hcju/proofwikiqa huggingface dataset:
=== $(1)$ implies $(2)$ === Assume that $E \in \Sigma$. It is clear that $x \notin \left\{{0, 1}\right\}$ implies $\chi_E^{-1} \left({x}\right) = \varnothing$. Hence Preimage of Union under Mapping and Characteristic Function Determined by 1-Fiber yield, for any $\alpha \in \R$: :$\left\{{x \in X: \chi_E \left({x}\right) \ge \alpha}\right\} = \begin{cases}\varnothing & \text{if $1 < \alpha$}\\ E & \text{if $0 < \alpha \le 1$}\\ X & \text{if $\alpha \le 0$}\end{cases}$ Since $\Sigma$ is a $\sigma$-algebra, $X \in \Sigma$. By assumption, also $E \in \Sigma$. Lastly, Sigma-Algebra Contains Empty Set ensures $\varnothing \in \Sigma$. This establishes $(4)$ of Characterization of Measurable Functions. Hence $\chi_E$ is $\Sigma$-measurable. {{qed|lemma}} === $(2)$ implies $(1)$ === Assume that $\chi_E$ is $\Sigma$-measurable. Since for all $x \in X$, it holds that: :$\chi_E \left({x}\right) > 0 \iff x \in E$ it follows that: :$E = \chi^{-1} \left({\left({0 \,.\,.\, +\infty}\right)}\right)$ and thus $E \in \Sigma$ as $\chi_E$ is $\Sigma$-measurable. {{qed}}
Let $f$ be written in the following form: :$f = \displaystyle \sum_{i \mathop = 1}^n a_i \chi_{S_i}$ where $a_i \in \R$ and the $S_i$ are $\Sigma$-measurable. Next, for each ordered $n$-tuple $b$ of zeroes and ones define: :$T_b \left({i}\right) := \begin{cases} S_i & \text{if $b \left({i}\right) = 0$}\\ X \setminus S_i & \text{if $b \left({i}\right) = 1$} \end{cases}$ and subsequently: :$T_b := \displaystyle \bigcap_{i \mathop = 1}^n T_b \left({i}\right)$ From Sigma-Algebra Closed under Intersection, $T_b \in \Sigma$ for all $b$. Also, the $T_b$ are pairwise disjoint, and furthermore: :$f = \displaystyle \sum_b a_b \chi_{T_b}$ where: :$a_b := \displaystyle \sum_{i \mathop = 1}^n b \left({i}\right) a_i$ {{finish|prove it, it's a messy business}} Now we have, for all $\lambda \in \R$: :$\left\{{x \in X: f \left({x}\right) > \lambda}\right\} = \displaystyle \bigcup \left\{{T_b: a_b > \lambda}\right\}$ which by Sigma-Algebra Closed under Union is a $\Sigma$-measurable set. From Characterization of Measurable Functions: $(5)$ it follows that $f$ is measurable. {{qed}}
By definition of identity mapping: :$\forall x \in S: \map {I_S} x = x$ So: :$x \mathrel \RR y \implies \map {I_S} x \mathrel \RR \map {I_S} y$ From Identity Mapping is Bijection, $I_S$ is a bijection. Hence: :$\map {I_S^{-1} } x = x$ So: :$x \mathrel \RR y \implies \map {I_S^{-1} } x \mathrel \RR \map {I_S^{-1} } y$ {{qed}}
=== Necessary Condition === Suppose that $f$ is a simple function, and that: :$\ds \forall x \in X: \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{S_i} } x$ Since each of the $\chi_{S_i}$ is a characteristic function, it can take only the values $0$ and $1$. Thus each summand can take two values. It follows immediately that $f$ can take at most $2^n$ different values. The conclusion follows from Simple Function is Measurable. {{qed|lemma}} === Sufficient Condition === Suppose that the image of $f$ is finite. Call the distinct values $f$ attains $y_1, \ldots, y_n$. For brevity, denote $\set {f = a}$ to mean $\set {x \in X: \map f x = a}$ (compare Set Definition by Predicate). Define for each $i$ with $1 \le i \le n$: :$B_i := \set {f = y_i}$ From Characterization of Measurable Functions $(2)$ and $(4)$, and Sigma-Algebra Closed under Intersection we obtain that: :$\set {f = y_i} = \set {f \ge y_i} \cap \set {f \le y_i} \in \Sigma$ Furthermore, since the $y_i$ are distinct, the $B_i$ are necessarily disjoint. It follows that: :$(1): \quad \map f x = \ds \sum_{i \mathop = 1}^n y_j \map {\chi_{B_j} } x$ As the $B_i$ are measurable, $f$ is shown to be a simple function. {{qed}}
