hcju/proofwikiqa huggingface dataset:
When $x,y,z \in \R$, then $(1)$ follows from Real Addition is Associative. From the definition of $+_{\overline{\R}}$, it follows that either expression in $(1)$ is defined precisely when at most one of $+\infty$ and $-\infty$ occurs. The case where neither occurs is already covered above; now assume that $+\infty$ occurs (the case with $-\infty$ is similar). By Extended Real Addition is Commutative, it suffices to consider the cases $x = +\infty$ and $y = +\infty$. Suppose $x = +\infty$. Then as $y \ne -\infty$ and $z \ne -\infty$, it follows that $\left({+\infty}\right) +_{\overline{\R}} y = +\infty$, and $\left({+\infty}\right) +_{\overline{\R}} z = +\infty$. Hence the right-hand side in $(1)$ equals $+\infty$. But $y +_{\overline{\R}} z \ne -\infty$ as well, and so the left-hand side also equals $+\infty$. Now suppose $y = +\infty$. Then, again, $x +_{\overline{\R}} \left({+\infty}\right) = +\infty$ and $\left({+\infty}\right) +_{\overline{\R}} z = +\infty$. The result follows by applying these two equalities in the appropriate order on both the left- and the right-hand side of $(1)$. Hence the result, from Proof by Cases. {{qed}} Category:Extended Real Numbers acaio0xh0iy7g3gdb3pf6vx8a0l9njy
Let $x, y \in \R$. Then from Real Multiplication is Commutative: :$x \times_{\overline \R} y = y \times_{\overline \R} x$ The remaining cases are explicitly imposed in the definition of $\times_{\overline \R}$. {{qed}} Category:Extended Real Numbers ers5dtbwrqxqxdxuunybf70906gsovl
When $x, y, z \in \R$, then $(1)$ follows from Real Multiplication is Associative. Next, the cases where at least one of $+\infty$ and $-\infty$ occurs need to be dealt with. {{ProofWanted|when someone thinks of a nice way to deal with the case distinctions, go ahead}} Category:Extended Real Numbers 42pagy9miv7xxla3wtfi3au46flcowq
Checking the axioms for a monoid in turn: === Closure === Immediate as $\cdot_{\overline \R}: \overline \R \times \overline \R \to \overline \R$ is a mapping. {{qed|lemma}} === Associativity === Proved on Extended Real Multiplication is Associative. {{qed|lemma}} === Identity === For all $x \in \R$, it holds that $1 \cdot_{\overline \R} x = x \cdot_{\overline \R} 1 = x$ by definition of $\cdot_{\overline \R}$ on $\R$. Furthermore, by definition, $1 \cdot_{\overline \R} \paren {+\infty} = \paren {+\infty} \cdot_{\overline \R} 1 = \paren {+\infty}$. Lastly $1 \cdot_{\overline \R} \paren {-\infty} = \paren {-\infty} \cdot_{\overline \R} 1 = \paren {-\infty}$. That is, $1 \in \overline \R$ is an identity for $\cdot_{\overline \R}$. {{qed|lemma}} Hence, satisfying all the axioms, $\paren {\overline \R, \cdot_{\overline \R} }$ is a monoid. {{qed}} Category:Extended Real Numbers Category:Examples of Monoids htl3b6zpo85s2okk8e80dj3u9wi27aw
By Extended Real Numbers under Multiplication form Monoid, $\left({\overline \R, \cdot_{\overline \R}}\right)$ is a monoid. By Extended Real Multiplication is Commutative, $\cdot_{\overline \R}$ is commutative. Hence $\left({\overline \R, \cdot_{\overline \R}}\right)$ is a commutative monoid. {{qed}} Category:Extended Real Numbers Category:Examples of Monoids rfaey9zmkxdwlog70vuju7j7u1ok4uw
By Characterization of Euclidean Borel Sigma-Algebra, $\mathcal B = \sigma \left({\mathcal{J}_{ho}^n}\right)$. Here, $\mathcal{J}_{ho}^n$ is the set of half-open $n$-rectangles, and $\sigma$ denotes generated $\sigma$-algebra. Now, for any half-open $n$-rectangle $\left[[{\mathbf a \,.\,.\, \mathbf b}\right))$, it is trivial that: :$\tau_{\mathbf x}^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) = \left[[{\mathbf a + \mathbf x \,.\,.\, \mathbf b + \mathbf x}\right))$ That is, the preimage of a half-open $n$-rectangle under $\tau_{\mathbf x}$ is again a half-open $n$-rectangle. In particular, since $\mathcal{J}_{ho}^n \subseteq \sigma \left({\mathcal{J}_{ho}^n}\right) = \mathcal B$, Mapping Measurable iff Measurable on Generator applies. Thus it follows that $\tau_{\mathbf x}$ is $\mathcal B \, / \, \mathcal B$-measurable. {{qed}}
