Does there exist a real number a given distance from each rational number?












4












$begingroup$



Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $xin mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$?




This problem was inspired by an easier version of the problem where we assume the stronger condition that $displaystylesum_{n=1}^infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too.




Let $Omega$ denote the set of $xin mathbb{R}$ that do not satisfy the given property; we claim that $Omega neq mathbb{R}$ and so an $xin mathbb{R}$ with given property exists. Indeed, $Omega={xinmathbb{R} | exists nin mathbb{N} mathrm{such that} |x-r_n| leq a_n}=displaystylebigcup_{n=1}^infty [r_n-a_n,r_n+a_n]$

whose Lebesgue measure $lambda(Omega)leq displaystylesum_{n=1}^infty lambda([r_n-a_n,r_n+a_n])=displaystylesum_{n=1}^infty 2a_n < infty$, so $Omega neq mathbb{R}$.




Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.










share|cite|improve this question









$endgroup$

















    4












    $begingroup$



    Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $xin mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$?




    This problem was inspired by an easier version of the problem where we assume the stronger condition that $displaystylesum_{n=1}^infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too.




    Let $Omega$ denote the set of $xin mathbb{R}$ that do not satisfy the given property; we claim that $Omega neq mathbb{R}$ and so an $xin mathbb{R}$ with given property exists. Indeed, $Omega={xinmathbb{R} | exists nin mathbb{N} mathrm{such that} |x-r_n| leq a_n}=displaystylebigcup_{n=1}^infty [r_n-a_n,r_n+a_n]$

    whose Lebesgue measure $lambda(Omega)leq displaystylesum_{n=1}^infty lambda([r_n-a_n,r_n+a_n])=displaystylesum_{n=1}^infty 2a_n < infty$, so $Omega neq mathbb{R}$.




    Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.










    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$



      Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $xin mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$?




      This problem was inspired by an easier version of the problem where we assume the stronger condition that $displaystylesum_{n=1}^infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too.




      Let $Omega$ denote the set of $xin mathbb{R}$ that do not satisfy the given property; we claim that $Omega neq mathbb{R}$ and so an $xin mathbb{R}$ with given property exists. Indeed, $Omega={xinmathbb{R} | exists nin mathbb{N} mathrm{such that} |x-r_n| leq a_n}=displaystylebigcup_{n=1}^infty [r_n-a_n,r_n+a_n]$

      whose Lebesgue measure $lambda(Omega)leq displaystylesum_{n=1}^infty lambda([r_n-a_n,r_n+a_n])=displaystylesum_{n=1}^infty 2a_n < infty$, so $Omega neq mathbb{R}$.




      Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.










      share|cite|improve this question









      $endgroup$





      Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $xin mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$?




      This problem was inspired by an easier version of the problem where we assume the stronger condition that $displaystylesum_{n=1}^infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too.




      Let $Omega$ denote the set of $xin mathbb{R}$ that do not satisfy the given property; we claim that $Omega neq mathbb{R}$ and so an $xin mathbb{R}$ with given property exists. Indeed, $Omega={xinmathbb{R} | exists nin mathbb{N} mathrm{such that} |x-r_n| leq a_n}=displaystylebigcup_{n=1}^infty [r_n-a_n,r_n+a_n]$

      whose Lebesgue measure $lambda(Omega)leq displaystylesum_{n=1}^infty lambda([r_n-a_n,r_n+a_n])=displaystylesum_{n=1}^infty 2a_n < infty$, so $Omega neq mathbb{R}$.




      Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.







      real-analysis sequences-and-series alternative-proof






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      AlephNullAlephNull

      2429




      2429






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).



          (At the same time, given a sequence $(a_i)_{iinmathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)



          For simplicity, let's look at $[0,infty)$ rather than $mathbb{R}$ (this doesn't make a substantive difference).





          The key point is the following picture: we chop $[0,infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[sum_{0<jle i}{1over j}, (sum_{0<jle i}{1over j})+{1over i+1})$. Note that each $B_k$ has "diameter" ${1over k+1}$, and hence if $qin B_k$ then the ball around $q$ with radius ${2over k+1}$ covers $B_k$.