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an relation isomorphism is also a bijection. By definition of composition of mappings: :$\map {\psi \circ \phi} x = \map \psi {\map \phi x}$ As $\phi$ is a relation isomorphism, we have: :$\forall x_1, y_1 \in S_1: x_1 \mathrel {\RR_1} y_1 \implies \map \phi {x_1} \mathrel {\RR_2} \map \phi {y_1}$ As $\psi$ is a relation isomorphism, we have: :$\forall x_2, y_2 \in S_2: x_2 \mathrel {\RR_2} y_2 \implies \map \psi {x_2} \mathrel {\RR_3} \map \psi {y_2}$ By setting $x_2 = \map \phi {x_1}, y_2 = \map \phi {y_1}$, it follows that: :$\forall x_1, y_1 \in S_1: x_1 \mathrel {\RR_1} y_1 \implies \map \psi {\map \phi {x_1} } \mathrel {\RR_3} \map \psi {\map \phi {y_1} }$ Similarly we can show that: :$\forall x_3, y_3 \in S_3: x_3 \mathrel {\RR_3} y_3 \implies \map {\phi^{-1} } {\map {\psi^{-1} } {x_3} } \mathrel {\RR_1} \map {\phi^{-1} } {\map {\psi^{-1} } {y_3} }$ Hence the result, by definition of a relation isomorphism. {{qed}}
We have $f + g = + \circ \innerprod f g \circ \Delta_X$, where: :$\Delta_X: X \to X \times X$ is the diagonal mapping on $X$ :$\innerprod f g: X \times X \to \R \times \R, \map {\innerprod f g} {x, y} := \tuple {\map f x, \map g y}$ :$+: \R \times \R \to \R$ is real addition. {{explain|What is the meaning of the notation $\innerprod f g$? Inner product?}} From this, we see that $+$ may be restricted to $\Img {\innerprod f g}$, the image of $\sequence {f, g}$. It is immediate that this image equals $\Img f \times \Img g$. By Measurable Function is Simple Function iff Finite Image Set, $\Img f$ and $\Img g$ are finite. Therefore, by Cardinality of Cartesian Product, $\Img {\sequence {f, g} }$ is also finite. Next, from Corollary to Image of Composite Mapping: :$\Img {\innerprod f g \circ \Delta_X} \subseteq \Img {\innerprod f g}$ whence the former is finite. Now $+$ is a surjection: :$+: \Img {\innerprod f g \circ \Delta_X} \to \Img {f + g}$ by Restriction of Mapping to Image is Surjection. By Cardinality of Surjection, this implies that $\Img {f + g}$ is finite. Whence Measurable Function is Simple Function iff Finite Image Set grants that $f + g$ is a simple function. {{qed}}
From Measurable Function is Simple Function iff Finite Image Set, it follows that there exist $x_1, \ldots, x_n$ and $y_1, \ldots y_m$ comprising the image of $f$ and $g$, respectively. But then it immediately follows that any value attained by $f \cdot g$ is of the form $x_i \cdot y_j$. Hence, there are at most $n \times m$ distinct values $f \cdot g$ can take. From Pointwise Product of Measurable Functions is Measurable, $f \cdot g$ is also measurable. Hence by Measurable Function is Simple Function iff Finite Image Set, $f \cdot g$ is a simple function. {{qed}}
Let $f$ have the following standard representation: :$f = \displaystyle \sum_{i \mathop = 0}^n a_i \chi_{E_i}$ Then we see that $f^+$ must satisfy: :$f^+ = \displaystyle \sum_{i \mathop = 0}^n \max \left\{{a_i, 0}\right\} \chi_{E_i}$ as the $E_i$ are disjoint, and $\chi_{E_i} \ge 0$ pointwise. Since all of the $E_i$ are measurable, it follows that $f^+$ is a simple function. {{qed}}
