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About the supremum and the infimum


Infimum and supremum of the set $ (-1)^n + 1/m : n,m in mathbb N cup -1$compact set always contains its supremum and infimumchecking the answer for infimum and supremum of a setFind the supremum and infimumSupremum and infimum of set with absolute value$S_1 = [ |a -b| : a in A , b in B]$ What is the infimum and supremum of this set?Supremum and infimum of $x, y ge 1 : fracxy3x + 2y + 1$Supremum and infimum of a set






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


I am learning about the supremum and infimum, I have however read a couple of things, that I did not understand.



1) is $A$ a two element containing set $A:=a,b$ then we use the notation



$a lor b := Sup(A)$ and $a land b := Inf(A) $



Why does this notation represent Sup of A and inf A? what do logical and and or have to do here with the lowest and highest bounds?



2) Let $A$ be a none-empty subset of $mathcal P(X) $ then follows



$cup A = Sup (A) $



$cap A = Inf (A) $



I also fail to see the relation. I am not even sure what $cup A$ means. Is it the unification of all none empty subsets of the bigger set $X$? or is it the unification of the single elements of the subset $A$? in either case, how does that produce the Sup/Inf?



Thanks!










share|cite|improve this question









$endgroup$




















    4












    $begingroup$


    I am learning about the supremum and infimum, I have however read a couple of things, that I did not understand.



    1) is $A$ a two element containing set $A:=a,b$ then we use the notation



    $a lor b := Sup(A)$ and $a land b := Inf(A) $



    Why does this notation represent Sup of A and inf A? what do logical and and or have to do here with the lowest and highest bounds?



    2) Let $A$ be a none-empty subset of $mathcal P(X) $ then follows



    $cup A = Sup (A) $



    $cap A = Inf (A) $



    I also fail to see the relation. I am not even sure what $cup A$ means. Is it the unification of all none empty subsets of the bigger set $X$? or is it the unification of the single elements of the subset $A$? in either case, how does that produce the Sup/Inf?



    Thanks!










    share|cite|improve this question









    $endgroup$
















      4












      4








      4





      $begingroup$


      I am learning about the supremum and infimum, I have however read a couple of things, that I did not understand.



      1) is $A$ a two element containing set $A:=a,b$ then we use the notation



      $a lor b := Sup(A)$ and $a land b := Inf(A) $



      Why does this notation represent Sup of A and inf A? what do logical and and or have to do here with the lowest and highest bounds?



      2) Let $A$ be a none-empty subset of $mathcal P(X) $ then follows



      $cup A = Sup (A) $



      $cap A = Inf (A) $



      I also fail to see the relation. I am not even sure what $cup A$ means. Is it the unification of all none empty subsets of the bigger set $X$? or is it the unification of the single elements of the subset $A$? in either case, how does that produce the Sup/Inf?



      Thanks!










      share|cite|improve this question









      $endgroup$




      I am learning about the supremum and infimum, I have however read a couple of things, that I did not understand.



      1) is $A$ a two element containing set $A:=a,b$ then we use the notation



      $a lor b := Sup(A)$ and $a land b := Inf(A) $



      Why does this notation represent Sup of A and inf A? what do logical and and or have to do here with the lowest and highest bounds?



      2) Let $A$ be a none-empty subset of $mathcal P(X) $ then follows



      $cup A = Sup (A) $



      $cap A = Inf (A) $



      I also fail to see the relation. I am not even sure what $cup A$ means. Is it the unification of all none empty subsets of the bigger set $X$? or is it the unification of the single elements of the subset $A$? in either case, how does that produce the Sup/Inf?



      Thanks!







      elementary-set-theory supremum-and-infimum






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      share|cite|improve this question




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      asked 8 hours ago









      MadSpaceMemerMadSpaceMemer

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          $begingroup$

          The notation $alor b$ for supremum (or join) and $aland b$ for infimum (or meet) stems from Boolean algebra. If we consider the formulas of propositional logic, and order them by entailment (i.e. $phi$ is smaller than $psi$ when $phivdashpsi$), then $lor$ is exactly the notion of a supremum, and $land$ that of infimum.



          Instead of using this notation just for Boolean algebras, the same symbols are used to describe the supremum and infimum for any kind of lattice. Depending on where your use of a lattice stems from, $land$ and $lor$ might have not much to do with logical conjunction or disjunction.




