Prove tht $f(x) = x^{2}$ is uniformly continuous on $bigcup_{n = 1}^{infty} [n, n + 1/n^{3}]$.












3












$begingroup$



How can I show that the function $f(x) = x^{2}$ is uniformly continuous on $bigcup_{n = 1}^{infty} [n, n + frac{1}{n^{3}}]$?




As $n to infty$, I know that we get only all of the integers; however, the result doesn't make sense to me. I'm aware of both the sequential and $epsilon-delta$ definitions of continuity, but I didn't get anywhere with them.






real-analysis analysis continuity epsilon-delta uniform-continuity







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$endgroup$












  • $begingroup$
    @SagnikSaha How is that false?
    $endgroup$
    – Migos
    Dec 14 '18 at 17:43










  • $begingroup$
    hi, i made a mistake. i meant to say uniformly continuous. not continuous.
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:43










  • $begingroup$
    The question in the title and the question in the body are different. Do you want to show continuity or uniform continuity?
    $endgroup$
    – saulspatz
    Dec 14 '18 at 17:48










  • $begingroup$
    Good question! Helps to build some intuition about uniform continuity. It isn't necessary for the set to be compact and/or the function to be "obviously uniform continuous".
    $endgroup$
    – ploosu2
    Dec 14 '18 at 19:31


















3












$begingroup$



How can I show that the function $f(x) = x^{2}$ is uniformly continuous on $bigcup_{n = 1}^{infty} [n, n + frac{1}{n^{3}}]$?




As $n to infty$, I know that we get only all of the integers; however, the result doesn't make sense to me. I'm aware of both the sequential and $epsilon-delta$ definitions of continuity, but I didn't get anywhere with them.






real-analysis analysis continuity epsilon-delta uniform-continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    @SagnikSaha How is that false?
    $endgroup$
    – Migos
    Dec 14 '18 at 17:43










  • $begingroup$
    hi, i made a mistake. i meant to say uniformly continuous. not continuous.
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:43










  • $begingroup$
    The question in the title and the question in the body are different. Do you want to show continuity or uniform continuity?
    $endgroup$
    – saulspatz
    Dec 14 '18 at 17:48










  • $begingroup$
    Good question! Helps to build some intuition about uniform continuity. It isn't necessary for the set to be compact and/or the function to be "obviously uniform continuous".
    $endgroup$
    – ploosu2
    Dec 14 '18 at 19:31
















3












3








3





$begingroup$



How can I show that the function $f(x) = x^{2}$ is uniformly continuous on $bigcup_{n = 1}^{infty} [n, n + frac{1}{n^{3}}]$?




As $n to infty$, I know that we get only all of the integers; however, the result doesn't make sense to me. I'm aware of both the sequential and $epsilon-delta$ definitions of continuity, but I didn't get anywhere with them.






real-analysis analysis continuity epsilon-delta uniform-continuity







share|cite|improve this question











$endgroup$





How can I show that the function $f(x) = x^{2}$ is uniformly continuous on $bigcup_{n = 1}^{infty} [n, n + frac{1}{n^{3}}]$?




As $n to infty$, I know that we get only all of the integers; however, the result doesn't make sense to me. I'm aware of both the sequential and $epsilon-delta$ definitions of continuity, but I didn't get anywhere with them.






real-analysis analysis continuity epsilon-delta uniform-continuity




real-analysis analysis continuity epsilon-delta uniform-continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 21:04









Batominovski

33.1k33293




33.1k33293










asked Dec 14 '18 at 17:22









stackofhay42stackofhay42

2437




2437












  • $begingroup$
    @SagnikSaha How is that false?
    $endgroup$
    – Migos
    Dec 14 '18 at 17:43










  • $begingroup$
    hi, i made a mistake. i meant to say uniformly continuous. not continuous.
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:43










  • $begingroup$
    The question in the title and the question in the body are different. Do you want to show continuity or uniform continuity?
    $endgroup$
    – saulspatz
    Dec 14 '18 at 17:48










  • $begingroup$
    Good question! Helps to build some intuition about uniform continuity. It isn't necessary for the set to be compact and/or the function to be "obviously uniform continuous".
    $endgroup$
    – ploosu2
    Dec 14 '18 at 19:31




















  • $begingroup$
    @SagnikSaha How is that false?
    $endgroup$
    – Migos
    Dec 14 '18 at 17:43










  • $begingroup$
    hi, i made a mistake. i meant to say uniformly continuous. not continuous.
    $endgroup$
    – stackofhay42
    Dec 14 '18 at 17:43










  • $begingroup$
    The question in the title and the question in the body are different. Do you want to show continuity or uniform continuity?
    $endgroup$
    – saulspatz
    Dec 14 '18 at 17:48










  • $begingroup$
    Good question! Helps to build some intuition about uniform continuity. It isn't necessary for the set to be compact and/or the function to be "obviously uniform continuous".
    $endgroup$
    – ploosu2
    Dec 14 '18 at 19:31


















