Nested sequence of closed sets












4












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Is it true that every nested sequence of non-empty closed sets $(I_n)$ (one such that $I_{n+1} subset I_n$) has a non-empty intersection?






real-analysis general-topology







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  • $begingroup$
    Closed sets where? In what topological space?
    $endgroup$
    – Asaf Karagila♦
    Sep 10 '13 at 7:55






  • 3




    $begingroup$
    I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =emptyset$ for large enough $n$.
    $endgroup$
    – Rhys
    Sep 10 '13 at 7:56


















4












$begingroup$


Is it true that every nested sequence of non-empty closed sets $(I_n)$ (one such that $I_{n+1} subset I_n$) has a non-empty intersection?






real-analysis general-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Closed sets where? In what topological space?
    $endgroup$
    – Asaf Karagila♦
    Sep 10 '13 at 7:55






  • 3




    $begingroup$
    I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =emptyset$ for large enough $n$.
    $endgroup$
    – Rhys
    Sep 10 '13 at 7:56
















4












4








4


1



$begingroup$


Is it true that every nested sequence of non-empty closed sets $(I_n)$ (one such that $I_{n+1} subset I_n$) has a non-empty intersection?






real-analysis general-topology







share|cite|improve this question











$endgroup$




Is it true that every nested sequence of non-empty closed sets $(I_n)$ (one such that $I_{n+1} subset I_n$) has a non-empty intersection?






real-analysis general-topology




real-analysis general-topology






share|cite|improve this question















share|cite|improve this question













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edited Sep 10 '13 at 7:58







user94055

















asked Sep 10 '13 at 7:51









user94055user94055

2313




2313












  • $begingroup$
    Closed sets where? In what topological space?
    $endgroup$
    – Asaf Karagila♦
    Sep 10 '13 at 7:55






  • 3




    $begingroup$
    I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =emptyset$ for large enough $n$.
    $endgroup$
    – Rhys
    Sep 10 '13 at 7:56




















  • $begingroup$
    Closed sets where? In what topological space?
    $endgroup$
    – Asaf Karagila♦
    Sep 10 '13 at 7:55






  • 3




    $begingroup$
    I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =emptyset$ for large enough $n$.
    $endgroup$
    – Rhys
    Sep 10 '13 at 7:56


















$begingroup$
Closed sets where? In what topological space?
$endgroup$
– Asaf Karagila♦
Sep 10 '13 at 7:55




$begingroup$
Closed sets where? In what topological space?
$endgroup$
– Asaf Karagila♦
Sep 10 '13 at 7:55




3




3




$begingroup$
I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =emptyset$ for large enough $n$.
$endgroup$
– Rhys
Sep 10 '13 at 7:56






$begingroup$
I added 'non-empty' to the question, because otherwise there is the trivial counter-example where $I_n =emptyset$ for large enough $n$.
$endgroup$
– Rhys
Sep 10 '13 at 7:56












3 Answers
3






active

oldest

votes


















4












$begingroup$

If $X$ is a compact space, this is true. $mathcal{F} = {I_alpha}_{alpha in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $mathcal{F}$ is nonempty, then $bigcap_{alpha in I} I_alpha$ is nonempty.



To prove this, suppose $bigcap_{alpha in I} I_alpha$ is empty. Then ${X - I_alpha}$ is an open cover of $X$. Since $X$ is compact, there exists $alpha_1, ..., alpha_n$ such that $bigcup_{j = 1}^n (X - I_{alpha_j}) = X$. Then $bigcap_{j = 1}^n I_{alpha_j} = emptyset$. This contradicts the assumption that any finite intersection of elements of $mathcal{F}$ is nonempty.