          Now fix any enumeration of rationals $E=(r_i)_{iinmathbb{N}}$, and pick a sequence $n_i$ ($iinmathbb{N}$) of naturals such that:




          • $n_i<n_{i+1}$, and


          • $r_{n_i}in B_i$.



          Such a sequence must exist since $B_icapmathbb{Q}$ is always infinite.



          Finally, let $A=(a_i)_{iinmathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2over i+1}$ for all $i$. The set $${(r_{n_i}-a_i, r_{n_i}+a_i): iinmathbb{N}}$$ covers $[0,infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.






          share|cite|improve this answer











          $endgroup$





















            3












            $begingroup$

            The answer is no. Consider the following open cover of $Bbb R$:



            $$mathcal U={ (H_{2n} , H_{2n+4}) | n in Bbb N_{ge 1}} cup { (-H_{2n+4} , -H_{2n}) | n in Bbb N_{ge 1}} cup { (-3, 3) }$$
            where $$H_n = sum_{j=1}^n frac 1j$$
            is the $n$-th harmonic number. Note that for all $varepsilon >0$ only finitely many elements $U in mathcal U$ satisfy $lambda (U) > varepsilon$.



            Let ${ U_{n} }_{n ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= lambda (U_n)$ converges to $0$.



            Let ${ r_n }_{n ge 1}$ be an enumeration of rational numbers such that for all $n$
            $$r_{n} in U_n$$
            holds. Recall that $a_n$ is rational, and it converges to $0$.



            Now, for all $x in Bbb R$ there exists $n ge 1$ such that $x in U_n$: thus $|x-r_n| < a_n$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
              $endgroup$
              – Noah Schweber
              4 hours ago











            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%2f3076002%2fdoes-there-exist-a-real-number-a-given-distance-from-each-rational-number%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).



            (At the same time, given a sequence $(a_i)_{iinmathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)



            For simplicity, let's look at $[0,infty)$ rather than $mathbb{R}$ (this doesn't make a substantive difference).





            The key point is the following picture: we chop $[0,infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[sum_{0<jle i}{1over j}, (sum_{0<jle i}{1over j})+{1over i+1})$. Note that each $B_k$ has "diameter" ${1over k+1}$, and hence if $qin B_k$ then the ball around $q$ with radius ${2over k+1}$ covers $B_k$.



            Now fix any enumeration of rationals $E=(r_i)_{iinmathbb{N}}$, and pick a sequence $n_i$ ($iinmathbb{N}$) of naturals such that:




            • $n_i<n_{i+1}$, and


            • $r_{n_i}in B_i$.



            Such a sequence must exist since $B_icapmathbb{Q}$ is always infinite.



            Finally, let $A=(a_i)_{iinmathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2over i+1}$ for all $i$. The set $${(r_{n_i}-a_i, r_{n_i}+a_i): iinmathbb{N}}$$ covers $[0,infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).



              (At the same time, given a sequence $(a_i)_{iinmathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)



              For simplicity, let's look at $[0,infty)$ rather than $mathbb{R}$ (this doesn't make a substantive difference).





              The key point is the following picture: we chop $[0,infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[sum_{0<jle i}{1over j}, (sum_{0<jle i}{1over j})+{1over i+1})$. Note that each $B_k$ has "diameter" ${1over k+1}$, and hence if $qin B_k$ then the ball around $q$ with radius ${2over k+1}$ covers $B_k$.



              Now fix any enumeration of rationals $E=(r_i)_{iinmathbb{N}}$, and pick a sequence $n_i$ ($iinmathbb{N}$) of naturals such that:




              • $n_i<n_{i+1}$, and


              • $r_{n_i}in B_i$.



              Such a sequence must exist since $B_icapmathbb{Q}$ is always infinite.



              Finally, let $A=(a_i)_{iinmathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2over i+1}$ for all $i$. The set $${(r_{n_i}-a_i, r_{n_i}+a_i): iinmathbb{N}}$$ covers $[0,infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).



                (At the same time, given a sequence $(a_i)_{iinmathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)



                For simplicity, let's look at $[0,infty)$ rather than $mathbb{R}$ (this doesn't make a substantive difference).