$begingroup$
@SagnikSaha How is that false?
$endgroup$
– Migos
Dec 14 '18 at 17:43




$begingroup$
@SagnikSaha How is that false?
$endgroup$
– Migos
Dec 14 '18 at 17:43












$begingroup$
hi, i made a mistake. i meant to say uniformly continuous. not continuous.
$endgroup$
– stackofhay42
Dec 14 '18 at 17:43




$begingroup$
hi, i made a mistake. i meant to say uniformly continuous. not continuous.
$endgroup$
– stackofhay42
Dec 14 '18 at 17:43












$begingroup$
The question in the title and the question in the body are different. Do you want to show continuity or uniform continuity?
$endgroup$
– saulspatz
Dec 14 '18 at 17:48




$begingroup$
The question in the title and the question in the body are different. Do you want to show continuity or uniform continuity?
$endgroup$
– saulspatz
Dec 14 '18 at 17:48












$begingroup$
Good question! Helps to build some intuition about uniform continuity. It isn't necessary for the set to be compact and/or the function to be "obviously uniform continuous".
$endgroup$
– ploosu2
Dec 14 '18 at 19:31






$begingroup$
Good question! Helps to build some intuition about uniform continuity. It isn't necessary for the set to be compact and/or the function to be "obviously uniform continuous".
$endgroup$
– ploosu2
Dec 14 '18 at 19:31












2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint



Let $epsilon>0$. You need to choose $delta$ so $|x-y|<delta$ implies $|x^2-y^2|<epsilon$, as $x,y$ range over your set. You can always make $delta$ smaller; choose $delta$ small enough so that $|x-y|<delta$ implies they are in the same interval $[n,n+n^{-3}]$.



The maximum value of $|x^2-y^2|$ for $x, yin [n,n+n^{-3}]$ is $(n+n^{-3})^2-n^2=2n^{-2}+n^{-6}$. We can choose $N$ large enough so $n>N$ implies this difference is less than $epsilon$. This allows us to restrict our attention to the intervals $[n,n+n^{-3}]$ where $n<N$. On those intervals, $f$ has a bounded derivative.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that for $x,yin [m,m+{1over m^3}]$ for some $m$ we have$$|x^2-y^2|le left(m+{1over m^3}right)^2-m^2={2over m^2}+{1over m^6}le {3over m^2}$$then for some $epsilon>0$ let $N=lfloorsqrt{3over epsilon}rfloor+1$ therefore we obtain $$text{for }x,y in [m,m+{1over m^3}], mge Nto |x^2-y^2|<epsilon$$and for $x,y in [m,m+{1over m^3}]$ , $ m< N$ let $delta_m(epsilon)>0$ be such that if $|x-y|<delta_m(epsilon)$ then $|x^2-y^2|<epsilon$ therefore we finally write the general $delta $ as$$delta= min_{1le mle lfloorsqrt{3over epsilon}rfloor}delta_m(epsilon)$$ which is a function of $epsilon$ and the proof is complete.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      Hint



      Let $epsilon>0$. You need to choose $delta$ so $|x-y|<delta$ implies $|x^2-y^2|<epsilon$, as $x,y$ range over your set. You can always make $delta$ smaller; choose $delta$ small enough so that $|x-y|<delta$ implies they are in the same interval $[n,n+n^{-3}]$.



      The maximum value of $|x^2-y^2|$ for $x, yin [n,n+n^{-3}]$ is $(n+n^{-3})^2-n^2=2n^{-2}+n^{-6}$. We can choose $N$ large enough so $n>N$ implies this difference is less than $epsilon$. This allows us to restrict our attention to the intervals $[n,n+n^{-3}]$ where $n<N$. On those intervals, $f$ has a bounded derivative.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Hint



        Let $epsilon>0$. You need to choose $delta$ so $|x-y|<delta$ implies $|x^2-y^2|<epsilon$, as $x,y$ range over your set. You can always make $delta$ smaller; choose $delta$ small enough so that $|x-y|<delta$ implies they are in the same interval $[n,n+n^{-3}]$.



        The maximum value of $|x^2-y^2|$ for $x, yin [n,n+n^{-3}]$ is $(n+n^{-3})^2-n^2=2n^{-2}+n^{-6}$. We can choose $N$ large enough so $n>N$ implies this difference is less than $epsilon$. This allows us to restrict our attention to the intervals $[n,n+n^{-3}]$ where $n<N$. On those intervals, $f$ has a bounded derivative.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Hint



          Let $epsilon>0$. You need to choose $delta$ so $|x-y|<delta$ implies $|x^2-y^2|<epsilon$, as $x,y$ range over your set. You can always make $delta$ smaller; choose $delta$ small enough so that $|x-y|<delta$ implies they are in the same interval $[n,n+n^{-3}]$.



