Is $[ sqrt 2, sqrt 3] cap mathbb{Q}$ an open subset of $mathbb{Q}$?












5












$begingroup$


Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.



I have some confusion in my mind that is



Is $K$ is an open subset of $mathbb{Q}$ ?



My attempt : my answer is No,



$K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



From this I can conclude that K is not open subset of $mathbb{Q}$



Is it True ?










share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.



    I have some confusion in my mind that is



    Is $K$ is an open subset of $mathbb{Q}$ ?



    My attempt : my answer is No,



    $K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



    From this I can conclude that K is not open subset of $mathbb{Q}$



    Is it True ?










    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$


      Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.



      I have some confusion in my mind that is



      Is $K$ is an open subset of $mathbb{Q}$ ?



      My attempt : my answer is No,



      $K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



      From this I can conclude that K is not open subset of $mathbb{Q}$



      Is it True ?










      share|cite|improve this question











      $endgroup$




      Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.



      I have some confusion in my mind that is



      Is $K$ is an open subset of $mathbb{Q}$ ?



      My attempt : my answer is No,



      $K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.



      From this I can conclude that K is not open subset of $mathbb{Q}$



      Is it True ?







      general-topology proof-verification compactness






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 6:07









      Asaf Karagila

      302k32429760




      302k32429760










      asked Dec 14 '18 at 13:36









      jasminejasmine

      1,679416




      1,679416






















          5 Answers
          5






          active

          oldest

          votes


















          11












          $begingroup$

          Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.






          share|cite|improve this answer









          $endgroup$





















            23












            $begingroup$

            No, that's wrong. The fact that a set is closed doesn't mean it is not open!



            In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.



            Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.






            share|cite|improve this answer











            $endgroup$









            • 4




              $begingroup$
              A set is not a door.
              $endgroup$
              – Arno
              Dec 14 '18 at 21:30






            • 5




              $begingroup$
              The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
              $endgroup$
              – Carmeister
              Dec 15 '18 at 3:58



















            4












            $begingroup$


            $K=[sqrt 2,sqrt 3]∩mathbb{Q}={qin mathbb{Q}|sqrt 2<q<sqrt 3}$ where$[sqrt 2,sqrt 3]$ is closed in R.



            From this I can conclude that K is not open subset of Q




            You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:




            1. K is an intersection between a closed set and a closed set.


            2. K is therefore closed.


            3. Therefore K is not open.



            The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.



            If we have the open ball topology, then since $sqrt2 <q$, we know that there is "space" between $sqrt 2$ and $q$, and similarly for $sqrt3$. So given any $q$, we can take $epsilon_1$ to be half the distance between $sqrt2$ and $q$, $epsilon_2$ to be half the distance between $sqrt3$ and $q$, and $epsilon$ to be the minimum of $epsilon_1$ and $epsilon_2$. Then everything withing $epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.






            share|cite|improve this answer











            $endgroup$





















              3












              $begingroup$

              A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.



              Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.






              share|cite|improve this answer









              $endgroup$





















                2












                $begingroup$

                With the metric $d(x,y)=|x-y|$ on $Bbb Q$ and the topology on $Bbb Q$ generated by $d$: For $qin K$ let $r(q)=min (q-sqrt 2,,sqrt 3 -q,).$ Let $K(q)={q'in Bbb Q: d(q',q)<r(q)}.$



                Then $K(q)$ is open in $Bbb Q$ and $qin K(q)subset K.$



                So $cup_{qin K}K(q)$ is open in $Bbb Q.$ And we have $K=cup_{qin K}
                ,{q}subset cup_{qin K},K(q)subset cup_{qin K},K=K,$
                so $K=cup_{qin K}K(q)$ is open in $Bbb Q.$



                An easily overlooked point about this Q:



                (i). Let $T$ be a topology on a set $X$ and let $Y subset X.$ The subspace topology $T|Y$ on $Y$ is defined as $T|Y={tcap Y:tin T}. $ If $B$ is a base (basis) for $T$ then $B|Y={bcap Y: bin B}$ is a base for $T|Y.$



