Understanding the Lim infimum and Lim supremum












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There are a bunch of questions I've gone through here that talk about how to solve a specific question. I am more curious on the intuition behind a particular problem. I am self studying probability theory, and am now in the analysis section.



I understand the general concepts of infinum and supremum:



$inf A_k = cap_{k=n}^infty A_k$



The intuition here is that the intersection of all possible of k-length subsequences will be the greatest lower bound.



The same goes for supremum:



$sup A_k = cup_{k=n}^infty A_k$



In this case the union of all possible k-length subsequences will be the least upper bound.



This problem however has me confused:



Check that $liminf_{n to infty} [0, frac{n}{n+1}) = limsup_{n to infty}[0, frac{n}{n+1}) = [0, 1)$



Setting out to try this I attempted to solve one part of it:



$liminf_{n to infty} [0, frac{n}{n+1})$



$lim_{n to infty} [inf_{k le n} [0, frac{n}{n+1})]$



$cup_{n = 1}^infty cap_{k=n}^infty [0, frac{n}{n+1})$



So I need to try to find the union of all intersections of the subsequences. This is where my intuition falls apart. Writing down some test sequences:



$k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



$k = 2: [0, frac{2}{3}) cap [0, frac{3}{4}) cap [0, frac{4}{5}) cap ...$



So now looking a bit deeper at $k = 1$



$[0, frac{1}{2}) cap [0, frac{2}{3})$



Would give me $[0, frac{1}{2})$ since its the only common subsequence right? Continuing down the line, it would appear eventually this repeated intersectioning would result in $[0, 1)$. This would mean that unioning all of the results of the intersectioning would also be $[0, 1)$ - right? Doing this for $k = 2$ reveals the same thing.



Am I proceeding through this question correctly? I don't have a professor available to ask...so I want to take the time to make sure I am doing this correctly.



Thank you!



EDIT: I've realized part of my interpretation was wrong:



For



$k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



This intersection will only ever result in $[0, frac{1}{2})$ because it is all the sequences here have in common. For $k = 2$ this value will be $[0, frac{2}{3})$. This continues on. Then, when you get to the unioning part you end up with:



$[0, frac{1}{2}) cup [0, frac{2}{3}) cup ...$



Which is the union of all of the k intersection results up to infinity.



etc etc. If you do this, it is obvious the union becomes $[0, 1)$ in the limit.



Is this more correct?










share|cite|improve this question











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    1












    $begingroup$


    There are a bunch of questions I've gone through here that talk about how to solve a specific question. I am more curious on the intuition behind a particular problem. I am self studying probability theory, and am now in the analysis section.



    I understand the general concepts of infinum and supremum:



    $inf A_k = cap_{k=n}^infty A_k$



    The intuition here is that the intersection of all possible of k-length subsequences will be the greatest lower bound.



    The same goes for supremum:



    $sup A_k = cup_{k=n}^infty A_k$



    In this case the union of all possible k-length subsequences will be the least upper bound.



    This problem however has me confused:



    Check that $liminf_{n to infty} [0, frac{n}{n+1}) = limsup_{n to infty}[0, frac{n}{n+1}) = [0, 1)$



    Setting out to try this I attempted to solve one part of it:



    $liminf_{n to infty} [0, frac{n}{n+1})$



    $lim_{n to infty} [inf_{k le n} [0, frac{n}{n+1})]$



    $cup_{n = 1}^infty cap_{k=n}^infty [0, frac{n}{n+1})$



    So I need to try to find the union of all intersections of the subsequences. This is where my intuition falls apart. Writing down some test sequences:



    $k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



    $k = 2: [0, frac{2}{3}) cap [0, frac{3}{4}) cap [0, frac{4}{5}) cap ...$



    So now looking a bit deeper at $k = 1$



    $[0, frac{1}{2}) cap [0, frac{2}{3})$



    Would give me $[0, frac{1}{2})$ since its the only common subsequence right? Continuing down the line, it would appear eventually this repeated intersectioning would result in $[0, 1)$. This would mean that unioning all of the results of the intersectioning would also be $[0, 1)$ - right? Doing this for $k = 2$ reveals the same thing.



    Am I proceeding through this question correctly? I don't have a professor available to ask...so I want to take the time to make sure I am doing this correctly.



    Thank you!



