Apostol Proof for Cantor Intersection Theorem












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$begingroup$


I am trying to understand the proof for the following theorem from Apostol:



enter image description here



Here is the proof:



enter image description here



I don't understand the parts underlined in red. For the first part, why is it trivial if each $Q$ is finite? For the second part, how does $Q$ contain all points but possibly a finite amount of points of $A$? Does it have something to do with the first condition $Q_{k+1}subseteq Q_{k}$?






real-analysis proof-explanation







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    0












    $begingroup$


    I am trying to understand the proof for the following theorem from Apostol:



    enter image description here



    Here is the proof:



    enter image description here



    I don't understand the parts underlined in red. For the first part, why is it trivial if each $Q$ is finite? For the second part, how does $Q$ contain all points but possibly a finite amount of points of $A$? Does it have something to do with the first condition $Q_{k+1}subseteq Q_{k}$?






    real-analysis proof-explanation







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      I am trying to understand the proof for the following theorem from Apostol:



      enter image description here



      Here is the proof:



      enter image description here



      I don't understand the parts underlined in red. For the first part, why is it trivial if each $Q$ is finite? For the second part, how does $Q$ contain all points but possibly a finite amount of points of $A$? Does it have something to do with the first condition $Q_{k+1}subseteq Q_{k}$?






      real-analysis proof-explanation







      share|cite|improve this question











      $endgroup$




      I am trying to understand the proof for the following theorem from Apostol:



      enter image description here



      Here is the proof:



      enter image description here



      I don't understand the parts underlined in red. For the first part, why is it trivial if each $Q$ is finite? For the second part, how does $Q$ contain all points but possibly a finite amount of points of $A$? Does it have something to do with the first condition $Q_{k+1}subseteq Q_{k}$?






      real-analysis proof-explanation




      real-analysis proof-explanation






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 19:00









      Andrés E. Caicedo

      65.6k8159250




      65.6k8159250










      asked Jan 10 at 18:36









      numericalorangenumericalorange

      1,775311




      1,775311






















          2 Answers
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          $begingroup$


          1. Suppose that $Q_i$ only has $n$ elements. Since $Q_isupset Q_{i+1}supset Q_{i+2}supsetcdots$, you have$$n=#Q_igeqslant#Q_{i+1}geqslant#Q_{i+2}geqslantcdots$$So, you have a decreasing sequence of elements of ${1,2,ldots,n}$. But then the sequence $#Q_i,#Q_{i+1},#Q_{i+2},ldots$ must becaom stable after some point. So, for some $Ngeqslant i$, you have $Q_N=Q_{N+1}=Q_{N+2}=cdots$ and therefore $bigcap_{ninmathbb N}Q_n=Q_Nneqemptyset$.

          2. Yes, it has to do with that. Note that $x_1$ belongs to every $Q_n$, that $x_2$ belongs to every $Q_n$ except perhaps for $Q_1$, then $x_3$ belongs to every $Q_n$ except perhaps for $Q_1$ and $Q_2$ and so on. So, for each $Q_n$, all but possibly a finite number of elements of $A$ belong to $Q_n$.






          share|cite|improve this answer









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            $begingroup$

            If at least one $Q$ is finite, say $|Q_n|=N$, then $|Q_k|$, $kge N$, is a non-decreasing sequence of positive integers. It follows that $|Q_k|$ is constant for $k$ large enough. But then $Q_k$ is also constant for these large $k$.




            By the nesting property, $x_jin Q_k$ for all $jge k$.






            share|cite|improve this answer









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              2 Answers
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              $begingroup$


              1. Suppose that $Q_i$ only has $n$ elements. Since $Q_isupset Q_{i+1}supset Q_{i+2}supsetcdots$, you have$$n=#Q_igeqslant#Q_{i+1}geqslant#Q_{i+2}geqslantcdots$$So, you have a decreasing sequence of elements of ${1,2,ldots,n}$. But then the sequence $#Q_i,#Q_{i+1},#Q_{i+2},ldots$ must becaom stable after some point. So, for some $Ngeqslant i$, you have $Q_N=Q_{N+1}=Q_{N+2}=cdots$ and therefore $bigcap_{ninmathbb N}Q_n=Q_Nneqemptyset$.

