Prove: A set containing limit points of a sequence is a closed set












3














The question:



Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.



What I have so far:


I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.



So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.



I'm at a loss on how to proceed from here.






real-analysis general-topology proof-verification







share|cite|improve this question




















  • 1




    (1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
    – DanielWainfleet
    Aug 7 '16 at 6:54












  • The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
    – martin.koeberl
    Mar 28 '17 at 13:40












  • The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
    – user247327
    Jan 2 '18 at 2:44
















3














The question:



Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.



What I have so far:


I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.



So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.



I'm at a loss on how to proceed from here.






real-analysis general-topology proof-verification







share|cite|improve this question




















  • 1




    (1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
    – DanielWainfleet
    Aug 7 '16 at 6:54












  • The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
    – martin.koeberl
    Mar 28 '17 at 13:40












  • The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
    – user247327
    Jan 2 '18 at 2:44














3












3








3







The question:



Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.



What I have so far:


I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.



So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.



I'm at a loss on how to proceed from here.






real-analysis general-topology proof-verification







share|cite|improve this question















The question:



Prove that a set, $S'$, containing the limit points of the sequence $S subset mathbb{R}$, is closed.



What I have so far:


I want to prove this by showing that the complement of $S'$ is open. In other words, $mathbb{R} backslash S' $ is open.



So take any limit point $x$ of $mathbb{R} backslash S' $. Clearly, $x$ is not a limit point of $S$.



I'm at a loss on how to proceed from here.






real-analysis general-topology proof-verification




real-analysis general-topology proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 12 '14 at 6:05









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked May 12 '14 at 0:24









user3025403

1488




1488








  • 1




    (1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
    – DanielWainfleet
    Aug 7 '16 at 6:54












  • The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
    – martin.koeberl
    Mar 28 '17 at 13:40












  • The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
    – user247327
    Jan 2 '18 at 2:44














  • 1




    (1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
    – DanielWainfleet
    Aug 7 '16 at 6:54












  • The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
    – martin.koeberl
    Mar 28 '17 at 13:40












  • The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
    – user247327
    Jan 2 '18 at 2:44








1




1




(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54






(1) Saying that a set $A$ contains the set $B$ means $Asupset B.$ ... (2). It is unclear whether you want to show that $P$ is closed or that $Pcup Q$ is closed, where $Q={s_n:nin N}$, and $P$ is the set of limit points of the sequence $S=(s_n)_{nin N}.$ (That is, $xin P$ iff ${n:s_nin (-r+x,r+x)}$ is infinite for every $r>0.$)
– DanielWainfleet
Aug 7 '16 at 6:54














The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40






The wording of this question is somehow unfortunate. Do you mean that $S^prime$ contains the limit points of one sequence $S$ or is $S^prime$ somehow quantified? Are you assuming that $S$ is a sequence in $S^prime$?
– martin.koeberl
Mar 28 '17 at 13:40














The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
– user247327
Jan 2 '18 at 2:44




The way you have stated this is not true! Take S to be the sequence 1/n, S' to be the open interval (-1, 1). S' contains the limit point of S but is not closed. What you meant to say is that "if a set contains all limit points of every sequence in S'. then S' is closed".
– user247327
Jan 2 '18 at 2:44










3 Answers
3






active

oldest

votes


















0














Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.






share|cite|improve this answer





















  • Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
    – Sayan Bandyapadhyay
    May 12 '14 at 1:01












  • @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
    – Zircht
    May 12 '14 at 1:05



















0














Another nice way to see this would be the following.
Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,



begin{align*}
x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
vdots \
x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
end{align*}
Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.






share|cite|improve this answer





























    0














    If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have



    $tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$



    Claim: The OP's $S'$ is equal to $bar {S}$.



    It is easy to see that $S' subset bar {S}$.



    Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval

    $tag 2 (r-frac{1}{n},r+frac{1}{n})$

    intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.



    Since $S' = bar {S}$, $S'$ is a closed set.






    share|cite|improve this answer





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f791041%2fprove-a-set-containing-limit-points-of-a-sequence-is-a-closed-set%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.






      share|cite|improve this answer





















      • Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
        – Sayan Bandyapadhyay
        May 12 '14 at 1:01












      • @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
        – Zircht
        May 12 '14 at 1:05
















      0














      Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.






      share|cite|improve this answer





















      • Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
        – Sayan Bandyapadhyay
        May 12 '14 at 1:01












      • @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
        – Zircht
        May 12 '14 at 1:05














      0












      0








      0






      Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.






      share|cite|improve this answer












      Since $x$ is not a limit point of $S$, there is an open neighborhood $U$ of $x$ such that $(Usetminus{x})cap S=emptyset$. Now prove $Ucap S'=emptyset$ and conclude $Usubset(Bbb Rsetminus S')$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered May 12 '14 at 0:39









