If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on...











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If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?



I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:



Since



$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$



Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)



$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$



as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?










share|cite|improve this question






















  • What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
    – Robert Israel
    Nov 20 at 20:27












  • @RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
    – Riley H
    Nov 20 at 20:31










  • Would there be any other changes if you used continuity at 2? @hamam_Abdallah
    – Riley H
    Nov 20 at 20:32






  • 2




    Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
    – Sangchul Lee
    Nov 20 at 20:35






  • 1




    For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
    – Sangchul Lee
    Nov 20 at 20:39

















up vote
2
down vote

favorite
1












If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?



I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:



Since



$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$



Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)



$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$



as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?










share|cite|improve this question






















  • What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
    – Robert Israel
    Nov 20 at 20:27












  • @RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
    – Riley H
    Nov 20 at 20:31










  • Would there be any other changes if you used continuity at 2? @hamam_Abdallah
    – Riley H
    Nov 20 at 20:32






  • 2




    Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
    – Sangchul Lee
    Nov 20 at 20:35






  • 1




    For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
    – Sangchul Lee
    Nov 20 at 20:39















up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?



I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:



Since



$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$



Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)



$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$



as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?










share|cite|improve this question













If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?



I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:



Since



$exists delta_1 s.t. |x-y|<delta_1 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [1,2]$ and $exists delta_2 s.t. |x-y|<delta_2 to |f(x)-f(y)|<frac{epsilon}{2}$ for $x,y in [2,5]$



Then set $delta_3 = min{delta_1,delta_2}$
and you get (by triangle inequality)



$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|leq |f(x)-f(3)| + |f(y)-f(3)|leqfrac{epsilon}{2}+frac{epsilon}{2}=epsilon$



as long as $|x-y|<delta_3$
What's wrong with this proof and how can I fix it?







real-analysis proof-verification epsilon-delta






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share|cite|improve this question











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asked Nov 20 at 20:24









Riley H

594




594












  • What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
    – Robert Israel
    Nov 20 at 20:27












  • @RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
    – Riley H
    Nov 20 at 20:31










  • Would there be any other changes if you used continuity at 2? @hamam_Abdallah
    – Riley H
    Nov 20 at 20:32






  • 2




    Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
    – Sangchul Lee
    Nov 20 at 20:35






  • 1




    For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
    – Sangchul Lee
    Nov 20 at 20:39




















  • What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
    – Robert Israel
    Nov 20 at 20:27












  • @RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
    – Riley H
    Nov 20 at 20:31










  • Would there be any other changes if you used continuity at 2? @hamam_Abdallah
    – Riley H
    Nov 20 at 20:32






  • 2




    Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
    – Sangchul Lee
    Nov 20 at 20:35






  • 1




    For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
    – Sangchul Lee
    Nov 20 at 20:39


















What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 at 20:27






What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small.
– Robert Israel
Nov 20 at 20:27














@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 at 20:31




@RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<epsilon$, right?
– Riley H
Nov 20 at 20:31












Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 at 20:32




Would there be any other changes if you used continuity at 2? @hamam_Abdallah
– Riley H
Nov 20 at 20:32




2




2




Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 at 20:35




Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous.
– Sangchul Lee
Nov 20 at 20:35




1




1




For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 at 20:39






For the standard proof, let $delta_1, delta_2$ be as in your setting and let $delta = min{delta_1, delta_2, 2}$. If $x,yin[1,5]$ satisfy $|x-y|<delta$, argue that either $x,yin[1,4]$ or $x,yin[2,5]$ holds, so that the defining property of $delta_i$'s kicks in.
– Sangchul Lee
Nov 20 at 20:39












1 Answer
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up vote
2
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accepted










hint



The uniform continuity at $[1,4]$ gives $delta_1$



the UC at $[4,5]$ will give $delta_2$.



the continuity at $x=4$ gives $delta_3$ such that



$$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$



Take $delta=min(delta_i,i=1,2,3).$



If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then



$|x-4|<delta_3$ and $|y-4|<delta_3$



thus



$$|f(x)-f(y)|=$$
$$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$






share|cite|improve this answer























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    active

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    up vote
    2
    down vote



    accepted










    hint



    The uniform continuity at $[1,4]$ gives $delta_1$



    the UC at $[4,5]$ will give $delta_2$.



    the continuity at $x=4$ gives $delta_3$ such that



    $$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$



    Take $delta=min(delta_i,i=1,2,3).$



    If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then



    $|x-4|<delta_3$ and $|y-4|<delta_3$



    thus



    $$|f(x)-f(y)|=$$
    $$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      hint



      The uniform continuity at $[1,4]$ gives $delta_1$



      the UC at $[4,5]$ will give $delta_2$.



      the continuity at $x=4$ gives $delta_3$ such that



      $$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$



      Take $delta=min(delta_i,i=1,2,3).$



      If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then



      $|x-4|<delta_3$ and $|y-4|<delta_3$



      thus



      $$|f(x)-f(y)|=$$
      $$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        hint



        The uniform continuity at $[1,4]$ gives $delta_1$



        the UC at $[4,5]$ will give $delta_2$.



        the continuity at $x=4$ gives $delta_3$ such that



        $$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$



        Take $delta=min(delta_i,i=1,2,3).$



        If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then



        $|x-4|<delta_3$ and $|y-4|<delta_3$



        thus



        $$|f(x)-f(y)|=$$
        $$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$






        share|cite|improve this answer














        hint



        The uniform continuity at $[1,4]$ gives $delta_1$



        the UC at $[4,5]$ will give $delta_2$.



        the continuity at $x=4$ gives $delta_3$ such that



        $$|x-4|<delta_3 implies |f(x)-f(4)|<frac{epsilon}{2}$$



        Take $delta=min(delta_i,i=1,2,3).$



        If $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then



        $|x-4|<delta_3$ and $|y-4|<delta_3$



        thus



        $$|f(x)-f(y)|=$$
        $$=|f(x)-f(4)+f(4)-f(y)|<epsilon$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 20 at 20:44

























        answered Nov 20 at 20:39









        hamam_Abdallah

        36.7k21533




        36.7k21533






























             

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