avewright/proofwiki-math huggingface dataset:
Theorem: Internal Direct Product Theorem/General Result

Let $G$ be a group whose identity is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.
Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}:
{{begin-itemize}}
{{item|(1):|$G {{=}} H_1 H_2 \cdots H_n$}}
{{item|(2):|$\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups}}
{{item|(3):|$\forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$}}
{{end-itemize}}
where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.

Proof:
By definition, $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}} the mapping:
:$\ds C: \prod_{k \mathop = 1}^n H_k \to G: \map C {h_1, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
is a group isomorphism from the cartesian product $\struct {H_1, \circ {\restriction_{H_1} } } \times \cdots \times \struct {H_n, \circ {\restriction_{H_n} } }$ onto $\struct {G, \circ}$. 
=== Necessary Condition ===
Let $G$ be the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$.
:$(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 H_2 \cdots H_n$.
:$(2): \quad$ From Internal Group Direct Product is Injective: General Result, $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups.
:$(3): \quad$ From Internal Group Direct Product Isomorphism:
:::: $\forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$
{{qed|lemma}}
=== Sufficient Condition ===
Now suppose the three conditions hold.
:$(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $C$ is surjective.
:$(2): \quad$ From Internal Group Direct Product is Injective: General Result, $C$ is injective.
:$(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism.
Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism.
So by definition, $G$ is the internal group direct product of $\sequence {H_k} _{1 \mathop \le k \mathop \le n}$.
{{qed}}
Theorem: Internal Direct Product Theorem/General Result

Let $G$ be a group whose identity is $e$.
Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.
Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}:
{{begin-itemize}}
{{item|(1):|$G {{=}} H_1 H_2 \cdots H_n$}}
{{item|(2):|$\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups}}
{{item|(3):|$\forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$}}
{{end-itemize}}
where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.

Proof:
It is to be shown that:
{{:Definition:Internal Group Direct Product/General Definition/Definition by Unique Expression}}
by definition of Internal Group Direct Product.
Condition $(3)$ already gives that $H_i$ is normal for all $i \in \set {1, 2, \ldots, n}$.
Condition $(1)$ gives us that each element $g$ of $G$ can be expressed in the form:
:$g = h_1 h_2 \dotsm h_n$ with $h_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$.
It is now shown that this expression is unique.
Suppose that:
:$g = h_1 h_2 \dotsm h_n = k_1 k_2 \dotsm k_n$
where $h_i, k_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$ and $h_j \ne k_j$ for at least one $j$.
Let $j$ be the largest integer such that $h_j \ne k_j$, so that $h_i = k_i$ for $i > j$.
Cancelling $h_i$ for $i > j$ gives:
:$h_i h_2 \dotsm h_j = k_1 k_2 \dotsm k_j$
and so:
:$h_j {k_j}^{-1} = \paren {h_1 h_2 \dotsm h_{j - 1} }^{-1} \paren {k_1 k_2 \dotsm k_{j - 1} } \in \paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j$
But by condition $(2)$:
:$\paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j = \set e$
by definition of independent subgroups.
Thus $h_j = k_j$, which contradicts our assertion that $h_j \ne k_j$.
Hence the decomposition is unique.
{{qed}}
Theorem: Structure Induced by Group Operation is Group

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is a group.

Proof:
Taking the group axioms in turn:
Theorem: Continuous Mapping is Measurable

Let $\struct {X, \tau}$ and $\struct {X', \tau'}$ be topological spaces.
Denote with $\map \BB {X, \tau}$ and $\map \BB {X', \tau'}$ their respective Borel $\sigma$-algebras.
Let $f: X \to X'$ be a continuous mapping.
Then $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable.

