Let $mathfrak{A}$ be a ordered field and $mathfrak{X}$ be the smallest subfield of $mathfrak{A}$. Is $X$...












1












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Let $mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be an ordered field and $mathfrak{X}=langle X,<,+,cdot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$.



It is well-known that $mathfrak{X}$ is uniquely isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$ where $Bbb Q$ is the set of rationals.




I would like to ask if $X$ is dense in $A$.




Thank you for your help!






ordered-fields







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    1












    $begingroup$


    Let $mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be an ordered field and $mathfrak{X}=langle X,<,+,cdot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$.



    It is well-known that $mathfrak{X}$ is uniquely isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$ where $Bbb Q$ is the set of rationals.




    I would like to ask if $X$ is dense in $A$.




    Thank you for your help!






    ordered-fields







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be an ordered field and $mathfrak{X}=langle X,<,+,cdot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$.



      It is well-known that $mathfrak{X}$ is uniquely isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$ where $Bbb Q$ is the set of rationals.




      I would like to ask if $X$ is dense in $A$.




      Thank you for your help!






      ordered-fields







      share|cite|improve this question









      $endgroup$




      Let $mathfrak{A}=langle A,<,+,cdot,0',1' rangle$ be an ordered field and $mathfrak{X}=langle X,<,+,cdot,0',1' rangle$ be the smallest subfield of $mathfrak{A}$.



      It is well-known that $mathfrak{X}$ is uniquely isomorphic to $langle Bbb Q,<,+,cdot,0,1 rangle$ where $Bbb Q$ is the set of rationals.




      I would like to ask if $X$ is dense in $A$.




      Thank you for your help!






      ordered-fields




      ordered-fields






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      asked Jan 9 at 4:20









      Le Anh DungLe Anh Dung

      1,0921521




      1,0921521






















          2 Answers
          2






          active

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          3












          $begingroup$

          Well, no. In brief, we can make an ordered field out of rational functions with real coefficients. A rational function is called positive if its value is positive for all sufficiently large positive $x$



          I'm looking up some details, give me a minute.



          Added: yes, I got it right, Proposition 18.1, pages 158-159 in Hartshorne, Geometry: Euclid and Beyond






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I gave you not only 1 minute but also +1 vote ^^
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:28












          • $begingroup$
            Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:33






          • 2




            $begingroup$
            @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
            $endgroup$
            – Will Jagy
            Jan 9 at 4:39



















          1












          $begingroup$

          No.




          1. There are ordered fields of arbitrary high cardinality.


          2. Every field has a countable subfield. (Therefore, its smallest subfield is countable.)


          3. If a totally ordered set has a countable dense subset, then its cardinality is at most $2^{aleph_0}$. (Call the ordered set $A$ and the countable dense subset $Q$; then the map $amapsto{qin Q:qlt a}$ is an injection from $A$ to the power set of $Q$.)



          Hence any ordered field of cardinality greater than $2^{aleph_0}$ is a counterexample.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your great insight!
            $endgroup$
            – Le Anh Dung
            Jan 9 at 5:10











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Well, no. In brief, we can make an ordered field out of rational functions with real coefficients. A rational function is called positive if its value is positive for all sufficiently large positive $x$



          I'm looking up some details, give me a minute.



          Added: yes, I got it right, Proposition 18.1, pages 158-159 in Hartshorne, Geometry: Euclid and Beyond






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I gave you not only 1 minute but also +1 vote ^^
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:28












          • $begingroup$
            Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:33






          • 2




            $begingroup$
            @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
            $endgroup$
            – Will Jagy
            Jan 9 at 4:39
















          3












          $begingroup$

          Well, no. In brief, we can make an ordered field out of rational functions with real coefficients. A rational function is called positive if its value is positive for all sufficiently large positive $x$



          I'm looking up some details, give me a minute.



          Added: yes, I got it right, Proposition 18.1, pages 158-159 in Hartshorne, Geometry: Euclid and Beyond






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I gave you not only 1 minute but also +1 vote ^^
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:28












          • $begingroup$
            Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:33






          • 2




            $begingroup$
            @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
            $endgroup$
            – Will Jagy
            Jan 9 at 4:39














          3












          3








          3





          $begingroup$

          Well, no. In brief, we can make an ordered field out of rational functions with real coefficients. A rational function is called positive if its value is positive for all sufficiently large positive $x$



          I'm looking up some details, give me a minute.



          Added: yes, I got it right, Proposition 18.1, pages 158-159 in Hartshorne, Geometry: Euclid and Beyond






          share|cite|improve this answer











          $endgroup$



          Well, no. In brief, we can make an ordered field out of rational functions with real coefficients. A rational function is called positive if its value is positive for all sufficiently large positive $x$



          I'm looking up some details, give me a minute.