    EDIT: I've realized part of my interpretation was wrong:



    For



    $k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



    This intersection will only ever result in $[0, frac{1}{2})$ because it is all the sequences here have in common. For $k = 2$ this value will be $[0, frac{2}{3})$. This continues on. Then, when you get to the unioning part you end up with:



    $[0, frac{1}{2}) cup [0, frac{2}{3}) cup ...$



    Which is the union of all of the k intersection results up to infinity.



    etc etc. If you do this, it is obvious the union becomes $[0, 1)$ in the limit.



    Is this more correct?










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      There are a bunch of questions I've gone through here that talk about how to solve a specific question. I am more curious on the intuition behind a particular problem. I am self studying probability theory, and am now in the analysis section.



      I understand the general concepts of infinum and supremum:



      $inf A_k = cap_{k=n}^infty A_k$



      The intuition here is that the intersection of all possible of k-length subsequences will be the greatest lower bound.



      The same goes for supremum:



      $sup A_k = cup_{k=n}^infty A_k$



      In this case the union of all possible k-length subsequences will be the least upper bound.



      This problem however has me confused:



      Check that $liminf_{n to infty} [0, frac{n}{n+1}) = limsup_{n to infty}[0, frac{n}{n+1}) = [0, 1)$



      Setting out to try this I attempted to solve one part of it:



      $liminf_{n to infty} [0, frac{n}{n+1})$



      $lim_{n to infty} [inf_{k le n} [0, frac{n}{n+1})]$



      $cup_{n = 1}^infty cap_{k=n}^infty [0, frac{n}{n+1})$



      So I need to try to find the union of all intersections of the subsequences. This is where my intuition falls apart. Writing down some test sequences:



      $k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



      $k = 2: [0, frac{2}{3}) cap [0, frac{3}{4}) cap [0, frac{4}{5}) cap ...$



      So now looking a bit deeper at $k = 1$



      $[0, frac{1}{2}) cap [0, frac{2}{3})$



      Would give me $[0, frac{1}{2})$ since its the only common subsequence right? Continuing down the line, it would appear eventually this repeated intersectioning would result in $[0, 1)$. This would mean that unioning all of the results of the intersectioning would also be $[0, 1)$ - right? Doing this for $k = 2$ reveals the same thing.



      Am I proceeding through this question correctly? I don't have a professor available to ask...so I want to take the time to make sure I am doing this correctly.



      Thank you!



      EDIT: I've realized part of my interpretation was wrong:



      For



      $k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



      This intersection will only ever result in $[0, frac{1}{2})$ because it is all the sequences here have in common. For $k = 2$ this value will be $[0, frac{2}{3})$. This continues on. Then, when you get to the unioning part you end up with:



      $[0, frac{1}{2}) cup [0, frac{2}{3}) cup ...$



      Which is the union of all of the k intersection results up to infinity.



      etc etc. If you do this, it is obvious the union becomes $[0, 1)$ in the limit.



      Is this more correct?










      share|cite|improve this question











      $endgroup$




      There are a bunch of questions I've gone through here that talk about how to solve a specific question. I am more curious on the intuition behind a particular problem. I am self studying probability theory, and am now in the analysis section.



      I understand the general concepts of infinum and supremum:



      $inf A_k = cap_{k=n}^infty A_k$



      The intuition here is that the intersection of all possible of k-length subsequences will be the greatest lower bound.



      The same goes for supremum:



      $sup A_k = cup_{k=n}^infty A_k$



      In this case the union of all possible k-length subsequences will be the least upper bound.



      This problem however has me confused:



      Check that $liminf_{n to infty} [0, frac{n}{n+1}) = limsup_{n to infty}[0, frac{n}{n+1}) = [0, 1)$



      Setting out to try this I attempted to solve one part of it:



      $liminf_{n to infty} [0, frac{n}{n+1})$



      $lim_{n to infty} [inf_{k le n} [0, frac{n}{n+1})]$



      $cup_{n = 1}^infty cap_{k=n}^infty [0, frac{n}{n+1})$



      So I need to try to find the union of all intersections of the subsequences. This is where my intuition falls apart. Writing down some test sequences:



      $k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



      $k = 2: [0, frac{2}{3}) cap [0, frac{3}{4}) cap [0, frac{4}{5}) cap ...$



      So now looking a bit deeper at $k = 1$



      $[0, frac{1}{2}) cap [0, frac{2}{3})$



      Would give me $[0, frac{1}{2})$ since its the only common subsequence right? Continuing down the line, it would appear eventually this repeated intersectioning would result in $[0, 1)$. This would mean that unioning all of the results of the intersectioning would also be $[0, 1)$ - right? Doing this for $k = 2$ reveals the same thing.



