Every compact metric space is complete












11














I need to prove that every compact metric space is complete. I think I need to use the following two facts:




  1. A set $K$ is compact if and only if every collection $mathcal{F}$ of closed subsets with finite intersection property has $bigcap{F:Finmathcal{F}}neqemptyset$.

  2. A metric space $(X,d)$ is complete if and only if for any sequence ${F_n}$ of non-empty closed sets with $F_1supset F_2supsetcdots$ and $text{diam}~F_nrightarrow0$, $bigcap_{n=1}^{infty}F_n$ contains a single point.


I do not know how to arrive at my result that every compact metric space is complete. Any help?



Thanks in advance.










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  • 1




    Closed subsets of compact spaces are compact. If $F={K_alpha}$ is a centered family of compact sets, $bigcap Fneq varnothing$. If, moreover, ${rm diam}K_nto 0$, it is easily seen $bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
    – Pedro Tamaroff
    Jan 5 '14 at 5:26












  • As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
    – Pedro Tamaroff
    Jan 5 '14 at 5:30


















11














I need to prove that every compact metric space is complete. I think I need to use the following two facts:




  1. A set $K$ is compact if and only if every collection $mathcal{F}$ of closed subsets with finite intersection property has $bigcap{F:Finmathcal{F}}neqemptyset$.

  2. A metric space $(X,d)$ is complete if and only if for any sequence ${F_n}$ of non-empty closed sets with $F_1supset F_2supsetcdots$ and $text{diam}~F_nrightarrow0$, $bigcap_{n=1}^{infty}F_n$ contains a single point.


I do not know how to arrive at my result that every compact metric space is complete. Any help?



Thanks in advance.










share|cite|improve this question




















  • 1




    Closed subsets of compact spaces are compact. If $F={K_alpha}$ is a centered family of compact sets, $bigcap Fneq varnothing$. If, moreover, ${rm diam}K_nto 0$, it is easily seen $bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
    – Pedro Tamaroff
    Jan 5 '14 at 5:26












  • As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
    – Pedro Tamaroff
    Jan 5 '14 at 5:30
















11












11








11


15





I need to prove that every compact metric space is complete. I think I need to use the following two facts:




  1. A set $K$ is compact if and only if every collection $mathcal{F}$ of closed subsets with finite intersection property has $bigcap{F:Finmathcal{F}}neqemptyset$.

  2. A metric space $(X,d)$ is complete if and only if for any sequence ${F_n}$ of non-empty closed sets with $F_1supset F_2supsetcdots$ and $text{diam}~F_nrightarrow0$, $bigcap_{n=1}^{infty}F_n$ contains a single point.


I do not know how to arrive at my result that every compact metric space is complete. Any help?



Thanks in advance.










share|cite|improve this question















I need to prove that every compact metric space is complete. I think I need to use the following two facts:




  1. A set $K$ is compact if and only if every collection $mathcal{F}$ of closed subsets with finite intersection property has $bigcap{F:Finmathcal{F}}neqemptyset$.

  2. A metric space $(X,d)$ is complete if and only if for any sequence ${F_n}$ of non-empty closed sets with $F_1supset F_2supsetcdots$ and $text{diam}~F_nrightarrow0$, $bigcap_{n=1}^{infty}F_n$ contains a single point.


I do not know how to arrive at my result that every compact metric space is complete. Any help?



Thanks in advance.







general-topology analysis metric-spaces compactness






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 14 '15 at 19:07









Asaf Karagila

301k32423755




301k32423755










asked Jan 5 '14 at 5:06









Abishanka Saha

7,79011022




7,79011022








  • 1




    Closed subsets of compact spaces are compact. If $F={K_alpha}$ is a centered family of compact sets, $bigcap Fneq varnothing$. If, moreover, ${rm diam}K_nto 0$, it is easily seen $bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
    – Pedro Tamaroff
    Jan 5 '14 at 5:26












  • As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
    – Pedro Tamaroff
    Jan 5 '14 at 5:30
















  • 1




    Closed subsets of compact spaces are compact. If $F={K_alpha}$ is a centered family of compact sets, $bigcap Fneq varnothing$. If, moreover, ${rm diam}K_nto 0$, it is easily seen $bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
    – Pedro Tamaroff
    Jan 5 '14 at 5:26












  • As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
    – Pedro Tamaroff
    Jan 5 '14 at 5:30










