If $f:[0,1] rightarrow mathbb{R}$ is continuous then it is uniformly continuous












1












$begingroup$


I do have a general proof for this problem. That is if there is a continuous function on a compact interval then the function is uniformly continuous.

But I'm wondering whether there is a different (more simpler one) for this particular domain.






real-analysis continuity metric-spaces uniform-continuity







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  • 7




    $begingroup$
    No, there is nothing peculiar about $[0,1]$ as far as this is concerned.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 23:57










  • $begingroup$
    You might be able to do something if you add that the function is of bounded variation.
    $endgroup$
    – Dunham
    Dec 20 '18 at 0:09










  • $begingroup$
    I am guessing that the general proof uses finite cover of a compact set. In that case, some extra assumption might ease the problem.
    $endgroup$
    – HumbleStudent
    Dec 20 '18 at 0:21






  • 1




    $begingroup$
    There is a proof in Michael Spivak's textbook Calculus which does not explicitly use the concept of compactness.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 0:22










  • $begingroup$
    see math.stackexchange.com/a/3065724/432081
    $endgroup$
    – CopyPasteIt
    Jan 11 at 15:07
















1












$begingroup$


I do have a general proof for this problem. That is if there is a continuous function on a compact interval then the function is uniformly continuous.

But I'm wondering whether there is a different (more simpler one) for this particular domain.






real-analysis continuity metric-spaces uniform-continuity







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    No, there is nothing peculiar about $[0,1]$ as far as this is concerned.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 23:57










  • $begingroup$
    You might be able to do something if you add that the function is of bounded variation.
    $endgroup$
    – Dunham
    Dec 20 '18 at 0:09










  • $begingroup$
    I am guessing that the general proof uses finite cover of a compact set. In that case, some extra assumption might ease the problem.
    $endgroup$
    – HumbleStudent
    Dec 20 '18 at 0:21






  • 1




    $begingroup$
    There is a proof in Michael Spivak's textbook Calculus which does not explicitly use the concept of compactness.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 0:22










  • $begingroup$
    see math.stackexchange.com/a/3065724/432081
    $endgroup$
    – CopyPasteIt
    Jan 11 at 15:07














1












1








1





$begingroup$


I do have a general proof for this problem. That is if there is a continuous function on a compact interval then the function is uniformly continuous.

But I'm wondering whether there is a different (more simpler one) for this particular domain.






real-analysis continuity metric-spaces uniform-continuity







share|cite|improve this question











$endgroup$




I do have a general proof for this problem. That is if there is a continuous function on a compact interval then the function is uniformly continuous.

But I'm wondering whether there is a different (more simpler one) for this particular domain.






real-analysis continuity metric-spaces uniform-continuity




real-analysis continuity metric-spaces uniform-continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 2:23









Luca Carai

31119




31119










asked Dec 19 '18 at 23:52









DD90DD90

2648




2648








  • 7




    $begingroup$
    No, there is nothing peculiar about $[0,1]$ as far as this is concerned.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 23:57










  • $begingroup$
    You might be able to do something if you add that the function is of bounded variation.
    $endgroup$
    – Dunham
    Dec 20 '18 at 0:09










  • $begingroup$
    I am guessing that the general proof uses finite cover of a compact set. In that case, some extra assumption might ease the problem.
    $endgroup$
    – HumbleStudent
    Dec 20 '18 at 0:21






  • 1




    $begingroup$
    There is a proof in Michael Spivak's textbook Calculus which does not explicitly use the concept of compactness.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 0:22










  • $begingroup$
    see math.stackexchange.com/a/3065724/432081
    $endgroup$
    – CopyPasteIt
    Jan 11 at 15:07














  • 7




    $begingroup$
    No, there is nothing peculiar about $[0,1]$ as far as this is concerned.
    $endgroup$
    – José Carlos Santos
    Dec 19 '18 at 23:57










  • $begingroup$
    You might be able to do something if you add that the function is of bounded variation.
    $endgroup$
    – Dunham
    Dec 20 '18 at 0:09










  • $begingroup$
    I am guessing that the general proof uses finite cover of a compact set. In that case, some extra assumption might ease the problem.
    $endgroup$
    – HumbleStudent
    Dec 20 '18 at 0:21






  • 1




    $begingroup$
    There is a proof in Michael Spivak's textbook Calculus which does not explicitly use the concept of compactness.
    $endgroup$
    – MathematicsStudent1122
    Dec 20 '18 at 0:22










  • $begingroup$
    see math.stackexchange.com/a/3065724/432081
    $endgroup$
    – CopyPasteIt
    Jan 11 at 15:07








7




7




$begingroup$
No, there is nothing peculiar about $[0,1]$ as far as this is concerned.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 23:57




$begingroup$
No, there is nothing peculiar about $[0,1]$ as far as this is concerned.
$endgroup$
– José Carlos Santos
Dec 19 '18 at 23:57












$begingroup$
You might be able to do something if you add that the function is of bounded variation.
$endgroup$
– Dunham
Dec 20 '18 at 0:09




$begingroup$
You might be able to do something if you add that the function is of bounded variation.
$endgroup$
– Dunham
Dec 20 '18 at 0:09












$begingroup$
I am guessing that the general proof uses finite cover of a compact set. In that case, some extra assumption might ease the problem.
$endgroup$
– HumbleStudent
Dec 20 '18 at 0:21




$begingroup$
I am guessing that the general proof uses finite cover of a compact set. In that case, some extra assumption might ease the problem.
$endgroup$
– HumbleStudent
Dec 20 '18 at 0:21




1




1




$begingroup$
There is a proof in Michael Spivak's textbook Calculus which does not explicitly use the concept of compactness.
$endgroup$
– MathematicsStudent1122
Dec 20 '18 at 0:22




