Nested sequence of half open intervals with non-empty intersection












1














Consider the sets $B_{2^n}^k = [frac{k}{2^n},frac{k+1}{2^n})$ with $n in mathbb{N}$ and $k in mathbb{Z}$.
Now we pick a sequence $(k_n)_{n in mathbb{N}}$ such that we get a nested sequence $B_{2^1}^{k_1} supset B_{2^2}^{k_2} supset dots$.
I suspect that the intersection $underset{{n in mathbb{N}}}{cap}B_{2^n}^{k_n} = {x }$, i.e. is a singleton.
As this sequence resembles the binary representation of a real number. For your intuition of this sequence: at each step we half the needle and pick either the right or the left part.



Problem is that I don't know how to prove it, as we cannot use the nested interval theorem (which only works for nested compact sets).






real-analysis general-topology real-numbers cantor-set







share|cite|improve this question





























    1














    Consider the sets $B_{2^n}^k = [frac{k}{2^n},frac{k+1}{2^n})$ with $n in mathbb{N}$ and $k in mathbb{Z}$.
    Now we pick a sequence $(k_n)_{n in mathbb{N}}$ such that we get a nested sequence $B_{2^1}^{k_1} supset B_{2^2}^{k_2} supset dots$.
    I suspect that the intersection $underset{{n in mathbb{N}}}{cap}B_{2^n}^{k_n} = {x }$, i.e. is a singleton.
    As this sequence resembles the binary representation of a real number. For your intuition of this sequence: at each step we half the needle and pick either the right or the left part.



    Problem is that I don't know how to prove it, as we cannot use the nested interval theorem (which only works for nested compact sets).






    real-analysis general-topology real-numbers cantor-set







    share|cite|improve this question



























      1












      1








      1


      0





      Consider the sets $B_{2^n}^k = [frac{k}{2^n},frac{k+1}{2^n})$ with $n in mathbb{N}$ and $k in mathbb{Z}$.
      Now we pick a sequence $(k_n)_{n in mathbb{N}}$ such that we get a nested sequence $B_{2^1}^{k_1} supset B_{2^2}^{k_2} supset dots$.
      I suspect that the intersection $underset{{n in mathbb{N}}}{cap}B_{2^n}^{k_n} = {x }$, i.e. is a singleton.
      As this sequence resembles the binary representation of a real number. For your intuition of this sequence: at each step we half the needle and pick either the right or the left part.



      Problem is that I don't know how to prove it, as we cannot use the nested interval theorem (which only works for nested compact sets).






      real-analysis general-topology real-numbers cantor-set







      share|cite|improve this question















      Consider the sets $B_{2^n}^k = [frac{k}{2^n},frac{k+1}{2^n})$ with $n in mathbb{N}$ and $k in mathbb{Z}$.
      Now we pick a sequence $(k_n)_{n in mathbb{N}}$ such that we get a nested sequence $B_{2^1}^{k_1} supset B_{2^2}^{k_2} supset dots$.
      I suspect that the intersection $underset{{n in mathbb{N}}}{cap}B_{2^n}^{k_n} = {x }$, i.e. is a singleton.
      As this sequence resembles the binary representation of a real number. For your intuition of this sequence: at each step we half the needle and pick either the right or the left part.



      Problem is that I don't know how to prove it, as we cannot use the nested interval theorem (which only works for nested compact sets).






      real-analysis general-topology real-numbers cantor-set




      real-analysis general-topology real-numbers cantor-set






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 at 12:48

























      asked Nov 27 at 12:45









      Jens Wagemaker

      540311




      540311






















          2 Answers
          2






          active

          oldest

          votes


















          0














          The intersection doesn't need to be a singleton. for example, taking $k_n=-1$ for all $ninmathbb N$, we see that:



          $$B_{2^n}^{k_n} = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right) = left[-frac{1}{2^n}, 0right)supsetleft[-frac{1}{2^{n+1}}, 0right) = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right)=B_{2^{n+1}}^{k_{n+1}}$$



          so your condition is met, however the intersection



          $$bigcap_{n=1}^infty B_{2^n}^{k_n} = bigcap_{n=1}^infty left[-frac{1}{2^n}, 0right)=emptyset$$



          is empty.




          What you can prove is that the intersection contains at most one element. The idea of the proof is that if it contains two distinct elements, you can reach a contradiction since these two elements $a,b$ must be elements of an interval with an arbitrarily small length, e.g. with a length of $frac{a+b}{2}$, which is impossible.






          share|cite|improve this answer































            0














            This is not necessarily true, by the same argument as usual counterexamples to the nested interval theorem without closedness:



            If you take $k := 2^n-1$, you get an empty intersection (intervals approach 1 from below, but it is not contained), corresponding to the binary representation $0.111... = 1$.






            share|cite|improve this answer





















              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%2f3015727%2fnested-sequence-of-half-open-intervals-with-non-empty-intersection%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









              0














              The intersection doesn't need to be a singleton. for example, taking $k_n=-1$ for all $ninmathbb N$, we see that:



              $$B_{2^n}^{k_n} = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right) = left[-frac{1}{2^n}, 0right)supsetleft[-frac{1}{2^{n+1}}, 0right) = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right)=B_{2^{n+1}}^{k_{n+1}}$$



              so your condition is met, however the intersection



              $$bigcap_{n=1}^infty B_{2^n}^{k_n} = bigcap_{n=1}^infty left[-frac{1}{2^n}, 0right)=emptyset$$



              is empty.