Let $f$ have the following standard representation: :$f = \displaystyle \sum_{i \mathop = 0}^n a_i \chi_{E_i}$ Then we see that $f^-$ must satisfy: :$f^- = \displaystyle \sum_{i \mathop = 0}^n \min \left\{{a_i, 0}\right\} \chi_{E_i}$ as the $E_i$ are disjoint, and $\chi_{E_i} \ge 0$ pointwise. Since all of the $E_i$ are measurable, it follows that $f^+$ is a simple function. {{qed}}
Let $x \in X$.  Consider the following four cases for the value of $f(x)$ in $\overline{\R}$: * $f \left({x}\right) = -\infty$ : By ordering on extended reals: ::$f^+ \left({x}\right) = \max \left\{{0, f \left({x}\right)}\right\} = \max \left\{{0, -\infty}\right\} = 0$ ::$f^- \left({x}\right) = - \min \left\{{0, f \left({x}\right)}\right\} = - \min \left\{{0, -\infty}\right\} = +\infty$ : By extended real subtraction, it thus follows that: ::$f^+ \left({x}\right) - f^- \left({x}\right) = 0 - \left({+\infty}\right) = - \infty = f \left({x}\right)$ * $f \left({x}\right) \in \left({-\infty \,.\,.\, 0}\right)$ : Since $f \left({x}\right) < 0$, it follows that: ::$f^+ \left({x}\right) = \max \left\{{0, f \left({x}\right)}\right\} = 0$ ::$f^- \left({x}\right) = - \min \left\{{0, f \left({x}\right)}\right\}= - f \left({x}\right)$ : Thence it immediately follows that: ::$f^+ \left({x}\right) - f^- \left({x}\right) = 0 - \left({- f \left({x}\right)}\right) = f \left({x}\right)$ * $f \left({x}\right) \in \left[{0 \,.\,.\, \infty}\right)$ : Since $f \left({x}\right) \geq 0$, we obtain: ::$f^+ \left({x}\right) = \max \left\{{0, f \left({x}\right)}\right\} = f \left({x}\right)$ ::$f^- \left({x}\right) = - \min \left\{{0, f \left({x}\right)}\right\} = 0$ : Then, these immediately imply: ::$f^+ \left({x}\right) - f^- \left({x}\right) = f \left({x}\right) - 0 = f \left({x}\right)$ * $f \left({x}\right) = +\infty$ : By ordering on extended reals: ::$f^+ \left({x}\right) = \max \left\{{0, f \left({x}\right)}\right\} = \max \left\{{0, +\infty}\right\} = +\infty$ ::$f^- \left({x}\right) = - \min \left\{{0, f \left({x}\right)}\right\}= - \min \left\{{0, +\infty}\right\} = 0$ : By extended real subtraction, it now follows that: ::$f^+ \left({x}\right) - f^- \left({x}\right) = +\infty - 0 = +\infty = f \left({x}\right)$ In all cases, $f^+ \left({x}\right) - f^- \left({x}\right) = f \left({x}\right)$. As $x\in X$ was arbitrary, we may conclude that: :$\forall x \in X: f^+ \left({x}\right) - f^- \left({x}\right) = f \left({x}\right)$ That is, we have shown: :$f = f^+ - f^-$ {{qed}}
Let $x \in X$. Suppose that $f \left({x}\right) \ge 0$, where $\ge$ signifies the extended real ordering. Then $\left\vert{f \left({x}\right)}\right\vert = f \left({x}\right)$, and: :$f^+ \left({x}\right) = \max \, \left\{{f \left({x}\right), 0}\right\} = f \left({x}\right)$ :$f^- \left({x}\right) = - \min \, \left\{{f \left({x}\right), 0}\right\} = 0$ Hence $f^+ \left({x}\right) + f^- \left({x}\right) = f \left({x}\right) = \left\vert{f \left({x}\right)}\right\vert$. Next, suppose that $f \left({x}\right) < 0$, again in the extended real ordering. Then $\left\vert{f \left({x}\right)}\right\vert = - f \left({x}\right)$, and: :$f^+ \left({x}\right) = \max \, \left\{{f \left({x}\right), 0}\right\} = 0$ :$f^- \left({x}\right) = - \min \, \left\{{f \left({x}\right), 0}\right\} = - f \left({x}\right)$ Hence $f^+ \left({x}\right) + f^- \left({x}\right) = - f \left({x}\right) = \left\vert{f \left({x}\right)}\right\vert$. Thus, for all $x \in X$: :$f^+ \left({x}\right) + f^- \left({x}\right) = \left\vert{f \left({x}\right)}\right\vert$ that is, $f^+ + f^- = \left\vert{f}\right\vert$. {{qed}}
