$$s:=\{(1/2n) - (1/(2m+1))\} n,m \in N$$
I know that the inf is $-1/3$ and the sup is $1/2$. I know to prove this I that I have to show they are the greatest lowerbound and least upperbound respectively but do I have to prove that $(1/2n)$ is decreasing and bounded $0$ and that $- (1/(2m+1)$ is increasing and bounded by $0$ first.
I'm unsure about what exactly I have to prove.
TO show $-1/3$ is the greatest lowerbound of the set you need the fact that $1/2n$ goes to $0$ as $n$ tends to $\infty$, of course all the elements in the set is greater than $-1/3$, now say $-1/3$ is not the greatest lowerbound, say $x$ is the greatest lowerbound, so $x>-1/3$ , so no term of the sequence must lie between $(-1/3,x)$ as $x$ is the greatest lowerbound
Now fix $m=1$, now as $1/2n$ goes to zero there is a term in the set which lies in the interval $(-1/3,x)$, hence a contradiction, so $-1/3$ is the greatest lowerbound.
Now to prove the least upperbound you need the fact that $−(1/(2m+1)$ is increasing and bounded by $0$.
I hope everything is clear to you now, proceed similarly for the least upperbound.