      Am I proceeding through this question correctly? I don't have a professor available to ask...so I want to take the time to make sure I am doing this correctly.



      Thank you!



      EDIT: I've realized part of my interpretation was wrong:



      For



      $k = 1: [0, frac{1}{2}) cap [0, frac{2}{3})cap [0, frac{3}{4}) cap ...$



      This intersection will only ever result in $[0, frac{1}{2})$ because it is all the sequences here have in common. For $k = 2$ this value will be $[0, frac{2}{3})$. This continues on. Then, when you get to the unioning part you end up with:



      $[0, frac{1}{2}) cup [0, frac{2}{3}) cup ...$



      Which is the union of all of the k intersection results up to infinity.



      etc etc. If you do this, it is obvious the union becomes $[0, 1)$ in the limit.



      Is this more correct?







      real-analysis






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      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 18 at 3:46









      Andrés E. Caicedo

      66.1k8160252




      66.1k8160252










      asked Jan 17 at 20:23









      CL40CL40

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          $begingroup$

          First, here is my mental picture:



          Imagine you have a set of friends $X$. Every day some of your friends come as guests to your house. In day $n$, the set of guests $A_n$ arrive.



          Then:



          The $limsup A_n$ are those friends that you, at any given day, are garanteed to see again, at some future day.



          The $liminf A_n$ are those friends that, on some day, stop going home, they are present on each following day.



          In this case, each set the sets keep "growing", $A_{n} subseteq A_{n+1}$. Since $frac{k}{k+1} to 1 $ We get:



          $$limsup A_n = [0,1) cap [0,1) cap [0,1) cap ... = [0,1) $$



          and



          $$liminf A_n = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$.



          If, instead $A_n = (-frac{1}{n}, frac{n}{n+1}) $, then $A_{1} notsubseteq A_{2}$ since $-1 < -frac{1}{2}$ and but we have:
          $$limsup A_n = (-1, 1) cap (-frac{1}{2}, 1) cap (-frac{1}{3},1) cap ... = [0,1) $$ since all "guests" below $0$ will leave forever at some point ($0$ we are garanteed to see again, on any present $n$, and $1$ will never come). Aslo note that this is an intersection of open sets that is not open (and also not closed), although every finite intersection is open. And



          $$liminf A_n = bigcup_{n=1}^{infty} bigcap_{k = n}^{infty} (-frac{1}{k}, frac{k}{k+1}) = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$ since all "guests" below $1$ will be forever present after some point. (and $1$ will never come)






          share|cite|improve this answer











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            1












            $begingroup$

            Let $A_k = [ 0, frac{k}{k+1})$.



            We have $A_k subset A_{k+1}$, for all $k > 0$. Hence, $bigcap_{k=n}^infty A_k = A_n$. Therefore, $lim inf_{n rightarrow infty} A_n = bigcup_{n=1}^infty (bigcap_{k=n}^infty A_k) = bigcup_{n=1}^infty A_n = [0,1)$.



            Now, $bigcup_{k=n}^infty A_k = bigcup_{k=n}^infty [0,frac{k}{k+1}) = [0,1)$. Hence, $lim sup_{n rightarrow infty} A_n = bigcap_{n=1}^infty (bigcup_{k=1}^infty A_k)= bigcap_{n=1}^infty [0,1) = [0,1)$.






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              2 Answers
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              2 Answers
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              $begingroup$

              First, here is my mental picture:



              Imagine you have a set of friends $X$. Every day some of your friends come as guests to your house. In day $n$, the set of guests $A_n$ arrive.



              Then:



              The $limsup A_n$ are those friends that you, at any given day, are garanteed to see again, at some future day.



              The $liminf A_n$ are those friends that, on some day, stop going home, they are present on each following day.



              In this case, each set the sets keep "growing", $A_{n} subseteq A_{n+1}$. Since $frac{k}{k+1} to 1 $ We get:



              $$limsup A_n = [0,1) cap [0,1) cap [0,1) cap ... = [0,1) $$



              and



              $$liminf A_n = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$.