Proof:
As $f$ is a continuous mapping, by definition, it holds that:
:$\forall S' \in \tau': \map {f^{-1} } {S'} \in \tau$
Now, by definition, $\map \BB {X, \tau} = \map \sigma \tau$, and so $\tau \subseteq \map \BB {X, \tau}$.
Also, $\map \BB {X', \tau'} = \map \sigma {\tau'}$.
Hence Mapping Measurable iff Measurable on Generator applies.
Therefore, $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable.
{{qed}}
Theorem: Composition of Measurable Mappings is Measurable

Let $\struct {X_1, \Sigma_1}$, $\struct {X_2, \Sigma_2}$ and $\struct {X_3, \Sigma_3}$ be measurable spaces.
Let $f: X_1 \to X_2$ be a $\Sigma_1 \, / \, \Sigma_2$-measurable mapping.
Let $g: X_2 \to X_3$ be a $\Sigma_2 \, / \, \Sigma_3$-measurable mapping.
Then their composition $g \circ f: X_1 \to X_3$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.

Proof:
Let $E_3 \in \Sigma_3$.
Then $\map {g^{-1} } {E_3} \in \Sigma_2$, and $\map {f^{-1} } {\map {g^{-1} } {E_3} } \in \Sigma_1$ as $f, g$ are measurable.
That is, $\map {\paren {g \circ f}^{-1} } {E_3} \in \Sigma_1$ for all $E_3 \in \Sigma_3$.
Hence, $g \circ f$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.
{{qed}}
Theorem: Restriction of Equivalence Relation is Equivalence

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an equivalence relation on $T$.

Proof:
Let $\RR$ be an equivalence relation on $S$.
Then by definition:
:$\RR$ is a reflexive relation on $S$
:$\RR$ is a symmetric relation on $S$
:$\RR$ is a transitive relation on $S$.
Then:
:from Restriction of Reflexive Relation is Reflexive, $\RR {\restriction_T}$ is a reflexive relation on $T$
:from Restriction of Symmetric Relation is Symmetric, $\RR {\restriction_T}$ is a symmetric relation on $T$
:from Restriction of Transitive Relation is Transitive, $\RR {\restriction_T}$ is a transitive relation on $T$
and so it follows by definition that $\RR {\restriction_T}$ is an equivalence relation on $T$.
{{qed}}
Theorem: Restriction of Reflexive Relation is Reflexive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a reflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a reflexive relation on $T$.

Proof:
Suppose $\RR$ is reflexive on $S$.
So:
:$\forall x \in S: \tuple {x, x} \in \RR$
We are given $T$ is a subset of $S$, so:
:$\forall x \in T: \tuple {x, x} \in \RR$
By definition of  restriction:
:$\forall x \in T: \tuple {x, x} \in \RR {\restriction_T}$
Hence $\RR {\restriction_T}$ is by definition reflexive on $T$.
{{qed}}
Theorem: Restriction of Symmetric Relation is Symmetric

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a symmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a symmetric relation on $T$.

Proof:
Suppose $\RR$ is symmetric on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
      | o = \in
      | r = \RR {\restriction_T}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, y}
      | o = \in
      | r = \paren {T \times T} \cap \RR
      | c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, y}
      | o = \in
      | r = T \times T
      | c = 
}}
{{eqn | lo= \land
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \in
      | r = T \times T
      | c = 
}}
{{eqn | lo= \land
      | l = \tuple {y, x}
      | o = \in
      | r = \RR
      | c = $\RR$ is symmetric on $S$
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \in
      | r = \paren {T \times T} \cap \RR
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \in
      | r = \RR {\restriction_T}
      | c = {{Defof|Restriction of Relation}}
}}
{{end-eqn}}
and so $\RR {\restriction_T}$ is symmetric on $T$.
{{qed}}
Theorem: Restriction of Transitive Relation is Transitive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a transitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a transitive relation on $T$.

Proof:
Suppose $\RR$ is transitive on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \in \RR$
So:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, z} }
      | o = \subseteq
      | r = \RR {\restriction_T}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \set {\tuple {x, y}, \tuple {y, z} }
      | o = \subseteq
      | r = \paren {T \times T} \cap \RR
      | c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, z}
      | o = \in
      | r = \paren {T \times T} \cap \RR
      | c = $\RR$ is transitive on $S$
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, z}
      | o = \in
      | r = \RR {\restriction_T}
      | c = {{Defof|Restriction of Relation}}
}}
{{end-eqn}}
Therefore, if $x, y, z \in T$, it follows that:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \in \RR {\restriction_T}$
and so by definition $\RR {\restriction_T}$ is a transitive relation on $T$.
{{qed}}
Theorem: Restriction of Asymmetric Relation is Asymmetric

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a asymmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a asymmetric relation on $T$.

