Is every real valued continuous function on $(0,1)$ is uniformly continuous?












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$begingroup$


Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?



I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:



Coming up with an example, a function that is continuous but not uniformly continuous



But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.






continuity







share|cite|improve this question











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  • 4




    $begingroup$
    Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:20










  • $begingroup$
    @Boshu: That does not work as I explained in the question.
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:46






  • 1




    $begingroup$
    Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:52










  • $begingroup$
    @Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:57










  • $begingroup$
    Will write a brief answer.
    $endgroup$
    – Boshu
    Dec 9 '18 at 7:05
















-1












$begingroup$


Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?



I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:



Coming up with an example, a function that is continuous but not uniformly continuous



But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.






continuity







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:20










  • $begingroup$
    @Boshu: That does not work as I explained in the question.
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:46






  • 1




    $begingroup$
    Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:52










  • $begingroup$
    @Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:57










  • $begingroup$
    Will write a brief answer.
    $endgroup$
    – Boshu
    Dec 9 '18 at 7:05














-1












-1








-1





$begingroup$


Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?



I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:



Coming up with an example, a function that is continuous but not uniformly continuous



But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.






continuity







share|cite|improve this question











$endgroup$




Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?



I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:



Coming up with an example, a function that is continuous but not uniformly continuous



But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.






continuity




continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 6:44







Sepide

















asked Dec 9 '18 at 6:16









SepideSepide

2788




2788








  • 4




    $begingroup$
    Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:20










  • $begingroup$
    @Boshu: That does not work as I explained in the question.
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:46






  • 1




    $begingroup$
    Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:52










  • $begingroup$
    @Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:57










  • $begingroup$
    Will write a brief answer.
    $endgroup$
    – Boshu
    Dec 9 '18 at 7:05














  • 4




    $begingroup$
    Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:20










  • $begingroup$
    @Boshu: That does not work as I explained in the question.
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:46






  • 1




    $begingroup$
    Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
    $endgroup$
    – Boshu
    Dec 9 '18 at 6:52










  • $begingroup$
    @Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
    $endgroup$
    – Sepide
    Dec 9 '18 at 6:57










  • $begingroup$
    Will write a brief answer.
    $endgroup$
    – Boshu
    Dec 9 '18 at 7:05








4




4




$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20




$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20












$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46




$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46




1




1




$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52




$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52












$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57




$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57












$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05




$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05










1 Answer
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$begingroup$

The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.



Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.






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    -1












    $begingroup$

    The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.



    Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.






    share|cite|improve this answer









    $endgroup$


















      -1












      $begingroup$

      The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.



      Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.






      share|cite|improve this answer









      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.



        Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.






        share|cite|improve this answer









        $endgroup$



        The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.



        Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 7:08









        BoshuBoshu

        705315




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