              If, instead $A_n = (-frac{1}{n}, frac{n}{n+1}) $, then $A_{1} notsubseteq A_{2}$ since $-1 < -frac{1}{2}$ and but we have:
              $$limsup A_n = (-1, 1) cap (-frac{1}{2}, 1) cap (-frac{1}{3},1) cap ... = [0,1) $$ since all "guests" below $0$ will leave forever at some point ($0$ we are garanteed to see again, on any present $n$, and $1$ will never come). Aslo note that this is an intersection of open sets that is not open (and also not closed), although every finite intersection is open. And



              $$liminf A_n = bigcup_{n=1}^{infty} bigcap_{k = n}^{infty} (-frac{1}{k}, frac{k}{k+1}) = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$ since all "guests" below $1$ will be forever present after some point. (and $1$ will never come)






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                First, here is my mental picture:



                Imagine you have a set of friends $X$. Every day some of your friends come as guests to your house. In day $n$, the set of guests $A_n$ arrive.



                Then:



                The $limsup A_n$ are those friends that you, at any given day, are garanteed to see again, at some future day.



                The $liminf A_n$ are those friends that, on some day, stop going home, they are present on each following day.



                In this case, each set the sets keep "growing", $A_{n} subseteq A_{n+1}$. Since $frac{k}{k+1} to 1 $ We get:



                $$limsup A_n = [0,1) cap [0,1) cap [0,1) cap ... = [0,1) $$



                and



                $$liminf A_n = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$.



                If, instead $A_n = (-frac{1}{n}, frac{n}{n+1}) $, then $A_{1} notsubseteq A_{2}$ since $-1 < -frac{1}{2}$ and but we have:
                $$limsup A_n = (-1, 1) cap (-frac{1}{2}, 1) cap (-frac{1}{3},1) cap ... = [0,1) $$ since all "guests" below $0$ will leave forever at some point ($0$ we are garanteed to see again, on any present $n$, and $1$ will never come). Aslo note that this is an intersection of open sets that is not open (and also not closed), although every finite intersection is open. And



                $$liminf A_n = bigcup_{n=1}^{infty} bigcap_{k = n}^{infty} (-frac{1}{k}, frac{k}{k+1}) = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$ since all "guests" below $1$ will be forever present after some point. (and $1$ will never come)






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  First, here is my mental picture:



                  Imagine you have a set of friends $X$. Every day some of your friends come as guests to your house. In day $n$, the set of guests $A_n$ arrive.



                  Then:



                  The $limsup A_n$ are those friends that you, at any given day, are garanteed to see again, at some future day.



                  The $liminf A_n$ are those friends that, on some day, stop going home, they are present on each following day.



                  In this case, each set the sets keep "growing", $A_{n} subseteq A_{n+1}$. Since $frac{k}{k+1} to 1 $ We get:



                  $$limsup A_n = [0,1) cap [0,1) cap [0,1) cap ... = [0,1) $$



                  and



                  $$liminf A_n = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$.



                  If, instead $A_n = (-frac{1}{n}, frac{n}{n+1}) $, then $A_{1} notsubseteq A_{2}$ since $-1 < -frac{1}{2}$ and but we have:
                  $$limsup A_n = (-1, 1) cap (-frac{1}{2}, 1) cap (-frac{1}{3},1) cap ... = [0,1) $$ since all "guests" below $0$ will leave forever at some point ($0$ we are garanteed to see again, on any present $n$, and $1$ will never come). Aslo note that this is an intersection of open sets that is not open (and also not closed), although every finite intersection is open. And



                  $$liminf A_n = bigcup_{n=1}^{infty} bigcap_{k = n}^{infty} (-frac{1}{k}, frac{k}{k+1}) = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$ since all "guests" below $1$ will be forever present after some point. (and $1$ will never come)






                  share|cite|improve this answer











                  $endgroup$



                  First, here is my mental picture:



                  Imagine you have a set of friends $X$. Every day some of your friends come as guests to your house. In day $n$, the set of guests $A_n$ arrive.



                  Then:



                  The $limsup A_n$ are those friends that you, at any given day, are garanteed to see again, at some future day.



                  The $liminf A_n$ are those friends that, on some day, stop going home, they are present on each following day.



                  In this case, each set the sets keep "growing", $A_{n} subseteq A_{n+1}$. Since $frac{k}{k+1} to 1 $ We get:



                  $$limsup A_n = [0,1) cap [0,1) cap [0,1) cap ... = [0,1) $$



                  and



                  $$liminf A_n = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$.



