Modules is Unitary Module

== Theorem ==
Let $R$ be a ring with unity whose unity is $1_R$.
Let $\family {\struct {M_i, +_i, \circ_i} }_{i \mathop \in I}$ be a family of unitary $R$-modules.
Let $\struct {M, +, \circ}$ be their direct product.
Then $\struct {M, +, \circ}$ is a unitary $R$-module.
== Proof ==
From Direct Product of Modules is Module, $M$ is an $R$-module.
It remains to verify that:
:$\forall x \in M: 1_R \circ x = x$
By the definition of direct product, $\circ$ is an $R$-action induced on $M$ by $\family {\circ_i}_{i \mathop \in I}$:
:$\forall r \in R: r \circ \family {m_i}_{i \mathop \in I} = \family {r \circ_i m_i}_{i \mathop \in I}$.
We have:
Category:Module Theory
Category:Direct Products
Direct Product of Vector Spaces is Vector Space

== Theorem ==
Let $K$ be a field.
Let $V_1, V_2, \ldots, V_n$ be $K$-vector spaces.
Let $\struct {V, + , \circ}_K$ be their direct product.
Then $\struct {V, + , \circ}_K$ is a $K$-vector space.
== Proof ==
This follows directly from Finite Direct Product of Modules is Module and the definition of vector space.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.5$
Category:Linear Algebra
Category:Direct Product of Vector Spaces
Direct Sum of Densely-Defined Linear Operators is Densely-Defined Linear Operator

== Theorem ==
Let $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$ be a sequence of Hilbert spaces over $\C$.
Let $\tuple {\HH, \innerprod \cdot \cdot}$ be the Hilbert space direct sum of $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$.
For each $n \in \N$, let $\tuple {\map D {T_n}, T_n}$ be a densely-defined linear operator. 
Define:
:$\ds \map D T = \set {\sequence {\phi_n}_{n \mathop \in \N} \in \HH : \sum_{n \mathop = 1}^\infty \norm {T_n \phi_n}^2_{\HH_n} < \infty}$
Define $T : \map D T \to \HH$ by:
:$\map T {\sequence {\phi_n}_{n \mathop \in \N} } = T_\phi$
for each $\phi = \sequence {\phi_n}_{n \mathop \in \N} \in \HH$, where $T_\phi \in \HH$ is such that:
:$\map {T_\phi} n = T_n \phi_n$
for each $n \in \N$.
Then $\tuple {\map D T, T}$ be a densely-defined.
== Proof ==
Let $\SS$ be the set of $f \in \HH$ such that:
:there exists $N \in \N$ such that:
::$\map f n = {\mathbf 0}_{\HH_n}$ for $n > N$.
For $\phi = \sequence {\phi_n}_{n \mathop \in \N} \in \SS$ and $N \in \N$ picked as a witness to the above condition, we have:
:$\ds \sum_{n \mathop = 1}^\infty \norm {T_n \phi_n}_{\HH_n}^2 = \sum_{n \mathop = 1}^N \norm {T_n \phi_n}_{\HH_n}^2 < \infty$ 
Hence $\phi \in \map D T$.
So $\SS \subseteq \map D T$.
From Everywhere Dense Subset of Countable Hilbert Space Direct Sum in terms of Everywhere Dense Subsets of Summands, $\SS$ is everywhere dense in $\HH$.
Hence from Set Closure Preserves Set Inclusion, $\map D T$ is everywhere dense in $\HH$.
It remains to show that $T$ is linear.
Let $\phi = \sequence {\phi_n}_{n \mathop \in \N} \in \map D T$, $\psi = \sequence {\psi_n}_{n \mathop \in \N} \in \map D T$ and $\lambda \in \C$.
Then, for each $n \in \N$, we have:
Hence we have:
:$T_{\phi + \lambda \psi} = T_\phi + \lambda T_\psi$
Hence:
:$\map T {\phi + \lambda \psi} = T \phi + \lambda T \psi$
So $T$ is linear.
Category:Direct Sums of Densely-Defined Linear Operators
Direct Sum of Identity Operators on Hilbert Spaces is Identity Operator

