Infimum of a union of sets











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Considering the set



$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $



I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.



Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?






real-analysis







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  • When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
    – Alex S
    Jan 29 '16 at 4:04






  • 3




    The set $(0,0]$ should be considered empty.
    – OnoL
    Jan 29 '16 at 4:05










  • I'm not sure if clopen sets of that form contain the point or not.
    – IntegrateThis
    Jan 29 '16 at 4:05










  • aw ok thus the source of my confusion.
    – IntegrateThis
    Jan 29 '16 at 4:05






  • 2




    When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
    – user238841
    Jan 29 '16 at 4:06















up vote
0
down vote

favorite












Considering the set



$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $



I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.



Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?






real-analysis







share|cite|improve this question
























  • When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
    – Alex S
    Jan 29 '16 at 4:04






  • 3




    The set $(0,0]$ should be considered empty.
    – OnoL
    Jan 29 '16 at 4:05










  • I'm not sure if clopen sets of that form contain the point or not.
    – IntegrateThis
    Jan 29 '16 at 4:05










  • aw ok thus the source of my confusion.
    – IntegrateThis
    Jan 29 '16 at 4:05






  • 2




    When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
    – user238841
    Jan 29 '16 at 4:06













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Considering the set



$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $



I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.



Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?






real-analysis







share|cite|improve this question















Considering the set



$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $



I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.



Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













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edited Nov 17 at 10:39









amWhy

191k27223437




191k27223437










asked Jan 29 '16 at 3:58









IntegrateThis

1,6971717




1,6971717












  • When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
    – Alex S
    Jan 29 '16 at 4:04






  • 3




    The set $(0,0]$ should be considered empty.
    – OnoL
    Jan 29 '16 at 4:05










  • I'm not sure if clopen sets of that form contain the point or not.
    – IntegrateThis
    Jan 29 '16 at 4:05










  • aw ok thus the source of my confusion.
    – IntegrateThis
    Jan 29 '16 at 4:05






  • 2




    When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
    – user238841
    Jan 29 '16 at 4:06


















  • When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
    – Alex S
    Jan 29 '16 at 4:04






  • 3




    The set $(0,0]$ should be considered empty.
    – OnoL
    Jan 29 '16 at 4:05










  • I'm not sure if clopen sets of that form contain the point or not.
    – IntegrateThis
    Jan 29 '16 at 4:05










  • aw ok thus the source of my confusion.
    – IntegrateThis
    Jan 29 '16 at 4:05






  • 2




    When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
    – user238841
    Jan 29 '16 at 4:06
















When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04




When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04




3




3




The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05




The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05












I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05




I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05












aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05




aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05




2




2




When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06




When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06










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Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.



Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.






share|cite|improve this answer





















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    Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.



    Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.



      Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.






      share|cite|improve this answer























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        0
        down vote










        up vote
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        down vote









        Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.



        Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.






        share|cite|improve this answer












        Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.



        Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 29 '16 at 4:32









        M10687

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