hcju/proofwikiqa huggingface dataset:
By definition of upper limit, we have: :$\displaystyle \limsup_{n \to \infty} f_n = \inf_{m \mathop \in \N} \ \sup_{n \ge m} f_n$ The result follows from combining: :Pointwise Supremum of Measurable Functions is Measurable :Pointwise Infimum of Measurable Functions is Measurable {{qed}}
By definition of limit inferior, we have: :$\displaystyle \liminf_{n \mathop \to \infty} f_n = \sup_{m \mathop \in \N} \ \inf_{n \mathop \ge m} f_n$ The result follows from combining: :Pointwise Infimum of Measurable Functions is Measurable :Pointwise Supremum of Measurable Functions is Measurable {{qed}}
From Convergence of Limsup and Liminf, it follows that: :$\displaystyle \lim_{n \to \infty} f_n = \limsup_{n \to \infty} f_n$ We have Pointwise Upper Limit of Measurable Functions is Measurable. Hence the result. {{qed}}
By Measurable Function Pointwise Limit of Simple Functions, we find sequences $\left({f_n}\right)_{n \in \N}, \left({g_n}\right)_{n \in \N}$ such that: :$\displaystyle f = \lim_{n \to \infty} f_n$ :$\displaystyle g = \lim_{n \to \infty} g_n$ where the limits are pointwise. It follows that for all $x \in X$: :$f \left({x}\right) + g \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right) + g_n \left({x}\right)$ {{explain|There needs to be a bunch of results establishing such equalities of convergence in function spaces}} so that we have the pointwise limit: :$\displaystyle f + g = \lim_{n \to \infty} f_n + g_n$ By Pointwise Sum of Simple Functions is Simple Function, $f + g$ is a pointwise limit of simple functions. By Simple Function is Measurable, $f + g$ is a pointwise limit of measurable functions. Hence $f + g$ is measurable, by Pointwise Limit of Measurable Functions is Measurable. {{qed}}
We have the apparent identity: :$f - g = f + \paren {-g}$ By Negative of Measurable Function is Measurable, $-g$ is a measurable function. Hence so is $f - g$, by Pointwise Sum of Measurable Functions is Measurable. {{qed}}
By Measurable Function Pointwise Limit of Simple Functions, we find sequences $\left({f_n}\right)_{n \in \N}, \left({g_n}\right)_{n \in \N}$ such that: :$\displaystyle f = \lim_{n \to \infty} f_n$ :$\displaystyle g = \lim_{n \to \infty} g_n$ where the limits are pointwise. It follows that for all $x \in X$: :$f \left({x}\right) \cdot g \left({x}\right) = \displaystyle \lim_{n \to \infty} f_n \left({x}\right) \cdot g_n \left({x}\right)$ {{explain|There needs to be a bunch of results establishing such equalities of convergence in function spaces}} so that we have the pointwise limit: :$\displaystyle f \cdot g = \lim_{n \to \infty} f_n \cdot g_n$ By Pointwise Product of Simple Functions is Simple Function, $f \cdot g$ is a pointwise limit of simple functions. By Simple Function is Measurable, $f \cdot g$ is a pointwise limit of measurable functions. Hence $f \cdot g$ is measurable, by Pointwise Limit of Measurable Functions is Measurable. {{qed}}
For all $x \in X$ and $a \in \R$, we have by Max yields Supremum of Parameters that: :$\max \left({f \left({x}\right), g \left({x}\right)}\right) \le a$ iff both $f \left({x}\right) \le a$ and $g \left({x}\right) \le a$. That is, for all $a \in \R$: :$\left\{{x \in X: \max \left({f \left({x}\right), g \left({x}\right)}\right) \le a}\right\} = \left\{{x \in X: f \left({x}\right) \le a}\right\} \cap \left\{{x \in X: g \left({x}\right) \le a}\right\}$ By Characterization of Measurable Functions: $(1) \implies (2)$, the two sets on the RHS are elements of $\Sigma$, i.e. measurable. By Sigma-Algebra Closed under Intersection, it follows that: :$\left\{{x \in X: \max \left({f \left({x}\right), g \left({x}\right)}\right) \le a}\right\} \in \Sigma$ Hence $\max \left({f, g}\right)$ is measurable, by Characterization of Measurable Functions: $(2) \implies (1)$. {{qed}}