1




1




Closed subsets of compact spaces are compact. If $F={K_alpha}$ is a centered family of compact sets, $bigcap Fneq varnothing$. If, moreover, ${rm diam}K_nto 0$, it is easily seen $bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
– Pedro Tamaroff
Jan 5 '14 at 5:26






Closed subsets of compact spaces are compact. If $F={K_alpha}$ is a centered family of compact sets, $bigcap Fneq varnothing$. If, moreover, ${rm diam}K_nto 0$, it is easily seen $bigcap F$ must consist of only one point. Nested sets are centered. Hence the result.
– Pedro Tamaroff
Jan 5 '14 at 5:26














As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
– Pedro Tamaroff
Jan 5 '14 at 5:30






As a side note, a metric space is compact iff it is complete and totally bounded. You have a part of the implication, so maybe you're interested in finishing things off. =)
– Pedro Tamaroff
Jan 5 '14 at 5:30












6 Answers
6






active

oldest

votes


















7














Let $langle F_nrangle_{ninBbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $operatorname{diam} F_nto 0$ as $ntoinfty$. You can easily check that if $m_1<m_2<cdots<m_k$ then
$$
F_{m_1}cap F_{m_2}capcdotscap F_{m_k} =F_{m_k}neq varnothing
$$

so $langle F_nrangle_{ninBbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $bigcap_{ninBbb{N}} F_n$ is not empty. Since
$$
operatorname{diam} bigcap_{ninBbb{N}} F_n le operatorname{diam} F_mto 0qquad text{as }> mtoinfty
$$

so $bigcap_{ninBbb{N}} F_n$ contains at most one point. So $bigcap_{ninBbb{N}} F_n$ is singleton.






share|cite|improve this answer































    18














    If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:



    Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}to x in X$.



    All that now remains to be shown is that $x_n to x$. Since $x_{n_k}to x$ there is $N_1$ with $n_k ge N_1$ implies $|x_{n_k}-x|<{varepsilonover 2}$. Let $N_2$ be such that $n,mge N_2$ implies $|x_n-x_m|<{varepsilon over 2}$.



    Then $n>N=max(N_1,N_2)$ implies
    $$ |x_n-x|le |x_n-x_N|+|x_N-x|<varepsilon$$



    Hence $X$ is complete.






    share|cite|improve this answer



















    • 2




      This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
      – Santiago Canez
      Apr 23 '14 at 13:00










    • @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
      – Student
      Apr 23 '14 at 13:39












    • @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
      – Student
      Apr 24 '14 at 10:10










    • @SantiagoCanez Do you find the edited answer more palatable?
      – Student
      Apr 24 '14 at 15:46










    • Yes, this looks great, thanks.
      – Santiago Canez
      Apr 24 '14 at 16:21



















    2














    Here's a not-so-fancy way:



    Let ${a_n}$ be a Cauchy sequence.



    If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).



    If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.






    share|cite|improve this answer































      1














      This follows from Heine-Borel (see the wiki page for the relevant proofs).






      share|cite|improve this answer

















      • 2




        How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
        – Pedro Tamaroff
        Jan 5 '14 at 5:29






      • 1




        @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
        – Student
        Apr 23 '14 at 12:10



















      1














      Let $X$ be a compact metric space and let ${p_n}$ be a Cauchy sequence in $X$. Then define $E_N$ as ${p_N, p_{N+1}, p_{N+2}, ldots}$. Let $overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well.



      By definition of Cauchy sequence, we have $lim_{Ntoinfty} text{diam } E_N = lim_{Ntoinfty} text{diam } overline{E_N} = 0$. Let $E = cap_{n=1}^infty overline{E_n}$. Because $E_N supset E_{N+1}$ and $overline{E_N} supset overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $lim_{Ntoinfty} text{diam } overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p in overline{E_N}$ for all $N$. Therefore $p in X$.



      For all $epsilon > 0$, there exists an $N$ such that $text{diam } overline{E_n} < epsilon$ for all $n > N$. Thus, $d(p,q) < epsilon$ for all $q in overline{E_n}$. Since $E_n subset overline{E_n}$, we have $d(p,q) < epsilon$ for all $q in E_n$. Therefore, ${p_n}$ converges to $p in X$.






      share|cite|improve this answer





























        0














        Hint: If you have a sequence wherein $d(a_{N},a_{k})<r$ for all $kgeq N$, then the limit of that sequence, if exist, must be in the closure of the open ball radius $r$ around $a_{N}$.






        share|cite|improve this answer





















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          6 Answers
          6






          active

          oldest

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          6 Answers
          6