For example (in your case) if $mathcal{F}$ consists of nonempty nested intervals, then $mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    No: take $I_n=[n,to)={xinBbb R:xge n}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $nge m$, and the intersection will be non-empty.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you use compactness to prove that?
      $endgroup$
      – user94055
      Sep 10 '13 at 7:54






    • 2




      $begingroup$
      @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
      $endgroup$
      – Brian M. Scott
      Sep 10 '13 at 7:57












    • $begingroup$
      @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
      $endgroup$
      – M.Sina
      Sep 10 '13 at 8:08








    • 1




      $begingroup$
      @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
      $endgroup$
      – Brian M. Scott
      Sep 10 '13 at 8:20






    • 1




      $begingroup$
      @BrianM.Scott: thanks. good night :).
      $endgroup$
      – M.Sina
      Sep 10 '13 at 8:27



















    2












    $begingroup$

    I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $mathbb{R}$ do not necessarily have a nonempty infinite intersection:



    $textbf{Claim 1}$: The set $A_{n} = (-infty, -n]$ is closed, $nin mathbb{N}$.



    $textit{proof}$: Let $x$ be a limit point for $A_{n}Rightarrow$ if we had $x>-n,$ choose $epsilon = x+nRightarrow x-epsilon=-n<displaystyle x-frac{epsilon}{2}Rightarrow
    V_{epsilon/2}(x)cap A_{n}=emptyset Rightarrow $     contradiction $Rightarrow xin A_{n}$.



    $textbf{Claim 2}$: $A_{n+1}subseteq A_{n}forall n$.



    $textit{Proof}$: $xin A_{n+1}Rightarrow xleq-n-1<-nRightarrow xin A_{n}$.



    $textbf{Claim 3}$: The sequence $A_{n}supset A_{n+1}, A_{n}=(-infty,-n] forall n$ has
    empty infinite intersection.



    $textit{Proof}: bigcap_{n=1}^{infty}A_{n}neq emptyset Rightarrow$ choose $xin
    bigcap_{n=1}^{infty}A_{n}Rightarrow xleq -nhspace{.5pc}forall nRightarrow -xgeq nhspace{.5pc}forall nRightarrow$     contradiction ($mathbb{N}$ is not bounded above)



    $textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.



    $underline{Legend}$:




    1. $V_{epsilon}(x)=(x-epsilon,x+epsilon)$

    2. Infinite intersection of $A_{n}=bigcap_{n=1}^{infty}A_{n}$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      4












      $begingroup$

      If $X$ is a compact space, this is true. $mathcal{F} = {I_alpha}_{alpha in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $mathcal{F}$ is nonempty, then $bigcap_{alpha in I} I_alpha$ is nonempty.



      To prove this, suppose $bigcap_{alpha in I} I_alpha$ is empty. Then ${X - I_alpha}$ is an open cover of $X$. Since $X$ is compact, there exists $alpha_1, ..., alpha_n$ such that $bigcup_{j = 1}^n (X - I_{alpha_j}) = X$. Then $bigcap_{j = 1}^n I_{alpha_j} = emptyset$. This contradicts the assumption that any finite intersection of elements of $mathcal{F}$ is nonempty.



      For example (in your case) if $mathcal{F}$ consists of nonempty nested intervals, then $mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        If $X$ is a compact space, this is true. $mathcal{F} = {I_alpha}_{alpha in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $mathcal{F}$ is nonempty, then $bigcap_{alpha in I} I_alpha$ is nonempty.



        To prove this, suppose $bigcap_{alpha in I} I_alpha$ is empty. Then ${X - I_alpha}$ is an open cover of $X$. Since $X$ is compact, there exists $alpha_1, ..., alpha_n$ such that $bigcup_{j = 1}^n (X - I_{alpha_j}) = X$. Then $bigcap_{j = 1}^n I_{alpha_j} = emptyset$. This contradicts the assumption that any finite intersection of elements of $mathcal{F}$ is nonempty.



        For example (in your case) if $mathcal{F}$ consists of nonempty nested intervals, then $mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          If $X$ is a compact space, this is true. $mathcal{F} = {I_alpha}_{alpha in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $mathcal{F}$ is nonempty, then $bigcap_{alpha in I} I_alpha$ is nonempty.



          To prove this, suppose $bigcap_{alpha in I} I_alpha$ is empty. Then ${X - I_alpha}$ is an open cover of $X$. Since $X$ is compact, there exists $alpha_1, ..., alpha_n$ such that $bigcup_{j = 1}^n (X - I_{alpha_j}) = X$. Then $bigcap_{j = 1}^n I_{alpha_j} = emptyset$. This contradicts the assumption that any finite intersection of elements of $mathcal{F}$ is nonempty.