For all $x \in X$ and $a \in \R$, we have by Min yields Infimum of Parameters that: :$a \le \min \left({f \left({x}\right), g \left({x}\right)}\right)$ iff both $a \le f \left({x}\right)$ and $a \le g \left({x}\right)$. That is, for all $a \in \R$: :$\left\{{x \in X: \min \left({f \left({x}\right), g \left({x}\right)}\right) \ge a}\right\} = \left\{{x \in X: f \left({x}\right) \ge a}\right\} \cap \left\{{x \in X: g \left({x}\right) \ge a}\right\}$ By Characterization of Measurable Functions: $(1) \implies (4)$, the two sets on the RHS are elements of $\Sigma$, i.e. measurable. By Sigma-Algebra Closed under Intersection, it follows that: :$\left\{{x \in X: \min \left({f \left({x}\right), g \left({x}\right)}\right) \ge a}\right\} \in \Sigma$ Hence $\min \left({f, g}\right)$ is measurable, by Characterization of Measurable Functions: $(4) \implies (1)$. {{qed}}
=== Necessary Condition === Suppose $f$ is measurable. By definition, its positive part $f^+$ equals the pointwise maximum: :$f^+ = \max \left\{{f, 0}\right\}$ where $0$ denotes the zero function. By Constant Function is Measurable, $0$ is a measurable function. Thus, by Pointwise Maximum of Measurable Functions is Measurable, $f^+$ is measurable. Subsequently, the negative part $f^-$ of $f$ is defined by means of a pointwise minimum: :$f^- = - \min \left\{{f, 0}\right\}$ By Pointwise Maximum of Measurable Functions is Measurable and Negative of Measurable Function is Measurable, $f^-$ is measurable as well. {{qed|lemma}} === Sufficient Condition === Suppose $f^+$ and $f^-$ are both measurable. By Difference of Positive and Negative Parts: :$f = f^+ - f^-$ and hence, by Pointwise Difference of Measurable Functions is Measurable, $f$ is also measurable. {{qed}}
From Pointwise Difference of Measurable Functions is Measurable, $f - g: X \to \overline \R$ is $\Sigma$-measurable. Now we have the following evident identities: :$\set {f < g} = \set {f - g < 0}$ :$\set {f \ge g} = \set {f - g \le 0}$ :$\set {f = g} = \set {f - g = 0}$ :$\set {f \ne g} = \set {f - g \ne 0}$ Subsequently, using the preimage under $f - g$, the latter may respectively be expressed as: :$\set {f - g < 0} = \paren {f - g}^{-1} \sqbrk {\openint {-\infty} 0}$ :$\set {f - g \ge 0} = \paren {f - g}^{-1} \sqbrk {\hointl {-\infty} 0}$ :$\set {f - g = 0} = \paren {f - g}^{-1} \sqbrk {\set 0}$ :$\set {f - g \ne 0} = \paren {f - g}^{-1} \sqbrk {\R \setminus \set 0}$ Now the sets: :$\openint {-\infty} 0$ :$\hointl {-\infty} 0$ :$\set 0$ :$\R \setminus \set 0$ are all open or closed in the Euclidean topology. Hence by definition of Borel $\sigma$-algebra, and by Closed Set Measurable in Borel Sigma-Algebra, they are $\map \BB \R$-measurable. Since $f - g$ is $\Sigma$-measurable, it follows that: :$\set {f < g}$ :$\set {f \le g}$ :$\set {f = g}$ :$\set {f \ne g}$ are all $\Sigma$-measurable sets. {{qed}}
