$X_nleq Y_n implies inf X_n leq inf Y_n$











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$x_nleq y_n forall n in N implies inf x_n leq inf y_n$



It is obvious from the definition of infimum and supremum, $sup x_n leq y_n$ and $inf x_n leq x_n leq y_n$. However I do not know how to use the definition to prove formally that $sup x_n leq inf y_n$ and conclude that
$inf x_n leq sup y_n$.






real-analysis sequences-and-series supremum-and-infimum







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  • $sup x_n le inf y_n$ does not hold. Try some sequences with $x_n = y_n$.
    – mschauer
    Nov 22 at 21:41










  • I would be careful with your statement. It is true that if $x_{n} leq c$ for some fixed real number $c$, then $sup x_{n} leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} leq y_{n}$ for all $n$, that doesn't mean $sup x_{n} leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - frac{2}{n}$ and $y_{n} = 1 - frac{1}{n}$. Clealry, $x_{n} leq y_{n}$ for all $n$. But $sup {x_{n}} = 1$ which is bigger than all $y_{n}$.
    – layman
    Nov 22 at 21:41










  • It is not true that $sup x_n leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$
    – Will M.
    Nov 22 at 21:41















up vote
1
down vote

favorite












$x_nleq y_n forall n in N implies inf x_n leq inf y_n$



It is obvious from the definition of infimum and supremum, $sup x_n leq y_n$ and $inf x_n leq x_n leq y_n$. However I do not know how to use the definition to prove formally that $sup x_n leq inf y_n$ and conclude that
$inf x_n leq sup y_n$.






real-analysis sequences-and-series supremum-and-infimum







share|cite|improve this question
























  • $sup x_n le inf y_n$ does not hold. Try some sequences with $x_n = y_n$.
    – mschauer
    Nov 22 at 21:41










  • I would be careful with your statement. It is true that if $x_{n} leq c$ for some fixed real number $c$, then $sup x_{n} leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} leq y_{n}$ for all $n$, that doesn't mean $sup x_{n} leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - frac{2}{n}$ and $y_{n} = 1 - frac{1}{n}$. Clealry, $x_{n} leq y_{n}$ for all $n$. But $sup {x_{n}} = 1$ which is bigger than all $y_{n}$.
    – layman
    Nov 22 at 21:41










  • It is not true that $sup x_n leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$
    – Will M.
    Nov 22 at 21:41













up vote
1
down vote

favorite









up vote
1
down vote

favorite











$x_nleq y_n forall n in N implies inf x_n leq inf y_n$



It is obvious from the definition of infimum and supremum, $sup x_n leq y_n$ and $inf x_n leq x_n leq y_n$. However I do not know how to use the definition to prove formally that $sup x_n leq inf y_n$ and conclude that
$inf x_n leq sup y_n$.






real-analysis sequences-and-series supremum-and-infimum







share|cite|improve this question















$x_nleq y_n forall n in N implies inf x_n leq inf y_n$



It is obvious from the definition of infimum and supremum, $sup x_n leq y_n$ and $inf x_n leq x_n leq y_n$. However I do not know how to use the definition to prove formally that $sup x_n leq inf y_n$ and conclude that
$inf x_n leq sup y_n$.






real-analysis sequences-and-series supremum-and-infimum




real-analysis sequences-and-series supremum-and-infimum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 21:37









Yadati Kiran

1,327418




1,327418










asked Nov 22 at 21:36









Pumpkin

4861417




4861417












  • $sup x_n le inf y_n$ does not hold. Try some sequences with $x_n = y_n$.
    – mschauer
    Nov 22 at 21:41










  • I would be careful with your statement. It is true that if $x_{n} leq c$ for some fixed real number $c$, then $sup x_{n} leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} leq y_{n}$ for all $n$, that doesn't mean $sup x_{n} leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - frac{2}{n}$ and $y_{n} = 1 - frac{1}{n}$. Clealry, $x_{n} leq y_{n}$ for all $n$. But $sup {x_{n}} = 1$ which is bigger than all $y_{n}$.
    – layman
    Nov 22 at 21:41










  • It is not true that $sup x_n leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$
    – Will M.
    Nov 22 at 21:41


















  • $sup x_n le inf y_n$ does not hold. Try some sequences with $x_n = y_n$.
    – mschauer
    Nov 22 at 21:41










  • I would be careful with your statement. It is true that if $x_{n} leq c$ for some fixed real number $c$, then $sup x_{n} leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} leq y_{n}$ for all $n$, that doesn't mean $sup x_{n} leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - frac{2}{n}$ and $y_{n} = 1 - frac{1}{n}$. Clealry, $x_{n} leq y_{n}$ for all $n$. But $sup {x_{n}} = 1$ which is bigger than all $y_{n}$.
    – layman
    Nov 22 at 21:41










  • It is not true that $sup x_n leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$
    – Will M.
    Nov 22 at 21:41
