          For example (in your case) if $mathcal{F}$ consists of nonempty nested intervals, then $mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.






          share|cite|improve this answer









          $endgroup$



          If $X$ is a compact space, this is true. $mathcal{F} = {I_alpha}_{alpha in I}$ is an arbitrary collection of nonempty closed sets such that any intersection of finitely many sets from $mathcal{F}$ is nonempty, then $bigcap_{alpha in I} I_alpha$ is nonempty.



          To prove this, suppose $bigcap_{alpha in I} I_alpha$ is empty. Then ${X - I_alpha}$ is an open cover of $X$. Since $X$ is compact, there exists $alpha_1, ..., alpha_n$ such that $bigcup_{j = 1}^n (X - I_{alpha_j}) = X$. Then $bigcap_{j = 1}^n I_{alpha_j} = emptyset$. This contradicts the assumption that any finite intersection of elements of $mathcal{F}$ is nonempty.



          For example (in your case) if $mathcal{F}$ consists of nonempty nested intervals, then $mathcal{F}$ has the finite intersection property and the above proof assured that the infinite intersection is nonempty.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 10 '13 at 8:10









          WilliamWilliam

          17.3k22256




          17.3k22256























              5












              $begingroup$

              No: take $I_n=[n,to)={xinBbb R:xge n}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $nge m$, and the intersection will be non-empty.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How do you use compactness to prove that?
                $endgroup$
                – user94055
                Sep 10 '13 at 7:54






              • 2




                $begingroup$
                @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 7:57












              • $begingroup$
                @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:08








              • 1




                $begingroup$
                @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 8:20






              • 1




                $begingroup$
                @BrianM.Scott: thanks. good night :).
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:27
















              5












              $begingroup$

              No: take $I_n=[n,to)={xinBbb R:xge n}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $nge m$, and the intersection will be non-empty.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How do you use compactness to prove that?
                $endgroup$
                – user94055
                Sep 10 '13 at 7:54






              • 2




                $begingroup$
                @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 7:57












              • $begingroup$
                @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:08








              • 1




                $begingroup$
                @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 8:20






              • 1




                $begingroup$
                @BrianM.Scott: thanks. good night :).
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:27














              5












              5








              5





              $begingroup$

              No: take $I_n=[n,to)={xinBbb R:xge n}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $nge m$, and the intersection will be non-empty.






              share|cite|improve this answer









              $endgroup$



              No: take $I_n=[n,to)={xinBbb R:xge n}$. If, however, some $I_m$ is compact, then $I_n$ is compact for all $nge m$, and the intersection will be non-empty.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Sep 10 '13 at 7:52









              Brian M. ScottBrian M. Scott

              459k38513916




              459k38513916












              • $begingroup$
                How do you use compactness to prove that?
                $endgroup$
                – user94055
                Sep 10 '13 at 7:54






              • 2




                $begingroup$
                @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 7:57












              • $begingroup$
                @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:08








              • 1




                $begingroup$
                @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 8:20






              • 1




                $begingroup$
                @BrianM.Scott: thanks. good night :).
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:27


















              • $begingroup$
                How do you use compactness to prove that?
                $endgroup$
                – user94055
                Sep 10 '13 at 7:54






              • 2




                $begingroup$
                @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 7:57












              • $begingroup$
                @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:08








              • 1




                $begingroup$
                @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
                $endgroup$
                – Brian M. Scott
                Sep 10 '13 at 8:20






              • 1




                $begingroup$
                @BrianM.Scott: thanks. good night :).
                $endgroup$
                – M.Sina
                Sep 10 '13 at 8:27
















              $begingroup$
              How do you use compactness to prove that?
              $endgroup$
              – user94055
              Sep 10 '13 at 7:54




              $begingroup$
              How do you use compactness to prove that?
              $endgroup$
              – user94055
              Sep 10 '13 at 7:54




              2




              2




              $begingroup$
              @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
              $endgroup$
              – Brian M. Scott
              Sep 10 '13 at 7:57






