Sequential compact set in $mathbb R^2$












1












$begingroup$


The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.



My approach is to show that I can find a subsequence that converges to a point in the set.



Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.



Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.






general-topology







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$endgroup$












  • $begingroup$
    I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
    $endgroup$
    – Don Thousand
    Jan 14 at 20:58












  • $begingroup$
    What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 22:00


















1












$begingroup$


The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.



My approach is to show that I can find a subsequence that converges to a point in the set.



Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.



Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.






general-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
    $endgroup$
    – Don Thousand
    Jan 14 at 20:58












  • $begingroup$
    What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 22:00
















1












1








1





$begingroup$


The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.



My approach is to show that I can find a subsequence that converges to a point in the set.



Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.



Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.






general-topology







share|cite|improve this question









$endgroup$




The problem : Show that the set of points in $mathbb R^2$ of the form $(x,sqrt{x})$ for
$x in [0, 2]$ is a sequentially compact subset of $mathbb R^2$.



My approach is to show that I can find a subsequence that converges to a point in the set.



Let $S$ = {$(x,sqrt{x}):x in [0,2]$}. First observe that $S$ is bounded below by $(0,0)$ and above by$(2,sqrt{2})$.So rewrite $S$ as the interval $[(0,0),(2,sqrt{2})]$. Let $(y_n)$ be a convergent sequence in $S$, so $y_n = (x,sqrt{x})$. Now since every subsequence of a convergent sequence has the same limit point, it follows that the subsequence has a limit point in S.



Would this be a good solution to the problem or am I leaving anything out?
Thank you for your feedback and hints.






general-topology




general-topology






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share|cite|improve this question











share|cite|improve this question




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asked Jan 14 at 20:53









AllorjaAllorja

789




789












  • $begingroup$
    I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
    $endgroup$
    – Don Thousand
    Jan 14 at 20:58












  • $begingroup$
    What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 22:00




















  • $begingroup$
    I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
    $endgroup$
    – Don Thousand
    Jan 14 at 20:58












  • $begingroup$
    What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
    $endgroup$
    – DanielWainfleet
    Jan 14 at 22:00


















$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58






$begingroup$
I would just note that $[0,2]$ is sequentially compact in $mathbb R$, and that $f : xto(x,sqrt x)$ is continuous and bijective, which completes your problem. The problem with your current approach is that you don't know $S=[(0,0),(2,sqrt2)]$.
$endgroup$
– Don Thousand
Jan 14 at 20:58














$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00






$begingroup$
What is an "interval" $[(0,0),(2,sqrt 2)]$ ? In $Bbb R$ the notation $[a,b]$ is a synonym for ${x: ale xle b}$.
$endgroup$
– DanielWainfleet
Jan 14 at 22:00












2 Answers
2






active

oldest

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1












$begingroup$

Let’s try to be more general.



You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.



You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.



So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.



The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.



We’re done.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      1












      $begingroup$

      Let’s try to be more general.



      You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.



      You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.



      So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.



      The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.



      We’re done.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Let’s try to be more general.



        You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.



        You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.



        So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.



        The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.



        We’re done.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Let’s try to be more general.



          You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.



          You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.



          So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.



          The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.



          We’re done.






          share|cite|improve this answer











          $endgroup$



          Let’s try to be more general.



          You have a compact interval $[a,b] subset mathbb R$ and a continuous map $f : [a,b] to mathbb R$. In your case $[a,b]=[0,2]$ and $f(x) = sqrt x$.



          You have to prove that $S= {(x,f(x)) ; x in [a,b]}$ is sequentially compact.



          So suppose that a sequence of points of $S$, namely $(P_n)=(x_n,f(x_n))$ converges to $(r,s) in mathbb R^2$. We have to prove that $(r,s) in S$.



          The coordinates of $(P_n)$ converge. So $x_n to r$ and $f(x_n) to s$. Continuity of $f$ implies that $f(x_n) to f(r)$. And unicity of a limit that $s=f(r)$. Which means that $(r,s)in S$.



          We’re done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 14 at 21:29

























          answered Jan 14 at 21:06









          mathcounterexamples.netmathcounterexamples.net

          26.5k22157




          26.5k22157























              0












              $begingroup$

              In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.






                  share|cite|improve this answer









                  $endgroup$



                  In $mathbb{R}^n$ compactness and sequential compactness are equivalent. So you just need to show that your subset is closed and bounded and that's trivial.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 21:04









                  GianniGianni

                  387




                  387






























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