          The maximum value of $|x^2-y^2|$ for $x, yin [n,n+n^{-3}]$ is $(n+n^{-3})^2-n^2=2n^{-2}+n^{-6}$. We can choose $N$ large enough so $n>N$ implies this difference is less than $epsilon$. This allows us to restrict our attention to the intervals $[n,n+n^{-3}]$ where $n<N$. On those intervals, $f$ has a bounded derivative.






          share|cite|improve this answer









          $endgroup$



          Hint



          Let $epsilon>0$. You need to choose $delta$ so $|x-y|<delta$ implies $|x^2-y^2|<epsilon$, as $x,y$ range over your set. You can always make $delta$ smaller; choose $delta$ small enough so that $|x-y|<delta$ implies they are in the same interval $[n,n+n^{-3}]$.



          The maximum value of $|x^2-y^2|$ for $x, yin [n,n+n^{-3}]$ is $(n+n^{-3})^2-n^2=2n^{-2}+n^{-6}$. We can choose $N$ large enough so $n>N$ implies this difference is less than $epsilon$. This allows us to restrict our attention to the intervals $[n,n+n^{-3}]$ where $n<N$. On those intervals, $f$ has a bounded derivative.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 17:57









          Mike EarnestMike Earnest

          24.1k22151




          24.1k22151























              1












              $begingroup$

              Note that for $x,yin [m,m+{1over m^3}]$ for some $m$ we have$$|x^2-y^2|le left(m+{1over m^3}right)^2-m^2={2over m^2}+{1over m^6}le {3over m^2}$$then for some $epsilon>0$ let $N=lfloorsqrt{3over epsilon}rfloor+1$ therefore we obtain $$text{for }x,y in [m,m+{1over m^3}], mge Nto |x^2-y^2|<epsilon$$and for $x,y in [m,m+{1over m^3}]$ , $ m< N$ let $delta_m(epsilon)>0$ be such that if $|x-y|<delta_m(epsilon)$ then $|x^2-y^2|<epsilon$ therefore we finally write the general $delta $ as$$delta= min_{1le mle lfloorsqrt{3over epsilon}rfloor}delta_m(epsilon)$$ which is a function of $epsilon$ and the proof is complete.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note that for $x,yin [m,m+{1over m^3}]$ for some $m$ we have$$|x^2-y^2|le left(m+{1over m^3}right)^2-m^2={2over m^2}+{1over m^6}le {3over m^2}$$then for some $epsilon>0$ let $N=lfloorsqrt{3over epsilon}rfloor+1$ therefore we obtain $$text{for }x,y in [m,m+{1over m^3}], mge Nto |x^2-y^2|<epsilon$$and for $x,y in [m,m+{1over m^3}]$ , $ m< N$ let $delta_m(epsilon)>0$ be such that if $|x-y|<delta_m(epsilon)$ then $|x^2-y^2|<epsilon$ therefore we finally write the general $delta $ as$$delta= min_{1le mle lfloorsqrt{3over epsilon}rfloor}delta_m(epsilon)$$ which is a function of $epsilon$ and the proof is complete.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note that for $x,yin [m,m+{1over m^3}]$ for some $m$ we have$$|x^2-y^2|le left(m+{1over m^3}right)^2-m^2={2over m^2}+{1over m^6}le {3over m^2}$$then for some $epsilon>0$ let $N=lfloorsqrt{3over epsilon}rfloor+1$ therefore we obtain $$text{for }x,y in [m,m+{1over m^3}], mge Nto |x^2-y^2|<epsilon$$and for $x,y in [m,m+{1over m^3}]$ , $ m< N$ let $delta_m(epsilon)>0$ be such that if $|x-y|<delta_m(epsilon)$ then $|x^2-y^2|<epsilon$ therefore we finally write the general $delta $ as$$delta= min_{1le mle lfloorsqrt{3over epsilon}rfloor}delta_m(epsilon)$$ which is a function of $epsilon$ and the proof is complete.






                  share|cite|improve this answer









                  $endgroup$



                  Note that for $x,yin [m,m+{1over m^3}]$ for some $m$ we have$$|x^2-y^2|le left(m+{1over m^3}right)^2-m^2={2over m^2}+{1over m^6}le {3over m^2}$$then for some $epsilon>0$ let $N=lfloorsqrt{3over epsilon}rfloor+1$ therefore we obtain $$text{for }x,y in [m,m+{1over m^3}], mge Nto |x^2-y^2|<epsilon$$and for $x,y in [m,m+{1over m^3}]$ , $ m< N$ let $delta_m(epsilon)>0$ be such that if $|x-y|<delta_m(epsilon)$ then $|x^2-y^2|<epsilon$ therefore we finally write the general $delta $ as$$delta= min_{1le mle lfloorsqrt{3over epsilon}rfloor}delta_m(epsilon)$$ which is a function of $epsilon$ and the proof is complete.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 21:11









                  Mostafa AyazMostafa Ayaz

                  15.7k3939




                  15.7k3939






























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