A question about metric space











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Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$



The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.




I know that:



The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.




But I could not prove, why $X$ is not compact by defined metric?



Please help me.










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  • You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 27 at 10:51















up vote
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down vote

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Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$



The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.




I know that:



The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.




But I could not prove, why $X$ is not compact by defined metric?



Please help me.










share|cite|improve this question
























  • You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 27 at 10:51













up vote
-1
down vote

favorite
1









up vote
-1
down vote

favorite
1






1





Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$



The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.




I know that:



The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.




But I could not prove, why $X$ is not compact by defined metric?



Please help me.










share|cite|improve this question















Let $$X = left{ (x_n)_{n=1}^{infty} mid x_n in mathbb{R}, quad 0 leq x_n leq 1 quad forall n in mathbb{N} right}.$$



The set $X$ equipped with metric $d(x,y) = sup { mid x_n - y_n mid : n in mathbb{N} }$.




I know that:



The metric space $Y$ is compact if and only if the every sequence in $Y$ has a convergent sub-sequence.




But I could not prove, why $X$ is not compact by defined metric?



Please help me.







real-analysis functional-analysis analysis






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edited Nov 22 at 14:57









Davide Giraudo

124k16150256




124k16150256










asked Nov 22 at 14:33









joe

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874












  • You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 27 at 10:51


















  • You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
    – 5xum
    Nov 27 at 10:51
















You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51




You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:51










2 Answers
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You can prove the space is not compact by finding a sequence without a convergent subsequence.



Hint:



Like a lot of times, you should try your search for a counterexample by looking at simple things.



Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.






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    That is not compact .
    Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
      3
      down vote













      You can prove the space is not compact by finding a sequence without a convergent subsequence.



      Hint:



      Like a lot of times, you should try your search for a counterexample by looking at simple things.



      Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.






      share|cite|improve this answer

























        up vote
        3
        down vote













        You can prove the space is not compact by finding a sequence without a convergent subsequence.



        Hint:



        Like a lot of times, you should try your search for a counterexample by looking at simple things.



        Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.






        share|cite|improve this answer























          up vote
          3
          down vote










          up vote
          3
          down vote









          You can prove the space is not compact by finding a sequence without a convergent subsequence.



          Hint:



          Like a lot of times, you should try your search for a counterexample by looking at simple things.



          Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.






          share|cite|improve this answer












          You can prove the space is not compact by finding a sequence without a convergent subsequence.



          Hint:



          Like a lot of times, you should try your search for a counterexample by looking at simple things.



          Think about very simple vectors in $X$. Very simple ones. With a whole lot of zeroes. Like, as many zeroes as you can get (i.e., with as few nonzero elements as possible). Also, to make them even simpler, have the nonzero element be something simple. Like a constant value. Some simple constant value, so it'll be easy to calculate with it.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 14:37









          5xum

          88.7k392160




          88.7k392160






















              up vote
              0
              down vote













              That is not compact .
              Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.






              share|cite|improve this answer

























                up vote
                0
                down vote













                That is not compact .
                Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  That is not compact .
                  Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.






                  share|cite|improve this answer












                  That is not compact .
                  Counterexample: $x_n:-n^{th}$ coordinate is 1 else $0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 14:50









                  Shubham

                  1,5721519




                  1,5721519






























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