A set $A = \left\{(-1)^n + \frac{1}{m} : n,m \in \Bbb N\right\} \cup \{-1\}$ is given.
a) I shall find out and justify what the supremum an infimum is.
b) Is the sup a maximum or the inf a minimum.
a) For the sup, I find out $n=2$ and $m=1$ because then I have $1+1=2$. For the inf, I have $\displaystyle \lim _{m\to\infty} (1/m) = 0$ and $n = \text{odd number} \implies 0-1=-1$.
Now my questions are: Is that correct? How can I now find out if there is a max or min? Can I say for the sup that $2$ and $1$ are elements of $\Bbb N$ so the sup is a max and for the inf same? And what about the $\{-1\}$ in the set, what does this $\{-1\}$ mean for the inf, sup, min and max?
Best Answer
Every element $x \in A$ is $x\leq2$(why?) Thus $2$ is an upper bound of $A$. Can you find a lower upper bound of $A$? The smallest of the upper bounds is the $\sup A$.
Every element $x \in A$ is $x\geq-1$(why?) Thus $-1$ is a lower bound of $A$. Can you find a larger lower bound of $A$? The largest of the lower bounds is the $\inf A$.
Edit: You got it right $\sup A=2$ and $\inf A=-1.$
To prove this you must show that $2$ is an upper bound of $A$ (i.e. $x\leq 2, \ \forall x \in A$ ) and that for any $\epsilon>0, \ 2-\epsilon$ is not an upper bound of $A$(i.e. there is an $x \in A$ with $x>2-\epsilon$) . SImilar with $\inf A$.