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

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          7














          Let $langle F_nrangle_{ninBbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $operatorname{diam} F_nto 0$ as $ntoinfty$. You can easily check that if $m_1<m_2<cdots<m_k$ then
          $$
          F_{m_1}cap F_{m_2}capcdotscap F_{m_k} =F_{m_k}neq varnothing
          $$

          so $langle F_nrangle_{ninBbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $bigcap_{ninBbb{N}} F_n$ is not empty. Since
          $$
          operatorname{diam} bigcap_{ninBbb{N}} F_n le operatorname{diam} F_mto 0qquad text{as }> mtoinfty
          $$

          so $bigcap_{ninBbb{N}} F_n$ contains at most one point. So $bigcap_{ninBbb{N}} F_n$ is singleton.






          share|cite|improve this answer




























            7














            Let $langle F_nrangle_{ninBbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $operatorname{diam} F_nto 0$ as $ntoinfty$. You can easily check that if $m_1<m_2<cdots<m_k$ then
            $$
            F_{m_1}cap F_{m_2}capcdotscap F_{m_k} =F_{m_k}neq varnothing
            $$

            so $langle F_nrangle_{ninBbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $bigcap_{ninBbb{N}} F_n$ is not empty. Since
            $$
            operatorname{diam} bigcap_{ninBbb{N}} F_n le operatorname{diam} F_mto 0qquad text{as }> mtoinfty
            $$

            so $bigcap_{ninBbb{N}} F_n$ contains at most one point. So $bigcap_{ninBbb{N}} F_n$ is singleton.






            share|cite|improve this answer


























              7












              7








              7






              Let $langle F_nrangle_{ninBbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $operatorname{diam} F_nto 0$ as $ntoinfty$. You can easily check that if $m_1<m_2<cdots<m_k$ then
              $$
              F_{m_1}cap F_{m_2}capcdotscap F_{m_k} =F_{m_k}neq varnothing
              $$

              so $langle F_nrangle_{ninBbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $bigcap_{ninBbb{N}} F_n$ is not empty. Since
              $$
              operatorname{diam} bigcap_{ninBbb{N}} F_n le operatorname{diam} F_mto 0qquad text{as }> mtoinfty
              $$

              so $bigcap_{ninBbb{N}} F_n$ contains at most one point. So $bigcap_{ninBbb{N}} F_n$ is singleton.






              share|cite|improve this answer














              Let $langle F_nrangle_{ninBbb{N}}$ be a descending sequence of nonempty closed sets satisfy that $operatorname{diam} F_nto 0$ as $ntoinfty$. You can easily check that if $m_1<m_2<cdots<m_k$ then
              $$
              F_{m_1}cap F_{m_2}capcdotscap F_{m_k} =F_{m_k}neq varnothing
              $$

              so $langle F_nrangle_{ninBbb{N}}$ satisfies finite intersection property. Since $(X,d)$ is a compact metric space, $bigcap_{ninBbb{N}} F_n$ is not empty. Since
              $$
              operatorname{diam} bigcap_{ninBbb{N}} F_n le operatorname{diam} F_mto 0qquad text{as }> mtoinfty
              $$

              so $bigcap_{ninBbb{N}} F_n$ contains at most one point. So $bigcap_{ninBbb{N}} F_n$ is singleton.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 26 at 0:29









              Robson

              771221




              771221










              answered Jan 5 '14 at 5:19









              Hanul Jeon

              17.5k42780




              17.5k42780























                  18














                  If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:



                  Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}to x in X$.



                  All that now remains to be shown is that $x_n to x$. Since $x_{n_k}to x$ there is $N_1$ with $n_k ge N_1$ implies $|x_{n_k}-x|<{varepsilonover 2}$. Let $N_2$ be such that $n,mge N_2$ implies $|x_n-x_m|<{varepsilon over 2}$.



                  Then $n>N=max(N_1,N_2)$ implies
                  $$ |x_n-x|le |x_n-x_N|+|x_N-x|<varepsilon$$



                  Hence $X$ is complete.






                  share|cite|improve this answer



















                  • 2




                    This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
                    – Santiago Canez
                    Apr 23 '14 at 13:00










                  • @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
                    – Student
                    Apr 23 '14 at 13:39












                  • @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
                    – Student
                    Apr 24 '14 at 10:10










                  • @SantiagoCanez Do you find the edited answer more palatable?
                    – Student
                    Apr 24 '14 at 15:46










                  • Yes, this looks great, thanks.
                    – Santiago Canez
                    Apr 24 '14 at 16:21
















                  18














                  If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:



                  Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}to x in X$.