Proof:
Suppose $\RR$ is asymmetric on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {x, y}
      | o = \in
      | r = \RR {\restriction_T}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, y}
      | o = \in
      | r = \paren {T \times T} \cap \RR
      | c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = Intersection is Subset
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \notin
      | r = \RR
      | c = $\RR$ is asymmetric on $S$
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \notin
      | r = \paren {T \times T} \cap \RR
      | c = Proof by Contraposition and Intersection is Subset
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \notin
      | r = \RR {\restriction_T}
      | c = {{Defof|Restriction of Relation}}
}}
{{end-eqn}}
and so $\RR {\restriction_T}$ is asymmetric on $T$.
{{qed}}
Theorem: Restriction of Antisymmetric Relation is Antisymmetric

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antisymmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antisymmetric relation on $T$.

Proof:
Suppose $\RR$ is antisymmetric on $S$.
Then:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, x} }
      | o = \subseteq
      | r = \RR {\restriction_T}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \set {\tuple {x, y}, \tuple {y, x} }
      | o = \subseteq
      | r = \paren {T \times T} \cap \RR
      | c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \set {\tuple {x, y}, \tuple {y, x} }
      | o = \subseteq
      | r = \RR
      | c = Intersection is Subset
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y
      | c = $\RR$ is Antisymmetric on $S$
}}
{{end-eqn}}
Thus $\RR {\restriction_T}$ is antisymmetric on $T$.
{{qed}}
Theorem: Restriction of Antireflexive Relation is Antireflexive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antireflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antireflexive relation on $T$.

Proof:
Suppose $\RR$ is antireflexive on $S$.
Then by definition of antireflexive:
:$\forall x \in S: \tuple {x, x} \notin \RR$
We are given  $T$ is a subset of $S$, so:
:$\forall x \in T: \tuple {x, x} \notin \RR$
We are given $\RR {\restriction_T} \subseteq T \times T$, so:
:$\forall x \in T: \tuple {x, x} \notin \RR \restriction_T$
Hence by definition, $\RR {\restriction_T}$ is antireflexive on $T$.
{{qed}}
Theorem: Restriction of Antitransitive Relation is Antitransitive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an antitransitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is an antitransitive relation on $T$.

Proof:
Suppose $\RR$ is antitransitive on $S$.
Then by definition:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR \implies \tuple {x, z} \notin \RR$
So:
{{begin-eqn}}
{{eqn | l = \set {\tuple {x, y}, \tuple {y, z} }
      | o = \subseteq
      | r = \RR {\restriction_T}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \set {\tuple {x, y}, \tuple {y, z} }
      | o = \subseteq
      | r = \paren {T \times T} \cap \RR
      | c = {{Defof|Restriction of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, z}
      | o = \notin
      | r = \paren {T \times T} \cap \RR
      | c = as $\RR$ is antitransitive on $S$
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, z}
      | o = \in
      | r = \RR {\restriction_T}
      | c = {{Defof|Restriction of Relation}}
}}
{{end-eqn}}
Therefore, if $x, y, z \in T$, it follows that:
:$\set {\tuple {x, y}, \tuple {y, z} } \subseteq \RR {\restriction_T} \implies \tuple {x, z} \notin \RR {\restriction_T}$
and so by definition $\RR {\restriction_T}$ is an antitransitive relation on $T$.
{{qed}}
Theorem: Relation is Total iff Union with Inverse is Trivial Relation

Let $\RR$ be a relation on $S$.
Then $\RR$ is a total relation {{iff}}:
:$\RR \cup \RR^{-1} = S \times S$
where:
:$\RR^{-1}$ is the inverse of $\RR$.
:$S \times S$ is the trivial relation on $S$.

