Is a set together with an operation always a relational structure?












0












$begingroup$


I'm reading Algebraic Methods in Philosophical Logic, an introductory book on Universal Algebra by by J. Michael Dunn, Gary Hardegree.

This book start its presentation by introducing the notions of "relational structure" and "operational structure" (An algebra).

An operation on a set is considered a particular case of a relation on that set,and so, an "operational structure" is a particular instance of a "relational structure".

However, an "operational structure" must have its carrier set closed under the operations of the structure. This might have confused me.

In my understanding an operation on a set is a kind of relation which combines,modifies, operates on some elements of the set.

This "definition" has nothing to do with the notion of closure, which we can consider an extra proprety of an operation on a set.

So given that the notion of "operational structure" requires a "closed operation". What structure is a set with an opertion, but not closed under that operation?

Can we say that is just a relational structure? (if we look at it through a "relational lens", in the sense that we consider that operation a relation)

Is there a similiar notion of "closure" for relations?

I'll try to illustrate my doubts with an example.
Let's take the set A={0,1,2,3} together with the operation of standatd integer addition. We can easily see that this is not an algebra, since A is not closed under integer addition.

Let's now consider our operation as a relation.

We still have our set A={0,1,2,3} but this time we a define a 3-place relation on A such that a1, a2, and, a3, are related if a3=a1+a2.

Is this considered a relational structure?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I'm reading Algebraic Methods in Philosophical Logic, an introductory book on Universal Algebra by by J. Michael Dunn, Gary Hardegree.

    This book start its presentation by introducing the notions of "relational structure" and "operational structure" (An algebra).

    An operation on a set is considered a particular case of a relation on that set,and so, an "operational structure" is a particular instance of a "relational structure".

    However, an "operational structure" must have its carrier set closed under the operations of the structure. This might have confused me.

    In my understanding an operation on a set is a kind of relation which combines,modifies, operates on some elements of the set.

    This "definition" has nothing to do with the notion of closure, which we can consider an extra proprety of an operation on a set.

    So given that the notion of "operational structure" requires a "closed operation". What structure is a set with an opertion, but not closed under that operation?

    Can we say that is just a relational structure? (if we look at it through a "relational lens", in the sense that we consider that operation a relation)

    Is there a similiar notion of "closure" for relations?

    I'll try to illustrate my doubts with an example.
    Let's take the set A={0,1,2,3} together with the operation of standatd integer addition. We can easily see that this is not an algebra, since A is not closed under integer addition.

    Let's now consider our operation as a relation.

    We still have our set A={0,1,2,3} but this time we a define a 3-place relation on A such that a1, a2, and, a3, are related if a3=a1+a2.

    Is this considered a relational structure?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm reading Algebraic Methods in Philosophical Logic, an introductory book on Universal Algebra by by J. Michael Dunn, Gary Hardegree.

      This book start its presentation by introducing the notions of "relational structure" and "operational structure" (An algebra).

      An operation on a set is considered a particular case of a relation on that set,and so, an "operational structure" is a particular instance of a "relational structure".

      However, an "operational structure" must have its carrier set closed under the operations of the structure. This might have confused me.

      In my understanding an operation on a set is a kind of relation which combines,modifies, operates on some elements of the set.

      This "definition" has nothing to do with the notion of closure, which we can consider an extra proprety of an operation on a set.

      So given that the notion of "operational structure" requires a "closed operation". What structure is a set with an opertion, but not closed under that operation?

      Can we say that is just a relational structure? (if we look at it through a "relational lens", in the sense that we consider that operation a relation)

      Is there a similiar notion of "closure" for relations?

      I'll try to illustrate my doubts with an example.
      Let's take the set A={0,1,2,3} together with the operation of standatd integer addition. We can easily see that this is not an algebra, since A is not closed under integer addition.

      Let's now consider our operation as a relation.

      We still have our set A={0,1,2,3} but this time we a define a 3-place relation on A such that a1, a2, and, a3, are related if a3=a1+a2.

      Is this considered a relational structure?










      share|cite|improve this question









      $endgroup$




      I'm reading Algebraic Methods in Philosophical Logic, an introductory book on Universal Algebra by by J. Michael Dunn, Gary Hardegree.

      This book start its presentation by introducing the notions of "relational structure" and "operational structure" (An algebra).

      An operation on a set is considered a particular case of a relation on that set,and so, an "operational structure" is a particular instance of a "relational structure".

      However, an "operational structure" must have its carrier set closed under the operations of the structure. This might have confused me.

      In my understanding an operation on a set is a kind of relation which combines,modifies, operates on some elements of the set.

      This "definition" has nothing to do with the notion of closure, which we can consider an extra proprety of an operation on a set.

      So given that the notion of "operational structure" requires a "closed operation". What structure is a set with an opertion, but not closed under that operation?