                  All that now remains to be shown is that $x_n to x$. Since $x_{n_k}to x$ there is $N_1$ with $n_k ge N_1$ implies $|x_{n_k}-x|<{varepsilonover 2}$. Let $N_2$ be such that $n,mge N_2$ implies $|x_n-x_m|<{varepsilon over 2}$.



                  Then $n>N=max(N_1,N_2)$ implies
                  $$ |x_n-x|le |x_n-x_N|+|x_N-x|<varepsilon$$



                  Hence $X$ is complete.






                  share|cite|improve this answer



















                  • 2




                    This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
                    – Santiago Canez
                    Apr 23 '14 at 13:00










                  • @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
                    – Student
                    Apr 23 '14 at 13:39












                  • @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
                    – Student
                    Apr 24 '14 at 10:10










                  • @SantiagoCanez Do you find the edited answer more palatable?
                    – Student
                    Apr 24 '14 at 15:46










                  • Yes, this looks great, thanks.
                    – Santiago Canez
                    Apr 24 '14 at 16:21














                  18












                  18








                  18






                  If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:



                  Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}to x in X$.



                  All that now remains to be shown is that $x_n to x$. Since $x_{n_k}to x$ there is $N_1$ with $n_k ge N_1$ implies $|x_{n_k}-x|<{varepsilonover 2}$. Let $N_2$ be such that $n,mge N_2$ implies $|x_n-x_m|<{varepsilon over 2}$.



                  Then $n>N=max(N_1,N_2)$ implies
                  $$ |x_n-x|le |x_n-x_N|+|x_N-x|<varepsilon$$



                  Hence $X$ is complete.






                  share|cite|improve this answer














                  If you do not wish to use the Heine-Borel theorem for metric spaces (as suggested in the answer by Igor Rivin) then here is another way of proving that a compact metric space is complete:



                  Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}to x in X$.



                  All that now remains to be shown is that $x_n to x$. Since $x_{n_k}to x$ there is $N_1$ with $n_k ge N_1$ implies $|x_{n_k}-x|<{varepsilonover 2}$. Let $N_2$ be such that $n,mge N_2$ implies $|x_n-x_m|<{varepsilon over 2}$.



                  Then $n>N=max(N_1,N_2)$ implies
                  $$ |x_n-x|le |x_n-x_N|+|x_N-x|<varepsilon$$



                  Hence $X$ is complete.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 24 '14 at 10:09

























                  answered Apr 23 '14 at 12:15









                  Student

                  664518




                  664518








                  • 2




                    This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
                    – Santiago Canez
                    Apr 23 '14 at 13:00










                  • @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
                    – Student
                    Apr 23 '14 at 13:39












                  • @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
                    – Student
                    Apr 24 '14 at 10:10










                  • @SantiagoCanez Do you find the edited answer more palatable?
                    – Student
                    Apr 24 '14 at 15:46










                  • Yes, this looks great, thanks.
                    – Santiago Canez
                    Apr 24 '14 at 16:21














                  • 2




                    This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
                    – Santiago Canez
                    Apr 23 '14 at 13:00










                  • @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
                    – Student
                    Apr 23 '14 at 13:39












                  • @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
                    – Student
                    Apr 24 '14 at 10:10










                  • @SantiagoCanez Do you find the edited answer more palatable?
                    – Student
                    Apr 24 '14 at 15:46










                  • Yes, this looks great, thanks.
                    – Santiago Canez
                    Apr 24 '14 at 16:21








                  2




                  2




                  This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
                  – Santiago Canez
                  Apr 23 '14 at 13:00




                  This is not yet correct, since you haven't yet shown that the original sequence converges. The result which you need to establish first is that if a Cauchy sequence has a convergent subsequence, then the original sequence converges.
                  – Santiago Canez
                  Apr 23 '14 at 13:00












                  @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
                  – Student
                  Apr 23 '14 at 13:39






                  @SantiagoCanez I believe this is a misunderstanding due to my clumsy writing. I will rewrite and improve the answer. Thank you for your comments.
                  – Student
                  Apr 23 '14 at 13:39














                  @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
                  – Student
                  Apr 24 '14 at 10:10




                  @SantiagoCanez I now included the proof of what I assumed and consider a triviality.
                  – Student
                  Apr 24 '14 at 10:10












                  @SantiagoCanez Do you find the edited answer more palatable?
                  – Student
                  Apr 24 '14 at 15:46




                  @SantiagoCanez Do you find the edited answer more palatable?
                  – Student
                  Apr 24 '14 at 15:46












                  Yes, this looks great, thanks.
                  – Santiago Canez
                  Apr 24 '14 at 16:21




                  Yes, this looks great, thanks.
                  – Santiago Canez
                  Apr 24 '14 at 16:21











                  2














                  Here's a not-so-fancy way:



                  Let ${a_n}$ be a Cauchy sequence.



