Algebras of the powerset monad











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Many authors treat algrebras of the powerset monad as a trivial example. It is really not trivial to me.



Can anyone help me with a detailed construction of such algebras. If you know of any article or book that explicitly construct such algebras which will be easy for a beginner to follow, please help out. Or if you can construct it as an answer here, I would really appreciate it.



Thank you.










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    up vote
    6
    down vote

    favorite












    Many authors treat algrebras of the powerset monad as a trivial example. It is really not trivial to me.



    Can anyone help me with a detailed construction of such algebras. If you know of any article or book that explicitly construct such algebras which will be easy for a beginner to follow, please help out. Or if you can construct it as an answer here, I would really appreciate it.



    Thank you.










    share|cite|improve this question
























      up vote
      6
      down vote

      favorite









      up vote
      6
      down vote

      favorite











      Many authors treat algrebras of the powerset monad as a trivial example. It is really not trivial to me.



      Can anyone help me with a detailed construction of such algebras. If you know of any article or book that explicitly construct such algebras which will be easy for a beginner to follow, please help out. Or if you can construct it as an answer here, I would really appreciate it.



      Thank you.










      share|cite|improve this question













      Many authors treat algrebras of the powerset monad as a trivial example. It is really not trivial to me.



      Can anyone help me with a detailed construction of such algebras. If you know of any article or book that explicitly construct such algebras which will be easy for a beginner to follow, please help out. Or if you can construct it as an answer here, I would really appreciate it.



      Thank you.







      category-theory monads






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      asked Nov 22 at 9:32









      Percy

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      718






















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          Algebras for the powerset monad are essentially sup lattices, that is complete ordered sets $S$ with a given operation $bigvee$ which takes a subset of $S$ to its least upper bound, and algebra maps are (weakly) monotone maps between them that preserve said operation.



          Let's call our monad $(T,mu,eta)$.



          An algebra for $T$ is a set $S$ with a map $h: TSto S$ : this will be interpreted as the lub map $bigvee$.



          Define $leq_h$ on $S$ as follows : for $x,yin S$, $h({x,y})in S$. Define $xleq_h y$ if and only if $h({x,y}) = y$. This is a well-defined relation, let's now prove that it's a partial order.



          It's clearly antisymmetric. To show that its reflexive, note that since $h:TSto S$ defines a $T$-algebra, we have $Sto^{eta_S} TSto^h S = Sto^{id_S} S$, and so $h({x,x}) = h({x}) = hcirceta_S(x) = x$, thus $xleq_h x$.



          We now have to prove that it's transitive : assume $xleq_h y, yleq_h z$. Note that ${x,y,z} = {x,y}cup{y,z}= mu ({{x,y}, {y,z}})$. Since $h$ makes $S$ into a $T$-algebra, we have that $hcirc mu_S = hcirc T(h)$. Thus $h({x,y,z}) = h({h({x,y}), h({y,z})})= h({y,z}) = z$.



          But also ${x,y,z}= {x}cup {y,z} = mu({{x}, {y,z}})$ and so we get $h({x,y,z}) = h({h({x}), h({y,z})}) = h({x,z})$. Thus $h({x,z}) = z$ and so $xleq_h z$.



          Therefore $leq_h$ defines an order on $S$. Let $Asubset S$ : we now want to prove that $h(A)$ is the least upper bound of $A$ for $leq_h$.



          Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_h$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map.



          Let $xin A$. Then $A=Acup{x}$ thus $h(A) = hcirc mu_S ({A,{x}}) = hcirc T(h) ({A, {x}}) = h({h(A), h({x})}) = h({h(A), x})$, therefore $xleq_h h(A)$ : $h(A)$ is an upper bound of $A$.



          Let $y$ be an upper bound of $A$. Then $h({y, h(A)}) = h({h({y}), h(A)}) = h({y}cup A)$. So it suffices to prove that if $z$ is the maximum of $B$, then $h(B) = z$.