=== Necessary Condition === Let $g$ be a $\sigma \left({f}\right) \, / \, \overline{\mathcal B}$-measurable function. We need to construct a measurable $\tilde g$ such that $g = \tilde g \circ f$. Let us proceed in the following fashion: * Establish the result for $g$ a characteristic function; * Establish the result for $g$ a simple function; * Establish the result for all $g$ So let $g = \chi_E$ be a characteristic function. By Characteristic Function Measurable iff Set Measurable, it follows that $E$ is $\sigma \left({f}\right)$-measurable. Thus there exists some $A \in \Sigma$ such that $E = f^{-1} \left({A}\right)$. Again by Characteristic Function Measurable iff Set Measurable, we have $\chi_A: Y \to \overline \R$ is measurable. It follows that $\chi_E = \chi_A \circ f$, and $\tilde g := \chi_A$ works. Now let $g = \displaystyle \sum_{i \mathop = 1}^n a_i \chi_{E_i}$ be a simple function. Let $A_i$ be associated to $E_i$ as above. Then we have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n a_i \chi_{E_i}       | r = \sum_{i \mathop = 1}^n a_i \left({\chi_{A_i} \circ f}\right)       | c = by the result for characteristic functions }} {{eqn | r = \left({\sum_{i \mathop = 1}^n a_i \chi_{A_i} }\right) \circ f       | c = Composition of Mappings is Linear }} {{end-eqn}} Now $\displaystyle \sum_{i \mathop = 1}^n a_i \chi_{A_i}$ is a simple function, hence measurable by Simple Function is Measurable. Therefore, it suffices as a choice for $\tilde g$. Next, let $g \ge 0$ be a measurable function. By Measurable Function Pointwise Limit of Simple Functions, we find simple functions $g_j$ such that: :$\displaystyle \lim_{j \to \infty} g_j = g$ Applying the previous step to each $g_j$, we find a sequence of $\tilde g_j$ satisfying: :$\displaystyle \lim_{j \to \infty} \tilde g_j \circ f = g$ From Composition with Pointwise Limit it follows that we have, putting $\tilde g := \displaystyle \lim_{j \to \infty} \tilde g_j$: :$\displaystyle \lim_{j \to \infty} \tilde g_j \circ f = \tilde g \circ f$ {{explain|Address why (if?!) $\tilde g$ is well defined}} An application of Pointwise Limit of Measurable Functions is Measurable yields $\tilde g$ measurable. Thus we have provided a suitable $\tilde g$ for every $g$, such that: :$g = \tilde g \circ f$ as desired. {{qed}} === Sufficient Condition === Suppose that such a $\tilde g$ exists. Note that $f$ is $\sigma \left({f}\right) \,/\, \Sigma$-measurable by definition of $\sigma \left({f}\right)$. The result follows immediately from Composition of Measurable Mappings is Measurable. {{qed}}
Let $E' \in \Sigma'$ be a $\Sigma'$-measurable set. Then by definition of preimage: :$\map {h^{-1} } {E'} = \set {x \in X: \map h x \in E'}$ Expanding the definition of $h$, this translates into: :$\map {h^{-1} } {E'} = \set {x \in E: \map f x \in E'} \cup \set {x \in \relcomp X E: \map g x \in E'}$ where $\complement$ denotes set complement. That is, we have: :$\map {h^{-1} } {E'} = \paren {E \cap \map {f^{-1} } {E'} } \cup \paren {\relcomp X E \cap \map {g^{-1} } {E'} }$ All sets on the {{RHS}} are $\Sigma$-measurable. By Sigma-Algebra Closed under Intersection and Sigma-Algebra Closed under Union, so is $\map {h^{-1} } {E'}$. Since $E' \in \Sigma'$ was arbitrary, $h$ is a $\Sigma \, / \, \Sigma'$-measurable mapping. {{qed}}
