Every ordered field has a countably infinite subset.












2














Is the following argument correct?



Theorem. Every ordered field has countably infinite subset.



Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
$$A = {1,1+1,1+1+1,dots}subseteq F$$
To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
$$cdots>1+1+1+1>1+1+1>1+1>1$$
appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.



$blacksquare$



Note:



Definition $textbf{1.1.5}$ is just the standard definition of a field.



Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that





  • $(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$


  • $(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$


Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that





  • $(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.


  • $(ii)$ If $x<y$ and $y<z$ then $x<z$.










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    2














    Is the following argument correct?



    Theorem. Every ordered field has countably infinite subset.



    Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
    $$A = {1,1+1,1+1+1,dots}subseteq F$$
    To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
    $$cdots>1+1+1+1>1+1+1>1+1>1$$
    appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.



    $blacksquare$



    Note:



    Definition $textbf{1.1.5}$ is just the standard definition of a field.



    Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that





    • $(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$


    • $(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$


    Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that





    • $(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.


    • $(ii)$ If $x<y$ and $y<z$ then $x<z$.










    share|cite|improve this question

























      2












      2








      2







      Is the following argument correct?



      Theorem. Every ordered field has countably infinite subset.



      Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
      $$A = {1,1+1,1+1+1,dots}subseteq F$$
      To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
      $$cdots>1+1+1+1>1+1+1>1+1>1$$
      appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.



      $blacksquare$



      Note:



      Definition $textbf{1.1.5}$ is just the standard definition of a field.



      Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that





      • $(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$


      • $(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$


      Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that





      • $(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.


      • $(ii)$ If $x<y$ and $y<z$ then $x<z$.










      share|cite|improve this question













      Is the following argument correct?



      Theorem. Every ordered field has countably infinite subset.



      Proof. Let $F$ be an arbitrary ordered field. From definition $textbf{1.1.5}$, we know that there exists a $1in F$ such that $1ne 0$, from the very same definition we also know that $x+yin F$, whenever $x,yin F$, implying
      $$A = {1,1+1,1+1+1,dots}subseteq F$$
      To prove the claim in question then it is sufficient to demonstrate that $A$ consists of distinct elements. Proposition $textbf{1.1.8}$ shows that $x^2>0$ whenever $xneq 0$, thus in particular $1>0$, definition $textbf{1.1.7}$ clause $(i)$ together with definition $textbf{1.1.1}$ clause $textit{(ii)}$ implies that
      $$cdots>1+1+1+1>1+1+1>1+1>1$$
      appealing then to definition $textbf{1.1.1}$ clause $textit{(i)}$ then implies that $sum_{i=1}^{j}1neqsum_{i=1}^{k} 1$ whenever $j>k$ consequently all elements of the set $A$ are distinct as required.



      $blacksquare$



      Note:



      Definition $textbf{1.1.5}$ is just the standard definition of a field.



      Definition $textbf{1.1.7}$: An field $F$ is said to be an ordered field if it is also an ordered set such that





      • $(i)$ For any $x,y,zin F$ $x<y$ implies $x+z<y+z$


      • $(ii)$ For any $x,yin F$, $x>0$ and $y>0$ implies $xy>0$


      Definition $textbf{1.1.1}$: An ordered set is a set $S$, together with a relation $<$ such that





      • $(i)$ For any $x,yin S$, exactly one of $x<y$,$x=y$ or $y<x$ holds.


      • $(ii)$ If $x<y$ and $y<z$ then $x<z$.







      real-analysis proof-verification






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      asked Nov 24 at 16:01









      Atif Farooq

      3,1422825




      3,1422825






















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          The argument is pretty solid, nice work!



          The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.






          share|cite|improve this answer





















          • Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
            – Atif Farooq
            Nov 24 at 16:32











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          1 Answer
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          active

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          active

          oldest

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          1














          The argument is pretty solid, nice work!



          The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.






          share|cite|improve this answer





















          • Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
            – Atif Farooq
            Nov 24 at 16:32
















          1














          The argument is pretty solid, nice work!



          The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.






          share|cite|improve this answer





















          • Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
            – Atif Farooq
            Nov 24 at 16:32














          1












          1








          1






          The argument is pretty solid, nice work!



          The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.






          share|cite|improve this answer












          The argument is pretty solid, nice work!



          The only thing I'd suggest to make the proof more complete would be to explicitly state your bijection between $A$ and the naturals. I know this is a simple one in this case, but there are many instances where the bijection may be harder to see, so it's good to get into a habit of writing it out.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 at 16:25









          Isaac Browne

          4,60731132




          4,60731132












          • Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
            – Atif Farooq
            Nov 24 at 16:32


















          • Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
            – Atif Farooq
            Nov 24 at 16:32
















          Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
          – Atif Farooq
          Nov 24 at 16:32




          Thank you very much for your reponse. I appreciate your comment concerning the explicit mention of the bijection aswell.
          – Atif Farooq
          Nov 24 at 16:32


















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