Find the lub and glb of $S= \left\{ \dfrac 32,-\dfrac 43,\dfrac 54,-\dfrac 65,\dfrac 76,-\dfrac 87,...\right\}$
My try:
\begin{align}
S &=\left\{ \dfrac 32,\dfrac 54,\dfrac 76,...\right\}\bigcup \left\{ -\dfrac 43,-\dfrac 65,-\dfrac 87,...\right\}\\
&= \left\{\dfrac{2n+1}{2n}:n\in\Bbb{N}\right\}\bigcup \left\{-\dfrac{2n+2}{2n+1}:n\in\Bbb{N}\right\}\\
&=\left\{1+\dfrac{1}{2n}:n\in\Bbb{N}\right\}\bigcup \left\{-1-\dfrac{1}{2n+1}:n\in\Bbb{N}\right\}\\
&= A \cup B
\end{align}
Now $\left\{1+\dfrac{1}{2n}\right\}_n$ is monotonically decreasing ans as $n\to\infty, \left\{1+\dfrac{1}{2n}\right\}\to 1 $. So for being monotonically decreasing, the maximum value is attained at $n=1$. So $\sup (A)=\dfrac 32$ and $\inf(A)=1$.
Also $\left\{-1-\dfrac{1}{2n+1}\right\}_n$ is monotonically increasing and $ \lim_ {n\to\infty} \left\{-1-\dfrac{1}{2n+1}\right\}=-1 $. So for being monotonically increasing, the minimum value is attained at $n=1$. So $\sup (B)=-1$ and $\inf(B)=-\dfrac 43$.
Now $\sup (S)= \max\{\sup (A), \sup(B)\}=\max\left\{\dfrac 32, -1\right\}=\dfrac 32$ and
$\inf (S)= \min\{\inf (A), \inf(B)\}=\max\left\{1,-\dfrac 43,\right\}=-\dfrac 43$
So is this correct? Or is there anything to improve? Is it still necessary to use the definition of being those numbers to be the supremum and infimum?