                  If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).



                  If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.






                  share|cite|improve this answer




























                    2














                    Here's a not-so-fancy way:



                    Let ${a_n}$ be a Cauchy sequence.



                    If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).



                    If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.






                    share|cite|improve this answer


























                      2












                      2








                      2






                      Here's a not-so-fancy way:



                      Let ${a_n}$ be a Cauchy sequence.



                      If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).



                      If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.






                      share|cite|improve this answer














                      Here's a not-so-fancy way:



                      Let ${a_n}$ be a Cauchy sequence.



                      If the set of values in the (image of the) sequence is finite, then use the Cauchy criterion to show that the sequence is eventually constant (and so converges).



                      If the set of values of the sequence is infinite, then use compactness to finite a limit point of this set. Use this limit point to construct a convergent subsequence of the original sequence. Then use the Cauchy criterion to show the original sequence converges to the same limit as the subsequence.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 5 '14 at 5:40

























                      answered Jan 5 '14 at 5:34









                      bryanj

                      2,4731027




                      2,4731027























                          1














                          This follows from Heine-Borel (see the wiki page for the relevant proofs).






                          share|cite|improve this answer

















                          • 2




                            How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
                            – Pedro Tamaroff
                            Jan 5 '14 at 5:29






                          • 1




                            @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
                            – Student
                            Apr 23 '14 at 12:10
















                          1














                          This follows from Heine-Borel (see the wiki page for the relevant proofs).






                          share|cite|improve this answer

















                          • 2




                            How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
                            – Pedro Tamaroff
                            Jan 5 '14 at 5:29






                          • 1




                            @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
                            – Student
                            Apr 23 '14 at 12:10














                          1












                          1








                          1






                          This follows from Heine-Borel (see the wiki page for the relevant proofs).






                          share|cite|improve this answer












                          This follows from Heine-Borel (see the wiki page for the relevant proofs).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 5 '14 at 5:10









                          Igor Rivin

                          15.9k11234




                          15.9k11234








                          • 2




                            How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
                            – Pedro Tamaroff
                            Jan 5 '14 at 5:29






                          • 1




                            @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
                            – Student
                            Apr 23 '14 at 12:10














                          • 2




                            How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
                            – Pedro Tamaroff
                            Jan 5 '14 at 5:29






                          • 1




                            @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
                            – Student
                            Apr 23 '14 at 12:10








                          2




                          2




                          How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
                          – Pedro Tamaroff
                          Jan 5 '14 at 5:29




                          How does this follow from Heine-Borel's theorem? Were you trying to answer another question?
                          – Pedro Tamaroff
                          Jan 5 '14 at 5:29




                          1




                          1




                          @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
                          – Student
                          Apr 23 '14 at 12:10




                          @PedroTamaroff The Heine-Borel for metric spaces says that a set is compact iff it is complete and totally bounded. So there is nothing to prove as the desired claim is an immediate consequence.
                          – Student
                          Apr 23 '14 at 12:10











                          1














                          Let $X$ be a compact metric space and let ${p_n}$ be a Cauchy sequence in $X$. Then define $E_N$ as ${p_N, p_{N+1}, p_{N+2}, ldots}$. Let $overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well.



                          By definition of Cauchy sequence, we have $lim_{Ntoinfty} text{diam } E_N = lim_{Ntoinfty} text{diam } overline{E_N} = 0$. Let $E = cap_{n=1}^infty overline{E_n}$. Because $E_N supset E_{N+1}$ and $overline{E_N} supset overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $lim_{Ntoinfty} text{diam } overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p in overline{E_N}$ for all $N$. Therefore $p in X$.



                          For all $epsilon > 0$, there exists an $N$ such that $text{diam } overline{E_n} < epsilon$ for all $n > N$. Thus, $d(p,q) < epsilon$ for all $q in overline{E_n}$. Since $E_n subset overline{E_n}$, we have $d(p,q) < epsilon$ for all $q in E_n$. Therefore, ${p_n}$ converges to $p in X$.