                The key point is the following picture: we chop $[0,infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[sum_{0<jle i}{1over j}, (sum_{0<jle i}{1over j})+{1over i+1})$. Note that each $B_k$ has "diameter" ${1over k+1}$, and hence if $qin B_k$ then the ball around $q$ with radius ${2over k+1}$ covers $B_k$.



                Now fix any enumeration of rationals $E=(r_i)_{iinmathbb{N}}$, and pick a sequence $n_i$ ($iinmathbb{N}$) of naturals such that:




                • $n_i<n_{i+1}$, and


                • $r_{n_i}in B_i$.



                Such a sequence must exist since $B_icapmathbb{Q}$ is always infinite.



                Finally, let $A=(a_i)_{iinmathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2over i+1}$ for all $i$. The set $${(r_{n_i}-a_i, r_{n_i}+a_i): iinmathbb{N}}$$ covers $[0,infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.






                share|cite|improve this answer











                $endgroup$



                No, this is not the case, and in fact is never the case: any enumeration of rationals whatsoever has an "unsatisfiable sequence" (that is, a sequence of reals tending to zero - indeed, strictly decreasing! - such that no real satisfies the corresponding requirement on distances from rationals).



                (At the same time, given a sequence $(a_i)_{iinmathbb{N}}$ it's easy to construct an enumeration of the rationals with respect to which that sequence is satisfiable - just play "keep away from $pi$," and make sure every rational gets tossed in eventually - so this is the strongest negative result we can hope for.)



                For simplicity, let's look at $[0,infty)$ rather than $mathbb{R}$ (this doesn't make a substantive difference).





                The key point is the following picture: we chop $[0,infty)$ into blocks $B_i$ such that the size of block $B_i$ goes to zero as $i$ goes to $infty$. One way to do this is to define the $B_i$s inductively by $B_0=[0,1)$ and $B_{i+1}=[sum_{0<jle i}{1over j}, (sum_{0<jle i}{1over j})+{1over i+1})$. Note that each $B_k$ has "diameter" ${1over k+1}$, and hence if $qin B_k$ then the ball around $q$ with radius ${2over k+1}$ covers $B_k$.



                Now fix any enumeration of rationals $E=(r_i)_{iinmathbb{N}}$, and pick a sequence $n_i$ ($iinmathbb{N}$) of naturals such that:




                • $n_i<n_{i+1}$, and


                • $r_{n_i}in B_i$.



                Such a sequence must exist since $B_icapmathbb{Q}$ is always infinite.



                Finally, let $A=(a_i)_{iinmathbb{N}}$ be any strictly descending sequence of rationals such that $a_{n_i}={2over i+1}$ for all $i$. The set $${(r_{n_i}-a_i, r_{n_i}+a_i): iinmathbb{N}}$$ covers $[0,infty)$ since each $(r_{n_i}-a_i, r_{n_i}+a_i)$ covers the corresponding $B_i$. So the sequence $A$ is unsatisfiable for the enumeration $E$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 4 hours ago

























                answered 4 hours ago









                Noah SchweberNoah Schweber

                122k10149284




                122k10149284























                    3












                    $begingroup$

                    The answer is no. Consider the following open cover of $Bbb R$:



                    $$mathcal U={ (H_{2n} , H_{2n+4}) | n in Bbb N_{ge 1}} cup { (-H_{2n+4} , -H_{2n}) | n in Bbb N_{ge 1}} cup { (-3, 3) }$$
                    where $$H_n = sum_{j=1}^n frac 1j$$
                    is the $n$-th harmonic number. Note that for all $varepsilon >0$ only finitely many elements $U in mathcal U$ satisfy $lambda (U) > varepsilon$.



                    Let ${ U_{n} }_{n ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= lambda (U_n)$ converges to $0$.



                    Let ${ r_n }_{n ge 1}$ be an enumeration of rational numbers such that for all $n$
                    $$r_{n} in U_n$$
                    holds. Recall that $a_n$ is rational, and it converges to $0$.



