Troubles by understanding arity of terms in A Course in Universal Algebra.












3














In A Course in Universal Algebra in definition 10.1 terms are introduced.



What puzzles me is the statement about arity:




"A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."




Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?



If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?



This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.



I really do not understand this definition.



If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?



I found another question concerning the definitions, but that was not enough to take away my lack of understanding.










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    3














    In A Course in Universal Algebra in definition 10.1 terms are introduced.



    What puzzles me is the statement about arity:




    "A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."




    Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?



    If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?



    This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.



    I really do not understand this definition.



    If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?



    I found another question concerning the definitions, but that was not enough to take away my lack of understanding.










    share|cite|improve this question



























      3












      3








      3


      2





      In A Course in Universal Algebra in definition 10.1 terms are introduced.



      What puzzles me is the statement about arity:




      "A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."




      Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?



      If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?



      This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.



      I really do not understand this definition.



      If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?



      I found another question concerning the definitions, but that was not enough to take away my lack of understanding.










      share|cite|improve this question















      In A Course in Universal Algebra in definition 10.1 terms are introduced.



      What puzzles me is the statement about arity:




      "A term $p$ is $n$-ary if the number of variables appearing explicitly in $p$ is $leq n$."




      Does that mean that the arity of a term $p$ is at most limited but not defined, and can take several values?



      If e.g. $p$ is a variable then can every integer $geq1$ be put forward as arity of $p$?



      This seems to be confirmed in the definition 10.2 (also troubling me) where for term $p(x_1,dots,x_n):=x_iin X$ a mapping $p^{mathbf A}:A^nto A$ is prescribed by $(a_1,dots,a_n)mapsto a_i$.



      I really do not understand this definition.



      If I just start with term $x$ wich is a variable then according to def. 10.2 $x^{mathbf A}$ must be some function $A^nto A$. But if so, then what is $n$ (can it take several values?) and how is the function prescribed?



      I found another question concerning the definitions, but that was not enough to take away my lack of understanding.







      soft-question definition universal-algebra






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      share|cite|improve this question








      edited 5 hours ago









      Shaun

      8,344113578




      8,344113578










      asked 5 hours ago









      drhab

      97k544128




      97k544128






















          2 Answers
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          2














          Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:



          Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:




          1. every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,

          2. if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.


          Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:




          1. if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.

          2. if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.


          The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:



          Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.






          share|cite|improve this answer





















          • Thank you. I will take a good look at this (and also the answer of Noah).
            – drhab
            8 mins ago



















          2














          I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.





          Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.



          Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^{vert Vvert}rightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.



          In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).






          share|cite|improve this answer





















          • Thank you Noah. Light starts shining, but (as always) I need time to digest.
            – drhab
            16 mins ago











          Your Answer





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          Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:



          Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:




          1. every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,

          2. if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.


          Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:




          1. if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.

          2. if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.


          The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:



          Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.






          share|cite|improve this answer





















          • Thank you. I will take a good look at this (and also the answer of Noah).
            – drhab
            8 mins ago
















          2














          Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:



          Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:




          1. every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,

          2. if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.


          Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:




          1. if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.

          2. if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.


          The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:



          Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.






          share|cite|improve this answer





















          • Thank you. I will take a good look at this (and also the answer of Noah).
            – drhab
            8 mins ago














          2












          2








          2






          Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:



          Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:




          1. every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,

          2. if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.


          Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:




          1. if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.

          2. if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.


          The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:



          Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.






          share|cite|improve this answer












          Personally, I wouldn't even define arity for a term; I would only define it for a term operation. Let me try to be more precise:



          Let $X$ be a set of symbols, called variables, and let $mathcal{F}$ be a set of operation symbols such that each operation symbol $f$ has an arity, $mathrm{ar}(f)$, that is a non-negative integer. The set $mathrm{T}(X)$ of all terms in $X$ is defined recursively:




          1. every variable $x$ or operation symbol $c$ with $mathrm{ar}(c)=0$ is in $mathrm{T}(X)$,

          2. if $t_1,dots,t_k$ are terms and $f$ is an operation symbol with $mathrm{ar}(f)=n$, then $f(t_1,dots,t_n)$ is in $mathrm{T}(X)$.