First, note that $f$ is well-defined, since if $x \in E_n$ and $x \in E_m$, we have that: :$f_n \left({x}\right) = f \left({x}\right) = f_m \left({x}\right)$ by $(1)$, since $x \in E_n \cap E_m$. Let $E' \in \Sigma'$. Then by definition of preimage, $f^{-1} \left({E'}\right) \subseteq X$, and hence: {{begin-eqn}} {{eqn | l = f^{-1} \left({E'}\right)       | r = X \cap h^{-1} \left({E'}\right)       | c = Intersection with Subset is Subset }} {{eqn | r = \left({\bigcup_{n \mathop \in \N} E_n}\right) \cap f^{-1} \left({E'}\right) }} {{eqn | r = \bigcup_{n \mathop \in \N} \left({E_n \cap f^{-1} \left({E'}\right)}\right)       | c = Intersection Distributes over Union: General Result }} {{eqn | r = \bigcup_{n \mathop \in \N} f_n^{-1} \left({E'}\right) }} {{end-eqn}} The last step we derive as follows. Whenever $x \in E_n$, we have by definition of $f$ that $f \left({x}\right) = f_n \left({x}\right)$. Hence, $x \in E_n \cap f^{-1} \left({E'}\right)$ implies $x \in f_n^{-1} \left({E'}\right)$. Conversely, if $x \in f_n^{-1} \left({E'}\right)$, i.e. $f_n \left({x}\right) \in E'$, then $x \in E_n$ since $E_n$ is the domain of $f_n$. By definition of $f$ then also $f \left({x}\right) = f_n \left({x}\right) \in E'$. Hence we have deduced that $E_n \cap f^{-1} \left({E'}\right) = f_n^{-1} \left({E'}\right)$. Since all $f_n$ are $\Sigma_{E_n} \, / \, \Sigma'$-measurable, it follows that for all $n \in \N$: :$f_n^{-1} \left({E'}\right) \in \Sigma$ By $\sigma$-algebra axiom $(3)$, it follows that: :$f^{-1} \left({E'}\right) = \displaystyle \bigcup_{n \mathop \in \N} f_n^{-1} \left({E'}\right) \in \Sigma$ Since $E' \in \Sigma'$ was arbitrary, it follows that $f$ is $\Sigma \, / \, \Sigma'$-measurable. {{qed}}
=== Necessary Condition === Suppose $g$ is a simple function. By Positive Part of Simple Function is Simple Function, so is $g^+$. By Negative Part of Simple Function is Simple Function, so is $g^-$. {{qed|lemma}} === Sufficient Condition === Suppose $g^+$ and $g^-$ are simple functions. From Difference of Positive and Negative Parts: :$g = g^+ - g^-$ Hence $g$ is simple, by Pointwise Difference of Simple Functions is Simple Function. {{finish}}
The sets $F_0, \ldots, F_m$ are pairwise disjoint, and: :$X = \displaystyle \bigcup_{j \mathop = 0}^m F_j$ From Characteristic Function of Disjoint Union, we have: :$\chi_X = \displaystyle \sum_{j \mathop = 0}^m \chi_{F_j}$ Remark that $\chi_X \left({x}\right) = 1$ for all $x \in X$, so that we have: {{begin-eqn}} {{eqn|l = f      |r = \sum_{i \mathop = 0}^n a_i \chi_{E_i} \cdot 1 }} {{eqn|r = \sum_{i \mathop = 0}^n a_i \chi_{E_i} \left({\sum_{j \mathop = 0}^m \chi_{F_j} }\right) }} {{eqn|r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^m a_i \chi_{E_i} \chi_{F_j} }} {{eqn|r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^m a_i \chi_{E_i \cap F_j}      |c = Characteristic Function of Intersection: Variant 1 }} {{end-eqn}} Repeating the argument with the $E_i$ and $F_j$ interchanged also yields: :$f = \displaystyle \sum_{j \mathop = 0}^m \sum_{i \mathop = 0}^n b_j \chi_{F_j \cap E_i}$ Now whenever $x \in E_i \cap F_j$, for some $i, j$, then since the $E_i$, $F_j$ are disjoint, we find: :$x \in E_{i'} \cap F_{j'}$ implies $i = i'$ and $j = j'$ Thus, evaluating both expressions for $f \left({x}\right)$ we find: :$a_i = f \left({x}\right) = b_j$ In conclusion, we have: :$a_i = b_j$ if $E_i \cap F_j \ne \varnothing$. Furthermore, we have for all $i$ that: :$\displaystyle E_i = E_i \cap X = E_i \cap \left({\bigcup_{j \mathop = 0}^m F_j}\right) = \bigcup_{j \mathop = 0}^m \left({E_i \cap F_j}\right)$ by Intersection Distributes over Union: General Result. Similarly, we obtain for all $j$: :$\displaystyle F_j = F_j \cap X = F_j \cap \left({\bigcup_{i \mathop = 0}^n E_i}\right) = \bigcup_{i \mathop = 0}^n \left({F_j \cap E_i}\right)$ With this knowledge, we compute: {{begin-eqn}} {{eqn|l = \sum_{i \mathop = 0}^n a_i \mu \left({E_i}\right)      |r = \sum_{i \mathop = 0}^n a_i \mu \left({\bigcup_{j \mathop = 0}^m \left({E_i \cap F_j}\right)}\right) }} {{eqn|r = \sum_{i \mathop = 0}^n a_i \sum_{j \mathop = 0}^m \mu \left({E_i \cap F_j}\right)      |c = Measure is Finitely Additive Function }} {{eqn|r = \sum_{i \mathop = 0}^n \sum_{j \mathop = 0}^m a_i \mu \left({E_i \cap F_j}\right)      |c = Summation is Linear }} {{eqn|r = \sum_{j \mathop = 0}^m \sum_{i \mathop = 0}^n b_j \mu \left({E_i \cap F_j}\right)      |c = If $a_i \ne b_j$ then $E_i \cap F_j = \varnothing$; $a_i \cdot 0 = b_j \cdot 0$ }} {{eqn|r = \sum_{j \mathop = 0}^m b_j \sum_{i \mathop = 0}^n \mu \left({E_i \cap F_j}\right)      |c = Summation is Linear }} {{eqn|r = \sum_{j \mathop = 0}^m b_j \mu \left({\bigcup_{i \mathop = 0}^n \left({E_i \cap F_j}\right)}\right)      |c = Measure is Finitely Additive Function }} {{eqn|r = \sum_{j \mathop = 0}^m b_j \mu \left({F_j}\right) }} {{end-eqn}} Hence the result. {{qed}}
Let $a_1 = 1$ and $E_1 = E$. As in the definition of standard representation, denote $a_0 = 0$ and $E_0 = X \setminus E_1$. Then for $x \in X$, we have: :$\chi_E \left({x}\right) = 0 \cdot \chi_{E_0} \left({x}\right) + 1 \cdot \chi_{E_1} \left({x}\right)$ since $E_1 = E$. Hence $\chi_E = a_0 \chi_{E_0} + a_1 \chi_{E_1}$ is a standard representation for $\chi_E$. Thus, by definition of $\mu$-integral: :$\displaystyle \int \chi_E \, \mathrm d \mu = a_0 \mu \left({E_0}\right) + a_1 \mu \left({E_1}\right) = \mu \left({E}\right)$ as desired. {{qed}}
Remark that $\lambda \cdot f$ is a positive simple function by Scalar Multiple of Simple Function is Simple Function. Let: :$f = \displaystyle \sum_{i \mathop = 0}^n a_i \chi_{E_i}$ be a standard representation for $f$. Then we also have, for all $x \in X$: {{begin-eqn}} {{eqn|l = \lambda \cdot f \left({x}\right)      |r = \lambda \sum_{i \mathop = 0}^n a_i \chi_{E_i} \left({x}\right) }} {{eqn|r = \sum_{i \mathop = 0}^n \left({\lambda a_i}\right) \chi_{E_i} \left({x}\right)      |c = Summation is Linear }} {{end-eqn}} and it is immediate from the definition that this yields a standard representation for $\lambda \cdot f$. Therefore, we have: {{begin-eqn}} {{eqn | l = \lambda \cdot I_\mu \left({f}\right)       | r = \lambda \sum_{i \mathop = 0}^n a_i \mu \left({E_i}\right)       | c = Definition of $\mu$-integration }} {{eqn | r = \sum_{i \mathop = 0}^n \left({\lambda a_i}\right) \mu \left({E_i}\right)       | c = Summation is Linear }} {{eqn | r = I_\mu \left({\lambda \cdot f}\right)       | c = Definition of $\mu$-integration }} {{end-eqn}} as desired. {{qed}}
Suppose $x \circ z \prec y \circ z$. By Invertible Element of Monoid is Cancellable, $z^{-1}$ is cancellable. Then from Strict Ordering Preserved under Product with Cancellable Element: :$x = \left({x \circ z}\right) \circ z^{-1} \prec \left({y \circ z}\right) \circ z^{-1} = y$ Likewise, if $z \circ x \prec z \circ y$: :$x = z^{-1} \circ \left({z \circ x}\right) \prec z^{-1} \circ \left({z \circ y}\right) = y$ {{qed}}