Let $g = h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}$. From the Internal Direct Product Theorem: General Result, $H_i$ and $H_j$ are normal in $G$. Hence $h_i \circ h_j \circ h_i^{-1} \in H_j$ and thus $g \in H_j$. Similarly, $g \in H_i$ and thus $g \in H_i \cap H_j$. But: {{begin-eqn}} {{eqn | l = H_i \cap H_j       | r = \set e       | c =  }} {{eqn | ll= \leadsto       | l = g       | r = h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}       | c =  }} {{eqn | r = e       | c =  }} {{eqn | ll= \leadsto       | l = h_i \circ h_j \circ h_i^{-1}       | r = h_j       | c =  }} {{eqn | ll= \leadsto       | l = h_i \circ h_j       | r = h_j \circ h_i       | c =  }} {{end-eqn}} {{qed}}
Verify the axioms for a $\sigma$-algebra in turn: === Axiom $(1)$ === As $f$ is a mapping, it is immediate that $f^{-1} \left({X'}\right) = X$. Also $X \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra. Hence $X' \in \Sigma'$. {{qed|lemma}} === Axiom $(2)$ === Let $S' \in \Sigma'$. By Preimage of Set Difference under Mapping and $f^{-1} \left({X'}\right) = X$: :$f^{-1} \left({X' \setminus S'}\right) = f^{-1} \left({X'}\right) \setminus f^{-1} \left({S'}\right)$ Now $f^{-1} \left({X'}\right) = X \in \Sigma$, and $f^{-1} \left({S'}\right) \in \Sigma$ was already assumed. As $\Sigma$ is a $\sigma$-algebra, it follows that $f^{-1} \left({X' \setminus S'}\right) \in \Sigma$ as well. Hence $X' \setminus S' \in \Sigma'$. {{qed|lemma}} === Axiom $(3)$ === Let $\left({S'_i}\right)_{i \mathop \in \N}$ be a sequence in $\Sigma'$. By Preimage of Union under Mapping: General Result, have: :$\displaystyle f^{-1} \left({\bigcup_{i \mathop \in \N} S'_i}\right) = \bigcup_{i \mathop \in \N} f^{-1} \left({S'_i}\right)$ By assumption, $S'_i \in \Sigma'$ for all $i \in \N$; hence $f^{-1} \left({S'_i}\right) \in \Sigma$ for all $i \in \N$. Now $\displaystyle \bigcup_{i \mathop \in \N} f^{-1} \left({S'_i}\right) \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra. Hence $\displaystyle \bigcup_{i \mathop \in \N} S'_i \in \Sigma'$. {{qed}}
=== Necessary Condition === Let $f$ be $\Sigma \, / \, \Sigma'$-measurable. By definition of generated $\sigma$-algebra $\mathcal{G}' \subseteq \Sigma'$. Hence, in particular, $f$ satisfies: :$\forall G' \in \mathcal{G}': f^{-1} \left({G'}\right) \in \Sigma$ {{qed|lemma}} === Sufficient Condition === Suppose that: :$\forall G' \in \mathcal{G}': f^{-1} \left({G'}\right) \in \Sigma$ Consider the pre-image $\sigma$-algebra $\Sigma''$ on $X'$. The supposition precisely states that $\mathcal{G}' \subseteq \Sigma''$. By definition of generated $\sigma$-algebra, $\sigma \left({\mathcal{G}'}\right) \subseteq \Sigma''$. By definition of $\Sigma''$, this precisely means: :$\forall E' \in \Sigma': f^{-1} \left({E'}\right) \in \Sigma$ That is, $f$ is $\Sigma \, / \, \Sigma'$-measurable. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === Let $f, g \in G^S$. As $\struct {G, \circ}$ is a group, it is closed by group axiom $G0$. From Closure of Pointwise Operation on Algebraic Structure it follows that $\struct {G^S, \oplus}$ is likewise closed. {{qed|lemma}} === $\text G 1$: Associativity === As $\struct {G, \circ}$ is a group, $\circ$ is associative. So from Structure Induced by Associative Operation is Associative, $\struct {G^S, \oplus}$ is also associative. {{qed|lemma}} === $\text G 2$: Identity === We have from Induced Structure Identity that the constant mapping $f_e: S \to T$ defined as: :$\forall x \in S: \map {f_e} x = e$ is the identity for $\struct {G^S, \oplus}$. {{qed|lemma}} === $\text G 3$: Inverses === Let $f \in G^S$. Let $f^* \in G^S$ be defined as follows: :$\forall f \in G^S: \forall x \in S: \map {f^*} x = \paren {\map f x}^{-1}$ From Induced Structure Inverse, $f^*$ is the inverse of $f$ for the pointwise operation $\oplus$ induced on $G^S$ by $\circ$. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\struct {G^S, \oplus}$ is a group. {{qed}}