                          share|cite|improve this answer


























                            1














                            Let $X$ be a compact metric space and let ${p_n}$ be a Cauchy sequence in $X$. Then define $E_N$ as ${p_N, p_{N+1}, p_{N+2}, ldots}$. Let $overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well.



                            By definition of Cauchy sequence, we have $lim_{Ntoinfty} text{diam } E_N = lim_{Ntoinfty} text{diam } overline{E_N} = 0$. Let $E = cap_{n=1}^infty overline{E_n}$. Because $E_N supset E_{N+1}$ and $overline{E_N} supset overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $lim_{Ntoinfty} text{diam } overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p in overline{E_N}$ for all $N$. Therefore $p in X$.



                            For all $epsilon > 0$, there exists an $N$ such that $text{diam } overline{E_n} < epsilon$ for all $n > N$. Thus, $d(p,q) < epsilon$ for all $q in overline{E_n}$. Since $E_n subset overline{E_n}$, we have $d(p,q) < epsilon$ for all $q in E_n$. Therefore, ${p_n}$ converges to $p in X$.






                            share|cite|improve this answer
























                              1












                              1








                              1






                              Let $X$ be a compact metric space and let ${p_n}$ be a Cauchy sequence in $X$. Then define $E_N$ as ${p_N, p_{N+1}, p_{N+2}, ldots}$. Let $overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well.



                              By definition of Cauchy sequence, we have $lim_{Ntoinfty} text{diam } E_N = lim_{Ntoinfty} text{diam } overline{E_N} = 0$. Let $E = cap_{n=1}^infty overline{E_n}$. Because $E_N supset E_{N+1}$ and $overline{E_N} supset overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $lim_{Ntoinfty} text{diam } overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p in overline{E_N}$ for all $N$. Therefore $p in X$.



                              For all $epsilon > 0$, there exists an $N$ such that $text{diam } overline{E_n} < epsilon$ for all $n > N$. Thus, $d(p,q) < epsilon$ for all $q in overline{E_n}$. Since $E_n subset overline{E_n}$, we have $d(p,q) < epsilon$ for all $q in E_n$. Therefore, ${p_n}$ converges to $p in X$.






                              share|cite|improve this answer












                              Let $X$ be a compact metric space and let ${p_n}$ be a Cauchy sequence in $X$. Then define $E_N$ as ${p_N, p_{N+1}, p_{N+2}, ldots}$. Let $overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well.



                              By definition of Cauchy sequence, we have $lim_{Ntoinfty} text{diam } E_N = lim_{Ntoinfty} text{diam } overline{E_N} = 0$. Let $E = cap_{n=1}^infty overline{E_n}$. Because $E_N supset E_{N+1}$ and $overline{E_N} supset overline{E_{N+1}}$ for all $N$, we have that $E$ is not empty. $E$ cannot have more than $1$ point because otherwise, $lim_{Ntoinfty} text{diam } overline{E_N} > 0$, which is a contradiction. Therefore $E$ contains exactly one point $p in overline{E_N}$ for all $N$. Therefore $p in X$.



                              For all $epsilon > 0$, there exists an $N$ such that $text{diam } overline{E_n} < epsilon$ for all $n > N$. Thus, $d(p,q) < epsilon$ for all $q in overline{E_n}$. Since $E_n subset overline{E_n}$, we have $d(p,q) < epsilon$ for all $q in E_n$. Therefore, ${p_n}$ converges to $p in X$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Apr 14 '15 at 19:05









                              Andy

                              11317




                              11317























                                  0














                                  Hint: If you have a sequence wherein $d(a_{N},a_{k})<r$ for all $kgeq N$, then the limit of that sequence, if exist, must be in the closure of the open ball radius $r$ around $a_{N}$.






                                  share|cite|improve this answer


























                                    0














                                    Hint: If you have a sequence wherein $d(a_{N},a_{k})<r$ for all $kgeq N$, then the limit of that sequence, if exist, must be in the closure of the open ball radius $r$ around $a_{N}$.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Hint: If you have a sequence wherein $d(a_{N},a_{k})<r$ for all $kgeq N$, then the limit of that sequence, if exist, must be in the closure of the open ball radius $r$ around $a_{N}$.






                                      share|cite|improve this answer












                                      Hint: If you have a sequence wherein $d(a_{N},a_{k})<r$ for all $kgeq N$, then the limit of that sequence, if exist, must be in the closure of the open ball radius $r$ around $a_{N}$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 5 '14 at 5:12









                                      Gina

                                      4,6071133




                                      4,6071133






























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