Is there a function $f: mathbb{R} to mathbb{R}$ such that $limlimits_{xto p}f(x)=infty$ for every $p in...












7












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I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.



I guess not, because what would its graph look like, but then again, I don't know how to prove it.










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closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:31










  • $begingroup$
    My intuition tells me that such function doesn't exist. I like to be proven wrong.
    $endgroup$
    – Math_QED
    Jan 8 at 8:40






  • 1




    $begingroup$
    @EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 8:44






  • 2




    $begingroup$
    No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
    $endgroup$
    – Song
    Jan 8 at 8:44










  • $begingroup$
    Hmmm, I see. Thanks for the insight @AndersKaseorg
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:45
















7












$begingroup$


I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.



I guess not, because what would its graph look like, but then again, I don't know how to prove it.










share|cite|improve this question











$endgroup$



closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:31










  • $begingroup$
    My intuition tells me that such function doesn't exist. I like to be proven wrong.
    $endgroup$
    – Math_QED
    Jan 8 at 8:40






  • 1




    $begingroup$
    @EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 8:44






  • 2




    $begingroup$
    No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
    $endgroup$
    – Song
    Jan 8 at 8:44










  • $begingroup$
    Hmmm, I see. Thanks for the insight @AndersKaseorg
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:45














7












7








7


2



$begingroup$


I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.



I guess not, because what would its graph look like, but then again, I don't know how to prove it.










share|cite|improve this question











$endgroup$




I was wondering if there is a function $f: mathbb{R} to mathbb{R}$ such that its limit at every point is infinite.



I guess not, because what would its graph look like, but then again, I don't know how to prove it.







limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 8:41









Did

248k23224463




248k23224463










asked Jan 8 at 8:09









user4201961user4201961

730411




730411




closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Holo, amWhy, Saad, user21820, José Carlos Santos Jan 12 at 11:41


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Holo, amWhy, Saad, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:31










  • $begingroup$
    My intuition tells me that such function doesn't exist. I like to be proven wrong.
    $endgroup$
    – Math_QED
    Jan 8 at 8:40






  • 1




    $begingroup$
    @EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 8:44






  • 2




    $begingroup$
    No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
    $endgroup$
    – Song
    Jan 8 at 8:44










  • $begingroup$
    Hmmm, I see. Thanks for the insight @AndersKaseorg
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:45


















  • $begingroup$
    While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:31










  • $begingroup$
    My intuition tells me that such function doesn't exist. I like to be proven wrong.
    $endgroup$
    – Math_QED
    Jan 8 at 8:40






  • 1




    $begingroup$
    @EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
    $endgroup$
    – Anders Kaseorg
    Jan 8 at 8:44






  • 2




    $begingroup$
    No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
    $endgroup$
    – Song
    Jan 8 at 8:44










  • $begingroup$
    Hmmm, I see. Thanks for the insight @AndersKaseorg
    $endgroup$
    – Eevee Trainer
    Jan 8 at 8:45
















$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31




$begingroup$
While I'm not sure how to properly formalize it, I think an issue is that you wouldn't have any $x$ in the domain $mathbb{R}$ for which $f$ is defined, probably. So in turn it wouldn't even meet the definition of a function at that point. Maybe the definition of a partial function, but usually we speak of "total" functions (for each $x$ in the domain of $f$ there exists a corresponding $f(x)$ in the codomain) in this context.
$endgroup$
– Eevee Trainer
Jan 8 at 8:31












$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40




$begingroup$
My intuition tells me that such function doesn't exist. I like to be proven wrong.
$endgroup$
– Math_QED
Jan 8 at 8:40




1




1




$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44




$begingroup$
@EeveeTrainer This isn’t about partial functions, merely discontinuous ones. There is no reason we can’t have $lim_{x to p} f(x) = infty$ while $f(p)$ takes a (finite) value.
$endgroup$
– Anders Kaseorg
Jan 8 at 8:44