As $f$ is a continuous mapping, by definition, it holds that: :$\forall S' \in \tau': f^{-1} \left({S'}\right) \in \tau$ Now, by definition, $\mathcal B \left({X, \tau}\right) = \sigma \left({\tau}\right)$, and so $\tau \subseteq \mathcal B \left({X, \tau}\right)$. Also, $\mathcal B \left({X', \tau'}\right) = \sigma \left({\tau'}\right)$. Hence Mapping Measurable iff Measurable on Generator applies. Therefore, $f$ is $\mathcal B \left({X, \tau}\right) \, / \, \mathcal B \left({X', \tau'}\right)$-measurable. {{qed}}
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}}
Let $\mathcal R$ be an equivalence relation on $S$. Then by definition: : $\mathcal R$ is a reflexive relation on $S$ : $\mathcal R$ is a symmetric relation on $S$ : $\mathcal R$ is a transitive relation on $S$. Then: : from Restriction of Reflexive Relation is Reflexive, $\mathcal R {\restriction_T}$ is a reflexive relation on $T$ : from Restriction of Symmetric Relation is Symmetric, $\mathcal R {\restriction_T}$ is a symmetric relation on $T$ : from Restriction of Transitive Relation is Transitive, $\mathcal R {\restriction_T}$ is a transitive relation on $T$ and so it follows by definition that $\mathcal R {\restriction_T}$ is an equivalence relation on $T$. {{qed}}
Suppose $\mathcal R$ is reflexive on $S$. Then: : $\forall x \in S: \left({x, x}\right) \in \mathcal R$ So: : $\forall x \in T: \left({x, x}\right) \in \mathcal R {\restriction_T}$ Thus $\mathcal R {\restriction_T}$ is reflexive on $T$. {{qed}}
Suppose $\mathcal R$ is symmetric on $S$. Then: {{begin-eqn}} {{eqn | l = \left({x, y}\right)       | o = \in       | r = \mathcal R {\restriction_T}       | c =  }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | o = \in       | r = \left({T \times T}\right) \cap \mathcal R       | c = Definition of Restriction of Relation }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | o = \in       | r = T \times T       | c =  }} {{eqn | lo= \land       | l = \left({x, y}\right)       | o = \in       | r = \mathcal R       | c = Definition of Set Intersection }} {{eqn | ll= \implies       | l = \left({y, x}\right)       | o = \in       | r = T \times T       | c =  }} {{eqn | lo= \land       | l = \left({y, x}\right)       | o = \in       | r = \mathcal R       | c = $\mathcal R$ is symmetric on $S$ }} {{eqn | ll= \implies       | l = \left({y, x}\right)       | o = \in       | r = \left({T \times T}\right) \cap \mathcal R       | c = Definition of Set Intersection }} {{eqn | ll= \implies       | l = \left({y, x}\right)       | o = \in       | r = \mathcal R {\restriction_T}       | c = Definition of Restriction of Relation }} {{end-eqn}} and so $\mathcal R {\restriction_T}$ is symmetric on $T$. {{qed}}
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}}
Suppose $\mathcal R$ is asymmetric on $S$. Then: {{begin-eqn}} {{eqn | l = \left({x, y}\right)       | o = \in       | r = \mathcal R \restriction_T       | c =  }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | o = \in       | r = \left({T \times T}\right) \cap \mathcal R       | c = {{Defof|Restriction of Relation}} }} {{eqn | ll= \implies       | l = \left({x, y}\right)       | o = \in       | r = \mathcal R       | c = Intersection is Subset }} {{eqn | ll= \implies       | l = \left({y, x}\right)       | o = \notin       | r = \mathcal R       | c = $\mathcal R$ is asymmetric on $S$ }} {{eqn | ll= \implies       | l = \left({y, x}\right)       | o = \notin       | r = \mathcal R \restriction_T       | c = {{Defof|Restriction of Relation}} }} {{end-eqn}} and so $\mathcal R \restriction_T$ is asymmetric on $T$. {{qed}}
