$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be it's bases?












-1












$begingroup$


$(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?



My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ {because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist}]



Can anyone please correct me if I go wrong anywhere?






general-topology







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    $begingroup$


    $(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?



    My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ {because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist}]



    Can anyone please correct me if I go wrong anywhere?






    general-topology







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      $(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?



      My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ {because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist}]



      Can anyone please correct me if I go wrong anywhere?






      general-topology







      share|cite|improve this question











      $endgroup$




      $(-sqrt 2, sqrt 2) cap mathbb Q$ in order topology. What will be its bases?



      My try: [ $(a,b)$ where $a,b in (-sqrt2, sqrt2) cap mathbb Q$ and $(-infty , b)$ , $(a, infty)$ {because supremum and infimum of $(-sqrt 2, sqrt 2) cap mathbb Q$ do not exist}]



      Can anyone please correct me if I go wrong anywhere?






      general-topology




      general-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jul 28 '18 at 6:48









      celtschk

      29.9k755101




      29.9k755101










      asked Jul 28 '18 at 1:54









      cmicmi

      1,109212




      1,109212






















          2 Answers
          2






          active

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          votes


















          2












          $begingroup$

          Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.



          I would put it this way:




          One basis for the order topology on $X:=left(-sqrt{2},sqrt{2}right)capBbb Q$ is $$bigl{(a,b)cap Xmid a,bin Xbigr}cupbigl{(-infty,b)cap Xmid bin Xbigr}cupbigl{(a,infty)cap Xmid ain Xbigr}.$$




          At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.



          Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.




          Added: As an exercise, you should try to prove that $bigl{(a,b)cap Xmid a,bin Xbigr}$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
            $endgroup$
            – Henning Makholm
            Jul 28 '18 at 2:17










          • $begingroup$
            @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
            $endgroup$
            – Cameron Buie
            Jul 28 '18 at 2:18






          • 1




            $begingroup$
            Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
            $endgroup$
            – Henning Makholm
            Jul 28 '18 at 2:21










          • $begingroup$
            @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
            $endgroup$
            – Cameron Buie
            Jul 28 '18 at 2:23



















          0












          $begingroup$

          Let r = $sqrt 2.$ A simple base is

          { (a,b) $cap$ Q : a,b in (-r,r) }.



          Exercise. Show a subbase is

          { (-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) }.



          In both definitions a,b can be futher restricted to rationals.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.



            I would put it this way:




            One basis for the order topology on $X:=left(-sqrt{2},sqrt{2}right)capBbb Q$ is $$bigl{(a,b)cap Xmid a,bin Xbigr}cupbigl{(-infty,b)cap Xmid bin Xbigr}cupbigl{(a,infty)cap Xmid ain Xbigr}.$$




            At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.



            Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.




            Added: As an exercise, you should try to prove that $bigl{(a,b)cap Xmid a,bin Xbigr}$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:17










            • $begingroup$
              @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:18






            • 1




              $begingroup$
              Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:21










            • $begingroup$
              @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:23
















            2












            $begingroup$

            Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.



            I would put it this way:




            One basis for the order topology on $X:=left(-sqrt{2},sqrt{2}right)capBbb Q$ is $$bigl{(a,b)cap Xmid a,bin Xbigr}cupbigl{(-infty,b)cap Xmid bin Xbigr}cupbigl{(a,infty)cap Xmid ain Xbigr}.$$




            At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.



            Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.




            Added: As an exercise, you should try to prove that $bigl{(a,b)cap Xmid a,bin Xbigr}$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:17










            • $begingroup$
              @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:18






            • 1




              $begingroup$
              Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:21










            • $begingroup$
              @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:23














            2












            2








            2





            $begingroup$

            Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.



            I would put it this way:




            One basis for the order topology on $X:=left(-sqrt{2},sqrt{2}right)capBbb Q$ is $$bigl{(a,b)cap Xmid a,bin Xbigr}cupbigl{(-infty,b)cap Xmid bin Xbigr}cupbigl{(a,infty)cap Xmid ain Xbigr}.$$




            At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.



            Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.




            Added: As an exercise, you should try to prove that $bigl{(a,b)cap Xmid a,bin Xbigr}$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).






            share|cite|improve this answer











            $endgroup$



            Looks like you mostly have the right idea! I will say that you're expressing it a bit awkwardly, though.