          Added: yes, I got it right, Proposition 18.1, pages 158-159 in Hartshorne, Geometry: Euclid and Beyond







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 9 at 4:29

























          answered Jan 9 at 4:26









          Will JagyWill Jagy

          102k5101199




          102k5101199












          • $begingroup$
            I gave you not only 1 minute but also +1 vote ^^
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:28












          • $begingroup$
            Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:33






          • 2




            $begingroup$
            @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
            $endgroup$
            – Will Jagy
            Jan 9 at 4:39


















          • $begingroup$
            I gave you not only 1 minute but also +1 vote ^^
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:28












          • $begingroup$
            Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
            $endgroup$
            – Le Anh Dung
            Jan 9 at 4:33






          • 2




            $begingroup$
            @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
            $endgroup$
            – Will Jagy
            Jan 9 at 4:39
















          $begingroup$
          I gave you not only 1 minute but also +1 vote ^^
          $endgroup$
          – Le Anh Dung
          Jan 9 at 4:28






          $begingroup$
          I gave you not only 1 minute but also +1 vote ^^
          $endgroup$
          – Le Anh Dung
          Jan 9 at 4:28














          $begingroup$
          Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
          $endgroup$
          – Le Anh Dung
          Jan 9 at 4:33




          $begingroup$
          Hi @Will. If $mathfrak{A}$ is a complete ordered field, then $X$ is dense in $A$. Is it right?
          $endgroup$
          – Le Anh Dung
          Jan 9 at 4:33




          2




          2




          $begingroup$
          @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
          $endgroup$
          – Will Jagy
          Jan 9 at 4:39




          $begingroup$
          @LeAnhDung en.wikipedia.org/wiki/Ordered_field " Any Dedekind-complete ordered field is isomorphic to the real numbers."
          $endgroup$
          – Will Jagy
          Jan 9 at 4:39











          1












          $begingroup$

          No.




          1. There are ordered fields of arbitrary high cardinality.


          2. Every field has a countable subfield. (Therefore, its smallest subfield is countable.)


          3. If a totally ordered set has a countable dense subset, then its cardinality is at most $2^{aleph_0}$. (Call the ordered set $A$ and the countable dense subset $Q$; then the map $amapsto{qin Q:qlt a}$ is an injection from $A$ to the power set of $Q$.)



          Hence any ordered field of cardinality greater than $2^{aleph_0}$ is a counterexample.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your great insight!
            $endgroup$
            – Le Anh Dung
            Jan 9 at 5:10
















          1












          $begingroup$

          No.




          1. There are ordered fields of arbitrary high cardinality.


          2. Every field has a countable subfield. (Therefore, its smallest subfield is countable.)


          3. If a totally ordered set has a countable dense subset, then its cardinality is at most $2^{aleph_0}$. (Call the ordered set $A$ and the countable dense subset $Q$; then the map $amapsto{qin Q:qlt a}$ is an injection from $A$ to the power set of $Q$.)



          Hence any ordered field of cardinality greater than $2^{aleph_0}$ is a counterexample.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your great insight!
            $endgroup$
            – Le Anh Dung
            Jan 9 at 5:10














          1












          1








          1





          $begingroup$

          No.




          1. There are ordered fields of arbitrary high cardinality.


          2. Every field has a countable subfield. (Therefore, its smallest subfield is countable.)


          3. If a totally ordered set has a countable dense subset, then its cardinality is at most $2^{aleph_0}$. (Call the ordered set $A$ and the countable dense subset $Q$; then the map $amapsto{qin Q:qlt a}$ is an injection from $A$ to the power set of $Q$.)



          Hence any ordered field of cardinality greater than $2^{aleph_0}$ is a counterexample.






          share|cite|improve this answer









          $endgroup$



          No.




          1. There are ordered fields of arbitrary high cardinality.


          2. Every field has a countable subfield. (Therefore, its smallest subfield is countable.)


          3. If a totally ordered set has a countable dense subset, then its cardinality is at most $2^{aleph_0}$. (Call the ordered set $A$ and the countable dense subset $Q$; then the map $amapsto{qin Q:qlt a}$ is an injection from $A$ to the power set of $Q$.)



          Hence any ordered field of cardinality greater than $2^{aleph_0}$ is a counterexample.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 4:41









          bofbof

          51k457120




          51k457120












          • $begingroup$
            Thank you for your great insight!
            $endgroup$
            – Le Anh Dung
            Jan 9 at 5:10


















          • $begingroup$
            Thank you for your great insight!
            $endgroup$
            – Le Anh Dung
            Jan 9 at 5:10
















          $begingroup$
          Thank you for your great insight!
          $endgroup$
          – Le Anh Dung
          Jan 9 at 5:10




          $begingroup$
          Thank you for your great insight!
          $endgroup$
          – Le Anh Dung
          Jan 9 at 5:10


















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