                  If, instead $A_n = (-frac{1}{n}, frac{n}{n+1}) $, then $A_{1} notsubseteq A_{2}$ since $-1 < -frac{1}{2}$ and but we have:
                  $$limsup A_n = (-1, 1) cap (-frac{1}{2}, 1) cap (-frac{1}{3},1) cap ... = [0,1) $$ since all "guests" below $0$ will leave forever at some point ($0$ we are garanteed to see again, on any present $n$, and $1$ will never come). Aslo note that this is an intersection of open sets that is not open (and also not closed), although every finite intersection is open. And



                  $$liminf A_n = bigcup_{n=1}^{infty} bigcap_{k = n}^{infty} (-frac{1}{k}, frac{k}{k+1}) = [0,frac{1}{2}) cup [0,frac{2}{3}) cup [0,frac{3}{4}) cup ... = [0,1) $$ since all "guests" below $1$ will be forever present after some point. (and $1$ will never come)







                  share|cite|improve this answer














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                  edited Jan 18 at 13:42

























                  answered Jan 18 at 12:02









                  JuliusL33tJuliusL33t

                  1,36811017




                  1,36811017























                      1












                      $begingroup$

                      Let $A_k = [ 0, frac{k}{k+1})$.



                      We have $A_k subset A_{k+1}$, for all $k > 0$. Hence, $bigcap_{k=n}^infty A_k = A_n$. Therefore, $lim inf_{n rightarrow infty} A_n = bigcup_{n=1}^infty (bigcap_{k=n}^infty A_k) = bigcup_{n=1}^infty A_n = [0,1)$.



                      Now, $bigcup_{k=n}^infty A_k = bigcup_{k=n}^infty [0,frac{k}{k+1}) = [0,1)$. Hence, $lim sup_{n rightarrow infty} A_n = bigcap_{n=1}^infty (bigcup_{k=1}^infty A_k)= bigcap_{n=1}^infty [0,1) = [0,1)$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Let $A_k = [ 0, frac{k}{k+1})$.



                        We have $A_k subset A_{k+1}$, for all $k > 0$. Hence, $bigcap_{k=n}^infty A_k = A_n$. Therefore, $lim inf_{n rightarrow infty} A_n = bigcup_{n=1}^infty (bigcap_{k=n}^infty A_k) = bigcup_{n=1}^infty A_n = [0,1)$.



                        Now, $bigcup_{k=n}^infty A_k = bigcup_{k=n}^infty [0,frac{k}{k+1}) = [0,1)$. Hence, $lim sup_{n rightarrow infty} A_n = bigcap_{n=1}^infty (bigcup_{k=1}^infty A_k)= bigcap_{n=1}^infty [0,1) = [0,1)$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let $A_k = [ 0, frac{k}{k+1})$.



                          We have $A_k subset A_{k+1}$, for all $k > 0$. Hence, $bigcap_{k=n}^infty A_k = A_n$. Therefore, $lim inf_{n rightarrow infty} A_n = bigcup_{n=1}^infty (bigcap_{k=n}^infty A_k) = bigcup_{n=1}^infty A_n = [0,1)$.



                          Now, $bigcup_{k=n}^infty A_k = bigcup_{k=n}^infty [0,frac{k}{k+1}) = [0,1)$. Hence, $lim sup_{n rightarrow infty} A_n = bigcap_{n=1}^infty (bigcup_{k=1}^infty A_k)= bigcap_{n=1}^infty [0,1) = [0,1)$.






                          share|cite|improve this answer











                          $endgroup$



                          Let $A_k = [ 0, frac{k}{k+1})$.



                          We have $A_k subset A_{k+1}$, for all $k > 0$. Hence, $bigcap_{k=n}^infty A_k = A_n$. Therefore, $lim inf_{n rightarrow infty} A_n = bigcup_{n=1}^infty (bigcap_{k=n}^infty A_k) = bigcup_{n=1}^infty A_n = [0,1)$.



                          Now, $bigcup_{k=n}^infty A_k = bigcup_{k=n}^infty [0,frac{k}{k+1}) = [0,1)$. Hence, $lim sup_{n rightarrow infty} A_n = bigcap_{n=1}^infty (bigcup_{k=1}^infty A_k)= bigcap_{n=1}^infty [0,1) = [0,1)$.







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                          edited Jan 17 at 22:32

























                          answered Jan 17 at 21:53









                          dkalocinskidkalocinski

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                          526






























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