          A power set algebra is a special kind of Boolean algebra, where we order the power set of some set by inclusion (i.e. if $A,Bsubset X$, then we say $A$ is less than $B$ if $Asubseteq B$). Translating supremum and infimum to sets, we see that they relate to union and intersection.



          $bigcup A$ means to take the union of all sets in $A$. For example, if we consider $mathcal P(1,2,3)$, and we let $A=1,1,2$, then $bigcup A=1cup1,2=1,2$. So the supremum (or join, or union) of the sets $1,1,2inmathcal P(1,2,3)$ is the set $1,2$.




          It is probably helpful to verify these claims by proving that the notions of conjunction / disjunction in propositional logic, or of union / intersection in a power set, satisfy the axioms for a supremum / infimum (i.e., that they form a least upper bound / greatest lower bound with respect to the ordering).






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            If A := 1 does it mean infimum = supremum = 1?
            $endgroup$
            – MadSpaceMemer
            7 hours ago






          • 1




            $begingroup$
            @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
            $endgroup$
            – Vsotvep
            7 hours ago













          Your Answer








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          1 Answer
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          active

          oldest

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          3













          $begingroup$

          The notation $alor b$ for supremum (or join) and $aland b$ for infimum (or meet) stems from Boolean algebra. If we consider the formulas of propositional logic, and order them by entailment (i.e. $phi$ is smaller than $psi$ when $phivdashpsi$), then $lor$ is exactly the notion of a supremum, and $land$ that of infimum.



          Instead of using this notation just for Boolean algebras, the same symbols are used to describe the supremum and infimum for any kind of lattice. Depending on where your use of a lattice stems from, $land$ and $lor$ might have not much to do with logical conjunction or disjunction.




          A power set algebra is a special kind of Boolean algebra, where we order the power set of some set by inclusion (i.e. if $A,Bsubset X$, then we say $A$ is less than $B$ if $Asubseteq B$). Translating supremum and infimum to sets, we see that they relate to union and intersection.



          $bigcup A$ means to take the union of all sets in $A$. For example, if we consider $mathcal P(1,2,3)$, and we let $A=1,1,2$, then $bigcup A=1cup1,2=1,2$. So the supremum (or join, or union) of the sets $1,1,2inmathcal P(1,2,3)$ is the set $1,2$.




          It is probably helpful to verify these claims by proving that the notions of conjunction / disjunction in propositional logic, or of union / intersection in a power set, satisfy the axioms for a supremum / infimum (i.e., that they form a least upper bound / greatest lower bound with respect to the ordering).






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            If A := 1 does it mean infimum = supremum = 1?
            $endgroup$
            – MadSpaceMemer
            7 hours ago






          • 1




            $begingroup$
            @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
            $endgroup$
            – Vsotvep
            7 hours ago















          3













          $begingroup$

          The notation $alor b$ for supremum (or join) and $aland b$ for infimum (or meet) stems from Boolean algebra. If we consider the formulas of propositional logic, and order them by entailment (i.e. $phi$ is smaller than $psi$ when $phivdashpsi$), then $lor$ is exactly the notion of a supremum, and $land$ that of infimum.



          Instead of using this notation just for Boolean algebras, the same symbols are used to describe the supremum and infimum for any kind of lattice. Depending on where your use of a lattice stems from, $land$ and $lor$ might have not much to do with logical conjunction or disjunction.




          A power set algebra is a special kind of Boolean algebra, where we order the power set of some set by inclusion (i.e. if $A,Bsubset X$, then we say $A$ is less than $B$ if $Asubseteq B$). Translating supremum and infimum to sets, we see that they relate to union and intersection.



          $bigcup A$ means to take the union of all sets in $A$. For example, if we consider $mathcal P(1,2,3)$, and we let $A=1,1,2$, then $bigcup A=1cup1,2=1,2$. So the supremum (or join, or union) of the sets $1,1,2inmathcal P(1,2,3)$ is the set $1,2$.




          It is probably helpful to verify these claims by proving that the notions of conjunction / disjunction in propositional logic, or of union / intersection in a power set, satisfy the axioms for a supremum / infimum (i.e., that they form a least upper bound / greatest lower bound with respect to the ordering).