      Zircht

      1,623813




      1,623813












      • Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
        – Sayan Bandyapadhyay
        May 12 '14 at 1:01












      • @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
        – Zircht
        May 12 '14 at 1:05


















      • Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
        – Sayan Bandyapadhyay
        May 12 '14 at 1:01












      • @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
        – Zircht
        May 12 '14 at 1:05
















      Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
      – Sayan Bandyapadhyay
      May 12 '14 at 1:01






      Here one thing to note is that $x$ is not just a limit point of $mathbb{R}setminus S'$, $x$ is any arbitrary point of $mathbb{R}setminus S'$. We need to prove for all such points.
      – Sayan Bandyapadhyay
      May 12 '14 at 1:01














      @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
      – Zircht
      May 12 '14 at 1:05




      @SayanBandyapadhyay Yes, I should have started with: "Let $xin Bbb Rsetminus S'$. Since...".
      – Zircht
      May 12 '14 at 1:05











      0














      Another nice way to see this would be the following.
      Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,



      begin{align*}
      x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
      x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
      x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
      vdots \
      x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
      end{align*}
      Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.






      share|cite|improve this answer


























        0














        Another nice way to see this would be the following.
        Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,



        begin{align*}
        x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
        x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
        x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
        vdots \
        x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
        end{align*}
        Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.






        share|cite|improve this answer
























          0












          0








          0






          Another nice way to see this would be the following.
          Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,



          begin{align*}
          x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
          x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
          x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
          vdots \
          x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
          end{align*}
          Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.






          share|cite|improve this answer












          Another nice way to see this would be the following.
          Suppose $xin bar{S'}$. then there exists a sequence of numbers in $S'$ which converges to $x$. Let that sequence be $(x_n)$. But again as $x_nin S'$, we have the following,



          begin{align*}
          x_{11}, &x_{12}, x_{13},cdots rightarrow x_1 && (text{a sequence converging to} x_1) \
          x_{21}, &x_{22}, x_{23},cdots rightarrow x_2 && (text{a sequence converging to} x_2) \
          x_{31}, &x_{32}, x_{33},cdots rightarrow x_3 && (text{a sequence converging to} x_3) \
          vdots \
          x_{n1}, &x_{n2}, x_{n3},cdots rightarrow x_n && (text{a sequence converging to} x_n)
          end{align*}
          Note that all the numbers $x_{ij}in S$. Now use a diagonal argument to get a sequence in $S$ converging to the limit, $x=lim_{nrightarrowinfty}x_n$. This shows that $xin S'$ by definition of $S'$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 28 '17 at 12:53









          Parish

          631315




          631315























              0














              If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have



              $tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$



              Claim: The OP's $S'$ is equal to $bar {S}$.



              It is easy to see that $S' subset bar {S}$.



              Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval

              $tag 2 (r-frac{1}{n},r+frac{1}{n})$

              intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.



              Since $S' = bar {S}$, $S'$ is a closed set.






              share|cite|improve this answer


























                0














                If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have



                $tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$



                Claim: The OP's $S'$ is equal to $bar {S}$.



                It is easy to see that $S' subset bar {S}$.



                Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval

                $tag 2 (r-frac{1}{n},r+frac{1}{n})$

                intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.



                Since $S' = bar {S}$, $S'$ is a closed set.






                share|cite|improve this answer
























                  0












                  0








                  0






                  If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have



                  $tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$



                  Claim: The OP's $S'$ is equal to $bar {S}$.



                  It is easy to see that $S' subset bar {S}$.



                  Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval

                  $tag 2 (r-frac{1}{n},r+frac{1}{n})$

                  intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.



                  Since $S' = bar {S}$, $S'$ is a closed set.






                  share|cite|improve this answer












                  If $X$ is any topological space and $S subset X$, then ${displaystyle {bar {S}}=Xbackslash [ (Xbackslash S)^{circ }]}$. Here, we have



                  $tag 1 {displaystyle {bar {S}}=mathbb Rbackslash [ (mathbb Rbackslash S)^{circ }]}$



                  Claim: The OP's $S'$ is equal to $bar {S}$.



                  It is easy to see that $S' subset bar {S}$.



                  Using constant sequences, we see that $S subset S'$. Now let $r in bar {S}backslash S$. By (1), for every $n ge 1$ the open interval

                  $tag 2 (r-frac{1}{n},r+frac{1}{n})$

                  intersects the set $S$, so we can select $s_n$ in both the interval (2) and $S$. The sequence $(s_n)$ is $S$ converges to $r$ so $r in S'$. Since both $S$ and its $bar {S}backslash S$ are contained in $S'$, $bar {S} subset S'$.



                  Since $S' = bar {S}$, $S'$ is a closed set.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 '18 at 3:44









                  CopyPasteIt

                  4,0651627




                  4,0651627






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f791041%2fprove-a-set-containing-limit-points-of-a-sequence-is-a-closed-set%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more