      Can we say that is just a relational structure? (if we look at it through a "relational lens", in the sense that we consider that operation a relation)

      Is there a similiar notion of "closure" for relations?

      I'll try to illustrate my doubts with an example.
      Let's take the set A={0,1,2,3} together with the operation of standatd integer addition. We can easily see that this is not an algebra, since A is not closed under integer addition.

      Let's now consider our operation as a relation.

      We still have our set A={0,1,2,3} but this time we a define a 3-place relation on A such that a1, a2, and, a3, are related if a3=a1+a2.

      Is this considered a relational structure?







      relations universal-algebra






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      asked Dec 31 '18 at 14:39









      Gabriele ScarlattiGabriele Scarlatti

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          $begingroup$

          I don't have a hardcopy, but my guess would be:




          • A relational structure is a set $A$ endowed with an n-ary relation subset of $A^n$ (short for $A times ldots times A$).

          • An operational structure additionally enforces the relation viewed as a binary relation subset of $A^{n-1} times A$ to be left-total and right-unique (i.e. representing a function $A^{n-1} to A$).


          With these definitions it is impossible to have a non-closed operation. In fact your first example $({0, 1, 2, 3}, +_{mathbb{N}})$ does not match either definition since the second component is not a relation on that set.



          However, usually in writings on math, you will see something like:




          (A, +), with $A$ being a set and $+$ being a binary function, is an algebra. [...] We consider the subalgebra induced by $B subseteq A$. Indeed that is a subalgebra since its operation $+$ is closed on $B$.




          Formally, this means that $+$ viewed as a ternary relation subset of $(A times A) times A$ restricted on $B times B$ is a relation subset of $(B times B) times B$. Equivalently in logical symbols:




          • $+ subseteq (A times A) times A$

          • $forall b_1, b_2 in B. notexists anotin B. (b_1, b_2, a) in +$


          Indeed note formally we can only speak of the subalgebra once we proved this closedness property. But then again the subalgebra's operation is trivially closed, so formally it doesn't make sense to speak of a closed or non-closed operational algebra with the definitions I guessed above.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
            $endgroup$
            – amrsa
            Dec 31 '18 at 17:59






          • 1




            $begingroup$
            @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
            $endgroup$
            – ComFreek
            Jan 1 at 7:14













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          $begingroup$

          I don't have a hardcopy, but my guess would be:




          • A relational structure is a set $A$ endowed with an n-ary relation subset of $A^n$ (short for $A times ldots times A$).

          • An operational structure additionally enforces the relation viewed as a binary relation subset of $A^{n-1} times A$ to be left-total and right-unique (i.e. representing a function $A^{n-1} to A$).


          With these definitions it is impossible to have a non-closed operation. In fact your first example $({0, 1, 2, 3}, +_{mathbb{N}})$ does not match either definition since the second component is not a relation on that set.



          However, usually in writings on math, you will see something like:




          (A, +), with $A$ being a set and $+$ being a binary function, is an algebra. [...] We consider the subalgebra induced by $B subseteq A$. Indeed that is a subalgebra since its operation $+$ is closed on $B$.




          Formally, this means that $+$ viewed as a ternary relation subset of $(A times A) times A$ restricted on $B times B$ is a relation subset of $(B times B) times B$. Equivalently in logical symbols:




          • $+ subseteq (A times A) times A$

          • $forall b_1, b_2 in B. notexists anotin B. (b_1, b_2, a) in +$


          Indeed note formally we can only speak of the subalgebra once we proved this closedness property. But then again the subalgebra's operation is trivially closed, so formally it doesn't make sense to speak of a closed or non-closed operational algebra with the definitions I guessed above.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
            $endgroup$
            – amrsa
            Dec 31 '18 at 17:59






          • 1




            $begingroup$
            @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
            $endgroup$
            – ComFreek
            Jan 1 at 7:14


















          0












          $begingroup$

          I don't have a hardcopy, but my guess would be:




          • A relational structure is a set $A$ endowed with an n-ary relation subset of $A^n$ (short for $A times ldots times A$).

          • An operational structure additionally enforces the relation viewed as a binary relation subset of $A^{n-1} times A$ to be left-total and right-unique (i.e. representing a function $A^{n-1} to A$).


          With these definitions it is impossible to have a non-closed operation. In fact your first example $({0, 1, 2, 3}, +_{mathbb{N}})$ does not match either definition since the second component is not a relation on that set.



          However, usually in writings on math, you will see something like:




          (A, +), with $A$ being a set and $+$ being a binary function, is an algebra. [...] We consider the subalgebra induced by $B subseteq A$. Indeed that is a subalgebra since its operation $+$ is closed on $B$.