Let $\preceq$ be a total ordering. Let $x \circ z \prec y \circ z$. But we have, by hypothesis: :$x \succeq y \implies x \circ z \succeq y \circ z$ which contradicts $x \circ z \prec y \circ z$. So $x \prec y$. Similarly for $z \circ x \prec z \circ y$. {{qed}}
From Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup: : $x \circ z \prec y \circ z \implies x \prec y$ From the definition of cancellable element: : $x \circ z = y \circ z \implies x = y$ {{qed}} Category:Ordered Semigroups oafce8s0w7tw3us87tqoozua1fopu57
{{tidy}} {{Proofread}} {{MissingLinks}} Since by definition $\displaystyle \sup _{n \mathop \in \N} f_n \ge f_m$ for all $m$, we have: :$\displaystyle \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \int f_m \rd \mu$ and hence the inequality holds for the supremum as well: :$\displaystyle \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \sup_{m \mathop \in \N} \int f_m \rd \mu$ To show the reverse inequality, since the integral of $\sup\limits_{n \mathop \in \N}f_n$ is defined as the supremum of the integrals of positive simple functions $s \le \sup\limits_{n \mathop \in \N}f_n$, we show that given a simple function $\displaystyle s = \sum_{i \mathop = 1}^k \lambda_i \chi_{E_i} \le \sup_{n \mathop \in \N} f_n$ (where $\lambda_i \in \closedint 0 {+\infty}$ and $E_i \in \Sigma$) we have: :$\displaystyle \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$ To show this, we use the fact that $\nu_s: \Sigma \to \closedint 0 {+\infty}$ defined by $\map {\nu_s} E = \displaystyle \int \chi_Es \rd \mu$ clearly defines a measure over $X$, because it is simply a linear combination (with positive coefficients) of the measures $\bigvalueat \mu {E_i}$. Now, if we fix $1 > \epsilon > 0$, we have that the sets: :$A_m = \set {x \in X: \map {f_m} x \ge \paren {1 - \epsilon} \map s x}$ form a cover of $X$ ($X = \displaystyle \bigcup_{m \mathop \in \N} A_m$) by definition of the supremum, because $\displaystyle s \le \sup_{n \mathop \in \N} f_n$. Furthermore, $A_m \uparrow X$ ($m \to \infty$). {{explain|the notation of the above statement: link to a page defining $A_m \uparrow X$ or replace with the house notational style.}} By definition of the $A_m$ we have that: :$\displaystyle \int f_m \rd \mu \ge \int \chi_{A_m} f_m \rd \mu \ge \paren {1 - \epsilon} \int \chi_{A_m} s \rd \mu = \paren {1 - \epsilon} \map {\nu_s} {A_m}$ where the first inequality follows from the fact that the $f_m$ are positive. Now, since the sequence $\sequence {f_n}_{n \mathop \in \N}$ increases monotonically, by the  increasing sequence of sets property given in any measure space, we have that taking the supremum of both sides yields: {{begin-eqn}} {{eqn | l = \sup_{m \mathop \in \N} \int f_m \rd \mu       | r = \lim_{m \mathop \to +\infty} \int f_m \rd \mu       | c =  }} {{eqn | o = \ge       | r = \lim_{m \mathop \to +\infty} \paren {1 - \epsilon} \map {\nu_s} {A_m}       | c =  }} {{eqn | r = \paren {1 - \epsilon} \map {\nu_s} X       | c =  }} {{eqn | r = \paren {1 - \epsilon} \int s \rd \mu       | c =  }} {{end-eqn}} Since $\epsilon$ was selected arbitrarily, we have that the desired inequality holds: :$\displaystyle \sup_{m \mathop \in \N} \int f_m \rd \mu \ge \int s \rd \mu$ {{qed}}
