If $X$ is a non empty closed subset of $mathbbR$ and $Y = [1,2]$ then $X+Y$ is closed.

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If $X$ is a non empty closed subset of $mathbbR$ and $Y = [1,2]$ then $X+Y$ is closed?



How to show this?



I know that sum of two closed set may not be closed. But how to prove this? Can anyone give me a hint?






real-analysis general-topology metric-spaces normed-spaces







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    up vote
    1
    down vote

    favorite












    If $X$ is a non empty closed subset of $mathbbR$ and $Y = [1,2]$ then $X+Y$ is closed?



    How to show this?



    I know that sum of two closed set may not be closed. But how to prove this? Can anyone give me a hint?






    real-analysis general-topology metric-spaces normed-spaces







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      If $X$ is a non empty closed subset of $mathbbR$ and $Y = [1,2]$ then $X+Y$ is closed?



      How to show this?



      I know that sum of two closed set may not be closed. But how to prove this? Can anyone give me a hint?






      real-analysis general-topology metric-spaces normed-spaces







      share|cite|improve this question















      If $X$ is a non empty closed subset of $mathbbR$ and $Y = [1,2]$ then $X+Y$ is closed?



      How to show this?



      I know that sum of two closed set may not be closed. But how to prove this? Can anyone give me a hint?






      real-analysis general-topology metric-spaces normed-spaces




      real-analysis general-topology metric-spaces normed-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 4 at 9:27









      José Carlos Santos

      122k16101186




      122k16101186










      asked Sep 4 at 7:54









      cmi

      946110




      946110




















          2 Answers
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          up vote
          3
          down vote



          accepted










          Suppose $x_n in X,y_n in [1,2]$ and $x_n +y_n to z$. Since $[1,2]$ is compact there is a subsequence $y_n_k$ which converges to some point $y$ in $[1,2]$. But $x_n_k+y_n_k to z$ so $x_n_k to z-y$. Since $X$ is closed, $z-y in X$. It follows that $z= lim (x_n_k+y_n_k) =(z-y)+y in X+Y$.






          share|cite|improve this answer




















          • Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
            – cmi
            Sep 4 at 8:01










          • @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
            – Kavi Rama Murthy
            Sep 4 at 8:10










          • In any topological group if you use nets instead of sequences. @cmi
            – Henno Brandsma
            Sep 4 at 8:50










          • Yes, you are right again.
            – Kavi Rama Murthy
            Sep 4 at 8:52

















          up vote
          1
          down vote













          And an answer to the second part of your question.
          Let $X=2,3,4,ldots$, $Y=y:y=-n+frac1n, n=2,3,4,ldots$. They are closed, but in their sum there is a set $frac1n:n=2,3,4, ldots$ but not $0$.






          share|cite|improve this answer




















          • I knew that. ……….
            – cmi
            Sep 4 at 13:54










          Your Answer




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          2 Answers
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          2 Answers
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          up vote
          3
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          accepted










          Suppose $x_n in X,y_n in [1,2]$ and $x_n +y_n to z$. Since $[1,2]$ is compact there is a subsequence $y_n_k$ which converges to some point $y$ in $[1,2]$. But $x_n_k+y_n_k to z$ so $x_n_k to z-y$. Since $X$ is closed, $z-y in X$. It follows that $z= lim (x_n_k+y_n_k) =(z-y)+y in X+Y$.






          share|cite|improve this answer




















          • Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
            – cmi
            Sep 4 at 8:01










          • @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
            – Kavi Rama Murthy
            Sep 4 at 8:10










          • In any topological group if you use nets instead of sequences. @cmi
            – Henno Brandsma
            Sep 4 at 8:50










          • Yes, you are right again.
            – Kavi Rama Murthy
            Sep 4 at 8:52














          up vote
          3
          down vote



          accepted










          Suppose $x_n in X,y_n in [1,2]$ and $x_n +y_n to z$. Since $[1,2]$ is compact there is a subsequence $y_n_k$ which converges to some point $y$ in $[1,2]$. But $x_n_k+y_n_k to z$ so $x_n_k to z-y$. Since $X$ is closed, $z-y in X$. It follows that $z= lim (x_n_k+y_n_k) =(z-y)+y in X+Y$.






          share|cite|improve this answer




















          • Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
            – cmi
            Sep 4 at 8:01










          • @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
            – Kavi Rama Murthy
            Sep 4 at 8:10










          • In any topological group if you use nets instead of sequences. @cmi
            – Henno Brandsma
            Sep 4 at 8:50