          But $B = displaystylebigcup_{xin B}{x,z} = mu_S({{x,z}mid xin B})$, thus $h(B) = hcirc mu_S({{x,z}mid xin B})= hcirc T(h) ({{x,z}mid xin B})= h({h({x,z})mid xin B}) = h({zmid xin B})$ because $h({x,z}) = z$ for $xin B$ by assumption, and thus $h(B) = h({z}) = z$ : for $B=Acup{y}, z=y$, this gets us $h(A)leq_h y$ : $h(A)$ is the least upper bound of $A$, and we are done.



          Of course my choice of sup lattices instead of inf lattices is arbitrary, but of course this comes from the fact that the category of sup lattices is isomorphic to the category of inf lattices, so there's no problem here.






          share|cite|improve this answer





















          • Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
            – Percy
            Nov 26 at 9:29










          • Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
            – Max
            Nov 26 at 10:37










          • This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
            – Percy
            Nov 27 at 13:16












          • Your computations don't make sense to me. Is that a different $h$ ?
            – Max
            Nov 27 at 15:27











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          up vote
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          accepted










          Algebras for the powerset monad are essentially sup lattices, that is complete ordered sets $S$ with a given operation $bigvee$ which takes a subset of $S$ to its least upper bound, and algebra maps are (weakly) monotone maps between them that preserve said operation.



          Let's call our monad $(T,mu,eta)$.



          An algebra for $T$ is a set $S$ with a map $h: TSto S$ : this will be interpreted as the lub map $bigvee$.



          Define $leq_h$ on $S$ as follows : for $x,yin S$, $h({x,y})in S$. Define $xleq_h y$ if and only if $h({x,y}) = y$. This is a well-defined relation, let's now prove that it's a partial order.



          It's clearly antisymmetric. To show that its reflexive, note that since $h:TSto S$ defines a $T$-algebra, we have $Sto^{eta_S} TSto^h S = Sto^{id_S} S$, and so $h({x,x}) = h({x}) = hcirceta_S(x) = x$, thus $xleq_h x$.



          We now have to prove that it's transitive : assume $xleq_h y, yleq_h z$. Note that ${x,y,z} = {x,y}cup{y,z}= mu ({{x,y}, {y,z}})$. Since $h$ makes $S$ into a $T$-algebra, we have that $hcirc mu_S = hcirc T(h)$. Thus $h({x,y,z}) = h({h({x,y}), h({y,z})})= h({y,z}) = z$.



          But also ${x,y,z}= {x}cup {y,z} = mu({{x}, {y,z}})$ and so we get $h({x,y,z}) = h({h({x}), h({y,z})}) = h({x,z})$. Thus $h({x,z}) = z$ and so $xleq_h z$.



          Therefore $leq_h$ defines an order on $S$. Let $Asubset S$ : we now want to prove that $h(A)$ is the least upper bound of $A$ for $leq_h$.



          Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_h$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map.



          Let $xin A$. Then $A=Acup{x}$ thus $h(A) = hcirc mu_S ({A,{x}}) = hcirc T(h) ({A, {x}}) = h({h(A), h({x})}) = h({h(A), x})$, therefore $xleq_h h(A)$ : $h(A)$ is an upper bound of $A$.



          Let $y$ be an upper bound of $A$. Then $h({y, h(A)}) = h({h({y}), h(A)}) = h({y}cup A)$. So it suffices to prove that if $z$ is the maximum of $B$, then $h(B) = z$.



          But $B = displaystylebigcup_{xin B}{x,z} = mu_S({{x,z}mid xin B})$, thus $h(B) = hcirc mu_S({{x,z}mid xin B})= hcirc T(h) ({{x,z}mid xin B})= h({h({x,z})mid xin B}) = h({zmid xin B})$ because $h({x,z}) = z$ for $xin B$ by assumption, and thus $h(B) = h({z}) = z$ : for $B=Acup{y}, z=y$, this gets us $h(A)leq_h y$ : $h(A)$ is the least upper bound of $A$, and we are done.