Proof:
=== Necessary Condition ===
Let $\RR$ be a total relation.
By definition of relation, both $\RR \subseteq S \times S$ and $\RR^{-1} \subseteq S \times S$.
So from Union is Smallest Superset (and indeed, trivially) $\RR \cup \RR^{-1} \subseteq S \times S$.
Let $\tuple {a, b} \in S \times S$.
As $\RR$ is total, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {b, a} \in \RR$
From the definition of inverse relation, this means that either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
That is:
:$\tuple {a, b} \in \RR \cup \RR^{-1}$
and so by definition of subset:
:$S \times S \subseteq \RR \cup \RR^{-1}$
Hence, by definition of set equality:
:$\RR \cup \RR^{-1} = S \times S$
{{qed|lemma}}
Theorem: Restriction of Connected Relation is Connected

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a connected relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is a connected relation on $T$.

Proof:
Suppose $\RR$ is connected on $S$.
That is:
:$\forall a, b \in S: a \ne b \implies \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
So:
{{begin-eqn}}
{{eqn | l = a, b
      | o = \in
      | r = T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tuple {a, b}
      | o = \in
      | r = T \times T
      | c = 
}}
{{eqn | lo= \land
      | l = \tuple {b, a}
      | o = \in
      | r = T \times T
      | c = {{Defof|Ordered Pair}} and {{Defof|Cartesian Product}}
}}
{{eqn | ll= \leadsto
      | l = \tuple {a, b}
      | o = \in
      | r = \paren {T \times T} \cap \RR
      | c = 
}}
{{eqn | lo= \lor
      | l = \tuple {b, a}
      | o = \in
      | r = \paren {T \times T} \cap \RR
      | c = as $\RR$ is connected on $S$
}}
{{eqn | ll= \leadsto
      | l = \tuple {a, b}
      | o = \in
      | r = R \restriction_T
      | c = 
}}
{{eqn | lo= \lor
      | l = \tuple {b, a}
      | o = \in
      | r = R {\restriction_T}
      | c = {{Defof|Restriction of Relation}}
}}
{{end-eqn}}
and so $\RR {\restriction_T}$ is connected on $T$.
{{qed}}
Theorem: Restriction of Ordering is Ordering

Let $S$ be a set.
Let $\preceq$ be an ordering on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is an ordering on $T$.

Proof:
Let $\preceq$ be an ordering on $S$.
Then, by definition:
:$\preceq$ is a reflexive relation on $S$
:$\preceq$ is an antisymmetric relation on $S$
:$\preceq$ is a transitive relation on $S$.
Then:
:from Restriction of Reflexive Relation is Reflexive, $\preceq \restriction_T$ is a reflexive relation on $T$
:from Restriction of Antisymmetric Relation is Antisymmetric, $\preceq \restriction_T$ is an antisymmetric relation on $T$
:from Restriction of Transitive Relation is Transitive, $\preceq \restriction_T$ is a transitive relation on $T$
and so it follows by definition that $\preceq \restriction_T$ is an ordering on $T$.
{{qed}}
Theorem: Properties of Relation Not Preserved by Restriction

If a relation is:
* serial,
* non-reflexive,
* non-symmetric,
* non-transitive or
* non-connected
it is impossible to state without further information whether or not any restriction of that relation has the same properties.

Proof:
=== Restriction of Serial Relation is Not Necessarily Serial ===
{{:Restriction of Serial Relation is Not Necessarily Serial}}
Theorem: Restriction of Serial Relation is Not Necessarily Serial

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a serial relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a serial relation on $T$.

Proof:
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a serial relation, as can be seen by definition.
Now let $T = \set a$.
Then:
:$\RR {\restriction_T} = \O$
So:
:$\not \exists y \in T: \tuple {a, y} \in \RR {\restriction_T}$
That is, $\RR {\restriction_T}$ is not a serial relation on $T$.
{{qed}}
Theorem: Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-reflexive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-reflexive relation on $T$.

Proof:
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {b, b} }$.
$\RR$ is a non-reflexive relation, as can be seen by definition:
:$\tuple {a, a} \notin \RR$
:$\tuple {b, b} \in \RR$
Now let $T = \set a$.
Then $\RR {\restriction_T} = \O$.
So:
:$\forall x \in T: \tuple {x, x} \notin \RR {\restriction_T}$
That is, $\RR {\restriction_T}$ is an antireflexive relation on $T$.
That is, specifically ''not'' a non-reflexive relation.
{{qed}}
Theorem: Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-symmetric relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \ \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-symmetric relation on $T$.