                    Now, for all $x in Bbb R$ there exists $n ge 1$ such that $x in U_n$: thus $|x-r_n| < a_n$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
                      $endgroup$
                      – Noah Schweber
                      4 hours ago
















                    3












                    $begingroup$

                    The answer is no. Consider the following open cover of $Bbb R$:



                    $$mathcal U={ (H_{2n} , H_{2n+4}) | n in Bbb N_{ge 1}} cup { (-H_{2n+4} , -H_{2n}) | n in Bbb N_{ge 1}} cup { (-3, 3) }$$
                    where $$H_n = sum_{j=1}^n frac 1j$$
                    is the $n$-th harmonic number. Note that for all $varepsilon >0$ only finitely many elements $U in mathcal U$ satisfy $lambda (U) > varepsilon$.



                    Let ${ U_{n} }_{n ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= lambda (U_n)$ converges to $0$.



                    Let ${ r_n }_{n ge 1}$ be an enumeration of rational numbers such that for all $n$
                    $$r_{n} in U_n$$
                    holds. Recall that $a_n$ is rational, and it converges to $0$.



                    Now, for all $x in Bbb R$ there exists $n ge 1$ such that $x in U_n$: thus $|x-r_n| < a_n$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
                      $endgroup$
                      – Noah Schweber
                      4 hours ago














                    3












                    3








                    3





                    $begingroup$

                    The answer is no. Consider the following open cover of $Bbb R$:



                    $$mathcal U={ (H_{2n} , H_{2n+4}) | n in Bbb N_{ge 1}} cup { (-H_{2n+4} , -H_{2n}) | n in Bbb N_{ge 1}} cup { (-3, 3) }$$
                    where $$H_n = sum_{j=1}^n frac 1j$$
                    is the $n$-th harmonic number. Note that for all $varepsilon >0$ only finitely many elements $U in mathcal U$ satisfy $lambda (U) > varepsilon$.



                    Let ${ U_{n} }_{n ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= lambda (U_n)$ converges to $0$.



                    Let ${ r_n }_{n ge 1}$ be an enumeration of rational numbers such that for all $n$
                    $$r_{n} in U_n$$
                    holds. Recall that $a_n$ is rational, and it converges to $0$.



                    Now, for all $x in Bbb R$ there exists $n ge 1$ such that $x in U_n$: thus $|x-r_n| < a_n$.






                    share|cite|improve this answer











                    $endgroup$



                    The answer is no. Consider the following open cover of $Bbb R$:



                    $$mathcal U={ (H_{2n} , H_{2n+4}) | n in Bbb N_{ge 1}} cup { (-H_{2n+4} , -H_{2n}) | n in Bbb N_{ge 1}} cup { (-3, 3) }$$
                    where $$H_n = sum_{j=1}^n frac 1j$$
                    is the $n$-th harmonic number. Note that for all $varepsilon >0$ only finitely many elements $U in mathcal U$ satisfy $lambda (U) > varepsilon$.



                    Let ${ U_{n} }_{n ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= lambda (U_n)$ converges to $0$.



                    Let ${ r_n }_{n ge 1}$ be an enumeration of rational numbers such that for all $n$
                    $$r_{n} in U_n$$
                    holds. Recall that $a_n$ is rational, and it converges to $0$.



                    Now, for all $x in Bbb R$ there exists $n ge 1$ such that $x in U_n$: thus $|x-r_n| < a_n$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 4 hours ago

























                    answered 4 hours ago









                    CrostulCrostul

                    27.7k22352




                    27.7k22352












                    • $begingroup$
                      Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
                      $endgroup$
                      – Noah Schweber
                      4 hours ago


















                    • $begingroup$
                      Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
                      $endgroup$
                      – Noah Schweber
                      4 hours ago
















                    $begingroup$
                    Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
                    $endgroup$
                    – Noah Schweber
                    4 hours ago




                    $begingroup$
                    Note that this just shows that there is an enumeration of rationals with an "unsatisfiable" sequence - it doesn't immediately imply that every enumeration of rationals has such an unsatisfiable sequence.
                    $endgroup$
                    – Noah Schweber
                    4 hours ago


















                    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%2f3076002%2fdoes-there-exist-a-real-number-a-given-distance-from-each-rational-number%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

                    Magento 2 controller redirect on button click in phtml file

                    Polycentropodidae