                (ii).Suppose $T$ is generated by a metric $d$ on $X,$ so that the set $B$ of open $d$-balls of $X$ is a base for $T$. So $B|Y$ is a base for $T|Y.$



                BUT in general $B|Y$ may NOT be the set of open $d$-balls of $Y.$ An open $d$-ball of $Y$ is $B_d^Y(y,r)={y'in Y: d(y',y)<r},$ for some $yin Y, $ which does belong to $B|Y,$ but there may be other members of $B|Y.$



                For example in your Q, with $X=Bbb R$ and $Y=Bbb Q$ and $d(u,v)=|u-v|,$ the set $K$ belongs to $B|Y$ but is not an open ball of $Bbb Q$ because $(sqrt 2 +sqrt 3)/2not in Bbb Q.$



                (iii). In the general case, for metric spacess it is a useful, widely used result that every member of $B|Y$ is a union of open $d$-balls of $Y,$ so the subspace topology $T|Y,$ as a subspace of $X,$ co-incides with the topology on $Y$ generated by the metric $d|_{Ytimes Y}.$






                share|cite|improve this answer









                $endgroup$













                  Your Answer





                  StackExchange.ifUsing("editor", function () {
                  return StackExchange.using("mathjaxEditing", function () {
                  StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
                  StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                  });
                  });
                  }, "mathjax-editing");

                  StackExchange.ready(function() {
                  var channelOptions = {
                  tags: "".split(" "),
                  id: "69"
                  };
                  initTagRenderer("".split(" "), "".split(" "), channelOptions);

                  StackExchange.using("externalEditor", function() {
                  // Have to fire editor after snippets, if snippets enabled
                  if (StackExchange.settings.snippets.snippetsEnabled) {
                  StackExchange.using("snippets", function() {
                  createEditor();
                  });
                  }
                  else {
                  createEditor();
                  }
                  });

                  function createEditor() {
                  StackExchange.prepareEditor({
                  heartbeatType: 'answer',
                  autoActivateHeartbeat: false,
                  convertImagesToLinks: true,
                  noModals: true,
                  showLowRepImageUploadWarning: true,
                  reputationToPostImages: 10,
                  bindNavPrevention: true,
                  postfix: "",
                  imageUploader: {
                  brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                  contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                  allowUrls: true
                  },
                  noCode: true, onDemand: true,
                  discardSelector: ".discard-answer"
                  ,immediatelyShowMarkdownHelp:true
                  });


                  }
                  });














                  draft saved

                  draft discarded


















                  StackExchange.ready(
                  function () {
                  StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039363%2fis-sqrt-2-sqrt-3-cap-mathbbq-an-open-subset-of-mathbbq%23new-answer', 'question_page');
                  }
                  );

                  Post as a guest















                  Required, but never shown

























                  5 Answers
                  5






                  active

                  oldest

                  votes








                  5 Answers
                  5






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  11












                  $begingroup$

                  Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.






                  share|cite|improve this answer









                  $endgroup$


















                    11












                    $begingroup$

                    Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.






                    share|cite|improve this answer









                    $endgroup$
















                      11












                      11








                      11





                      $begingroup$

                      Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.






                      share|cite|improve this answer









                      $endgroup$



                      Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 14 '18 at 13:38









                      José Carlos SantosJosé Carlos Santos

                      154k22123226




                      154k22123226























                          23












                          $begingroup$

                          No, that's wrong. The fact that a set is closed doesn't mean it is not open!



                          In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.



                          Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.






                          share|cite|improve this answer











                          $endgroup$









                          • 4




                            $begingroup$
                            A set is not a door.
                            $endgroup$
                            – Arno
                            Dec 14 '18 at 21:30






                          • 5




                            $begingroup$
                            The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
                            $endgroup$
                            – Carmeister
                            Dec 15 '18 at 3:58
















                          23












                          $begingroup$

                          No, that's wrong. The fact that a set is closed doesn't mean it is not open!



                          In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.



                          Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.