          Now, let $X_k={x_1,dots,x_k}$ and let $mathbf{A}$ be an algebra of type $mathcal{F}$. To every term $t$ in $mathrm{T}(X_k)$ we can define a $k$-ary operation $t^A$ on $A$ as follows:




          1. if $t=x_i$ or $t=c$ with $mathrm{ar}(c)=0$, then $t^A:A^kto A$ is defined by $t^A(a_1,dots,a_k)=a_i$ or $t^A(a_1,dots,a_k)=c^A$, respectively.

          2. if $t=f(t_1,dots,t_n)$, then $t^A(a_1,dots,a_k)=f^A(t_1^A(a_1,dots,a_k),dots,t_n^A(a_1,dots,a_k))$.


          The important thing to notice is that terms are just formal strings of symbols whereas when we assign meaning to a term in a specific algebra that is how we arrive at term operation. Based on this, I feel no reason to define arity for a term and the arity of a term operation is just the arity of the operation. I will conclude with an example:



          Let $mathcal{F}={+,-,0}$. Then $t=(x_1+(-x_2))+x_3$ is in $mathrm{T}(X_3)$, but it is also in $mathrm{T}(X_k)$ for any $kgeq 3$. So, given an algebra $mathbf{A}$ of type $mathcal{F}$, we can define the term operation $t^A:A^3to A$ by $t(a_1,a_2,a_3)=(a_1+(-a_2))+a_3$ or we could define it $t^A:A^kto A$ for any $kgeq 3$ by $t(a_1,dots,a_k)=(a_1+(-a_2))+a_3$. This first term operation has arity equal to 3 and the second one has arity equal to $k$, but there is no reason to define arity for $t$.







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          answered 51 mins ago









          Eran

          1,198818




          1,198818












          • Thank you. I will take a good look at this (and also the answer of Noah).
            – drhab
            8 mins ago


















          • Thank you. I will take a good look at this (and also the answer of Noah).
            – drhab
            8 mins ago
















          Thank you. I will take a good look at this (and also the answer of Noah).
          – drhab
          8 mins ago




          Thank you. I will take a good look at this (and also the answer of Noah).
          – drhab
          8 mins ago











          2














          I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.





          Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.



          Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^{vert Vvert}rightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.



          In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).






          share|cite|improve this answer





















          • Thank you Noah. Light starts shining, but (as always) I need time to digest.
            – drhab
            16 mins ago
















          2














          I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.





          Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.



          Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^{vert Vvert}rightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.



          In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).






          share|cite|improve this answer





















          • Thank you Noah. Light starts shining, but (as always) I need time to digest.
            – drhab
            16 mins ago














          2












          2








          2






          I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.





          Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.



          Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^{vert Vvert}rightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.



          In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).






          share|cite|improve this answer












          I think you've got it (ugly as it seems). The point is that a term on its own isn't a "complete" object, and so it doesn't have a unique arity.





          Certainly there's no problem talking about the arity of a map from $A^n$ to $A$ (namely, just $n$). The point is that a single term can have multiple different interpretations as functions, and these yield different arities. You point this out yourself.



          Specifically, for any term $p$ and any set $V$ of variables such that every variable occurring in $p$ is in $V$, we can view $p$ as a map $p_V: A^{vert Vvert}rightarrow A$. Note that the translation $(p,V)mapsto p_V$ is completely straightforward; this is why one will (sadly) often conflate $p$ and $p_V$ when $V$ is clear from context.



          In light of this, it isn't too weird to allow a term to have multiple arities - the point being that in some sense a term on its own is incomplete (for spiritual similarity, one might argue that the graph of a function is incomplete since it doesn't tell you the codomain).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 47 mins ago









          Noah Schweber

          120k10146278




          120k10146278












          • Thank you Noah. Light starts shining, but (as always) I need time to digest.
            – drhab
            16 mins ago


















          • Thank you Noah. Light starts shining, but (as always) I need time to digest.
            – drhab
            16 mins ago
















          Thank you Noah. Light starts shining, but (as always) I need time to digest.
          – drhab
          16 mins ago




          Thank you Noah. Light starts shining, but (as always) I need time to digest.
          – drhab
          16 mins ago


















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