If $f$ is continuous $displaystylelim_{x to infty}f(x)=infty$ iff $f^{-1}(K)$ is compact for every compact $K...












1












$begingroup$



Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:



(i) $displaystylelim_{x to infty}f(x)=infty$;



(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.




My attempt.



(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.



I want to conclude that $f(x_{n}) to infty$. How can I ensure it?



(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.






real-analysis continuity compactness







share|cite|improve this question









$endgroup$












  • $begingroup$
    How do you define $xtoinfty$ exactly? With the one-point compactification?
    $endgroup$
    – SmileyCraft
    Jan 6 at 3:55








  • 1




    $begingroup$
    Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 3:58












  • $begingroup$
    @SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
    $endgroup$
    – Lucas Corrêa
    Jan 6 at 4:02










  • $begingroup$
    @ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
    $endgroup$
    – mathcounterexamples.net
    Jan 6 at 4:02






  • 1




    $begingroup$
    @LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 4:06
















1












$begingroup$



Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:



(i) $displaystylelim_{x to infty}f(x)=infty$;



(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.




My attempt.



(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.



I want to conclude that $f(x_{n}) to infty$. How can I ensure it?



(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.






real-analysis continuity compactness







share|cite|improve this question









$endgroup$












  • $begingroup$
    How do you define $xtoinfty$ exactly? With the one-point compactification?
    $endgroup$
    – SmileyCraft
    Jan 6 at 3:55








  • 1




    $begingroup$
    Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 3:58












  • $begingroup$
    @SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
    $endgroup$
    – Lucas Corrêa
    Jan 6 at 4:02










  • $begingroup$
    @ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
    $endgroup$
    – mathcounterexamples.net
    Jan 6 at 4:02






  • 1




    $begingroup$
    @LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 4:06














1












1








1





$begingroup$



Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:



(i) $displaystylelim_{x to infty}f(x)=infty$;



(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.




My attempt.



(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.



I want to conclude that $f(x_{n}) to infty$. How can I ensure it?



(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.






real-analysis continuity compactness







share|cite|improve this question









$endgroup$





Let $f: mathbb{R}^{m}to mathbb{R}^{n}$ be a continuous function. Prove that the following statements are equivalents:



(i) $displaystylelim_{x to infty}f(x)=infty$;



(ii) $f^{-1}(K)$ is compact for every compact $K subset mathbb{R}^{n}$.




My attempt.



(i) $Longrightarrow$ (ii): If $K subset mathbb{R}^{n}$ is compact, $f^{-1}(K)$ is closed, since $f$ is continuous. If $f^{-1}(K)$ is unbounded, there is a sequence $(x_{n}) subset f^{-1}(K)$ such that $x_{n} to infty$.



I want to conclude that $f(x_{n}) to infty$. How can I ensure it?



(ii) $Longrightarrow$ (i): I dont have a good idea. Can someone help me? I want just a hint, not a solution.






real-analysis continuity compactness




real-analysis continuity compactness






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 6 at 3:49









Lucas CorrêaLucas Corrêa

1,5351421




1,5351421












  • $begingroup$
    How do you define $xtoinfty$ exactly? With the one-point compactification?
    $endgroup$
    – SmileyCraft
    Jan 6 at 3:55








  • 1




    $begingroup$
    Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 3:58












  • $begingroup$
    @SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
    $endgroup$
    – Lucas Corrêa
    Jan 6 at 4:02










  • $begingroup$
    @ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
    $endgroup$
    – mathcounterexamples.net
    Jan 6 at 4:02






  • 1




    $begingroup$
    @LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 4:06


















  • $begingroup$
    How do you define $xtoinfty$ exactly? With the one-point compactification?
    $endgroup$
    – SmileyCraft
    Jan 6 at 3:55








  • 1




    $begingroup$
    Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 3:58












  • $begingroup$
    @SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
    $endgroup$
    – Lucas Corrêa
    Jan 6 at 4:02










  • $begingroup$
    @ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
    $endgroup$
    – mathcounterexamples.net
    Jan 6 at 4:02






  • 1




    $begingroup$
    @LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
    $endgroup$
    – ItsJustLogicBro
    Jan 6 at 4:06
















$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55






$begingroup$
How do you define $xtoinfty$ exactly? With the one-point compactification?
$endgroup$
– SmileyCraft
Jan 6 at 3:55






1




1




$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58






$begingroup$
Consider the function $f(x)=e^x$ and the compact set $K=[-1,1]$. Then $lim f(x)=infty$ but $f^{-1}(K)=(-infty,0]$. So either this problem isn't correct or I am losing my mind.
$endgroup$
– ItsJustLogicBro
Jan 6 at 3:58