              $begingroup$
              @user94055: For $nge m$ let $U_n=Bbb Rsetminus I_n$. Show that if $bigcap_{nge m}I_n=varnothing$, then ${U_n:nge m}$ is an open cover of $I_m$ with no finite subcover. You may find this post from Dan Ma’s Topology Blog helpful.
              $endgroup$
              – Brian M. Scott
              Sep 10 '13 at 7:57














              $begingroup$
              @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
              $endgroup$
              – M.Sina
              Sep 10 '13 at 8:08






              $begingroup$
              @BrianM.Scott: Can i ask you to help me about this two questions 1, 2, Thanks.
              $endgroup$
              – M.Sina
              Sep 10 '13 at 8:08






              1




              1




              $begingroup$
              @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
              $endgroup$
              – Brian M. Scott
              Sep 10 '13 at 8:20




              $begingroup$
              @M.Sina: I did see your questions, but I’m probably going to need to refresh my memory a bit before answering. I’m about to head off to bed now, but I’ll take a look tomorrow.
              $endgroup$
              – Brian M. Scott
              Sep 10 '13 at 8:20




              1




              1




              $begingroup$
              @BrianM.Scott: thanks. good night :).
              $endgroup$
              – M.Sina
              Sep 10 '13 at 8:27




              $begingroup$
              @BrianM.Scott: thanks. good night :).
              $endgroup$
              – M.Sina
              Sep 10 '13 at 8:27











              2












              $begingroup$

              I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $mathbb{R}$ do not necessarily have a nonempty infinite intersection:



              $textbf{Claim 1}$: The set $A_{n} = (-infty, -n]$ is closed, $nin mathbb{N}$.



              $textit{proof}$: Let $x$ be a limit point for $A_{n}Rightarrow$ if we had $x>-n,$ choose $epsilon = x+nRightarrow x-epsilon=-n<displaystyle x-frac{epsilon}{2}Rightarrow
              V_{epsilon/2}(x)cap A_{n}=emptyset Rightarrow $               contradiction $Rightarrow xin A_{n}$.



              $textbf{Claim 2}$: $A_{n+1}subseteq A_{n}forall n$.



              $textit{Proof}$: $xin A_{n+1}Rightarrow xleq-n-1<-nRightarrow xin A_{n}$.



              $textbf{Claim 3}$: The sequence $A_{n}supset A_{n+1}, A_{n}=(-infty,-n] forall n$ has
              empty infinite intersection.



              $textit{Proof}: bigcap_{n=1}^{infty}A_{n}neq emptyset Rightarrow$ choose $xin
              bigcap_{n=1}^{infty}A_{n}Rightarrow xleq -nhspace{.5pc}forall nRightarrow -xgeq nhspace{.5pc}forall nRightarrow$               contradiction ($mathbb{N}$ is not bounded above)



              $textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.



              $underline{Legend}$:




              1. $V_{epsilon}(x)=(x-epsilon,x+epsilon)$

              2. Infinite intersection of $A_{n}=bigcap_{n=1}^{infty}A_{n}$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $mathbb{R}$ do not necessarily have a nonempty infinite intersection:



                $textbf{Claim 1}$: The set $A_{n} = (-infty, -n]$ is closed, $nin mathbb{N}$.



                $textit{proof}$: Let $x$ be a limit point for $A_{n}Rightarrow$ if we had $x>-n,$ choose $epsilon = x+nRightarrow x-epsilon=-n<displaystyle x-frac{epsilon}{2}Rightarrow
                V_{epsilon/2}(x)cap A_{n}=emptyset Rightarrow $                 contradiction $Rightarrow xin A_{n}$.



                $textbf{Claim 2}$: $A_{n+1}subseteq A_{n}forall n$.



                $textit{Proof}$: $xin A_{n+1}Rightarrow xleq-n-1<-nRightarrow xin A_{n}$.



                $textbf{Claim 3}$: The sequence $A_{n}supset A_{n+1}, A_{n}=(-infty,-n] forall n$ has
                empty infinite intersection.



