Determine the least upper and the greatest lower bounds of the following sets:

$a)$ A={${\frac{m} {n} : m,n \in N ,m<2n}$}

$b) $B={${n^{1/2} -[n^{1/2}] : n \in N}$}

my attempts : for option a) if m= n=1 then greatest lower bound will be 1 and if m = 2n ,the the least upper bound will be 2

for option b) here greatest lower bound will be 0... that is $n^{1/2} -n^{1/2}$=0 here i don't know the what is least upper bounds

Pliz verified its and tell me the solution

thanks in advance


a) Supremum will be 2. As choose any $\epsilon >0$ there exist a rational number such that $m-\epsilon <m/n<2$ And Infimum will be 0. As by Archimedean property for any $\epsilon>0$ there exist $m\in N$ such that $0<1/m<\epsilon$

b)note that $0\leq b<1$ for all $b\in B$ so Infimum is 0 as 0 is the element of B for any square number. Now Supremum is 1(we show that to you) Now choose $\epsilon>0$ and let $y=1-\epsilon$ and $1>y>0$ consider any integer $m$ such that $m>\frac{y^2}{2(1-y)}$ so then $y<\sqrt{m^2+2m}-m$ now note that, $\sqrt{m^2+2m}-m =\sqrt{m^2+2m}-[\sqrt{m^2+2m}]<1$

so given any $y>0$ we find a number $m'$ in $B$ such that $1-\epsilon<m'<1$ so Supremum is 1.

To understand the solution, you must know the definition of Supremum and Infimum.

  • $\begingroup$ thanks a lots karambir.kd $\endgroup$ – user476275 Mar 16 '18 at 3:37

Consider (a). All elements of $A$ are between $0$ and $2$. Now think about limits.

  • $\begingroup$ h0w zero come? as natural number doesnot contain 0 @ncmthsadist $\endgroup$ – user476275 Mar 16 '18 at 1:43
  • 1
    $\begingroup$ between 0 and 2, m/n<2 $\endgroup$ – NickC Mar 16 '18 at 1:52

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