$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02




$begingroup$
@SmileyCraft, my book has not yet exactly defined $x to infty$. When we work with infinite limits, we use the standard definition.
$endgroup$
– Lucas Corrêa
Jan 6 at 4:02












$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02




$begingroup$
@ItsJustLogicBro $limlimits_{x to -infty } e^x =0$. So this isn’t a counterexample.
$endgroup$
– mathcounterexamples.net
Jan 6 at 4:02




1




1




$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06




$begingroup$
@LucasCorrêa I suspect that someone is simply abusing notation, and we really do want $|x|rightarrowinfty$.
$endgroup$
– ItsJustLogicBro
Jan 6 at 4:06










3 Answers
3






active

oldest

votes


















2












$begingroup$

To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.



ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.



    I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
      $endgroup$
      – Lucas Corrêa
      Jan 6 at 12:58



















    1












    $begingroup$

    Regarding (ii) $Longrightarrow$ (i)



    Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.



    The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.



    Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.



      ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.



        ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.



          ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.






          share|cite|improve this answer











          $endgroup$



          To complete your argument for $(i)$ $|x_n|rightarrow infty$ implies that $f(x_n)rightarrow infty$ by hypothesis.



          ii) implies i)Suppose that there exists a sequence $x_n$ such that $lim_n|x_n|=infty$ and $f(x_n)$ is bounded, $f(x_n)$ is contained in a closed ball $B$ which is compact, so $f^{-1}(B)=C$ is compact, and bounded by hypothesis, but $x_nin C$ contradiction since $x_n$ is not bounded.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 4:05

























          answered Jan 6 at 3:59









          Tsemo AristideTsemo Aristide

          59.6k11446




          59.6k11446























              1












              $begingroup$

              The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.



              I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
                $endgroup$
                – Lucas Corrêa
                Jan 6 at 12:58
















              1












              $begingroup$

              The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.



              I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
                $endgroup$
                – Lucas Corrêa
                Jan 6 at 12:58














              1












              1








              1





              $begingroup$

              The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.



              I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.






              share|cite|improve this answer









              $endgroup$



              The definition of $x_ntoinfty$ I am used to is that for every compact set $K$ there exists $N$ such that for all $n>N$ we have $x_nnotin K$. So then the statement $$lim_{xtoinfty}f(x)=infty$$ would literally mean that for every compact set $K$ there exists compact set $L$ such that for all $xnotin L$ we have $f(x)notin K$, so $f^{-1}(K)subseteq L$.



              I would like to mention how this elegant definition can be made precise with topology. In topology we say that $x_nto x$ if for every open set $U$ containing $x$ there exists $N$ such that for all $n>N$ we have $x_nin U$. There is also a thing in topology called the one-point compactification of a topological space $X$. The idea is to add a "point at infinity", so you get $Xcup{infty}$, and the open sets containing $infty$ are exactly the complements of the compact sets in $X$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 6 at 4:18









              SmileyCraftSmileyCraft

              3,749519




              3,749519












              • $begingroup$
                Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
                $endgroup$
                – Lucas Corrêa
                Jan 6 at 12:58


















              • $begingroup$
                Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
                $endgroup$
                – Lucas Corrêa
                Jan 6 at 12:58
















              $begingroup$
              Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
              $endgroup$
              – Lucas Corrêa
              Jan 6 at 12:58




              $begingroup$
              Its a very clever proof for bounded! I didnt know this definition, is that a topological property?
              $endgroup$
              – Lucas Corrêa
              Jan 6 at 12:58











              1












              $begingroup$

              Regarding (ii) $Longrightarrow$ (i)



              Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.



              The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.



              Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Regarding (ii) $Longrightarrow$ (i)



                Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.



                The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.



                Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Regarding (ii) $Longrightarrow$ (i)



                  Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.



                  The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.



                  Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.






                  share|cite|improve this answer











                  $endgroup$



                  Regarding (ii) $Longrightarrow$ (i)



                  Take $p in mathbb N$. You want to find $M>0$ such that for $Vert x Vert >M$ you have $Vert f(x) Vert >p$.



                  The closed ball $K_p$centered on zero with radius $p$ is a compact subset of $mathbb R^n$. $f^{-1}(mathbb R^n setminus K_p) = f^{-1}(mathbb R^n)setminus f^{-1}(K_p)$. And $f^{-1}(K_p)$ is by hypothesis a compact subset of $mathbb R^n$, therefore bounded and included in a closed ball centered on zero with a radius $M>0$.



                  Where are done: the image of a point $x$ with a norm $>M$ is such that $Vert f(x)Vert >p$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 6 at 9:09

























                  answered Jan 6 at 4:12









                  mathcounterexamples.netmathcounterexamples.net

                  27k22158




                  27k22158






























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