Show that if $f(-1)0$, then there exist $-1<a<1$ such that $f(a)=0$












0












$begingroup$



Let $f:mathbb{R}tomathbb{R}$ continuous. Show that if $f(-1)<0$ and $f(1)>0$, then there exist $-1<a<1$ such that $f(a)=0$.




My approach: Suppose that $f(x)neq 0$ for all $xin (-1,1)$. Then $$f((-1,1))=(f(-1),0)cup(0,f(1))qquadqquad(*)$$



Furthermore, $f$ is continuous and the image of connected set is connected, then (*) is a contradiction.



Therefore, must be exist some $ain(-1,1)$ such that $f(a)=0$. This is correct? Can give me some hint. Thanks!






real-analysis continuity connectedness







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$endgroup$








  • 1




    $begingroup$
    Would it not just be sufficient to use the intermediate value theorem?
    $endgroup$
    – Eevee Trainer
    Dec 23 '18 at 23:16










  • $begingroup$
    I know these theorem, but I want to demonstrate this problem without using the IVT.
    $endgroup$
    – user570343
    Dec 23 '18 at 23:18










  • $begingroup$
    Using the intermediate value theorem is a bit like assuming what you are to prove: the statement of the question basically is the intermediate value theorem.
    $endgroup$
    – Winther
    Dec 23 '18 at 23:41
















0












$begingroup$



Let $f:mathbb{R}tomathbb{R}$ continuous. Show that if $f(-1)<0$ and $f(1)>0$, then there exist $-1<a<1$ such that $f(a)=0$.




My approach: Suppose that $f(x)neq 0$ for all $xin (-1,1)$. Then $$f((-1,1))=(f(-1),0)cup(0,f(1))qquadqquad(*)$$



Furthermore, $f$ is continuous and the image of connected set is connected, then (*) is a contradiction.



Therefore, must be exist some $ain(-1,1)$ such that $f(a)=0$. This is correct? Can give me some hint. Thanks!






real-analysis continuity connectedness







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Would it not just be sufficient to use the intermediate value theorem?
    $endgroup$
    – Eevee Trainer
    Dec 23 '18 at 23:16










  • $begingroup$
    I know these theorem, but I want to demonstrate this problem without using the IVT.
    $endgroup$
    – user570343
    Dec 23 '18 at 23:18










  • $begingroup$
    Using the intermediate value theorem is a bit like assuming what you are to prove: the statement of the question basically is the intermediate value theorem.
    $endgroup$
    – Winther
    Dec 23 '18 at 23:41














0












0








0





$begingroup$



Let $f:mathbb{R}tomathbb{R}$ continuous. Show that if $f(-1)<0$ and $f(1)>0$, then there exist $-1<a<1$ such that $f(a)=0$.




My approach: Suppose that $f(x)neq 0$ for all $xin (-1,1)$. Then $$f((-1,1))=(f(-1),0)cup(0,f(1))qquadqquad(*)$$



Furthermore, $f$ is continuous and the image of connected set is connected, then (*) is a contradiction.



Therefore, must be exist some $ain(-1,1)$ such that $f(a)=0$. This is correct? Can give me some hint. Thanks!






real-analysis continuity connectedness







share|cite|improve this question











$endgroup$





Let $f:mathbb{R}tomathbb{R}$ continuous. Show that if $f(-1)<0$ and $f(1)>0$, then there exist $-1<a<1$ such that $f(a)=0$.




My approach: Suppose that $f(x)neq 0$ for all $xin (-1,1)$. Then $$f((-1,1))=(f(-1),0)cup(0,f(1))qquadqquad(*)$$



Furthermore, $f$ is continuous and the image of connected set is connected, then (*) is a contradiction.



Therefore, must be exist some $ain(-1,1)$ such that $f(a)=0$. This is correct? Can give me some hint. Thanks!






real-analysis continuity connectedness




real-analysis continuity connectedness






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 23 '18 at 23:27

























asked Dec 23 '18 at 23:11







user570343















  • 1




    $begingroup$
    Would it not just be sufficient to use the intermediate value theorem?
    $endgroup$
    – Eevee Trainer
    Dec 23 '18 at 23:16










  • $begingroup$
    I know these theorem, but I want to demonstrate this problem without using the IVT.
    $endgroup$
    – user570343
    Dec 23 '18 at 23:18










  • $begingroup$
    Using the intermediate value theorem is a bit like assuming what you are to prove: the statement of the question basically is the intermediate value theorem.
    $endgroup$
    – Winther
    Dec 23 '18 at 23:41














  • 1




    $begingroup$
    Would it not just be sufficient to use the intermediate value theorem?
    $endgroup$
    – Eevee Trainer
    Dec 23 '18 at 23:16










  • $begingroup$
    I know these theorem, but I want to demonstrate this problem without using the IVT.
    $endgroup$
    – user570343
    Dec 23 '18 at 23:18










  • $begingroup$
    Using the intermediate value theorem is a bit like assuming what you are to prove: the statement of the question basically is the intermediate value theorem.
    $endgroup$
    – Winther
    Dec 23 '18 at 23:41