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}}
Suppose $\mathcal R$ is antireflexive on $S$. Then: : $\forall x \in S: \left({x, x}\right) \notin \mathcal R$ So: : $\forall x \in T: \left({x, x}\right) \notin \mathcal R \restriction_T$ Thus $\mathcal R \restriction_T$ is antireflexive on $T$. {{qed}}
Suppose $\mathcal R$ is antitransitive on $S$. Then by definition: : $\left\{ {\left({x, y}\right), \left({y, z}\right)}\right\} \subseteq \mathcal R \implies \left({x, z}\right) \notin \mathcal R$ So: {{begin-eqn}} {{eqn | l=\left\{ {\left({x, y}\right), \left({y, z}\right)}\right\}       | o=\subseteq       | r=\mathcal R \restriction_T       | c= }} {{eqn | ll=\implies       | l=\left\{ {\left({x, y}\right), \left({y, z}\right)}\right\}       | o=\subseteq       | r=\left({T \times T}\right) \cap \mathcal R       | c=by definition of restriction of relation }} {{eqn | ll=\implies       | l=\left({x, z}\right)       | o=\notin       | r=\left({T \times T}\right) \cap \mathcal R       | c=as $\mathcal R$ is antitransitive on $S$ }} {{eqn | ll=\implies       | l=\left({x, z}\right)       | o=\in       | r=\mathcal R \restriction_T       | c=by definition of restriction of relation }} {{end-eqn}} Therefore, if $x, y, z \in T$, it follows that: : $\left\{ {\left({x, y}\right), \left({y, z}\right)}\right\} \subseteq \mathcal R \restriction_T \implies \left({x, z}\right) \notin \mathcal R \restriction_T$ and so by definition $\mathcal R \restriction_T$ is an antitransitive relation on $T$. {{qed}}
=== Necessary Condition === Let $\mathcal R$ be a total relation. By definition of relation, both $\mathcal R \subseteq S \times S$ and $\mathcal R^{-1} \subseteq S \times S$. So from Union is Smallest Superset (and indeed, trivially) $\mathcal R \cup \mathcal R^{-1} \subseteq S \times S$. Let $\left({a, b}\right) \in S \times S$. As $\mathcal R$ is total, either: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({b, a}\right) \in \mathcal R$ From the definition of inverse relation, this means that either: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({a, b}\right) \in \mathcal R^{-1}$ That is: :$\left({a, b}\right) \in \mathcal R \cup \mathcal R^{-1}$ and so by definition of subset: :$S \times S \subseteq \mathcal R \cup \mathcal R^{-1}$ Hence, by definition of set equality: :$\mathcal R \cup \mathcal R^{-1} = S \times S$ {{qed|lemma}} === Sufficient Condition === Let $\mathcal R \cup \mathcal R^{-1} = S \times S$. Let $\left({a, b}\right) \in S \times S$. Then by definition of set union: :$\left({a, b}\right) \in \mathcal R$ or: :$\left({a, b}\right) \in \mathcal R^{-1}$ That is, by definition of inverse relation: :$\left({a, b}\right) \in R$ or: :$\left({b, a}\right) \in R$ So by definition $\mathcal R$ is total. {{Qed}}
Suppose $\mathcal R$ is connected on $S$. That is: :$\forall a, b \in S: a \ne b \implies \left({a, b}\right) \in \mathcal R \lor \left({b, a}\right) \in \mathcal R$ So: {{begin-eqn}} {{eqn | l=a, b       | o=\in       | r=T       | c= }} {{eqn | ll=\implies       | l=\left({a, b}\right)       | o=\in       | r=T \times T       | c= }} {{eqn | lo=\land       | l=\left({b, a}\right)       | o=\in       | r=T \times T       | c=by definition of ordered pair and cartesian product }} {{eqn | ll=\implies       | l=\left({a, b}\right)       | o=\in       | r=\left({T \times T}\right) \cap \mathcal R       | c= }} {{eqn | lo=\lor       | l=\left({b, a}\right)       | o=\in       | r=\left({T \times T}\right) \cap \mathcal R       | c=as $\mathcal R$ is connected on $S$ }} {{eqn | ll=\implies       | l=\left({a, b}\right)       | o=\in       | r=R \restriction_T       | c= }} {{eqn | lo=\lor       | l=\left({b, a}\right)       | o=\in       | r=R \restriction_T       | c=by definition of restriction of relation }} {{end-eqn}} and so $\mathcal R \restriction_T$ is connected on $T$. {{qed}}