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            If A := 1 does it mean infimum = supremum = 1?
            $endgroup$
            – MadSpaceMemer
            7 hours ago






          • 1




            $begingroup$
            @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
            $endgroup$
            – Vsotvep
            7 hours ago













          3














          3










          3







          $begingroup$

          The notation $alor b$ for supremum (or join) and $aland b$ for infimum (or meet) stems from Boolean algebra. If we consider the formulas of propositional logic, and order them by entailment (i.e. $phi$ is smaller than $psi$ when $phivdashpsi$), then $lor$ is exactly the notion of a supremum, and $land$ that of infimum.



          Instead of using this notation just for Boolean algebras, the same symbols are used to describe the supremum and infimum for any kind of lattice. Depending on where your use of a lattice stems from, $land$ and $lor$ might have not much to do with logical conjunction or disjunction.




          A power set algebra is a special kind of Boolean algebra, where we order the power set of some set by inclusion (i.e. if $A,Bsubset X$, then we say $A$ is less than $B$ if $Asubseteq B$). Translating supremum and infimum to sets, we see that they relate to union and intersection.



          $bigcup A$ means to take the union of all sets in $A$. For example, if we consider $mathcal P(1,2,3)$, and we let $A=1,1,2$, then $bigcup A=1cup1,2=1,2$. So the supremum (or join, or union) of the sets $1,1,2inmathcal P(1,2,3)$ is the set $1,2$.




          It is probably helpful to verify these claims by proving that the notions of conjunction / disjunction in propositional logic, or of union / intersection in a power set, satisfy the axioms for a supremum / infimum (i.e., that they form a least upper bound / greatest lower bound with respect to the ordering).






          share|cite|improve this answer











          $endgroup$



          The notation $alor b$ for supremum (or join) and $aland b$ for infimum (or meet) stems from Boolean algebra. If we consider the formulas of propositional logic, and order them by entailment (i.e. $phi$ is smaller than $psi$ when $phivdashpsi$), then $lor$ is exactly the notion of a supremum, and $land$ that of infimum.



          Instead of using this notation just for Boolean algebras, the same symbols are used to describe the supremum and infimum for any kind of lattice. Depending on where your use of a lattice stems from, $land$ and $lor$ might have not much to do with logical conjunction or disjunction.




          A power set algebra is a special kind of Boolean algebra, where we order the power set of some set by inclusion (i.e. if $A,Bsubset X$, then we say $A$ is less than $B$ if $Asubseteq B$). Translating supremum and infimum to sets, we see that they relate to union and intersection.



          $bigcup A$ means to take the union of all sets in $A$. For example, if we consider $mathcal P(1,2,3)$, and we let $A=1,1,2$, then $bigcup A=1cup1,2=1,2$. So the supremum (or join, or union) of the sets $1,1,2inmathcal P(1,2,3)$ is the set $1,2$.




          It is probably helpful to verify these claims by proving that the notions of conjunction / disjunction in propositional logic, or of union / intersection in a power set, satisfy the axioms for a supremum / infimum (i.e., that they form a least upper bound / greatest lower bound with respect to the ordering).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          VsotvepVsotvep

          2,0135 silver badges13 bronze badges




          2,0135 silver badges13 bronze badges














          • $begingroup$
            If A := 1 does it mean infimum = supremum = 1?
            $endgroup$
            – MadSpaceMemer
            7 hours ago






          • 1




            $begingroup$
            @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
            $endgroup$
            – Vsotvep
            7 hours ago
















          • $begingroup$
            If A := 1 does it mean infimum = supremum = 1?
            $endgroup$
            – MadSpaceMemer
            7 hours ago






          • 1




            $begingroup$
            @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
            $endgroup$
            – Vsotvep
            7 hours ago















          $begingroup$
          If A := 1 does it mean infimum = supremum = 1?
          $endgroup$
          – MadSpaceMemer
          7 hours ago




          $begingroup$
          If A := 1 does it mean infimum = supremum = 1?
          $endgroup$
          – MadSpaceMemer
          7 hours ago




          1




          1




          $begingroup$
          @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
          $endgroup$
          – Vsotvep
          7 hours ago




          $begingroup$
          @MadSpaceMemer Yes, it does. Similar to how the supremum and infimum of the set of numbers $1$ is $1$. In fact, the supremum and infimum of any single-element set are equal.
          $endgroup$
          – Vsotvep
          7 hours ago

















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