1




1




$begingroup$
Would it not just be sufficient to use the intermediate value theorem?
$endgroup$
– Eevee Trainer
Dec 23 '18 at 23:16




$begingroup$
Would it not just be sufficient to use the intermediate value theorem?
$endgroup$
– Eevee Trainer
Dec 23 '18 at 23:16












$begingroup$
I know these theorem, but I want to demonstrate this problem without using the IVT.
$endgroup$
– user570343
Dec 23 '18 at 23:18




$begingroup$
I know these theorem, but I want to demonstrate this problem without using the IVT.
$endgroup$
– user570343
Dec 23 '18 at 23:18












$begingroup$
Using the intermediate value theorem is a bit like assuming what you are to prove: the statement of the question basically is the intermediate value theorem.
$endgroup$
– Winther
Dec 23 '18 at 23:41




$begingroup$
Using the intermediate value theorem is a bit like assuming what you are to prove: the statement of the question basically is the intermediate value theorem.
$endgroup$
– Winther
Dec 23 '18 at 23:41










3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint: the only connected subsets of $Bbb{R}$ are intervals (closed, open, half-closed/half-open and possibly with infinite endpoints). The image $f[[-1, 1]]$ of the closed interval $[-1, 1]$ under your function $f$ is connected and includes $f(-1) < 0$ and $f(1) > 0$, so it is an interval containing a negative number and a positive number. Can you take it from there?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohh great! Thanks, But, why my "solution" is not correct?
    $endgroup$
    – user570343
    Dec 23 '18 at 23:28










  • $begingroup$
    Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
    $endgroup$
    – Rob Arthan
    Dec 23 '18 at 23:31












  • $begingroup$
    I don't see what implies that $f(-1)>f(-1/2)$?? I
    $endgroup$
    – user570343
    Dec 23 '18 at 23:35










  • $begingroup$
    If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
    $endgroup$
    – Rob Arthan
    Dec 23 '18 at 23:44





















0












$begingroup$

Just because $f(x) neq 0$ on $x in (-1,1)$ does not mean that $0$ is the only omitted value, which you assert via "$(f(-1),0) cup (0,f(1))$"${} = (f(-1),f(1)) smallsetminus {0}$. My age (in years) is continuous and has never been $-1$, but this does not mean that $-1$ is the only number my age can never be.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    (*) is not correct. You are assuming that all values of $f$ are between $f(-1)$ and $f(1)$ which may not be true. Use: $f(-1,1)=[f(-1,1)cap (-infty,0)]cup [f(-1,1)cap (0,infty)]$.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint: the only connected subsets of $Bbb{R}$ are intervals (closed, open, half-closed/half-open and possibly with infinite endpoints). The image $f[[-1, 1]]$ of the closed interval $[-1, 1]$ under your function $f$ is connected and includes $f(-1) < 0$ and $f(1) > 0$, so it is an interval containing a negative number and a positive number. Can you take it from there?






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Ohh great! Thanks, But, why my "solution" is not correct?
        $endgroup$
        – user570343
        Dec 23 '18 at 23:28










      • $begingroup$
        Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:31












      • $begingroup$
        I don't see what implies that $f(-1)>f(-1/2)$?? I
        $endgroup$
        – user570343
        Dec 23 '18 at 23:35










      • $begingroup$
        If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:44


















      1












      $begingroup$

      Hint: the only connected subsets of $Bbb{R}$ are intervals (closed, open, half-closed/half-open and possibly with infinite endpoints). The image $f[[-1, 1]]$ of the closed interval $[-1, 1]$ under your function $f$ is connected and includes $f(-1) < 0$ and $f(1) > 0$, so it is an interval containing a negative number and a positive number. Can you take it from there?






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Ohh great! Thanks, But, why my "solution" is not correct?
        $endgroup$
        – user570343
        Dec 23 '18 at 23:28










      • $begingroup$
        Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:31












      • $begingroup$
        I don't see what implies that $f(-1)>f(-1/2)$?? I
        $endgroup$
        – user570343
        Dec 23 '18 at 23:35










      • $begingroup$
        If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:44
















      1












      1








      1





      $begingroup$

      Hint: the only connected subsets of $Bbb{R}$ are intervals (closed, open, half-closed/half-open and possibly with infinite endpoints). The image $f[[-1, 1]]$ of the closed interval $[-1, 1]$ under your function $f$ is connected and includes $f(-1) < 0$ and $f(1) > 0$, so it is an interval containing a negative number and a positive number. Can you take it from there?






      share|cite|improve this answer











      $endgroup$



      Hint: the only connected subsets of $Bbb{R}$ are intervals (closed, open, half-closed/half-open and possibly with infinite endpoints). The image $f[[-1, 1]]$ of the closed interval $[-1, 1]$ under your function $f$ is connected and includes $f(-1) < 0$ and $f(1) > 0$, so it is an interval containing a negative number and a positive number. Can you take it from there?