          Of course my choice of sup lattices instead of inf lattices is arbitrary, but of course this comes from the fact that the category of sup lattices is isomorphic to the category of inf lattices, so there's no problem here.






          share|cite|improve this answer





















          • Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
            – Percy
            Nov 26 at 9:29










          • Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
            – Max
            Nov 26 at 10:37










          • This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
            – Percy
            Nov 27 at 13:16












          • Your computations don't make sense to me. Is that a different $h$ ?
            – Max
            Nov 27 at 15:27















          up vote
          8
          down vote



          accepted










          Algebras for the powerset monad are essentially sup lattices, that is complete ordered sets $S$ with a given operation $bigvee$ which takes a subset of $S$ to its least upper bound, and algebra maps are (weakly) monotone maps between them that preserve said operation.



          Let's call our monad $(T,mu,eta)$.



          An algebra for $T$ is a set $S$ with a map $h: TSto S$ : this will be interpreted as the lub map $bigvee$.



          Define $leq_h$ on $S$ as follows : for $x,yin S$, $h({x,y})in S$. Define $xleq_h y$ if and only if $h({x,y}) = y$. This is a well-defined relation, let's now prove that it's a partial order.



          It's clearly antisymmetric. To show that its reflexive, note that since $h:TSto S$ defines a $T$-algebra, we have $Sto^{eta_S} TSto^h S = Sto^{id_S} S$, and so $h({x,x}) = h({x}) = hcirceta_S(x) = x$, thus $xleq_h x$.



          We now have to prove that it's transitive : assume $xleq_h y, yleq_h z$. Note that ${x,y,z} = {x,y}cup{y,z}= mu ({{x,y}, {y,z}})$. Since $h$ makes $S$ into a $T$-algebra, we have that $hcirc mu_S = hcirc T(h)$. Thus $h({x,y,z}) = h({h({x,y}), h({y,z})})= h({y,z}) = z$.



          But also ${x,y,z}= {x}cup {y,z} = mu({{x}, {y,z}})$ and so we get $h({x,y,z}) = h({h({x}), h({y,z})}) = h({x,z})$. Thus $h({x,z}) = z$ and so $xleq_h z$.



          Therefore $leq_h$ defines an order on $S$. Let $Asubset S$ : we now want to prove that $h(A)$ is the least upper bound of $A$ for $leq_h$.



          Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_h$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map.



          Let $xin A$. Then $A=Acup{x}$ thus $h(A) = hcirc mu_S ({A,{x}}) = hcirc T(h) ({A, {x}}) = h({h(A), h({x})}) = h({h(A), x})$, therefore $xleq_h h(A)$ : $h(A)$ is an upper bound of $A$.



          Let $y$ be an upper bound of $A$. Then $h({y, h(A)}) = h({h({y}), h(A)}) = h({y}cup A)$. So it suffices to prove that if $z$ is the maximum of $B$, then $h(B) = z$.



          But $B = displaystylebigcup_{xin B}{x,z} = mu_S({{x,z}mid xin B})$, thus $h(B) = hcirc mu_S({{x,z}mid xin B})= hcirc T(h) ({{x,z}mid xin B})= h({h({x,z})mid xin B}) = h({zmid xin B})$ because $h({x,z}) = z$ for $xin B$ by assumption, and thus $h(B) = h({z}) = z$ : for $B=Acup{y}, z=y$, this gets us $h(A)leq_h y$ : $h(A)$ is the least upper bound of $A$, and we are done.



          Of course my choice of sup lattices instead of inf lattices is arbitrary, but of course this comes from the fact that the category of sup lattices is isomorphic to the category of inf lattices, so there's no problem here.






          share|cite|improve this answer





















          • Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
            – Percy
            Nov 26 at 9:29










          • Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
            – Max
            Nov 26 at 10:37










          • This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
            – Percy
            Nov 27 at 13:16












          • Your computations don't make sense to me. Is that a different $h$ ?
            – Max
            Nov 27 at 15:27













          up vote
          8
          down vote



          accepted







          up vote
          8
          down vote



          accepted






          Algebras for the powerset monad are essentially sup lattices, that is complete ordered sets $S$ with a given operation $bigvee$ which takes a subset of $S$ to its least upper bound, and algebra maps are (weakly) monotone maps between them that preserve said operation.