$begingroup$
There is a proof in Michael Spivak's textbook Calculus which does not explicitly use the concept of compactness.
$endgroup$
– MathematicsStudent1122
Dec 20 '18 at 0:22












$begingroup$
see math.stackexchange.com/a/3065724/432081
$endgroup$
– CopyPasteIt
Jan 11 at 15:07




$begingroup$
see math.stackexchange.com/a/3065724/432081
$endgroup$
– CopyPasteIt
Jan 11 at 15:07










1 Answer
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$begingroup$

Wether there is a different or a simpler proof is not the same question. You can find proof for any closed interval of $mathbb{R}$ that do not explicitly call compactness ( an example is to be found in Michael Spivak's textbook Calculus, as @MathematicsStudent1122 stated in his comment, see page 142 of https://notendur.hi.is/mbh6/html/_downloads/Calculus%20-%20Michael%20Spivak.pdf). But all these proofs are based on the ability to explicitly exhibit a finite open cover for intervals of $mathbb{R}$, which in the end, is just as strong as calling compactness.



If f is differentiable, then the mean-value theorem will definitely give you a shortcut, but once again, the ability to find an upper bound for the derivative will be based on the compactness of $[0,1]$, wether explicitly or implicitly.



As to know if the above proofs will be simpler, there are some cases in mathematics (more frequent than you would think) where getting more general makes simpler proof. I reckon this is one of these cases. What we can say however, is that not calling explicitly compactness is gonna use less advanced knowledge with regard with a basic cursus in mathematics.






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    $begingroup$

    Wether there is a different or a simpler proof is not the same question. You can find proof for any closed interval of $mathbb{R}$ that do not explicitly call compactness ( an example is to be found in Michael Spivak's textbook Calculus, as @MathematicsStudent1122 stated in his comment, see page 142 of https://notendur.hi.is/mbh6/html/_downloads/Calculus%20-%20Michael%20Spivak.pdf). But all these proofs are based on the ability to explicitly exhibit a finite open cover for intervals of $mathbb{R}$, which in the end, is just as strong as calling compactness.



    If f is differentiable, then the mean-value theorem will definitely give you a shortcut, but once again, the ability to find an upper bound for the derivative will be based on the compactness of $[0,1]$, wether explicitly or implicitly.



    As to know if the above proofs will be simpler, there are some cases in mathematics (more frequent than you would think) where getting more general makes simpler proof. I reckon this is one of these cases. What we can say however, is that not calling explicitly compactness is gonna use less advanced knowledge with regard with a basic cursus in mathematics.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Wether there is a different or a simpler proof is not the same question. You can find proof for any closed interval of $mathbb{R}$ that do not explicitly call compactness ( an example is to be found in Michael Spivak's textbook Calculus, as @MathematicsStudent1122 stated in his comment, see page 142 of https://notendur.hi.is/mbh6/html/_downloads/Calculus%20-%20Michael%20Spivak.pdf). But all these proofs are based on the ability to explicitly exhibit a finite open cover for intervals of $mathbb{R}$, which in the end, is just as strong as calling compactness.



      If f is differentiable, then the mean-value theorem will definitely give you a shortcut, but once again, the ability to find an upper bound for the derivative will be based on the compactness of $[0,1]$, wether explicitly or implicitly.



      As to know if the above proofs will be simpler, there are some cases in mathematics (more frequent than you would think) where getting more general makes simpler proof. I reckon this is one of these cases. What we can say however, is that not calling explicitly compactness is gonna use less advanced knowledge with regard with a basic cursus in mathematics.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Wether there is a different or a simpler proof is not the same question. You can find proof for any closed interval of $mathbb{R}$ that do not explicitly call compactness ( an example is to be found in Michael Spivak's textbook Calculus, as @MathematicsStudent1122 stated in his comment, see page 142 of https://notendur.hi.is/mbh6/html/_downloads/Calculus%20-%20Michael%20Spivak.pdf). But all these proofs are based on the ability to explicitly exhibit a finite open cover for intervals of $mathbb{R}$, which in the end, is just as strong as calling compactness.



        If f is differentiable, then the mean-value theorem will definitely give you a shortcut, but once again, the ability to find an upper bound for the derivative will be based on the compactness of $[0,1]$, wether explicitly or implicitly.



        As to know if the above proofs will be simpler, there are some cases in mathematics (more frequent than you would think) where getting more general makes simpler proof. I reckon this is one of these cases. What we can say however, is that not calling explicitly compactness is gonna use less advanced knowledge with regard with a basic cursus in mathematics.






        share|cite|improve this answer









        $endgroup$



        Wether there is a different or a simpler proof is not the same question. You can find proof for any closed interval of $mathbb{R}$ that do not explicitly call compactness ( an example is to be found in Michael Spivak's textbook Calculus, as @MathematicsStudent1122 stated in his comment, see page 142 of https://notendur.hi.is/mbh6/html/_downloads/Calculus%20-%20Michael%20Spivak.pdf). But all these proofs are based on the ability to explicitly exhibit a finite open cover for intervals of $mathbb{R}$, which in the end, is just as strong as calling compactness.



        If f is differentiable, then the mean-value theorem will definitely give you a shortcut, but once again, the ability to find an upper bound for the derivative will be based on the compactness of $[0,1]$, wether explicitly or implicitly.



        As to know if the above proofs will be simpler, there are some cases in mathematics (more frequent than you would think) where getting more general makes simpler proof. I reckon this is one of these cases. What we can say however, is that not calling explicitly compactness is gonna use less advanced knowledge with regard with a basic cursus in mathematics.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 3:23









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