                          share|cite|improve this answer











                          $endgroup$









                          • 4




                            $begingroup$
                            A set is not a door.
                            $endgroup$
                            – Arno
                            Dec 14 '18 at 21:30






                          • 5




                            $begingroup$
                            The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
                            $endgroup$
                            – Carmeister
                            Dec 15 '18 at 3:58














                          23












                          23








                          23





                          $begingroup$

                          No, that's wrong. The fact that a set is closed doesn't mean it is not open!



                          In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.



                          Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.






                          share|cite|improve this answer











                          $endgroup$



                          No, that's wrong. The fact that a set is closed doesn't mean it is not open!



                          In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.



                          Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 15 '18 at 12:15

























                          answered Dec 14 '18 at 13:38









                          YankoYanko

                          6,6841728




                          6,6841728








                          • 4




                            $begingroup$
                            A set is not a door.
                            $endgroup$
                            – Arno
                            Dec 14 '18 at 21:30






                          • 5




                            $begingroup$
                            The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
                            $endgroup$
                            – Carmeister
                            Dec 15 '18 at 3:58














                          • 4




                            $begingroup$
                            A set is not a door.
                            $endgroup$
                            – Arno
                            Dec 14 '18 at 21:30






                          • 5




                            $begingroup$
                            The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
                            $endgroup$
                            – Carmeister
                            Dec 15 '18 at 3:58








                          4




                          4




                          $begingroup$
                          A set is not a door.
                          $endgroup$
                          – Arno
                          Dec 14 '18 at 21:30




                          $begingroup$
                          A set is not a door.
                          $endgroup$
                          – Arno
                          Dec 14 '18 at 21:30




                          5




                          5




                          $begingroup$
                          The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
                          $endgroup$
                          – Carmeister
                          Dec 15 '18 at 3:58




                          $begingroup$
                          The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580
                          $endgroup$
                          – Carmeister
                          Dec 15 '18 at 3:58











                          4












                          $begingroup$


                          $K=[sqrt 2,sqrt 3]∩mathbb{Q}={qin mathbb{Q}|sqrt 2<q<sqrt 3}$ where$[sqrt 2,sqrt 3]$ is closed in R.



                          From this I can conclude that K is not open subset of Q




                          You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:




                          1. K is an intersection between a closed set and a closed set.


                          2. K is therefore closed.


                          3. Therefore K is not open.



                          The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.



                          If we have the open ball topology, then since $sqrt2 <q$, we know that there is "space" between $sqrt 2$ and $q$, and similarly for $sqrt3$. So given any $q$, we can take $epsilon_1$ to be half the distance between $sqrt2$ and $q$, $epsilon_2$ to be half the distance between $sqrt3$ and $q$, and $epsilon$ to be the minimum of $epsilon_1$ and $epsilon_2$. Then everything withing $epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.






                          share|cite|improve this answer











                          $endgroup$


















                            4












                            $begingroup$


                            $K=[sqrt 2,sqrt 3]∩mathbb{Q}={qin mathbb{Q}|sqrt 2<q<sqrt 3}$ where$[sqrt 2,sqrt 3]$ is closed in R.



                            From this I can conclude that K is not open subset of Q




                            You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:




                            1. K is an intersection between a closed set and a closed set.


                            2. K is therefore closed.


                            3. Therefore K is not open.



                            The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.



                            If we have the open ball topology, then since $sqrt2 <q$, we know that there is "space" between $sqrt 2$ and $q$, and similarly for $sqrt3$. So given any $q$, we can take $epsilon_1$ to be half the distance between $sqrt2$ and $q$, $epsilon_2$ to be half the distance between $sqrt3$ and $q$, and $epsilon$ to be the minimum of $epsilon_1$ and $epsilon_2$. Then everything withing $epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.






                            share|cite|improve this answer











                            $endgroup$
















                              4












                              4








                              4





                              $begingroup$


                              $K=[sqrt 2,sqrt 3]∩mathbb{Q}={qin mathbb{Q}|sqrt 2<q<sqrt 3}$ where$[sqrt 2,sqrt 3]$ is closed in R.



                              From this I can conclude that K is not open subset of Q




                              You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:




                              1. K is an intersection between a closed set and a closed set.


                              2. K is therefore closed.


                              3. Therefore K is not open.



                              The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.