          Formally, this means that $+$ viewed as a ternary relation subset of $(A times A) times A$ restricted on $B times B$ is a relation subset of $(B times B) times B$. Equivalently in logical symbols:




          • $+ subseteq (A times A) times A$

          • $forall b_1, b_2 in B. notexists anotin B. (b_1, b_2, a) in +$


          Indeed note formally we can only speak of the subalgebra once we proved this closedness property. But then again the subalgebra's operation is trivially closed, so formally it doesn't make sense to speak of a closed or non-closed operational algebra with the definitions I guessed above.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
            $endgroup$
            – amrsa
            Dec 31 '18 at 17:59






          • 1




            $begingroup$
            @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
            $endgroup$
            – ComFreek
            Jan 1 at 7:14
















          0












          0








          0





          $begingroup$

          I don't have a hardcopy, but my guess would be:




          • A relational structure is a set $A$ endowed with an n-ary relation subset of $A^n$ (short for $A times ldots times A$).

          • An operational structure additionally enforces the relation viewed as a binary relation subset of $A^{n-1} times A$ to be left-total and right-unique (i.e. representing a function $A^{n-1} to A$).


          With these definitions it is impossible to have a non-closed operation. In fact your first example $({0, 1, 2, 3}, +_{mathbb{N}})$ does not match either definition since the second component is not a relation on that set.



          However, usually in writings on math, you will see something like:




          (A, +), with $A$ being a set and $+$ being a binary function, is an algebra. [...] We consider the subalgebra induced by $B subseteq A$. Indeed that is a subalgebra since its operation $+$ is closed on $B$.




          Formally, this means that $+$ viewed as a ternary relation subset of $(A times A) times A$ restricted on $B times B$ is a relation subset of $(B times B) times B$. Equivalently in logical symbols:




          • $+ subseteq (A times A) times A$

          • $forall b_1, b_2 in B. notexists anotin B. (b_1, b_2, a) in +$


          Indeed note formally we can only speak of the subalgebra once we proved this closedness property. But then again the subalgebra's operation is trivially closed, so formally it doesn't make sense to speak of a closed or non-closed operational algebra with the definitions I guessed above.






          share|cite|improve this answer









          $endgroup$



          I don't have a hardcopy, but my guess would be:




          • A relational structure is a set $A$ endowed with an n-ary relation subset of $A^n$ (short for $A times ldots times A$).

          • An operational structure additionally enforces the relation viewed as a binary relation subset of $A^{n-1} times A$ to be left-total and right-unique (i.e. representing a function $A^{n-1} to A$).


          With these definitions it is impossible to have a non-closed operation. In fact your first example $({0, 1, 2, 3}, +_{mathbb{N}})$ does not match either definition since the second component is not a relation on that set.



          However, usually in writings on math, you will see something like:




          (A, +), with $A$ being a set and $+$ being a binary function, is an algebra. [...] We consider the subalgebra induced by $B subseteq A$. Indeed that is a subalgebra since its operation $+$ is closed on $B$.




          Formally, this means that $+$ viewed as a ternary relation subset of $(A times A) times A$ restricted on $B times B$ is a relation subset of $(B times B) times B$. Equivalently in logical symbols:




          • $+ subseteq (A times A) times A$

          • $forall b_1, b_2 in B. notexists anotin B. (b_1, b_2, a) in +$


          Indeed note formally we can only speak of the subalgebra once we proved this closedness property. But then again the subalgebra's operation is trivially closed, so formally it doesn't make sense to speak of a closed or non-closed operational algebra with the definitions I guessed above.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 15:04









          ComFreekComFreek

          5521411




          5521411












          • $begingroup$
            I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
            $endgroup$
            – amrsa
            Dec 31 '18 at 17:59






          • 1




            $begingroup$
            @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
            $endgroup$
            – ComFreek
            Jan 1 at 7:14




















          • $begingroup$
            I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
            $endgroup$
            – amrsa
            Dec 31 '18 at 17:59






          • 1




            $begingroup$
            @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
            $endgroup$
            – ComFreek
            Jan 1 at 7:14


















          $begingroup$
          I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
          $endgroup$
          – amrsa
          Dec 31 '18 at 17:59




          $begingroup$
          I think what the OP is asking for is that if there is a name for the situation in which $f(x,y)$ exists for some pairs $(x,y) in A^2$ but not for all of them (here $f$ stands in prefix notation for a generic candidate for an operation). As you put it, this is an algebra if $f(x,y)$ exists for all $(x,y)$. But in some situations, the concept of partial algebra is also considered, and that's the one that applies to the example of the set ${0,1,2,3}$ with the natural addition. Here, $+$ is a partial operation.
          $endgroup$
          – amrsa
          Dec 31 '18 at 17:59




          1




          1




          $begingroup$
          @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
          $endgroup$
          – ComFreek
          Jan 1 at 7:14






          $begingroup$
          @amrsa Thanks, I didn't consider making $f$ just partial on those input arguments on which it "breaks out of the set $A$". Apparently, closed $Leftrightarrow$ total operation and non-closed $Leftrightarrow$ partial operation then.
          $endgroup$
          – ComFreek
          Jan 1 at 7:14




















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