By Simple Function has Standard Representation, $f$ has a standard representation, say: :$(1): \quad f = \displaystyle \sum_{k \mathop = 0}^n a_k \chi_{E_k}$ Then, for all $x \in X$: :$\left\vert{f}\right\vert \left({x}\right) = \displaystyle \left\vert{ \sum_{k \mathop = 0}^n a_k \chi_{E_k} \left({x}\right) }\right\vert$ by definition of pointwise absolute value. The fact that $(1)$ forms a standard representation ensures that for every $x \in X$, precisely one $k$ has $x \in E_k$. Now suppose that $x \in E_l$.  Then $\chi_{E_l} \left({x}\right) = 0$ {{iff}} $k \ne l$ by definition of characteristic function. It follows that $\left\vert{f}\right\vert \left({x}\right) = \left\vert{a_l \cdot 1}\right\vert = \left\vert{a_l}\right\vert$. Now define $g: X \to \R$ by: $g \left({x}\right) := \displaystyle \sum_{k \mathop = 0}^n \left\vert{a_k}\right\vert \chi_{E_k} \left({x}\right)$ By construction, $g$ is a simple function, and for $x \in E_l$, $g \left({x}\right) = \left\vert{a_l}\right\vert$. Thus, since every $x$ is in $E_l$ for precisely one $l$, it is showed that $g = \left\vert{f}\right\vert$. As $g$ was a simple function, it follows that $\left\vert{f}\right\vert$ is, too. {{qed}}
By Sum of Positive and Negative Parts, we have: :$\left\vert{f}\right\vert = f^+ + f^-$ We also have that Positive Part of Simple Function is Simple Function and Negative Part of Simple Function is Simple Function. Hence $\left\vert{f}\right\vert$ is a pointwise sum of simple functions. The result follows from Pointwise Sum of Simple Functions is Simple Function. {{qed}}
We use mathematical induction. The base case $n = 2$ is proved in Kuratowski Formalization of Ordered Pair, and the induction step follows directly from the definition of an ordered tuple. {{qed}} {{finish}} Category:Cartesian Product 5aqtynhmlw7t2z7lkrvk4nv37eqx2c9
{{WLOG}} it is necessary to prove only that if $\mathcal R_1$ is reflexive then $\mathcal R_2$ is reflexive. Let $\phi: S \to T$ be a relation isomorphism. Let $y \in T$. Let $x = \map {\phi^{-1} } y$. As $\phi$ is a bijection it follows from Inverse Element of Bijection that: :$y = \map \phi x$ As $\mathcal R_1$ is a reflexive relation it follows that: :$x \mathrel {\mathcal R_1} x$ As $\phi$ is a relation isomorphism it follows that: :$\map \phi x \mathrel {\mathcal R_2} \map \phi x$ Hence the result. {{qed}}
A mapping is a specific kind of relation. Hence, Inverse of Composite Relation applies, and it follows that: :$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$ {{qed}}
Let the cardinality of a set $S$ be denoted $\card S$. === Necessary Condition === Suppose $S$ is infinite. We use mathematical induction on $n$. For all $n \in \N$, let $\map P n$ be the proposition: :There exists a subset of $S$ whose cardinality is $n$. ==== Basis for the Induction ==== From Empty Set is Subset of All Sets: :$\O \subseteq S$ From Cardinality of Empty Set, its cardinality is $0$. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :There exists a subset of $S$ whose cardinality is $k$. Then we need to show: :There exists a subset of $S$ whose cardinality is $k+1$. ==== Induction Step ==== This is our induction step: By the induction hypothesis, there exists $T \subseteq S$ such that $\card T = k$. First, note that $S \ne T$, otherwise by definition of infinite $S$ would be finite. So $T$ is therefore a proper subset of $S$. So: {{begin-eqn}} {{eqn | l = S \setminus T       | o = \ne       | r = \O       | c = Set Difference with Proper Subset }} {{eqn | l = \exists x       | o = \in       | r = S \setminus T       | c = {{Defof|Empty Set}} }} {{eqn | ll= \leadsto       | l = \set x       | o = \subseteq       | r = S \setminus T       | c = {{Defof|Singleton}} }} {{eqn | ll= \leadsto       | l = \set x       | o = \subseteq       | r = S       | c = Set Difference is Subset: $S \setminus T \subseteq S$ }} {{eqn | ll= \leadsto       | l = T \cup \set x       | o = \subseteq       | r = S       | c = $T \subseteq S$ and $\set x \subseteq S$, and Union is Smallest Superset }} {{end-eqn}} As $x \notin T$ it follows that: :$\set x \cap T = \O$ Thus by Cardinality of Set Union: :$\card {T \cup \set x} = k + 1$ That is, $T \cup \set x$ is a subset of $S$ whose cardinality is $k + 1$. This is the set whose existence was to be be proved. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :For all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$. {{qed|lemma}} === Sufficient Condition === Suppose that for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$. {{AimForCont}} that $S$ is finite. Let $N = \card S$. As $N \in \N$ it follows that $N + 1 \in \N$. By hypothesis, there exists a subset $T \subseteq S$ whose cardinality is $N + 1$. From Cardinality of Subset of Finite Set, $\card S \ge \card T$. But then $\card S = N \ge N + 1 = \card T$, which contradicts the fact that $N < N + 1$. From this contradiction it follows that $S$ can not be finite. {{qed}}
Let $\phi: S \to T$ be a relation isomorphism. By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism. Thus WLOG it suffices to prove only that if $\mathcal R_1$ is symmetric, then also $\mathcal R_2$ is symmetric. So, suppose $\mathcal R_1$ is a symmetric relation. Let $y_1, y_2 \in T$ such that $y_1 \mathrel {\mathcal R_2} y_2$. Let $x_1 = \phi^{-1} \left({y_1}\right)$ and $x_2 = \phi^{-1} \left({y_2}\right)$. As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \phi \left({x_1}\right)$ and $y_2 = \phi \left({x_2}\right)$. As $\phi^{-1}$ is a relation isomorphism it follows that: :$x_1 = \phi^{-1} \left({y_1}\right) \mathrel {\mathcal R_1} \phi^{-1} \left({y_2}\right) = x_2$ As $\mathcal R_1$ is a symmetric relation it follows that: :$x_2 \mathrel {\mathcal R_1} x_1$ As $\phi$ is a relation isomorphism it follows that: :$y_2 = \phi \left({x_2}\right) \mathrel {\mathcal R_2} \phi \left({x_1}\right) = y_1$ So: :$y_1 \mathrel {\mathcal R_2} y_2 \implies y_2 \mathrel {\mathcal R_2} y_1$ and so by definition $\mathcal R_2$ is a symmetric relation. Hence the result. {{qed}}
{{AimForCont}} $S \setminus T = \O$. Then: :$\not \exists x \in S: x \notin T$ By De Morgan's laws: :$\forall x \in S: x \in T$ By definition of subset: :$S \subseteq T$ By definition of proper subset, we have that $T \subseteq S$ such that $T \ne S$. But we have $T \subseteq S$ and $S \subseteq T$. So by definition of set equality: :$S = T$ From this contradiction it follows that: :$S \setminus T \ne \O$ {{qed}} Category:Set Difference Category:Subsets 5r0nemxsix8lymqklsuqmwslr2kxwhd