                              If we have the open ball topology, then since $sqrt2 <q$, we know that there is "space" between $sqrt 2$ and $q$, and similarly for $sqrt3$. So given any $q$, we can take $epsilon_1$ to be half the distance between $sqrt2$ and $q$, $epsilon_2$ to be half the distance between $sqrt3$ and $q$, and $epsilon$ to be the minimum of $epsilon_1$ and $epsilon_2$. Then everything withing $epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.






                              share|cite|improve this answer











                              $endgroup$




                              $K=[sqrt 2,sqrt 3]∩mathbb{Q}={qin mathbb{Q}|sqrt 2<q<sqrt 3}$ where$[sqrt 2,sqrt 3]$ is closed in R.



                              From this I can conclude that K is not open subset of Q




                              You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:




                              1. K is an intersection between a closed set and a closed set.


                              2. K is therefore closed.


                              3. Therefore K is not open.



                              The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.



                              If we have the open ball topology, then since $sqrt2 <q$, we know that there is "space" between $sqrt 2$ and $q$, and similarly for $sqrt3$. So given any $q$, we can take $epsilon_1$ to be half the distance between $sqrt2$ and $q$, $epsilon_2$ to be half the distance between $sqrt3$ and $q$, and $epsilon$ to be the minimum of $epsilon_1$ and $epsilon_2$. Then everything withing $epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 14 '18 at 21:07









                              Yanko

                              6,6841728




                              6,6841728










                              answered Dec 14 '18 at 15:56









                              AcccumulationAcccumulation

                              6,8342618




                              6,8342618























                                  3












                                  $begingroup$

                                  A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.



                                  Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    3












                                    $begingroup$

                                    A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.



                                    Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      3












                                      3








                                      3





                                      $begingroup$

                                      A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.



                                      Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.






                                      share|cite|improve this answer









                                      $endgroup$



                                      A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.



                                      Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 14 '18 at 13:38









                                      ArthurArthur

                                      112k7108191




                                      112k7108191























                                          2












                                          $begingroup$

                                          With the metric $d(x,y)=|x-y|$ on $Bbb Q$ and the topology on $Bbb Q$ generated by $d$: For $qin K$ let $r(q)=min (q-sqrt 2,,sqrt 3 -q,).$ Let $K(q)={q'in Bbb Q: d(q',q)<r(q)}.$



                                          Then $K(q)$ is open in $Bbb Q$ and $qin K(q)subset K.$



                                          So $cup_{qin K}K(q)$ is open in $Bbb Q.$ And we have $K=cup_{qin K}
                                          ,{q}subset cup_{qin K},K(q)subset cup_{qin K},K=K,$
                                          so $K=cup_{qin K}K(q)$ is open in $Bbb Q.$



                                          An easily overlooked point about this Q:



                                          (i). Let $T$ be a topology on a set $X$ and let $Y subset X.$ The subspace topology $T|Y$ on $Y$ is defined as $T|Y={tcap Y:tin T}. $ If $B$ is a base (basis) for $T$ then $B|Y={bcap Y: bin B}$ is a base for $T|Y.$



                                          (ii).Suppose $T$ is generated by a metric $d$ on $X,$ so that the set $B$ of open $d$-balls of $X$ is a base for $T$. So $B|Y$ is a base for $T|Y.$



                                          BUT in general $B|Y$ may NOT be the set of open $d$-balls of $Y.$ An open $d$-ball of $Y$ is $B_d^Y(y,r)={y'in Y: d(y',y)<r},$ for some $yin Y, $ which does belong to $B|Y,$ but there may be other members of $B|Y.$



                                          For example in your Q, with $X=Bbb R$ and $Y=Bbb Q$ and $d(u,v)=|u-v|,$ the set $K$ belongs to $B|Y$ but is not an open ball of $Bbb Q$ because $(sqrt 2 +sqrt 3)/2not in Bbb Q.$



                                          (iii). In the general case, for metric spacess it is a useful, widely used result that every member of $B|Y$ is a union of open $d$-balls of $Y,$ so the subspace topology $T|Y,$ as a subspace of $X,$ co-incides with the topology on $Y$ generated by the metric $d|_{Ytimes Y}.$






