Natural numbers as a subset of integer numbers: $mathbb{N}subsetmathbb{Z}$.












1












$begingroup$


Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
.



Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
as sets, preserving the sum and product operations?






number-theory elementary-set-theory integers natural-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
    .



    Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
    as sets, preserving the sum and product operations?






    number-theory elementary-set-theory integers natural-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
      .



      Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
      as sets, preserving the sum and product operations?






      number-theory elementary-set-theory integers natural-numbers







      share|cite|improve this question











      $endgroup$




      Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
      .



      Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
      as sets, preserving the sum and product operations?






      number-theory elementary-set-theory integers natural-numbers




      number-theory elementary-set-theory integers natural-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 4:20









      Andrés E. Caicedo

      65.2k8158247




      65.2k8158247










      asked Dec 4 '18 at 19:48









      Dr PotatoDr Potato

      394




      394






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Yes (and the same argument goes for the other number sets).



          Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.



          Now define
          $$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
          where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.



          Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
          $$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$



          [Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
            $endgroup$
            – dbx
            Dec 4 '18 at 20:04










          • $begingroup$
            @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
            $endgroup$
            – Clive Newstead
            Dec 4 '18 at 21:00












          • $begingroup$
            Although it answers the question, I am looking for something more elegant.
            $endgroup$
            – Dr Potato
            Dec 4 '18 at 22:22










          • $begingroup$
            @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 2:37








          • 1




            $begingroup$
            @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 16:22













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          1 Answer
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          1 Answer
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          oldest

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          active

          oldest

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          0












          $begingroup$

          Yes (and the same argument goes for the other number sets).



          Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.



          Now define
          $$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
          where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.



          Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
          $$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$



          [Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
            $endgroup$
            – dbx
            Dec 4 '18 at 20:04










          • $begingroup$
            @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
            $endgroup$
            – Clive Newstead
            Dec 4 '18 at 21:00












          • $begingroup$
            Although it answers the question, I am looking for something more elegant.
            $endgroup$
            – Dr Potato
            Dec 4 '18 at 22:22










          • $begingroup$
            @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 2:37








          • 1




            $begingroup$
            @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 16:22


















          0












          $begingroup$

          Yes (and the same argument goes for the other number sets).



          Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.



          Now define
          $$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
          where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.



          Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
          $$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$



          [Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
            $endgroup$
            – dbx
            Dec 4 '18 at 20:04










          • $begingroup$
            @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
            $endgroup$
            – Clive Newstead
            Dec 4 '18 at 21:00












          • $begingroup$
            Although it answers the question, I am looking for something more elegant.
            $endgroup$
            – Dr Potato
            Dec 4 '18 at 22:22










          • $begingroup$
            @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 2:37








          • 1




            $begingroup$
            @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 16:22
















          0












          0








          0





          $begingroup$

          Yes (and the same argument goes for the other number sets).



          Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.



          Now define
          $$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
          where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.



          Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
          $$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$



          [Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]






          share|cite|improve this answer











          $endgroup$



          Yes (and the same argument goes for the other number sets).



          Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.



          Now define
          $$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
          where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.



          Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
          $$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$



          [Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 19:59

























          answered Dec 4 '18 at 19:54









          Clive NewsteadClive Newstead

          51.3k474135




          51.3k474135












          • $begingroup$
            I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
            $endgroup$
            – dbx
            Dec 4 '18 at 20:04










          • $begingroup$
            @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
            $endgroup$
            – Clive Newstead
            Dec 4 '18 at 21:00












          • $begingroup$
            Although it answers the question, I am looking for something more elegant.
            $endgroup$
            – Dr Potato
            Dec 4 '18 at 22:22










          • $begingroup$
            @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 2:37








          • 1




            $begingroup$
            @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 16:22




















          • $begingroup$
            I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
            $endgroup$
            – dbx
            Dec 4 '18 at 20:04










          • $begingroup$
            @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
            $endgroup$
            – Clive Newstead
            Dec 4 '18 at 21:00












          • $begingroup$
            Although it answers the question, I am looking for something more elegant.
            $endgroup$
            – Dr Potato
            Dec 4 '18 at 22:22










          • $begingroup$
            @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 2:37








          • 1




            $begingroup$
            @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
            $endgroup$
            – Clive Newstead
            Dec 5 '18 at 16:22


















          $begingroup$
          I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
          $endgroup$
          – dbx
          Dec 4 '18 at 20:04




          $begingroup$
          I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
          $endgroup$
          – dbx
          Dec 4 '18 at 20:04












          $begingroup$
          @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
          $endgroup$
          – Clive Newstead
          Dec 4 '18 at 21:00






          $begingroup$
          @dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
          $endgroup$
          – Clive Newstead
          Dec 4 '18 at 21:00














          $begingroup$
          Although it answers the question, I am looking for something more elegant.
          $endgroup$
          – Dr Potato
          Dec 4 '18 at 22:22




          $begingroup$
          Although it answers the question, I am looking for something more elegant.
          $endgroup$
          – Dr Potato
          Dec 4 '18 at 22:22












          $begingroup$
          @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
          $endgroup$
          – Clive Newstead
          Dec 5 '18 at 2:37






          $begingroup$
          @DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
          $endgroup$
          – Clive Newstead
          Dec 5 '18 at 2:37






          1




          1




          $begingroup$
          @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
          $endgroup$
          – Clive Newstead
          Dec 5 '18 at 16:22






          $begingroup$
          @dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
          $endgroup$
          – Clive Newstead
          Dec 5 '18 at 16:22




















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