Show that there exists $phi :Xto mathbb{R}$ which is continuous and not bounded.












4












$begingroup$


Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$



Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). I have shown that $f$ is continous. Now suppose that $X$ is not complete, then we want to show that there exists $phi :Xto mathbb{R}$ which is continuous and not bounded.



I am guessing that $phi = f,$ but I am not able to argue why $f$ is not bounded. Any ideas will be much appreciated.






real-analysis







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$endgroup$












  • $begingroup$
    Terminology: A sequence is not called convergent unless it has a limit. But you are correct in saying that $f$ is continuous.
    $endgroup$
    – DanielWainfleet
    Nov 25 '18 at 19:58
















4












$begingroup$


Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$



Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). I have shown that $f$ is continous. Now suppose that $X$ is not complete, then we want to show that there exists $phi :Xto mathbb{R}$ which is continuous and not bounded.



I am guessing that $phi = f,$ but I am not able to argue why $f$ is not bounded. Any ideas will be much appreciated.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Terminology: A sequence is not called convergent unless it has a limit. But you are correct in saying that $f$ is continuous.
    $endgroup$
    – DanielWainfleet
    Nov 25 '18 at 19:58














4












4








4





$begingroup$


Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$



Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). I have shown that $f$ is continous. Now suppose that $X$ is not complete, then we want to show that there exists $phi :Xto mathbb{R}$ which is continuous and not bounded.



I am guessing that $phi = f,$ but I am not able to argue why $f$ is not bounded. Any ideas will be much appreciated.






real-analysis







share|cite|improve this question









$endgroup$




Let $X$ be a metric space and $(a_n)$ a Cauchy sequence in $X.$



Let $f(x) = lim_{nto infty}d(a_n,x)$ (the limit exists since $d(a_n,x)$ is a Cauchy sequence and hence convergent). I have shown that $f$ is continous. Now suppose that $X$ is not complete, then we want to show that there exists $phi :Xto mathbb{R}$ which is continuous and not bounded.



I am guessing that $phi = f,$ but I am not able to argue why $f$ is not bounded. Any ideas will be much appreciated.






real-analysis




real-analysis






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 '18 at 18:02









Hello_WorldHello_World

4,11621731




4,11621731












  • $begingroup$
    Terminology: A sequence is not called convergent unless it has a limit. But you are correct in saying that $f$ is continuous.
    $endgroup$
    – DanielWainfleet
    Nov 25 '18 at 19:58


















  • $begingroup$
    Terminology: A sequence is not called convergent unless it has a limit. But you are correct in saying that $f$ is continuous.
    $endgroup$
    – DanielWainfleet
    Nov 25 '18 at 19:58
















$begingroup$
Terminology: A sequence is not called convergent unless it has a limit. But you are correct in saying that $f$ is continuous.
$endgroup$
– DanielWainfleet
Nov 25 '18 at 19:58




$begingroup$
Terminology: A sequence is not called convergent unless it has a limit. But you are correct in saying that $f$ is continuous.
$endgroup$
– DanielWainfleet
Nov 25 '18 at 19:58










2 Answers
2






active

oldest

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1












$begingroup$

First things first: you assume that $X$ is not complete, so you know there exists a Cauchy sequence $(a_{n})$ in $X$ with no limit in $X.$ So you should fix such a sequence, and use that in your definition of $f. $



Now, your function $f$ might be bounded, namely if the metric space $X$ is bounded (for example, take $X$ to be the rational numbers between $0$ and $1;$ then $f$ is bounded below by $0$ and bounded above by $1$).



So now the question becomes: can you use $f$ to define a different function $phicolon Xtomathbb{R}$ which is unbounded, but which is also continuous?



Hint:




Due to your choice of $(a_{n}),$ you know that $f$ never takes the value $0.$ Use this to your advantage.







share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Do you know that metric space completions exist? If so, then consider $tilde{X}$, the metric space completion of $X$, and the metric $tilde{d}$ on $tilde{X}$ which coincides with $d$ on elements of $X$. Since $X$ was not complete, there exists a point $tilde{x} in tilde{X}$ which is not in $X$, and a Cauchy sequence $(a_n)$ of elements of $X$ which converges to $tilde{x}$. Then the function $f(x) = lim_{n to infty} tilde{d}(a_n, x)$ is continuous, as you have shown. It should also take on non-zero values for any $x in X$ (why?). I think the restriction of $1/f$ to $X$ will be your desired function.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      1












      $begingroup$

      First things first: you assume that $X$ is not complete, so you know there exists a Cauchy sequence $(a_{n})$ in $X$ with no limit in $X.$ So you should fix such a sequence, and use that in your definition of $f. $



      Now, your function $f$ might be bounded, namely if the metric space $X$ is bounded (for example, take $X$ to be the rational numbers between $0$ and $1;$ then $f$ is bounded below by $0$ and bounded above by $1$).



      So now the question becomes: can you use $f$ to define a different function $phicolon Xtomathbb{R}$ which is unbounded, but which is also continuous?