2




2




$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44




$begingroup$
No, there does not exist such $f$. In fact, the set $$ {xinmathbb{R};|;lim_{yto x}f(y)=infty} $$ is a countable subset of $mathbb{R}$ (hence cannot be the whole real line.) You can see this earlier post math.stackexchange.com/questions/3060529/….
$endgroup$
– Song
Jan 8 at 8:44












$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45




$begingroup$
Hmmm, I see. Thanks for the insight @AndersKaseorg
$endgroup$
– Eevee Trainer
Jan 8 at 8:45










2 Answers
2






active

oldest

votes


















9












$begingroup$

If $f$ is such a function and $M>0$ then
$$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
But
$$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
$$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    Let us show that such a function $f$ does not exist.



    Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.



    Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.



    Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.






    share|cite|improve this answer











    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9












      $begingroup$

      If $f$ is such a function and $M>0$ then
      $$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
      But
      $$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
      and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
      $$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
      So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
      So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.






      share|cite|improve this answer











      $endgroup$


















        9












        $begingroup$

        If $f$ is such a function and $M>0$ then
        $$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
        But
        $$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
        and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
        $$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
        So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
        So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.






        share|cite|improve this answer











        $endgroup$
















          9












          9








          9





          $begingroup$

          If $f$ is such a function and $M>0$ then
          $$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
          But
          $$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
          and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
          $$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
          So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
          So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.






          share|cite|improve this answer











          $endgroup$



          If $f$ is such a function and $M>0$ then
          $$ forall x_0in ! [0,1] qquad exists delta_{x_0} qquad 0<|x-x_0|<delta_{x_0} implies f(x)>M.$$
          But
          $$bigcup_{x_0in [0,1]} B(x_0,delta_{x_0})setminus {x_0}supseteq [0,1]$$
          and $[0,1]$ is compact So there exists $x_1,ldots , x_n$ such that
          $$B(x_1,delta_{x_1})cup ldots cup B(x_n,delta_{x_n}) setminus {x_1,dotsc,x_n}supseteq [0,1]$$
          So $f(x)>M$ for all $xin [0,1]setminus{x_1,ldots,x_n } $.
          So $ { x: f(x)le M } $ is at most countable and $$bigcuplimits_{Min mathbb{N}} { x:f(x)le M }$$ is at most countable. Since $[0,1]$ is uncountable, this is absurd hence no such $f$ exists.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 9:46









          SvanN

          2,0661422




          2,0661422










          answered Jan 8 at 8:30









          DarmanDarman

          538112




          538112























              6












              $begingroup$

              Let us show that such a function $f$ does not exist.



              Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.



              Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.



              Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                Let us show that such a function $f$ does not exist.



                Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.



                Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.



                Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  Let us show that such a function $f$ does not exist.



                  Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.



                  Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.



                  Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.






                  share|cite|improve this answer











                  $endgroup$



                  Let us show that such a function $f$ does not exist.



                  Choose any function $f:mathbb Rtomathbb R$ and, for every positive integer $n$, consider the set $$A_n={xin[0,1],;,|f(x)|leqslant n}$$ then, since $f(x)$ is finite for every $x$, $$bigcup_nA_n=[0,1]$$ in particular, there exists some $n$ such that $A_n$ is infinite.



                  Pick a sequence $(x_k)$ such that $x_kne x_j$ for every $kne j$ and $x_k$ is in $A_n$ for every $k$. Then the set $X={x_k,;,kgeqslant0}$ is infinite and such that $Xsubseteq A_n$. Since $[0,1]$ is compact, $X$ has a limit point $p$ in $[0,1]$. Assume without loss of generality that $limlimits_{ktoinfty}x_k=p$ and that $x_kne p$ for every $k$.



                  Now, $|f(x_k)|leqslant n$ for every $k$ hence $limlimits_{xto p}|f(x)|=+infty$ is impossible because $limlimits_{ktoinfty}x_k=p$ with $x_kne p$ for every $k$ and $limsuplimits_{ktoinfty}|f(x_k)|leqslant n$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 12 at 9:56

























                  answered Jan 8 at 9:32









                  DidDid

                  248k23224463




                  248k23224463















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