Nested sequence of half open intervals with non-empty intersection











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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).






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    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


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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






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      edited Nov 27 at 12:48

























      asked Nov 27 at 12:45









      Jens Wagemaker

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          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






























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            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





















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              2 Answers
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              active

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              2 Answers
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              active

              oldest

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              active

              oldest

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              active

              oldest

              votes








              up vote
              0
              down vote



              accepted










              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



























                up vote
                0
                down vote



                accepted










                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

























                  up vote
                  0
                  down vote



                  accepted







                  up vote
                  0
                  down vote



                  accepted






                  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

                  89.4k393161




                  89.4k393161






















                      up vote
                      0
                      down vote













                      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

























                        up vote
                        0
                        down vote













                        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























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          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






























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