            I would put it this way:




            One basis for the order topology on $X:=left(-sqrt{2},sqrt{2}right)capBbb Q$ is $$bigl{(a,b)cap Xmid a,bin Xbigr}cupbigl{(-infty,b)cap Xmid bin Xbigr}cupbigl{(a,infty)cap Xmid ain Xbigr}.$$




            At that point, you must justify that this is a basis for the given topology. It sounds like you have some results you can use to help you accomplish that.



            Let me know if you have any trouble justifying the claims, or seeing the many differences between your version and mine.




            Added: As an exercise, you should try to prove that $bigl{(a,b)cap Xmid a,bin Xbigr}$ is a basis for the topology on $X,$ all on its own. This is the main reason why I emphasized the word "One" in my rewording above. Another is that the entire topology also comprises a basis (though it's much less straightforward to describe).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 6 at 19:11

























            answered Jul 28 '18 at 2:16









            Cameron BuieCameron Buie

            85.1k771155




            85.1k771155








            • 1




              $begingroup$
              It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:17










            • $begingroup$
              @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:18






            • 1




              $begingroup$
              Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:21










            • $begingroup$
              @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:23














            • 1




              $begingroup$
              It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:17










            • $begingroup$
              @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:18






            • 1




              $begingroup$
              Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
              $endgroup$
              – Henning Makholm
              Jul 28 '18 at 2:21










            • $begingroup$
              @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
              $endgroup$
              – Cameron Buie
              Jul 28 '18 at 2:23








            1




            1




            $begingroup$
            It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
            $endgroup$
            – Henning Makholm
            Jul 28 '18 at 2:17




            $begingroup$
            It probably deserves mention that the last two terms in that union can be omitted -- the first one is itself a basis.
            $endgroup$
            – Henning Makholm
            Jul 28 '18 at 2:17












            $begingroup$
            @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
            $endgroup$
            – Cameron Buie
            Jul 28 '18 at 2:18




            $begingroup$
            @Henning: Fair point. I thought I'd minimize the "burden of proof" (pun intended) on the OP.
            $endgroup$
            – Cameron Buie
            Jul 28 '18 at 2:18




            1




            1




            $begingroup$
            Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
            $endgroup$
            – Henning Makholm
            Jul 28 '18 at 2:21




            $begingroup$
            Also, I suspect the OP may be working in a context where, say, $(a,infty)$ doesn't necessarily mean a real half-interval, but an interval of whatever ordered set we're working in -- here, $X$. If that is the case, you explicit intersections with $X$ are not needed.
            $endgroup$
            – Henning Makholm
            Jul 28 '18 at 2:21












            $begingroup$
            @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
            $endgroup$
            – Cameron Buie
            Jul 28 '18 at 2:23




            $begingroup$
            @Henning: I wondered about that, too. I thought it better to be redundant, just in case, rather than assume too much.
            $endgroup$
            – Cameron Buie
            Jul 28 '18 at 2:23











            0












            $begingroup$

            Let r = $sqrt 2.$ A simple base is

            { (a,b) $cap$ Q : a,b in (-r,r) }.



            Exercise. Show a subbase is

            { (-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) }.



            In both definitions a,b can be futher restricted to rationals.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let r = $sqrt 2.$ A simple base is

              { (a,b) $cap$ Q : a,b in (-r,r) }.



              Exercise. Show a subbase is

              { (-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) }.



              In both definitions a,b can be futher restricted to rationals.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let r = $sqrt 2.$ A simple base is

                { (a,b) $cap$ Q : a,b in (-r,r) }.



                Exercise. Show a subbase is

                { (-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) }.



                In both definitions a,b can be futher restricted to rationals.






                share|cite|improve this answer









                $endgroup$



                Let r = $sqrt 2.$ A simple base is

                { (a,b) $cap$ Q : a,b in (-r,r) }.



                Exercise. Show a subbase is

                { (-r,a) $cap$ Q, (a,r) $cap$ Q : a in (-r,r) }.



                In both definitions a,b can be futher restricted to rationals.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jul 28 '18 at 5:24









                William ElliotWilliam Elliot

                7,5472720




                7,5472720






























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