$sup x_n le inf y_n$ does not hold. Try some sequences with $x_n = y_n$.
– mschauer
Nov 22 at 21:41




$sup x_n le inf y_n$ does not hold. Try some sequences with $x_n = y_n$.
– mschauer
Nov 22 at 21:41












I would be careful with your statement. It is true that if $x_{n} leq c$ for some fixed real number $c$, then $sup x_{n} leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} leq y_{n}$ for all $n$, that doesn't mean $sup x_{n} leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - frac{2}{n}$ and $y_{n} = 1 - frac{1}{n}$. Clealry, $x_{n} leq y_{n}$ for all $n$. But $sup {x_{n}} = 1$ which is bigger than all $y_{n}$.
– layman
Nov 22 at 21:41




I would be careful with your statement. It is true that if $x_{n} leq c$ for some fixed real number $c$, then $sup x_{n} leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} leq y_{n}$ for all $n$, that doesn't mean $sup x_{n} leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - frac{2}{n}$ and $y_{n} = 1 - frac{1}{n}$. Clealry, $x_{n} leq y_{n}$ for all $n$. But $sup {x_{n}} = 1$ which is bigger than all $y_{n}$.
– layman
Nov 22 at 21:41












It is not true that $sup x_n leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$
– Will M.
Nov 22 at 21:41




It is not true that $sup x_n leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$
– Will M.
Nov 22 at 21:41










2 Answers
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Note that
$$
inf x_nleq x_nle y_n
$$
for all $n$. So $inf x_n$ is a lower bound for $y_n$ whence
$$
inf x_nleq inf y_n.
$$






share|cite|improve this answer




























    up vote
    1
    down vote













    Assume by contradiction $$inf x_n=x>inf y_n=y$$therefore$$forall 0<epsilon <x-y,exists
    Nqquad forall n>Nto 0le y_n-y<epsilon<x-y$$    which means that the exists $nin Bbb N$ such that $$yle y_n<xle x_n$$or equivalently $$y_n<x_n$$which is a contradiction. Therefore $xle y$ and the proof is complete $blacksquare$






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
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      up vote
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      down vote



      accepted










      Note that
      $$
      inf x_nleq x_nle y_n
      $$
      for all $n$. So $inf x_n$ is a lower bound for $y_n$ whence
      $$
      inf x_nleq inf y_n.
      $$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Note that
        $$
        inf x_nleq x_nle y_n
        $$
        for all $n$. So $inf x_n$ is a lower bound for $y_n$ whence
        $$
        inf x_nleq inf y_n.
        $$






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Note that
          $$
          inf x_nleq x_nle y_n
          $$
          for all $n$. So $inf x_n$ is a lower bound for $y_n$ whence
          $$
          inf x_nleq inf y_n.
          $$






          share|cite|improve this answer












          Note that
          $$
          inf x_nleq x_nle y_n
          $$
          for all $n$. So $inf x_n$ is a lower bound for $y_n$ whence
          $$
          inf x_nleq inf y_n.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 22:00









          Foobaz John

          20.3k41250




          20.3k41250






















              up vote
              1
              down vote













              Assume by contradiction $$inf x_n=x>inf y_n=y$$therefore$$forall 0<epsilon <x-y,exists
              Nqquad forall n>Nto 0le y_n-y<epsilon<x-y$$              which means that the exists $nin Bbb N$ such that $$yle y_n<xle x_n$$or equivalently $$y_n<x_n$$which is a contradiction. Therefore $xle y$ and the proof is complete $blacksquare$






              share|cite|improve this answer

























                up vote
                1
                down vote













                Assume by contradiction $$inf x_n=x>inf y_n=y$$therefore$$forall 0<epsilon <x-y,exists
                Nqquad forall n>Nto 0le y_n-y<epsilon<x-y$$                which means that the exists $nin Bbb N$ such that $$yle y_n<xle x_n$$or equivalently $$y_n<x_n$$which is a contradiction. Therefore $xle y$ and the proof is complete $blacksquare$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Assume by contradiction $$inf x_n=x>inf y_n=y$$therefore$$forall 0<epsilon <x-y,exists
                  Nqquad forall n>Nto 0le y_n-y<epsilon<x-y$$                  which means that the exists $nin Bbb N$ such that $$yle y_n<xle x_n$$or equivalently $$y_n<x_n$$which is a contradiction. Therefore $xle y$ and the proof is complete $blacksquare$






                  share|cite|improve this answer












                  Assume by contradiction $$inf x_n=x>inf y_n=y$$therefore$$forall 0<epsilon <x-y,exists
                  Nqquad forall n>Nto 0le y_n-y<epsilon<x-y$$                  which means that the exists $nin Bbb N$ such that $$yle y_n<xle x_n$$or equivalently $$y_n<x_n$$which is a contradiction. Therefore $xle y$ and the proof is complete $blacksquare$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 at 21:45









                  Mostafa Ayaz

                  13.5k3836




                  13.5k3836






























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