                                          share|cite|improve this answer









                                          $endgroup$


















                                            2












                                            $begingroup$

                                            With the metric $d(x,y)=|x-y|$ on $Bbb Q$ and the topology on $Bbb Q$ generated by $d$: For $qin K$ let $r(q)=min (q-sqrt 2,,sqrt 3 -q,).$ Let $K(q)={q'in Bbb Q: d(q',q)<r(q)}.$



                                            Then $K(q)$ is open in $Bbb Q$ and $qin K(q)subset K.$



                                            So $cup_{qin K}K(q)$ is open in $Bbb Q.$ And we have $K=cup_{qin K}
                                            ,{q}subset cup_{qin K},K(q)subset cup_{qin K},K=K,$
                                            so $K=cup_{qin K}K(q)$ is open in $Bbb Q.$



                                            An easily overlooked point about this Q:



                                            (i). Let $T$ be a topology on a set $X$ and let $Y subset X.$ The subspace topology $T|Y$ on $Y$ is defined as $T|Y={tcap Y:tin T}. $ If $B$ is a base (basis) for $T$ then $B|Y={bcap Y: bin B}$ is a base for $T|Y.$



                                            (ii).Suppose $T$ is generated by a metric $d$ on $X,$ so that the set $B$ of open $d$-balls of $X$ is a base for $T$. So $B|Y$ is a base for $T|Y.$



                                            BUT in general $B|Y$ may NOT be the set of open $d$-balls of $Y.$ An open $d$-ball of $Y$ is $B_d^Y(y,r)={y'in Y: d(y',y)<r},$ for some $yin Y, $ which does belong to $B|Y,$ but there may be other members of $B|Y.$



                                            For example in your Q, with $X=Bbb R$ and $Y=Bbb Q$ and $d(u,v)=|u-v|,$ the set $K$ belongs to $B|Y$ but is not an open ball of $Bbb Q$ because $(sqrt 2 +sqrt 3)/2not in Bbb Q.$



                                            (iii). In the general case, for metric spacess it is a useful, widely used result that every member of $B|Y$ is a union of open $d$-balls of $Y,$ so the subspace topology $T|Y,$ as a subspace of $X,$ co-incides with the topology on $Y$ generated by the metric $d|_{Ytimes Y}.$






                                            share|cite|improve this answer









                                            $endgroup$
















                                              2












                                              2








                                              2





                                              $begingroup$

                                              With the metric $d(x,y)=|x-y|$ on $Bbb Q$ and the topology on $Bbb Q$ generated by $d$: For $qin K$ let $r(q)=min (q-sqrt 2,,sqrt 3 -q,).$ Let $K(q)={q'in Bbb Q: d(q',q)<r(q)}.$



                                              Then $K(q)$ is open in $Bbb Q$ and $qin K(q)subset K.$



                                              So $cup_{qin K}K(q)$ is open in $Bbb Q.$ And we have $K=cup_{qin K}
                                              ,{q}subset cup_{qin K},K(q)subset cup_{qin K},K=K,$
                                              so $K=cup_{qin K}K(q)$ is open in $Bbb Q.$



                                              An easily overlooked point about this Q:



                                              (i). Let $T$ be a topology on a set $X$ and let $Y subset X.$ The subspace topology $T|Y$ on $Y$ is defined as $T|Y={tcap Y:tin T}. $ If $B$ is a base (basis) for $T$ then $B|Y={bcap Y: bin B}$ is a base for $T|Y.$



                                              (ii).Suppose $T$ is generated by a metric $d$ on $X,$ so that the set $B$ of open $d$-balls of $X$ is a base for $T$. So $B|Y$ is a base for $T|Y.$



                                              BUT in general $B|Y$ may NOT be the set of open $d$-balls of $Y.$ An open $d$-ball of $Y$ is $B_d^Y(y,r)={y'in Y: d(y',y)<r},$ for some $yin Y, $ which does belong to $B|Y,$ but there may be other members of $B|Y.$