                $textit{Proof}: bigcap_{n=1}^{infty}A_{n}neq emptyset Rightarrow$ choose $xin
                bigcap_{n=1}^{infty}A_{n}Rightarrow xleq -nhspace{.5pc}forall nRightarrow -xgeq nhspace{.5pc}forall nRightarrow$                 contradiction ($mathbb{N}$ is not bounded above)



                $textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.



                $underline{Legend}$:




                1. $V_{epsilon}(x)=(x-epsilon,x+epsilon)$

                2. Infinite intersection of $A_{n}=bigcap_{n=1}^{infty}A_{n}$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $mathbb{R}$ do not necessarily have a nonempty infinite intersection:



                  $textbf{Claim 1}$: The set $A_{n} = (-infty, -n]$ is closed, $nin mathbb{N}$.



                  $textit{proof}$: Let $x$ be a limit point for $A_{n}Rightarrow$ if we had $x>-n,$ choose $epsilon = x+nRightarrow x-epsilon=-n<displaystyle x-frac{epsilon}{2}Rightarrow
                  V_{epsilon/2}(x)cap A_{n}=emptyset Rightarrow $                   contradiction $Rightarrow xin A_{n}$.



                  $textbf{Claim 2}$: $A_{n+1}subseteq A_{n}forall n$.



                  $textit{Proof}$: $xin A_{n+1}Rightarrow xleq-n-1<-nRightarrow xin A_{n}$.



                  $textbf{Claim 3}$: The sequence $A_{n}supset A_{n+1}, A_{n}=(-infty,-n] forall n$ has
                  empty infinite intersection.



                  $textit{Proof}: bigcap_{n=1}^{infty}A_{n}neq emptyset Rightarrow$ choose $xin
                  bigcap_{n=1}^{infty}A_{n}Rightarrow xleq -nhspace{.5pc}forall nRightarrow -xgeq nhspace{.5pc}forall nRightarrow$                   contradiction ($mathbb{N}$ is not bounded above)



                  $textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.



                  $underline{Legend}$:




                  1. $V_{epsilon}(x)=(x-epsilon,x+epsilon)$

                  2. Infinite intersection of $A_{n}=bigcap_{n=1}^{infty}A_{n}$






                  share|cite|improve this answer









                  $endgroup$



                  I'm studying intro to analysis and I got that a nested sequence of closed and nonempty subsets of $mathbb{R}$ do not necessarily have a nonempty infinite intersection:



                  $textbf{Claim 1}$: The set $A_{n} = (-infty, -n]$ is closed, $nin mathbb{N}$.



                  $textit{proof}$: Let $x$ be a limit point for $A_{n}Rightarrow$ if we had $x>-n,$ choose $epsilon = x+nRightarrow x-epsilon=-n<displaystyle x-frac{epsilon}{2}Rightarrow
                  V_{epsilon/2}(x)cap A_{n}=emptyset Rightarrow $                   contradiction $Rightarrow xin A_{n}$.



                  $textbf{Claim 2}$: $A_{n+1}subseteq A_{n}forall n$.



                  $textit{Proof}$: $xin A_{n+1}Rightarrow xleq-n-1<-nRightarrow xin A_{n}$.



                  $textbf{Claim 3}$: The sequence $A_{n}supset A_{n+1}, A_{n}=(-infty,-n] forall n$ has
                  empty infinite intersection.



                  $textit{Proof}: bigcap_{n=1}^{infty}A_{n}neq emptyset Rightarrow$ choose $xin
                  bigcap_{n=1}^{infty}A_{n}Rightarrow xleq -nhspace{.5pc}forall nRightarrow -xgeq nhspace{.5pc}forall nRightarrow$                   contradiction ($mathbb{N}$ is not bounded above)



                  $textbf{Conclusion}$: By claims 1,2, and 3, we have a nested sequence of closed sets with empty infinite intersection.



                  $underline{Legend}$:




                  1. $V_{epsilon}(x)=(x-epsilon,x+epsilon)$

                  2. Infinite intersection of $A_{n}=bigcap_{n=1}^{infty}A_{n}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 25 '18 at 19:15









                  JayJay

                  296




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