=== Necessary Condition === Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an order isomorphism. Then by definition $\phi$ is a bijection and so a surjection. Suppose $x \mathop{\prec_1} y$. That is: :$x \mathop{\preceq_1} y$ :$x \ne y$ Then: :$x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$ as $\phi$ is an order isomorphism. But as $\phi$ is a bijection it is also an injection. Thus: : $\phi \left({x}\right) = \phi \left({y}\right) \implies x = y$ and so it follows that: :$x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\prec_2} \phi \left({y}\right)$ {{qed|lemma}} === Sufficient Condition === Suppose $\phi$ is a mapping which satisfies the conditions: : $(1): \quad \phi$ is a surjection : $(2): \quad \forall x, y \in S: x \mathop{\prec_1} y \implies \phi \left({x}\right) \mathop{\prec_2} \phi \left({y}\right)$ From $(2)$ and Strictly Increasing Mapping is Increasing we have: :$x \mathop{\preceq_1} y \implies \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$ Now suppose $\phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$. Suppose $y \mathop{\prec_1} x$. Then from $(2)$ it would follow that $\phi \left({y}\right) \mathop{\prec_2} \phi \left({x}\right)$. So it is not the case that $y \mathop{\prec_1} x$. So from the Trichotomy Law $x \mathop{\preceq_1} y$. Thus it follows that: : $x \mathop{\preceq_1} y \iff \phi \left({x}\right) \mathop{\preceq_2} \phi \left({y}\right)$ It follows from Order Isomorphism is Surjective Order Embedding that $\phi$ is an order isomorphism. {{qed}}
Let $\phi: S \to T$ be a relation isomorphism. By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism. {{WLOG}}, it is therefore sufficient to prove only that if $\RR_1$ is transitive, then $\RR_2$ is also transitive. So, suppose $\RR_1$ is a transitive relation. Let $y_1, y_2, y_3 \in T$ such that $y_1 \mathrel {\RR_2} y_2$ and $y_2 \mathrel {\RR_2} y_3$. Let $x_1 = \map {\phi^{-1} } {y_1}$, $x_2 = \map {\phi^{-1} } {y_2}$ and $x_3 = \map {\phi^{-1} } {y_3}$. As $\phi$ is a bijection it follows from Inverse Element of Bijection that: :$y_1 = \map \phi {x_1}$ :$y_2 = \map \phi {x_2}$ :$y_3 = \map \phi {x_3}$ As $\phi^{-1}$ is a relation isomorphism it follows that: :$x_1 = \map {\phi^{-1} } {y_1} \mathrel {\RR_1} \map {\phi^{-1} } {y_2} = x_2$ :$x_2 = \map {\phi^{-1} } {y_2} \mathrel {\RR_1} \map {\phi^{-1} } {y_3} = x_3$ As $\RR_1$ is a transitive relation it follows that: :$x_1 \mathrel {\RR_1} x_3$ As $\phi$ is a relation isomorphism it follows that: :$y_1 = \map \phi {x_1} \mathrel {\RR_2} \map \phi {x_3} = y_3$ Hence if $y_1 \mathrel {\RR_2} y_2$ and $y_2 \mathrel {\RR_2} y_3$, then also: :$y_1 \mathrel {\RR_2} y_3$ Hence, $\RR_2$ is a transitive relation, by definition. Hence the result. {{qed}}
Let $\phi: S \to T$ be a relation isomorphism. By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism. {{WLOG}}, it therefore suffices to prove only that if $\mathcal R_1$ is antisymmetric, then also $\mathcal R_2$ is antisymmetric. So, suppose $\mathcal R_1$ is an antisymmetric relation. Let $y_1, y_2 \in T$ such that both $y_1 \mathrel {\mathcal R_2}, y_2$ and $y_2 \mathrel {\mathcal R_2} y_1$. Let $x_1 = \phi^{-1} \left({y_1}\right)$ and $x_2 = \phi^{-1} \left({y_2}\right)$. As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \phi \left({x_1}\right)$ and $y_2 = \phi \left({x_2}\right)$. As $\phi^{-1}$ is a relation isomorphism it follows that: : $x_1 = \phi^{-1} \left({y_1}\right) \mathrel {\mathcal R_1} \phi^{-1} \left({y_2}\right) = x_2$ : $x_2 = \phi^{-1} \left({y_2}\right) \mathrel {\mathcal R_1} \phi^{-1} \left({y_1}\right) = x_1$ As $\mathcal R_1$ is an antisymmetric relation it follows that $x_1 = x_2$. As $\phi$ is a bijection it follows that: : $y_1 = \phi \left({x_1}\right) = \phi \left({x_2}\right) = y_2$ Hence $y_1 = y_2$ and so by definition, $\mathcal R_2$ is an antisymmetric relation. Hence the result. {{qed}}
Let $\phi: S \to T$ be a relation isomorphism. By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism. {{WLOG}}, it thus is necessary to prove only that if $\RR_1$ is an equivalence relation then $\RR_2$ is an equivalence relation. So, suppose $\RR_1$ is an equivalence relation. By definition: :$(1): \quad \RR_1$ is reflexive :$(2): \quad \RR_1$ is symmetric :$(3): \quad \RR_1$ is transitive. It follows that: :From Relation Isomorphism Preserves Reflexivity, $\RR_2$ is reflexive. :From Relation Isomorphism Preserves Symmetry, $\RR_2$ is symmetric. :From Relation Isomorphism Preserves Transitivity, $\RR_2$ is transitive. So by definition $\RR_2$ is an equivalence relation. Hence the result. {{qed}}
Let $a \in \R$; for all $i \in I$, we have by Characterization of Measurable Functions that: :$\left\{{f_i > a}\right\} \in \Sigma$ and as $\Sigma$ is a $\sigma$-algebra and $I$ is countable, also: :$\displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \in \Sigma$ We will now show that: :$\displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\} = \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\}$ First, observe that for all $i \in I$: :$f_i \left({x}\right) \le \displaystyle \sup_{i \mathop \in I} f_i \left({x}\right)$ and hence: :$\left\{{f_i > a}\right\} \subseteq \displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$ From Union is Smallest Superset: Family of Sets: :$\displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \subseteq \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$ Next, suppose that: :$x \notin \displaystyle \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\}$ Then, by definition of union: :$\forall i \in I: f_i \left({x}\right) \le a$ which is to say that $a$ is an upper bound for the $f_i \left({x}\right)$. Hence, by definition of supremum, it follows that: :$\displaystyle \sup_{i \mathop \in I} f_i \left({x}\right) \le a$ and therefore: :$x \notin \displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\}$ Thus, we have shown: :$\displaystyle \left\{{\sup_{i \mathop \in I} f_i > a}\right\} = \bigcup_{i \mathop \in I} \left\{{f_i > a}\right\} \in \Sigma$ {{explain|There must be a result covering this type of set equality}} and by Characterization of Measurable Functions, it follows that $\displaystyle \sup_{i \mathop \in I} f_i$ is measurable. {{qed}}
From Infimum as Supremum, we have the Equality of Mappings: :$\displaystyle \inf_{i \mathop \in I} f_i = - \left({\sup_{i \mathop \in I} \, \left({- f_i}\right)}\right)$ Now, from Negative of Measurable Function is Measurable and Pointwise Supremum of Measurable Functions is Measurable, it follows that: :$\displaystyle - \left({\sup_{i \mathop \in I} \, \left({- f_i}\right)}\right)$ is a measurable function. Hence the result. {{qed}}