Proving $B={f:fin C_{[0,1]}text{ and }:d(f,0)leqslant 1}$ is not compact












1












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Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.



Prove that $B$ is not compact.




Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.



Question:



Is my strategy right? If so, how should I find the sequence of functions?



Thanks in advance!










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    $begingroup$



    Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
    Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.



    Prove that $B$ is not compact.




    Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.



    Question:



    Is my strategy right? If so, how should I find the sequence of functions?



    Thanks in advance!










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
      Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.



      Prove that $B$ is not compact.




      Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.



      Question:



      Is my strategy right? If so, how should I find the sequence of functions?



      Thanks in advance!










      share|cite|improve this question











      $endgroup$





      Let $(C_{[0,1]},d)$ be the metric space defined with the supremum metric.
      Let $B={f:fin C_{[0,1]}text{ and}:d(f,0)leqslant 1}$ where $0$ denotes the constant function form $[0,1]$ into $mathbb{R}$.



      Prove that $B$ is not compact.




      Since we are working on a metric space I can use the compactness definition that every convergent sequence has a convergent subsequence so I am trying to find a function that acts as a counterexample to prove the space is not compact.



      Question:



      Is my strategy right? If so, how should I find the sequence of functions?



      Thanks in advance!







      general-topology metric-spaces






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 22:10









      Bernard

      123k741116




      123k741116










      asked Dec 16 '18 at 21:55









      Pedro GomesPedro Gomes

      1,9252721




      1,9252721






















          1 Answer
          1






          active

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          votes


















          3












          $begingroup$

          You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 21:59










          • $begingroup$
            Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:00










          • $begingroup$
            So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 22:21










          • $begingroup$
            The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
            $endgroup$
            – Matematleta
            Dec 16 '18 at 22:27










          • $begingroup$
            @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:27











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 21:59










          • $begingroup$
            Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:00










          • $begingroup$
            So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 22:21










          • $begingroup$
            The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
            $endgroup$
            – Matematleta
            Dec 16 '18 at 22:27










          • $begingroup$
            @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:27
















          3












          $begingroup$

          You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 21:59










          • $begingroup$
            Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:00










          • $begingroup$
            So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 22:21










          • $begingroup$
            The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
            $endgroup$
            – Matematleta
            Dec 16 '18 at 22:27










          • $begingroup$
            @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:27














          3












          3








          3





          $begingroup$

          You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.






          share|cite|improve this answer











          $endgroup$



          You can indeed use a sequence, e.g. $f_n(x) = x^n$ is a good candidate.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 12:30

























          answered Dec 16 '18 at 21:57









          Henno BrandsmaHenno Brandsma

          112k348121




          112k348121












          • $begingroup$
            Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 21:59










          • $begingroup$
            Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:00










          • $begingroup$
            So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 22:21










          • $begingroup$
            The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
            $endgroup$
            – Matematleta
            Dec 16 '18 at 22:27










          • $begingroup$
            @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:27


















          • $begingroup$
            Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 21:59










          • $begingroup$
            Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:00










          • $begingroup$
            So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
            $endgroup$
            – Pedro Gomes
            Dec 16 '18 at 22:21










          • $begingroup$
            The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
            $endgroup$
            – Matematleta
            Dec 16 '18 at 22:27










          • $begingroup$
            @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
            $endgroup$
            – Henno Brandsma
            Dec 16 '18 at 22:27
















          $begingroup$
          Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
          $endgroup$
          – Pedro Gomes
          Dec 16 '18 at 21:59




          $begingroup$
          Could you please explain me how should I use those functions? I am not seeing how to apply the sequence subsequence definition of compactness, so I think I need to see an example. Thanks for your answer!
          $endgroup$
          – Pedro Gomes
          Dec 16 '18 at 21:59












          $begingroup$
          Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
          $endgroup$
          – Henno Brandsma
          Dec 16 '18 at 22:00




          $begingroup$
          Use that a supremum metric (sub)sequence limit is in particular a pointwise one. @PedroGomes
          $endgroup$
          – Henno Brandsma
          Dec 16 '18 at 22:00












          $begingroup$
          So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
          $endgroup$
          – Pedro Gomes
          Dec 16 '18 at 22:21




          $begingroup$
          So if I pick $x^n$, then for $n>N d(x^n,0)=sup|x^n-0|<epsilon$ for $epsilon>x^N$. However I am not seeing how to relate to the compactness concept.
          $endgroup$
          – Pedro Gomes
          Dec 16 '18 at 22:21












          $begingroup$
          The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
          $endgroup$
          – Matematleta
          Dec 16 '18 at 22:27




          $begingroup$
          The point is that these sequences do not converge to a $continuous$ function. i.e. they are sequences in $C$ with no convergent subsequences (in $C!$)
          $endgroup$
          – Matematleta
          Dec 16 '18 at 22:27












          $begingroup$
          @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
          $endgroup$
          – Henno Brandsma
          Dec 16 '18 at 22:27




          $begingroup$
          @PedroGomes The pointwise limit of $x^n$ is $0$ on $[0,1)$ and $1$ on $1$. The sup limit of a sequence of continuous functions is continuous.
          $endgroup$
          – Henno Brandsma
          Dec 16 '18 at 22:27


















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