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 23 '18 at 23:35

























      answered Dec 23 '18 at 23:27









      Rob ArthanRob Arthan

      29.3k42966




      29.3k42966












      • $begingroup$
        Ohh great! Thanks, But, why my "solution" is not correct?
        $endgroup$
        – user570343
        Dec 23 '18 at 23:28










      • $begingroup$
        Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:31












      • $begingroup$
        I don't see what implies that $f(-1)>f(-1/2)$?? I
        $endgroup$
        – user570343
        Dec 23 '18 at 23:35










      • $begingroup$
        If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:44




















      • $begingroup$
        Ohh great! Thanks, But, why my "solution" is not correct?
        $endgroup$
        – user570343
        Dec 23 '18 at 23:28










      • $begingroup$
        Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:31












      • $begingroup$
        I don't see what implies that $f(-1)>f(-1/2)$?? I
        $endgroup$
        – user570343
        Dec 23 '18 at 23:35










      • $begingroup$
        If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
        $endgroup$
        – Rob Arthan
        Dec 23 '18 at 23:44


















      $begingroup$
      Ohh great! Thanks, But, why my "solution" is not correct?
      $endgroup$
      – user570343
      Dec 23 '18 at 23:28




      $begingroup$
      Ohh great! Thanks, But, why my "solution" is not correct?
      $endgroup$
      – user570343
      Dec 23 '18 at 23:28












      $begingroup$
      Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
      $endgroup$
      – Rob Arthan
      Dec 23 '18 at 23:31






      $begingroup$
      Because you are making unjustified assumptions about the behaviour of $f$. E.g. you could have $f(-1) > f(-1/2)$.
      $endgroup$
      – Rob Arthan
      Dec 23 '18 at 23:31














      $begingroup$
      I don't see what implies that $f(-1)>f(-1/2)$?? I
      $endgroup$
      – user570343
      Dec 23 '18 at 23:35




      $begingroup$
      I don't see what implies that $f(-1)>f(-1/2)$?? I
      $endgroup$
      – user570343
      Dec 23 '18 at 23:35












      $begingroup$
      If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
      $endgroup$
      – Rob Arthan
      Dec 23 '18 at 23:44






      $begingroup$
      If you had $f(-1) > f(-1/2)$ then your equation (*) might be false. All you need to say is that the image $f[[-1, 1]]$ is connected and includes the point $0 in [f(-1), f(1)]$. Your equation for $f[(-1, 1)]$ need not hold. (I am writing $f[X]$ for the image of the set $X$ under the function $f$ where you are writing $f((X))$.)
      $endgroup$
      – Rob Arthan
      Dec 23 '18 at 23:44













      0












      $begingroup$

      Just because $f(x) neq 0$ on $x in (-1,1)$ does not mean that $0$ is the only omitted value, which you assert via "$(f(-1),0) cup (0,f(1))$"${} = (f(-1),f(1)) smallsetminus {0}$. My age (in years) is continuous and has never been $-1$, but this does not mean that $-1$ is the only number my age can never be.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Just because $f(x) neq 0$ on $x in (-1,1)$ does not mean that $0$ is the only omitted value, which you assert via "$(f(-1),0) cup (0,f(1))$"${} = (f(-1),f(1)) smallsetminus {0}$. My age (in years) is continuous and has never been $-1$, but this does not mean that $-1$ is the only number my age can never be.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Just because $f(x) neq 0$ on $x in (-1,1)$ does not mean that $0$ is the only omitted value, which you assert via "$(f(-1),0) cup (0,f(1))$"${} = (f(-1),f(1)) smallsetminus {0}$. My age (in years) is continuous and has never been $-1$, but this does not mean that $-1$ is the only number my age can never be.






          share|cite|improve this answer









          $endgroup$



          Just because $f(x) neq 0$ on $x in (-1,1)$ does not mean that $0$ is the only omitted value, which you assert via "$(f(-1),0) cup (0,f(1))$"${} = (f(-1),f(1)) smallsetminus {0}$. My age (in years) is continuous and has never been $-1$, but this does not mean that $-1$ is the only number my age can never be.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 23:19









          Eric TowersEric Towers

          32.7k22370




          32.7k22370























              0












              $begingroup$

              (*) is not correct. You are assuming that all values of $f$ are between $f(-1)$ and $f(1)$ which may not be true. Use: $f(-1,1)=[f(-1,1)cap (-infty,0)]cup [f(-1,1)cap (0,infty)]$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                (*) is not correct. You are assuming that all values of $f$ are between $f(-1)$ and $f(1)$ which may not be true. Use: $f(-1,1)=[f(-1,1)cap (-infty,0)]cup [f(-1,1)cap (0,infty)]$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  (*) is not correct. You are assuming that all values of $f$ are between $f(-1)$ and $f(1)$ which may not be true. Use: $f(-1,1)=[f(-1,1)cap (-infty,0)]cup [f(-1,1)cap (0,infty)]$.






                  share|cite|improve this answer









                  $endgroup$



                  (*) is not correct. You are assuming that all values of $f$ are between $f(-1)$ and $f(1)$ which may not be true. Use: $f(-1,1)=[f(-1,1)cap (-infty,0)]cup [f(-1,1)cap (0,infty)]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 23 '18 at 23:30









                  Kavi Rama MurthyKavi Rama Murthy

                  61.6k42262




                  61.6k42262






























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                      Mont Emei

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                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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