Supremum of a set with sequence

Question

I am trying to solve this practice question to prepare for my exam but I am getting stuck in the final part. Consider the sequence $(z_{n})$ given by

\begin{equation} z_{n}=(-1)^{n}+\frac{1}{n}, \;\; n\in\mathbb{N_{+}}=\{1,2,3,...\} \end{equation} Then let $A\subset\mathbb{R}$ be a subset with the properties $\inf A=-2$ and $\sup A = 1$. Let \begin{equation} B=\{az_{n}|a\in A,n\in\mathbb{N_{+}\} } \end{equation} So now I want to find $\sup B$ and prove this. Finding the $\sup B$ I did using the sequence and the infimum and supremum of $A$ making me claim that $\sup B=2$. Now I want to prove this using the alternative characterization of supremum. Namely, \begin{equation} (i) \text{ for all }b\in B, \text{ we have that } b\leq 2 \\ (ii) \text{ for all }\epsilon>0, \text{ there exists } b\in B, \text{ such that } b>2-\epsilon. \end{equation} Proving the first property is not too difficult as $az_{n}\in(-3,2)$ so we can simply say that: let $b\in B$, then $b\in (-3,2)$ so indeed $b\leq 2$.

Now proving $(ii)$ is where I get stuck. My intuition tells me that I should use the definition of infimum and/or supremum of $A$. We know these so we are able to say that \begin{equation} \text{ for all }\epsilon>0, \text{ there exists }a_{1}\in A, \text{ such that }a_{1}>1-\epsilon \\ \text{ and that for all }\epsilon>0,\text{ there exists }a_{2}\in A, \text{ such that }a_{2}<-2+\epsilon \end{equation} I believe that too complete my proof I should choose one of these two $a_{1}$ or $a_{2}$ and then be able to construct the inequality that I want. But for some reason I can't seem to find it, so I am beginning to wonder if I should be using these definitions or not.

Answer 1

You do have $\sup B=2$; besides $\inf B=-3$. Take $b\in B$ with $b\geqslant0$. Then $b=az_n$ for some $a\in A$ and some $n\in\Bbb N$. But then $b\leqslant2$, since:

• if $a,z_n\geqslant0$, then $a\leqslant1$ and $z_n\leqslant\frac32$, and therefore $b=az_n\leqslant\frac32$;
• if $a,z_n\leqslant0$, then $a\geqslant-2$ and $z_n>-1$, and therefore $b=az_n<2$; and $\frac32<2$.

So, $\sup B\leqslant2$. Also, if $b\leqslant0$, then $b\geqslant-3$, since

• if $a\leqslant0\leqslant z_n$, then $a\geqslant-2$ and $z_n\leqslant\frac32$, and therefore $b=az_n\geqslant-3$;
• if $z_n\leqslant0\leqslant a$, then $a\leqslant1$ and $z_n>-1$, and therefore $b=az_n>-1$; and $-3<-1$.

So, $\inf B\geqslant-3$.

On the other hand, since $\inf A=-2$, there is some sequence $(a_n)_{n\in\Bbb N}$ of elements of $A$ such that $\lim_{n\to\infty}a_n=-2$. So, $\lim_{n\to\infty}a_nz_{2n-1}=2$, and therefore $\sup B=2$.

Finally, since $\sup A=1$, there is some sequence $(a_n)_{n\in\Bbb N}$ of elements of $A$ such that $\lim_{n\to\infty}a_n=-2$. So, $\lim_{n\to\infty}a_nz_2=-3$, and therefore $\inf B=-3$.