      Hint:




      Due to your choice of $(a_{n}),$ you know that $f$ never takes the value $0.$ Use this to your advantage.







      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        First things first: you assume that $X$ is not complete, so you know there exists a Cauchy sequence $(a_{n})$ in $X$ with no limit in $X.$ So you should fix such a sequence, and use that in your definition of $f. $



        Now, your function $f$ might be bounded, namely if the metric space $X$ is bounded (for example, take $X$ to be the rational numbers between $0$ and $1;$ then $f$ is bounded below by $0$ and bounded above by $1$).



        So now the question becomes: can you use $f$ to define a different function $phicolon Xtomathbb{R}$ which is unbounded, but which is also continuous?



        Hint:




        Due to your choice of $(a_{n}),$ you know that $f$ never takes the value $0.$ Use this to your advantage.







        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          First things first: you assume that $X$ is not complete, so you know there exists a Cauchy sequence $(a_{n})$ in $X$ with no limit in $X.$ So you should fix such a sequence, and use that in your definition of $f. $



          Now, your function $f$ might be bounded, namely if the metric space $X$ is bounded (for example, take $X$ to be the rational numbers between $0$ and $1;$ then $f$ is bounded below by $0$ and bounded above by $1$).



          So now the question becomes: can you use $f$ to define a different function $phicolon Xtomathbb{R}$ which is unbounded, but which is also continuous?



          Hint:




          Due to your choice of $(a_{n}),$ you know that $f$ never takes the value $0.$ Use this to your advantage.







          share|cite|improve this answer











          $endgroup$



          First things first: you assume that $X$ is not complete, so you know there exists a Cauchy sequence $(a_{n})$ in $X$ with no limit in $X.$ So you should fix such a sequence, and use that in your definition of $f. $



          Now, your function $f$ might be bounded, namely if the metric space $X$ is bounded (for example, take $X$ to be the rational numbers between $0$ and $1;$ then $f$ is bounded below by $0$ and bounded above by $1$).



          So now the question becomes: can you use $f$ to define a different function $phicolon Xtomathbb{R}$ which is unbounded, but which is also continuous?



          Hint:




          Due to your choice of $(a_{n}),$ you know that $f$ never takes the value $0.$ Use this to your advantage.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 '18 at 19:38

























          answered Nov 25 '18 at 19:13









          Will RWill R

          6,57731429




          6,57731429























              0












              $begingroup$

              Do you know that metric space completions exist? If so, then consider $tilde{X}$, the metric space completion of $X$, and the metric $tilde{d}$ on $tilde{X}$ which coincides with $d$ on elements of $X$. Since $X$ was not complete, there exists a point $tilde{x} in tilde{X}$ which is not in $X$, and a Cauchy sequence $(a_n)$ of elements of $X$ which converges to $tilde{x}$. Then the function $f(x) = lim_{n to infty} tilde{d}(a_n, x)$ is continuous, as you have shown. It should also take on non-zero values for any $x in X$ (why?). I think the restriction of $1/f$ to $X$ will be your desired function.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Do you know that metric space completions exist? If so, then consider $tilde{X}$, the metric space completion of $X$, and the metric $tilde{d}$ on $tilde{X}$ which coincides with $d$ on elements of $X$. Since $X$ was not complete, there exists a point $tilde{x} in tilde{X}$ which is not in $X$, and a Cauchy sequence $(a_n)$ of elements of $X$ which converges to $tilde{x}$. Then the function $f(x) = lim_{n to infty} tilde{d}(a_n, x)$ is continuous, as you have shown. It should also take on non-zero values for any $x in X$ (why?). I think the restriction of $1/f$ to $X$ will be your desired function.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Do you know that metric space completions exist? If so, then consider $tilde{X}$, the metric space completion of $X$, and the metric $tilde{d}$ on $tilde{X}$ which coincides with $d$ on elements of $X$. Since $X$ was not complete, there exists a point $tilde{x} in tilde{X}$ which is not in $X$, and a Cauchy sequence $(a_n)$ of elements of $X$ which converges to $tilde{x}$. Then the function $f(x) = lim_{n to infty} tilde{d}(a_n, x)$ is continuous, as you have shown. It should also take on non-zero values for any $x in X$ (why?). I think the restriction of $1/f$ to $X$ will be your desired function.






                  share|cite|improve this answer









                  $endgroup$



                  Do you know that metric space completions exist? If so, then consider $tilde{X}$, the metric space completion of $X$, and the metric $tilde{d}$ on $tilde{X}$ which coincides with $d$ on elements of $X$. Since $X$ was not complete, there exists a point $tilde{x} in tilde{X}$ which is not in $X$, and a Cauchy sequence $(a_n)$ of elements of $X$ which converges to $tilde{x}$. Then the function $f(x) = lim_{n to infty} tilde{d}(a_n, x)$ is continuous, as you have shown. It should also take on non-zero values for any $x in X$ (why?). I think the restriction of $1/f$ to $X$ will be your desired function.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 '18 at 18:10









                  Bartosz MalmanBartosz Malman

                  8161620




                  8161620






























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