              2. Yes, it has to do with that. Note that $x_1$ belongs to every $Q_n$, that $x_2$ belongs to every $Q_n$ except perhaps for $Q_1$, then $x_3$ belongs to every $Q_n$ except perhaps for $Q_1$ and $Q_2$ and so on. So, for each $Q_n$, all but possibly a finite number of elements of $A$ belong to $Q_n$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$


                1. Suppose that $Q_i$ only has $n$ elements. Since $Q_isupset Q_{i+1}supset Q_{i+2}supsetcdots$, you have$$n=#Q_igeqslant#Q_{i+1}geqslant#Q_{i+2}geqslantcdots$$So, you have a decreasing sequence of elements of ${1,2,ldots,n}$. But then the sequence $#Q_i,#Q_{i+1},#Q_{i+2},ldots$ must becaom stable after some point. So, for some $Ngeqslant i$, you have $Q_N=Q_{N+1}=Q_{N+2}=cdots$ and therefore $bigcap_{ninmathbb N}Q_n=Q_Nneqemptyset$.

                2. Yes, it has to do with that. Note that $x_1$ belongs to every $Q_n$, that $x_2$ belongs to every $Q_n$ except perhaps for $Q_1$, then $x_3$ belongs to every $Q_n$ except perhaps for $Q_1$ and $Q_2$ and so on. So, for each $Q_n$, all but possibly a finite number of elements of $A$ belong to $Q_n$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$


                  1. Suppose that $Q_i$ only has $n$ elements. Since $Q_isupset Q_{i+1}supset Q_{i+2}supsetcdots$, you have$$n=#Q_igeqslant#Q_{i+1}geqslant#Q_{i+2}geqslantcdots$$So, you have a decreasing sequence of elements of ${1,2,ldots,n}$. But then the sequence $#Q_i,#Q_{i+1},#Q_{i+2},ldots$ must becaom stable after some point. So, for some $Ngeqslant i$, you have $Q_N=Q_{N+1}=Q_{N+2}=cdots$ and therefore $bigcap_{ninmathbb N}Q_n=Q_Nneqemptyset$.

                  2. Yes, it has to do with that. Note that $x_1$ belongs to every $Q_n$, that $x_2$ belongs to every $Q_n$ except perhaps for $Q_1$, then $x_3$ belongs to every $Q_n$ except perhaps for $Q_1$ and $Q_2$ and so on. So, for each $Q_n$, all but possibly a finite number of elements of $A$ belong to $Q_n$.






                  share|cite|improve this answer









                  $endgroup$




                  1. Suppose that $Q_i$ only has $n$ elements. Since $Q_isupset Q_{i+1}supset Q_{i+2}supsetcdots$, you have$$n=#Q_igeqslant#Q_{i+1}geqslant#Q_{i+2}geqslantcdots$$So, you have a decreasing sequence of elements of ${1,2,ldots,n}$. But then the sequence $#Q_i,#Q_{i+1},#Q_{i+2},ldots$ must becaom stable after some point. So, for some $Ngeqslant i$, you have $Q_N=Q_{N+1}=Q_{N+2}=cdots$ and therefore $bigcap_{ninmathbb N}Q_n=Q_Nneqemptyset$.

                  2. Yes, it has to do with that. Note that $x_1$ belongs to every $Q_n$, that $x_2$ belongs to every $Q_n$ except perhaps for $Q_1$, then $x_3$ belongs to every $Q_n$ except perhaps for $Q_1$ and $Q_2$ and so on. So, for each $Q_n$, all but possibly a finite number of elements of $A$ belong to $Q_n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 18:49









                  José Carlos SantosJosé Carlos Santos

                  165k22132235




                  165k22132235























                      1












                      $begingroup$

                      If at least one $Q$ is finite, say $|Q_n|=N$, then $|Q_k|$, $kge N$, is a non-decreasing sequence of positive integers. It follows that $|Q_k|$ is constant for $k$ large enough. But then $Q_k$ is also constant for these large $k$.




                      By the nesting property, $x_jin Q_k$ for all $jge k$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        If at least one $Q$ is finite, say $|Q_n|=N$, then $|Q_k|$, $kge N$, is a non-decreasing sequence of positive integers. It follows that $|Q_k|$ is constant for $k$ large enough. But then $Q_k$ is also constant for these large $k$.




                        By the nesting property, $x_jin Q_k$ for all $jge k$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          If at least one $Q$ is finite, say $|Q_n|=N$, then $|Q_k|$, $kge N$, is a non-decreasing sequence of positive integers. It follows that $|Q_k|$ is constant for $k$ large enough. But then $Q_k$ is also constant for these large $k$.




                          By the nesting property, $x_jin Q_k$ for all $jge k$.






                          share|cite|improve this answer









                          $endgroup$



                          If at least one $Q$ is finite, say $|Q_n|=N$, then $|Q_k|$, $kge N$, is a non-decreasing sequence of positive integers. It follows that $|Q_k|$ is constant for $k$ large enough. But then $Q_k$ is also constant for these large $k$.




                          By the nesting property, $x_jin Q_k$ for all $jge k$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 10 at 18:49









                          Hagen von EitzenHagen von Eitzen

                          282k23272505




                          282k23272505






























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