              What you can prove is that the intersection contains at most one element. The idea of the proof is that if it contains two distinct elements, you can reach a contradiction since these two elements $a,b$ must be elements of an interval with an arbitrarily small length, e.g. with a length of $frac{a+b}{2}$, which is impossible.






              share|cite|improve this answer




























                0














                The intersection doesn't need to be a singleton. for example, taking $k_n=-1$ for all $ninmathbb N$, we see that:



                $$B_{2^n}^{k_n} = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right) = left[-frac{1}{2^n}, 0right)supsetleft[-frac{1}{2^{n+1}}, 0right) = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right)=B_{2^{n+1}}^{k_{n+1}}$$



                so your condition is met, however the intersection



                $$bigcap_{n=1}^infty B_{2^n}^{k_n} = bigcap_{n=1}^infty left[-frac{1}{2^n}, 0right)=emptyset$$



                is empty.




                What you can prove is that the intersection contains at most one element. The idea of the proof is that if it contains two distinct elements, you can reach a contradiction since these two elements $a,b$ must be elements of an interval with an arbitrarily small length, e.g. with a length of $frac{a+b}{2}$, which is impossible.






                share|cite|improve this answer


























                  0












                  0








                  0






                  The intersection doesn't need to be a singleton. for example, taking $k_n=-1$ for all $ninmathbb N$, we see that:



                  $$B_{2^n}^{k_n} = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right) = left[-frac{1}{2^n}, 0right)supsetleft[-frac{1}{2^{n+1}}, 0right) = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right)=B_{2^{n+1}}^{k_{n+1}}$$



                  so your condition is met, however the intersection



                  $$bigcap_{n=1}^infty B_{2^n}^{k_n} = bigcap_{n=1}^infty left[-frac{1}{2^n}, 0right)=emptyset$$



                  is empty.




                  What you can prove is that the intersection contains at most one element. The idea of the proof is that if it contains two distinct elements, you can reach a contradiction since these two elements $a,b$ must be elements of an interval with an arbitrarily small length, e.g. with a length of $frac{a+b}{2}$, which is impossible.






                  share|cite|improve this answer














                  The intersection doesn't need to be a singleton. for example, taking $k_n=-1$ for all $ninmathbb N$, we see that:



                  $$B_{2^n}^{k_n} = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right) = left[-frac{1}{2^n}, 0right)supsetleft[-frac{1}{2^{n+1}}, 0right) = left[frac{k_n}{2^n}, frac{k_n+1}{2^n}right)=B_{2^{n+1}}^{k_{n+1}}$$



                  so your condition is met, however the intersection



                  $$bigcap_{n=1}^infty B_{2^n}^{k_n} = bigcap_{n=1}^infty left[-frac{1}{2^n}, 0right)=emptyset$$



                  is empty.




                  What you can prove is that the intersection contains at most one element. The idea of the proof is that if it contains two distinct elements, you can reach a contradiction since these two elements $a,b$ must be elements of an interval with an arbitrarily small length, e.g. with a length of $frac{a+b}{2}$, which is impossible.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 27 at 13:02

























                  answered Nov 27 at 12:51









                  5xum

                  89.5k393161




                  89.5k393161























                      0














                      This is not necessarily true, by the same argument as usual counterexamples to the nested interval theorem without closedness:



                      If you take $k := 2^n-1$, you get an empty intersection (intervals approach 1 from below, but it is not contained), corresponding to the binary representation $0.111... = 1$.






                      share|cite|improve this answer


























                        0














                        This is not necessarily true, by the same argument as usual counterexamples to the nested interval theorem without closedness:



                        If you take $k := 2^n-1$, you get an empty intersection (intervals approach 1 from below, but it is not contained), corresponding to the binary representation $0.111... = 1$.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          This is not necessarily true, by the same argument as usual counterexamples to the nested interval theorem without closedness:



                          If you take $k := 2^n-1$, you get an empty intersection (intervals approach 1 from below, but it is not contained), corresponding to the binary representation $0.111... = 1$.






                          share|cite|improve this answer












                          This is not necessarily true, by the same argument as usual counterexamples to the nested interval theorem without closedness:



                          If you take $k := 2^n-1$, you get an empty intersection (intervals approach 1 from below, but it is not contained), corresponding to the binary representation $0.111... = 1$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 27 at 12:53









                          Gnampfissimo

                          18011




                          18011






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f3015727%2fnested-sequence-of-half-open-intervals-with-non-empty-intersection%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

                              How do I know what Microsoft account the skydrive app is syncing to?

                              Image
                              up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
                              Read more

                              Grease: Live!

                              Image
                              A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
                              Read more

                              When does type information flow backwards in C++?

                              Image
                              up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
                              Read more