Proof:
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, b} }$.
$\RR$ is a non-symmetric relation, as can be seen by definition.
Now let $T = \set b$.
Then $\RR {\restriction_T} \ = \set {\tuple {b, b} }$.
So:
:$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} \implies \tuple {y, x} \in \RR {\restriction_T}$
as can be seen by setting $x = y = b$.
So $\RR {\restriction_T}$ is a symmetric relation on $T$.
That is, $\RR {\restriction_T}$ is not a non-symmetric relation on $T$.
{{qed}}
Theorem: Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a non-transitive relation on $S$.
Let $T \subseteq S$ be a subset of $S$.
Let $\RR {\restriction_T} \subseteq T \times T$ be the restriction of $\RR$ to $T$.
Then $\RR {\restriction_T}$ is not necessarily a non-transitive relation on $T$.

Proof:
Proof by Counterexample:
Let $S = \set {a, b}$.
Let $\RR = \set {\tuple {a, b}, \tuple {b, a}, \tuple {b, b} }$.
$\RR$ is a non-transitive relation, as can be seen by definition.
Now let $T = \set b$.
Then:
:$\RR {\restriction_T} = \set {\tuple {b, b} }$
So:
:$\forall x, y \in T: \tuple {x, y} \in \RR {\restriction_T} \land \tuple {y, z} \in \RR {\restriction_T} \implies \tuple {y, z} \in \RR {\restriction_T}$
as can be seen by setting $x = y = z = b$.
So $\RR {\restriction_T}$ is a transitive relation on $T$.
That is, $\RR {\restriction_T}$ is not a non-transitive relation on $T$.
{{qed}}
Theorem: Existence and Uniqueness of Sigma-Algebra Generated by Collection of Mappings

Let $I$ be an indexing set.
Let $\family {\struct {X_i, \Sigma_i} }_{i \mathop \in I}$ be a family of measurable spaces.
Let $X$ be a set.
Let $\family {f_i: X \to X_i}_{i \mathop \in I}$ be a family of mappings.
Then $\map \sigma {f_i: i \in I}$, the $\sigma$-algebra generated by $\family {f_i}_{i \mathop \in I}$, exists and is unique.

Proof:
By Characterization of Sigma-Algebra Generated by Collection of Mappings:
:$\ds \map \sigma {f_i: i \in I} = \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where the second is a $\sigma$-algebra generated by a collection of subsets.
The result follows from applying Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets.
{{qed}}
Theorem: Characterization of Sigma-Algebra Generated by Collection of Mappings

Let $\struct {X_i, \Sigma_i}$ be measurable spaces, with $i \in I$ for some index set $I$.
Let $X$ be a set, and let, for $i \in I$, $f_i: X \to X_i$ be a mapping.
Then:
:$\map \sigma {f_i: i \in I} = \map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
where:
:$\map \sigma {f_i: i \in I}$ is the $\sigma$-algebra generated by $\family {f_i}_{i \mathop \in I}$
:$\map \sigma {\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$ is the $\sigma$-algebra generated by $\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i}$
:$\map {f_i^{-1} } {\Sigma_i}$ denotes the pre-image $\sigma$-algebra on $X$ by $f$

Proof:
For each $i \in I$, one has by definition of generated $\sigma$-algebra:
:$\ds \map {f_i^{-1} } {\Sigma_i} \subseteq \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} }$
which shows that each of the $f_i$ is measurable.
Next, suppose that $\Sigma$ is a $\sigma$-algebra such that each of the $f_i$ is $\Sigma \,/\, \Sigma_i$-measurable.
Then for all $i \in I$, one has:
:$\map {f_i^{-1} } {\Sigma_i} \subseteq \Sigma$
and hence by Union is Smallest Superset: Family of Sets:
:$\ds \bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} \subseteq \Sigma$
Finally, by Generated Sigma-Algebra Preserves Subset, it follows that:
:$\ds \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} } \subseteq \Sigma$
Thus:
:$\ds \map \sigma {\bigcup_{i \mathop \in I} \map {f_i^{-1} } {\Sigma_i} } = \map \sigma {f_i : i \in I}$
by definition of the latter.
{{qed}}