          Let's call our monad $(T,mu,eta)$.



          An algebra for $T$ is a set $S$ with a map $h: TSto S$ : this will be interpreted as the lub map $bigvee$.



          Define $leq_h$ on $S$ as follows : for $x,yin S$, $h({x,y})in S$. Define $xleq_h y$ if and only if $h({x,y}) = y$. This is a well-defined relation, let's now prove that it's a partial order.



          It's clearly antisymmetric. To show that its reflexive, note that since $h:TSto S$ defines a $T$-algebra, we have $Sto^{eta_S} TSto^h S = Sto^{id_S} S$, and so $h({x,x}) = h({x}) = hcirceta_S(x) = x$, thus $xleq_h x$.



          We now have to prove that it's transitive : assume $xleq_h y, yleq_h z$. Note that ${x,y,z} = {x,y}cup{y,z}= mu ({{x,y}, {y,z}})$. Since $h$ makes $S$ into a $T$-algebra, we have that $hcirc mu_S = hcirc T(h)$. Thus $h({x,y,z}) = h({h({x,y}), h({y,z})})= h({y,z}) = z$.



          But also ${x,y,z}= {x}cup {y,z} = mu({{x}, {y,z}})$ and so we get $h({x,y,z}) = h({h({x}), h({y,z})}) = h({x,z})$. Thus $h({x,z}) = z$ and so $xleq_h z$.



          Therefore $leq_h$ defines an order on $S$. Let $Asubset S$ : we now want to prove that $h(A)$ is the least upper bound of $A$ for $leq_h$.



          Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_h$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map.



          Let $xin A$. Then $A=Acup{x}$ thus $h(A) = hcirc mu_S ({A,{x}}) = hcirc T(h) ({A, {x}}) = h({h(A), h({x})}) = h({h(A), x})$, therefore $xleq_h h(A)$ : $h(A)$ is an upper bound of $A$.



          Let $y$ be an upper bound of $A$. Then $h({y, h(A)}) = h({h({y}), h(A)}) = h({y}cup A)$. So it suffices to prove that if $z$ is the maximum of $B$, then $h(B) = z$.



          But $B = displaystylebigcup_{xin B}{x,z} = mu_S({{x,z}mid xin B})$, thus $h(B) = hcirc mu_S({{x,z}mid xin B})= hcirc T(h) ({{x,z}mid xin B})= h({h({x,z})mid xin B}) = h({zmid xin B})$ because $h({x,z}) = z$ for $xin B$ by assumption, and thus $h(B) = h({z}) = z$ : for $B=Acup{y}, z=y$, this gets us $h(A)leq_h y$ : $h(A)$ is the least upper bound of $A$, and we are done.



          Of course my choice of sup lattices instead of inf lattices is arbitrary, but of course this comes from the fact that the category of sup lattices is isomorphic to the category of inf lattices, so there's no problem here.






          share|cite|improve this answer












          Algebras for the powerset monad are essentially sup lattices, that is complete ordered sets $S$ with a given operation $bigvee$ which takes a subset of $S$ to its least upper bound, and algebra maps are (weakly) monotone maps between them that preserve said operation.



          Let's call our monad $(T,mu,eta)$.



          An algebra for $T$ is a set $S$ with a map $h: TSto S$ : this will be interpreted as the lub map $bigvee$.



          Define $leq_h$ on $S$ as follows : for $x,yin S$, $h({x,y})in S$. Define $xleq_h y$ if and only if $h({x,y}) = y$. This is a well-defined relation, let's now prove that it's a partial order.



          It's clearly antisymmetric. To show that its reflexive, note that since $h:TSto S$ defines a $T$-algebra, we have $Sto^{eta_S} TSto^h S = Sto^{id_S} S$, and so $h({x,x}) = h({x}) = hcirceta_S(x) = x$, thus $xleq_h x$.



          We now have to prove that it's transitive : assume $xleq_h y, yleq_h z$. Note that ${x,y,z} = {x,y}cup{y,z}= mu ({{x,y}, {y,z}})$. Since $h$ makes $S$ into a $T$-algebra, we have that $hcirc mu_S = hcirc T(h)$. Thus $h({x,y,z}) = h({h({x,y}), h({y,z})})= h({y,z}) = z$.



