A Singleton in a metric space is closed.

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There is another proof on math stack exchange, but I would like to ask if the following reasoning makes any sense:



It contains all of its limit points, namely the "singleton"(?) sequence, the sequence with only one element, namely the element in the singleton. Or, is this not in line with the definition of a sequence? Must a sequence have infinitely many elements?






real-analysis metric-spaces







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    up vote
    1
    down vote

    favorite












    There is another proof on math stack exchange, but I would like to ask if the following reasoning makes any sense:



    It contains all of its limit points, namely the "singleton"(?) sequence, the sequence with only one element, namely the element in the singleton. Or, is this not in line with the definition of a sequence? Must a sequence have infinitely many elements?






    real-analysis metric-spaces







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      There is another proof on math stack exchange, but I would like to ask if the following reasoning makes any sense:



      It contains all of its limit points, namely the "singleton"(?) sequence, the sequence with only one element, namely the element in the singleton. Or, is this not in line with the definition of a sequence? Must a sequence have infinitely many elements?






      real-analysis metric-spaces







      share|cite|improve this question















      There is another proof on math stack exchange, but I would like to ask if the following reasoning makes any sense:



      It contains all of its limit points, namely the "singleton"(?) sequence, the sequence with only one element, namely the element in the singleton. Or, is this not in line with the definition of a sequence? Must a sequence have infinitely many elements?






      real-analysis metric-spaces




      real-analysis metric-spaces






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      edited 1 hour ago









      Asaf Karagila♦

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      asked 5 hours ago









      Rafael Vergnaud

      19213




      19213




















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          Your reasoning is correct, but the justification should be rephrased. Rather than a sequence with one element, you should instead be referring to a constant sequence. A sequence is not a set of elements but a ''list of points'' given by a function $x_n : mathbbN to X$



          With this terminology, you should say that the only sequence in $x$ is the constant sequence $(x_n)$ with $x_n = x$ for all $n$. Thus, the limit of every convergent sequence in $x$ is an element of $x$.






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          • Thank you! ...filler characters....
            – Rafael Vergnaud
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          • @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
            – Brahadeesh
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          up vote
          4
          down vote













          Your reasoning is correct, but the justification should be rephrased. Rather than a sequence with one element, you should instead be referring to a constant sequence. A sequence is not a set of elements but a ''list of points'' given by a function $x_n : mathbbN to X$



          With this terminology, you should say that the only sequence in $x$ is the constant sequence $(x_n)$ with $x_n = x$ for all $n$. Thus, the limit of every convergent sequence in $x$ is an element of $x$.






          share|cite|improve this answer




















          • Thank you! ...filler characters....
            – Rafael Vergnaud
            5 hours ago











          • @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
            – Brahadeesh
            1 hour ago















          up vote
          4
          down vote













          Your reasoning is correct, but the justification should be rephrased. Rather than a sequence with one element, you should instead be referring to a constant sequence. A sequence is not a set of elements but a ''list of points'' given by a function $x_n : mathbbN to X$



          With this terminology, you should say that the only sequence in $x$ is the constant sequence $(x_n)$ with $x_n = x$ for all $n$. Thus, the limit of every convergent sequence in $x$ is an element of $x$.






          share|cite|improve this answer




















          • Thank you! ...filler characters....
            – Rafael Vergnaud
            5 hours ago











          • @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
            – Brahadeesh
            1 hour ago













          up vote
          4
          down vote










          up vote
          4
          down vote









          Your reasoning is correct, but the justification should be rephrased. Rather than a sequence with one element, you should instead be referring to a constant sequence. A sequence is not a set of elements but a ''list of points'' given by a function $x_n : mathbbN to X$



          With this terminology, you should say that the only sequence in $x$ is the constant sequence $(x_n)$ with $x_n = x$ for all $n$. Thus, the limit of every convergent sequence in $x$ is an element of $x$.






          share|cite|improve this answer












          Your reasoning is correct, but the justification should be rephrased. Rather than a sequence with one element, you should instead be referring to a constant sequence. A sequence is not a set of elements but a ''list of points'' given by a function $x_n : mathbbN to X$



          With this terminology, you should say that the only sequence in $x$ is the constant sequence $(x_n)$ with $x_n = x$ for all $n$. Thus, the limit of every convergent sequence in $x$ is an element of $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          rolandcyp

          6116




          6116











          • Thank you! ...filler characters....
            – Rafael Vergnaud
            5 hours ago











          • @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
            – Brahadeesh
            1 hour ago

















          • Thank you! ...filler characters....
            – Rafael Vergnaud
            5 hours ago











          • @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
            – Brahadeesh
            1 hour ago
















          Thank you! ...filler characters....
          – Rafael Vergnaud
          5 hours ago





          Thank you! ...filler characters....
          – Rafael Vergnaud
          5 hours ago













          @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
          – Brahadeesh
          1 hour ago





          @RafaelVergnaud if you found the answer useful you should consider accepting it. Upvoting and accepting answers are better forms of thanking a user for help, than only commenting. :)
          – Brahadeesh
          1 hour ago


















           

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