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

          32.6k22370




          32.6k22370























              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

                  58.5k42161




                  58.5k42161






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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                      Gyllenstierna

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