                                              For example in your Q, with $X=Bbb R$ and $Y=Bbb Q$ and $d(u,v)=|u-v|,$ the set $K$ belongs to $B|Y$ but is not an open ball of $Bbb Q$ because $(sqrt 2 +sqrt 3)/2not in Bbb Q.$



                                              (iii). In the general case, for metric spacess it is a useful, widely used result that every member of $B|Y$ is a union of open $d$-balls of $Y,$ so the subspace topology $T|Y,$ as a subspace of $X,$ co-incides with the topology on $Y$ generated by the metric $d|_{Ytimes Y}.$






                                              share|cite|improve this answer









                                              $endgroup$



                                              With the metric $d(x,y)=|x-y|$ on $Bbb Q$ and the topology on $Bbb Q$ generated by $d$: For $qin K$ let $r(q)=min (q-sqrt 2,,sqrt 3 -q,).$ Let $K(q)={q'in Bbb Q: d(q',q)<r(q)}.$



                                              Then $K(q)$ is open in $Bbb Q$ and $qin K(q)subset K.$



                                              So $cup_{qin K}K(q)$ is open in $Bbb Q.$ And we have $K=cup_{qin K}
                                              ,{q}subset cup_{qin K},K(q)subset cup_{qin K},K=K,$
                                              so $K=cup_{qin K}K(q)$ is open in $Bbb Q.$



                                              An easily overlooked point about this Q:



                                              (i). Let $T$ be a topology on a set $X$ and let $Y subset X.$ The subspace topology $T|Y$ on $Y$ is defined as $T|Y={tcap Y:tin T}. $ If $B$ is a base (basis) for $T$ then $B|Y={bcap Y: bin B}$ is a base for $T|Y.$



                                              (ii).Suppose $T$ is generated by a metric $d$ on $X,$ so that the set $B$ of open $d$-balls of $X$ is a base for $T$. So $B|Y$ is a base for $T|Y.$



                                              BUT in general $B|Y$ may NOT be the set of open $d$-balls of $Y.$ An open $d$-ball of $Y$ is $B_d^Y(y,r)={y'in Y: d(y',y)<r},$ for some $yin Y, $ which does belong to $B|Y,$ but there may be other members of $B|Y.$



                                              For example in your Q, with $X=Bbb R$ and $Y=Bbb Q$ and $d(u,v)=|u-v|,$ the set $K$ belongs to $B|Y$ but is not an open ball of $Bbb Q$ because $(sqrt 2 +sqrt 3)/2not in Bbb Q.$



                                              (iii). In the general case, for metric spacess it is a useful, widely used result that every member of $B|Y$ is a union of open $d$-balls of $Y,$ so the subspace topology $T|Y,$ as a subspace of $X,$ co-incides with the topology on $Y$ generated by the metric $d|_{Ytimes Y}.$







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 14 '18 at 15:22









                                              DanielWainfleetDanielWainfleet

                                              34.6k31648




                                              34.6k31648






























                                                  draft saved

                                                  draft discarded




















































                                                  Thanks for contributing an answer to Mathematics Stack Exchange!


                                                  • Please be sure to answer the question. Provide details and share your research!

                                                  But avoid



                                                  • Asking for help, clarification, or responding to other answers.

                                                  • Making statements based on opinion; back them up with references or personal experience.


                                                  Use MathJax to format equations. MathJax reference.


                                                  To learn more, see our tips on writing great answers.




                                                  draft saved


                                                  draft discarded














                                                  StackExchange.ready(
                                                  function () {
                                                  StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3039363%2fis-sqrt-2-sqrt-3-cap-mathbbq-an-open-subset-of-mathbbq%23new-answer', 'question_page');
                                                  }
                                                  );

                                                  Post as a guest















                                                  Required, but never shown





















































                                                  Required, but never shown














                                                  Required, but never shown












                                                  Required, but never shown







                                                  Required, but never shown

































                                                  Required, but never shown














                                                  Required, but never shown












                                                  Required, but never shown







                                                  Required, but never shown







                                                  Popular posts from this blog

                                                  Bressuire

                                                  Cabo Verde

                                                  Gyllenstierna