          • Yes, you are right again.
            – Kavi Rama Murthy
            Sep 4 at 8:52












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Suppose $x_n in X,y_n in [1,2]$ and $x_n +y_n to z$. Since $[1,2]$ is compact there is a subsequence $y_n_k$ which converges to some point $y$ in $[1,2]$. But $x_n_k+y_n_k to z$ so $x_n_k to z-y$. Since $X$ is closed, $z-y in X$. It follows that $z= lim (x_n_k+y_n_k) =(z-y)+y in X+Y$.






          share|cite|improve this answer












          Suppose $x_n in X,y_n in [1,2]$ and $x_n +y_n to z$. Since $[1,2]$ is compact there is a subsequence $y_n_k$ which converges to some point $y$ in $[1,2]$. But $x_n_k+y_n_k to z$ so $x_n_k to z-y$. Since $X$ is closed, $z-y in X$. It follows that $z= lim (x_n_k+y_n_k) =(z-y)+y in X+Y$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 4 at 7:58









          Kavi Rama Murthy

          26.1k31437




          26.1k31437











          • Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
            – cmi
            Sep 4 at 8:01










          • @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
            – Kavi Rama Murthy
            Sep 4 at 8:10










          • In any topological group if you use nets instead of sequences. @cmi
            – Henno Brandsma
            Sep 4 at 8:50










          • Yes, you are right again.
            – Kavi Rama Murthy
            Sep 4 at 8:52
















          • Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
            – cmi
            Sep 4 at 8:01










          • @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
            – Kavi Rama Murthy
            Sep 4 at 8:10










          • In any topological group if you use nets instead of sequences. @cmi
            – Henno Brandsma
            Sep 4 at 8:50










          • Yes, you are right again.
            – Kavi Rama Murthy
            Sep 4 at 8:52















          Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
          – cmi
          Sep 4 at 8:01




          Can I use this technic if $X$ and $Y$ were subsets of any NLS and one is compact and another is closed?@Kavi Rama Murthy
          – cmi
          Sep 4 at 8:01












          @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
          – Kavi Rama Murthy
          Sep 4 at 8:10




          @cmi Yes, the same argument works for the sum of a closed set and a compact set in any NLS.
          – Kavi Rama Murthy
          Sep 4 at 8:10












          In any topological group if you use nets instead of sequences. @cmi
          – Henno Brandsma
          Sep 4 at 8:50




          In any topological group if you use nets instead of sequences. @cmi
          – Henno Brandsma
          Sep 4 at 8:50












          Yes, you are right again.
          – Kavi Rama Murthy
          Sep 4 at 8:52




          Yes, you are right again.
          – Kavi Rama Murthy
          Sep 4 at 8:52










          up vote
          1
          down vote













          And an answer to the second part of your question.
          Let $X=2,3,4,ldots$, $Y=y:y=-n+frac1n, n=2,3,4,ldots$. They are closed, but in their sum there is a set $frac1n:n=2,3,4, ldots$ but not $0$.






          share|cite|improve this answer




















          • I knew that. ……….
            – cmi
            Sep 4 at 13:54














          up vote
          1
          down vote













          And an answer to the second part of your question.
          Let $X=2,3,4,ldots$, $Y=y:y=-n+frac1n, n=2,3,4,ldots$. They are closed, but in their sum there is a set $frac1n:n=2,3,4, ldots$ but not $0$.






          share|cite|improve this answer




















          • I knew that. ……….
            – cmi
            Sep 4 at 13:54












          up vote
          1
          down vote










          up vote
          1
          down vote









          And an answer to the second part of your question.
          Let $X=2,3,4,ldots$, $Y=y:y=-n+frac1n, n=2,3,4,ldots$. They are closed, but in their sum there is a set $frac1n:n=2,3,4, ldots$ but not $0$.






          share|cite|improve this answer












          And an answer to the second part of your question.
          Let $X=2,3,4,ldots$, $Y=y:y=-n+frac1n, n=2,3,4,ldots$. They are closed, but in their sum there is a set $frac1n:n=2,3,4, ldots$ but not $0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 4 at 8:23









          Przemysław Scherwentke

          11.8k52751




          11.8k52751











          • I knew that. ……….
            – cmi
            Sep 4 at 13:54
















          • I knew that. ……….
            – cmi
            Sep 4 at 13:54















          I knew that. ……….
          – cmi
          Sep 4 at 13:54




          I knew that. ……….
          – cmi
          Sep 4 at 13:54

















           

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