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\GF$.
For each $i \in I$, let $I_{\map B {\HH_i} } : \HH_i \to \HH_i$ be the identity mapping.
Let:
:$\ds \HH = \bigoplus_{i \mathop \in I} \HH_i$
be the Hilbert space direct sum of $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ with norm $\norm {\, \cdot \,}_\HH$.
Let $I_{\map B \HH} : \HH \to \HH$ be the identity mapping.
Then:
:$\ds I_{\map B \HH} = \bigoplus_{i \mathop \in I} I_{\map B {\HH_i} }$
where $\ds \bigoplus_{i \mathop \in I} I_{\map B {\HH_i} }$ is the direct sum of $\family {I_{\map B {\HH_i} } }_{i \mathop \in I}$. 
== Proof ==
Let $f \in \HH$. 
Then we have, for each $i \in I$:
:$\ds \map {\paren {\bigoplus_{i \mathop \in I} I_{\map B {\HH_i} } f} } i = I_{\map B {\HH_i} } \map f i = \map f i$
Hence: 
:$\ds \bigoplus_{i \mathop \in I} I_{\map B {\HH_i} } f = f = I_{\map B \HH} f$
for each $f \in \HH$.
Hence the result.
Category:Hilbert Space Direct Sums
Direct Sum of Modules is Module

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $I$ be an indexing set.
Let $\family {M_i}_{i \mathop \in I}$ be a family of $A$-modules indexed by $I$.
Let $\ds M = \bigoplus_{i \mathop \in I} M_i$ be their direct sum.
Then $M$ is an $A$-module.
== Proof ==
== Also see ==
* Elements of Finite Support form Submagma of Direct Product
Category:Module Theory
Direct Sum of Representations of C*-Algebra is Representation

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra. 
Let $\family {\tuple {\pi_i, \HH_i} }_{i \mathop \in I}$ be an $I$-index family of representations of $\struct {A, \ast, \norm {\, \cdot \,} }$.
Let:
:$\ds \HH = \bigoplus_{i \mathop \in I} \HH_i$
be the Hilbert space direct sum of $\sequence {\struct {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ with norm $\norm {\, \cdot \,}_\HH$.
Let:
:$\ds \pi = \bigoplus_{i \mathop \in I} \pi_i$
be the direct sum of $\family {\pi_i}_{i \mathop \in I}$. 
Then $\tuple {\pi, \HH}$ be a representation of $A$.
== Proof ==
Let $a, b \in A$ and $\lambda \in \C$.
so $\pi$ is a linear transformation.
We also have:
Hence $\pi$ is an algebra homomorphism.
We also have:
so $\pi$ is a $\ast$-algebra homomorphism.
Finally, if $A$ is unital, we have:
Hence $\tuple {\pi, \HH}$ is a representation of $A$.
== Sources ==
* $\text {VIII}.5.7$
Category:Hilbert Space Direct Sums
Category:Representations of C*-Algebras
Direct Sum of Subspace and Orthocomplement

== Theorem ==
Let $H$ be a Hilbert space.
Let $M$ be a closed linear subspace of $H$.
Denote by $M^\perp$ its orthocomplement.
Then the direct sum $M \oplus M^\perp$ is isomorphic to $H$.
== Proof ==
Assert that $U: M \oplus M^\perp \to H: \left({m, m^\perp}\right) \mapsto m + m^\perp$ is an isomorphism.
According to the definition of isomorphism, it is sufficient to check that $U$ is surjective and that:
:$\left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H = \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}$
First the surjectivity:
Denote by $P$ the orthogonal projection on $M$.
Then for any $h \in H$, have $h = Ph + \left({h - Ph}\right)$.
By definition of $P$, $Ph \in M$.
Furthermore, by Orthogonal Projection onto Orthocomplement, $h - Ph \in M^\perp$.
It follows that $h = U \left({Ph, h - Ph}\right)$, showing that $U$ is a surjection.
It remains to check that $U$ preserves the inner product:
Hence $U$ is an isomorphism, and so $M \oplus M^\perp$ is isomorphic to $H$.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Exercise $3$
Category:Hilbert Spaces
Category:Orthocomplements
Directed Set has Strict Successors iff Unbounded Above

== Theorem ==
Let $\struct {S, \le}$ be a directed set.
Then every element of $S$ has a strict successor in $S$ $S$ has no upper bound in $S$.
== Proof ==
=== Necessary Condition ===
Suppose that each element of $S$ has a strict successor in $S$.
If $x$ is any element of $S$, then $x$ has a strict successor.
Therefore, $x$ is not an upper bound of $S$.
=== Sufficient Condition ===
Suppose $S$ has no upper bound in $S$.
Let $x \in S$.
Since $S$ has no upper bound, $x$ is not an upper bound of $S$.
Thus for some $p \in S$, $p \not\le x$.
By the definition of a directed set, there is a $y \in S$ such that $p \le y$ and $x \le y$.
Since $p \not\le x$, it follows that $x \ne y$.
Hence, since we already know that $x \le y$, in fact $x < y$.
Thus $y$ is a strict successor of $x$.
Category:Order Theory