exchange max and limit












4












$begingroup$


Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?



I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?



Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.



Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.






real-analysis







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$endgroup$












  • $begingroup$
    You are correct.
    $endgroup$
    – martini
    Mar 27 '15 at 12:32










  • $begingroup$
    @martini Thanks!
    $endgroup$
    – simonzack
    Mar 27 '15 at 12:33






  • 1




    $begingroup$
    Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
    $endgroup$
    – 5xum
    Mar 27 '15 at 12:38
















4












$begingroup$


Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?



I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?



Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.



Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are correct.
    $endgroup$
    – martini
    Mar 27 '15 at 12:32










  • $begingroup$
    @martini Thanks!
    $endgroup$
    – simonzack
    Mar 27 '15 at 12:33






  • 1




    $begingroup$
    Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
    $endgroup$
    – 5xum
    Mar 27 '15 at 12:38














4












4








4


1



$begingroup$


Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?



I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?



Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.



Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.






real-analysis







share|cite|improve this question









$endgroup$




Let $lim_{n rightarrow infty} a_n = a$ and $lim_{n rightarrow infty} b_n
= b$ exist, then is it true that $lim_{n rightarrow infty} max { a_n, b_n } = max { a, b }$?



I couldn't find this on wiki, but it seems correct. Here's my proof, can somebody check it?



Suppose $a > b$. Let $epsilon^{prime} > 0$. Take $0 < epsilon < min {
epsilon^{prime}, (a - b) / 2 }$. Then there is an $N$ s.t. $n > N
Rightarrow | a - a_n | < epsilon wedge | b - b_n | < epsilon Rightarrow
a_n - b_n > (a - epsilon) - (b + epsilon) > 0 Rightarrow | max { a_n, b_n
} - a | = | a_n - a | < epsilon < epsilon^{prime}$.



Suppose $c := a = b$. Let $epsilon > 0$. Then there is an $N$ s.t. $n >
N Rightarrow | c - a_n | < epsilon wedge | c - b_n | < epsilon Rightarrow
| c - max { a_n, b_n } | < epsilon$.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 27 '15 at 12:30









simonzacksimonzack

728619




728619












  • $begingroup$
    You are correct.
    $endgroup$
    – martini
    Mar 27 '15 at 12:32










  • $begingroup$
    @martini Thanks!
    $endgroup$
    – simonzack
    Mar 27 '15 at 12:33






  • 1




    $begingroup$
    Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
    $endgroup$
    – 5xum
    Mar 27 '15 at 12:38


















  • $begingroup$
    You are correct.
    $endgroup$
    – martini
    Mar 27 '15 at 12:32










  • $begingroup$
    @martini Thanks!
    $endgroup$
    – simonzack
    Mar 27 '15 at 12:33






  • 1




    $begingroup$
    Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
    $endgroup$
    – 5xum
    Mar 27 '15 at 12:38
















$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32




$begingroup$
You are correct.
$endgroup$
– martini
Mar 27 '15 at 12:32












$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33




$begingroup$
@martini Thanks!
$endgroup$
– simonzack
Mar 27 '15 at 12:33




1




1




$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38




$begingroup$
Very nicely proven. I would just change the third paragraph from "suppose $a>b$" into "suppose $aneq b$. Then, without loss of generality, $a>b$", but that's almost nitpicking.
$endgroup$
– 5xum
Mar 27 '15 at 12:38










2 Answers
2






active

oldest

votes


















1












$begingroup$

For posterity: On the set of real numbers, $max$ and $min$ are given by
$$
max(a, b) = frac{a + b + |b - a|}{2},qquad
min(a, b) = frac{a + b - |b - a|}{2}.
$$
(Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")



These functions are (almost obviously) continuous as functions of two variables, so
$$
lim_{n to infty} max(a_{n}, b_{n})
= maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
= max(a, b),
$$
and similarly for $min$.



In fact, continuity guarantees stronger "double limit" assertions:
$$
max(a, b)
= lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
= lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
$$
etc.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.



    To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      For posterity: On the set of real numbers, $max$ and $min$ are given by
      $$
      max(a, b) = frac{a + b + |b - a|}{2},qquad
      min(a, b) = frac{a + b - |b - a|}{2}.
      $$
      (Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")



      These functions are (almost obviously) continuous as functions of two variables, so
      $$
      lim_{n to infty} max(a_{n}, b_{n})
      = maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
      = max(a, b),
      $$
      and similarly for $min$.



      In fact, continuity guarantees stronger "double limit" assertions:
      $$
      max(a, b)
      = lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
      = lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
      $$
      etc.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For posterity: On the set of real numbers, $max$ and $min$ are given by
        $$
        max(a, b) = frac{a + b + |b - a|}{2},qquad
        min(a, b) = frac{a + b - |b - a|}{2}.
        $$
        (Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")



        These functions are (almost obviously) continuous as functions of two variables, so
        $$
        lim_{n to infty} max(a_{n}, b_{n})
        = maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
        = max(a, b),
        $$
        and similarly for $min$.



        In fact, continuity guarantees stronger "double limit" assertions:
        $$
        max(a, b)
        = lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
        = lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
        $$
        etc.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For posterity: On the set of real numbers, $max$ and $min$ are given by
          $$
          max(a, b) = frac{a + b + |b - a|}{2},qquad
          min(a, b) = frac{a + b - |b - a|}{2}.
          $$
          (Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")



          These functions are (almost obviously) continuous as functions of two variables, so
          $$
          lim_{n to infty} max(a_{n}, b_{n})
          = maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
          = max(a, b),
          $$
          and similarly for $min$.



          In fact, continuity guarantees stronger "double limit" assertions:
          $$
          max(a, b)
          = lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
          = lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
          $$
          etc.






          share|cite|improve this answer









          $endgroup$



          For posterity: On the set of real numbers, $max$ and $min$ are given by
          $$
          max(a, b) = frac{a + b + |b - a|}{2},qquad
          min(a, b) = frac{a + b - |b - a|}{2}.
          $$
          (Proof: "Go to the midpoint $frac{1}{2}(a + b)$, then step to the right or left by half the distance between the numbers, $frac{1}{2}|b - a|$.")



          These functions are (almost obviously) continuous as functions of two variables, so
          $$
          lim_{n to infty} max(a_{n}, b_{n})
          = maxbigl(lim_{n to infty} a_{n}, lim_{n to infty} b_{n}bigr)
          = max(a, b),
          $$
          and similarly for $min$.



          In fact, continuity guarantees stronger "double limit" assertions:
          $$
          max(a, b)
          = lim_{m to infty} lim_{n to infty} max(a_{m}, b_{n})
          = lim_{n to infty} lim_{m to infty} max(a_{m}, b_{n}),
          $$
          etc.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 27 '15 at 14:16









          Andrew D. HwangAndrew D. Hwang

          52.9k447113




          52.9k447113























              0












              $begingroup$

              A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.



              To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.



                To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.



                  To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.






                  share|cite|improve this answer









                  $endgroup$



                  A kind of overkill: fix $K$ compact (in this case $K={1,2}$) and consider the Banach space $(mathcal{C}(K),|cdot|_infty)$. Since the norm is continuous, then $lim_n |f_n|_infty=|lim_n f_n|_infty$ for each sequence $(f_n)$ in $mathcal{C}(K)$, provided the limits exist.



                  To conclude the proof, use that $x_n to x$ iff $x_n^+ to x^+$ and $x_n^- to x^-$, where $a=a^+-a^-$ is the unique decomposition into positive and negative part.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 21:30









                  Paolo LeonettiPaolo Leonetti

                  11.5k21550




                  11.5k21550






























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