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}}
=== Restriction of Serial Relation is Not Necessarily Serial === {{:Restriction of Serial Relation is Not Necessarily Serial}} === Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive === {{:Restriction of Non-Reflexive Relation is Not Necessarily Non-Reflexive}} === Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric === {{:Restriction of Non-Symmetric Relation is Not Necessarily Non-Symmetric}} === Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive === {{:Restriction of Non-Transitive Relation is Not Necessarily Non-Transitive}} === Restriction of Non-Connected Relation is Not Necessarily Non-Connected === {{:Restriction of Non-Connected Relation is Not Necessarily Non-Connected}}
Proof by Counterexample: Let $S = \set {a, b}$. Let $\mathcal R = \set {\tuple {a, b}, \tuple {b, b} }$. $\mathcal R$ is a serial relation, as can be seen by definition. Now let $T = \set a$. Then $\mathcal R {\restriction_T} = \O$. So $\not \exists y \in T: \tuple {a, y} \in \mathcal R {\restriction_T}$. That is, $\mathcal R {\restriction_T}$ is not a serial relation on $T$. {{qed}}
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}}
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}}
Proof by Counterexample: Let $S = \left\{{a, b}\right\}$. Let $\mathcal R = \left\{{\left({a, b}\right), \left({b, a}\right), \left({b, b}\right)}\right\}$. $\mathcal R$ is a non-transitive relation, as can be seen by definition. Now let $T = \left\{{b}\right\}$. Then $\mathcal R \restriction_T \ = \left\{{\left({b, b}\right)}\right\}$. So: : $\forall x, y \in T: \left({x, y}\right) \in \mathcal R \restriction_T \land \left({y, z}\right) \in \mathcal R \restriction_T \implies \left({y, z}\right) \in \mathcal R \restriction_T$ as can be seen by setting $x = y = z = b$. So $\mathcal R \restriction_T$ is a transitive relation on $T$. That is, $\mathcal R \restriction_T$ is not a non-transitive relation on $T$. {{qed}}
By Characterization of Sigma-Algebra Generated by Collection of Mappings, we have that: :$\sigma \left({f_i: i \in I}\right) = \sigma \left({\displaystyle \bigcup_{i \mathop \in I} f_i^{-1} \left({\Sigma_i}\right)}\right)$ 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}}
For each $i \in I$, one has by definition of generated $\sigma$-algebra: :$\displaystyle f_i^{-1} \left({\Sigma_i}\right) \subseteq \bigcup_{i \mathop \in I} f_i^{-1} \left({\Sigma_i}\right) \subseteq \sigma \left({\bigcup_{i \mathop \in I} f_i^{-1} \left({\Sigma_i}\right)}\right)$ 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: :$f_i^{-1} \left({\Sigma_i}\right) \subseteq \Sigma$ and hence by Union is Smallest Superset: Family of Sets: :$\displaystyle \bigcup_{i \mathop \in I} f_i^{-1} \left({\Sigma_i}\right) \subseteq \Sigma$ Finally, by Generated Sigma-Algebra Preserves Subset, it follows that: :$\displaystyle \sigma \left({\bigcup_{i \mathop \in I} f_i^{-1} \left({\Sigma_i}\right)}\right) \subseteq \Sigma$ Thus: :$\displaystyle \sigma \left({\bigcup_{i \mathop \in I} f_i^{-1} \left({\Sigma_i}\right)}\right) = \sigma \left({f_i : i \in I}\right)$ by definition of the latter. {{qed}}
Proof by Counterexample: Let $S = \set {a, b}$. Let $\RR = \set {\tuple {a, a}, \tuple {b, b} }$. $\RR$ is a non-connected relation, as can be seen by definition: neither $a \mathrel \RR b$ nor $b \mathrel \RR a$. Now let $T = \set a$. Then $\RR {\restriction_T} = \set {\tuple {a, a} }$. Then $\RR {\restriction_T}$ is trivially connected on $T$. {{qed}}