          But also ${x,y,z}= {x}cup {y,z} = mu({{x}, {y,z}})$ and so we get $h({x,y,z}) = h({h({x}), h({y,z})}) = h({x,z})$. Thus $h({x,z}) = z$ and so $xleq_h z$.



          Therefore $leq_h$ defines an order on $S$. Let $Asubset S$ : we now want to prove that $h(A)$ is the least upper bound of $A$ for $leq_h$.



          Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_h$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map.



          Let $xin A$. Then $A=Acup{x}$ thus $h(A) = hcirc mu_S ({A,{x}}) = hcirc T(h) ({A, {x}}) = h({h(A), h({x})}) = h({h(A), x})$, therefore $xleq_h h(A)$ : $h(A)$ is an upper bound of $A$.



          Let $y$ be an upper bound of $A$. Then $h({y, h(A)}) = h({h({y}), h(A)}) = h({y}cup A)$. So it suffices to prove that if $z$ is the maximum of $B$, then $h(B) = z$.



          But $B = displaystylebigcup_{xin B}{x,z} = mu_S({{x,z}mid xin B})$, thus $h(B) = hcirc mu_S({{x,z}mid xin B})= hcirc T(h) ({{x,z}mid xin B})= h({h({x,z})mid xin B}) = h({zmid xin B})$ because $h({x,z}) = z$ for $xin B$ by assumption, and thus $h(B) = h({z}) = z$ : for $B=Acup{y}, z=y$, this gets us $h(A)leq_h y$ : $h(A)$ is the least upper bound of $A$, and we are done.



          Of course my choice of sup lattices instead of inf lattices is arbitrary, but of course this comes from the fact that the category of sup lattices is isomorphic to the category of inf lattices, so there's no problem here.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 10:54









          Max

          12.5k11040




          12.5k11040












          • Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
            – Percy
            Nov 26 at 9:29










          • Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
            – Max
            Nov 26 at 10:37










          • This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
            – Percy
            Nov 27 at 13:16












          • Your computations don't make sense to me. Is that a different $h$ ?
            – Max
            Nov 27 at 15:27


















          • Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
            – Percy
            Nov 26 at 9:29










          • Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
            – Max
            Nov 26 at 10:37










          • This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
            – Percy
            Nov 27 at 13:16












          • Your computations don't make sense to me. Is that a different $h$ ?
            – Max
            Nov 27 at 15:27
















          Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
          – Percy
          Nov 26 at 9:29




          Thank you @Max . It is very clear now. Does this paragraph cater for the converse? "Once we prove this, it will be clear that the desired result is true : indeed the definition of $leq_{h}$ proves that a $T$-algebra map preserves the order, and the least upper bounds, and it will be clear that a map that preserves least upper bounds will be a $T$-algebra map."
          – Percy
          Nov 26 at 9:29












          Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
          – Max
          Nov 26 at 10:37




          Yes (though obviously saying "it's clear" is not a proof, but I'm saying that it's easy enough that you should be able to do this exercise)
          – Max
          Nov 26 at 10:37












          This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
          – Percy
          Nov 27 at 13:16






          This is what I came out with as the converse: Let $h$ be a map preserving least upper bounds. Let $xin S$, then $h(eta_{S}(x))=h({x})=x$. Let ${A_{i}:iin I}in mathcal{P}^{2}(S)$, then $h(mu_{S}({A_{i}:iin I}))=h(bigcup_{iin I}A_{i})$. Let $x=h(bigcup_{iin I}A_{i})$. On the other hand, $h(mathcal{P}(h({A_{i}:iin I})))=h({h(A_{i}):iin I})$. How can I show that $h({h(A_{i}):iin I})=x$?
          – Percy
          Nov 27 at 13:16














          Your computations don't make sense to me. Is that a different $h$ ?
          – Max
          Nov 27 at 15:27




          Your computations don't make sense to me. Is that a different $h$ ?
          – Max
          Nov 27 at 15:27


















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