=== Necessary Condition === Let $\phi: \left({S, \circ, \preceq}\right) \to \left({T, *, \preccurlyeq}\right)$ be an ordered semigroup isomorphism. Then by definition: : $\phi$ is a semigroup isomorphism from the semigroup $\left({S, \circ}\right)$ to the semigroup $\left({T, *}\right)$ : $\phi$ is an order isomorphism from the ordered set $\left({S, \preceq}\right)$ to the ordered set $\left({T, \preccurlyeq}\right)$. A semigroup isomorphism is by definition: : A semigroup homomorphism which is: : A monomorphism and an epimorphism. From Order Isomorphism is Surjective Order Embedding, an order isomorphism is an order embedding which is also a surjection. Putting this all together, we see that an ordered semigroup isomorphism is: : A monomorphism : An order embedding : A surjection. An ordered semigroup monomorphism is by definition:  : A monomorphism which is also : An order embedding Hence $\phi$ is: : An ordered semigroup monomorphism : A surjection. {{qed|lemma}} === Sufficient Condition === Let $\phi$ be: : An ordered semigroup monomorphism : A surjection. By definition, that means $\phi$ be: : A monomorphism : An order embedding : A surjection. From Order Isomorphism is Surjective Order Embedding, an order isomorphism is an order embedding which is also a surjection. A semigroup isomorphism is by definition: : A semigroup homomorphism which is: : A semigroup monomorphism and an semigroup epimorphism. Thus a semigroup monomorphism which is also a surjection is a semigroup isomorphism. So $\phi$ is: : A semigroup isomorphism from the semigroup $\left({S, \circ}\right)$ to the semigroup $\left({T, *}\right)$ : An order isomorphism from the ordered set $\left({S, \preceq}\right)$ to the ordered set $\left({T, \preccurlyeq}\right)$. {{qed}}
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup monomorphism. Then $\phi$ is an injection into $\struct {T, *, \preccurlyeq}$ by definition. From Restriction of Mapping to Image is Surjection, a mapping from a set to the image of that mapping is a surjection. Thus the surjective restriction of $\phi$ onto $S'$ is an ordered semigroup monomorphism which is also a surjection. Hence the result from Ordered Semigroup Isomorphism is Surjective Monomorphism. {{qed}}
Both $\left({T, \preceq_T}\right)$ and $\left({T', \preceq_{T'}}\right)$ are order completions for $\left({S, \preceq_S}\right)$. Hence they both satisfy condition $(4)$ (and also $(1)$, $(2)$ and $(3)$). Thus, applying condition $(4)$ to $\left({T, \preceq_T}\right)$ (with respect to $\left({T', \preceq_{T'}}\right)$), obtain a unique order-preserving mapping $\phi: T' \to T$. Applying $(4)$ to $\left({T', \preceq_{T'}}\right)$ (with respect to $\left({T, \preceq_T}\right)$) gives also a unique order-preserving mapping $\psi: T \to T'$. By Composite of Order-Preserving Mappings is Order-Preserving, their composites $\psi \circ \phi: T' \to T'$ and $\phi \circ \psi: T \to T$ are also order-preserving. Now applying $(4)$ to $\left({T, \preceq_T}\right)$ (with respect to itself), it follows that $\phi \circ \psi$ is unique. Now from Identity Mapping is Order Isomorphism, the identity mapping $\operatorname{id}_T: T \to T$ is also order-preserving. Thus, uniqueness of $\phi \circ \psi$ implies that $\phi \circ \psi = \operatorname{id}_T$. Similarly, it follows that $\psi \circ \phi = \operatorname{id}_{T'}$. It follows that $\psi: T \to T'$ is a bijection from Bijection iff Left and Right Inverse. Thus, $\psi$ is an order-preserving bijection whose inverse is also order-preserving. That is, $\psi$ is an order isomorphism. Its uniqueness was already remarked above. {{qed}} Category:Order Theory sfexel14u95lq4kg46ovitl61vyknd1