it sounds like a silly question, but I'm having trouble to prove that
$$ \sup\left(\left\{\left. \frac1n-1 \right|n \text{ is odd}\right\}\right)=0 $$
I'm stuck in proving that $$ \forall \varepsilon >0 \exists a\in \left\{\left. \frac1n-1 \right|n \text{ is odd}\right\} a>0-\varepsilon $$
thanks in advance
Since it is unclear whether $n$ can be negative or not, we assume that it can be.
However, if $n$ is negative, $\frac 1 n$ is negative and hence $\frac 1 n - 1 \lt -1$.
Now say that $N$ is positive. Notice if $a \gt b \gt 0$, $\frac 1 a \lt \frac 1 b$.
Since $1$ is the smallest positive odd number, $\max S$ is achieved at $N=1$, which is $0$.
Now we will show that $\max S = \sup S$.
Suppose that it is not true. Then there are some number $K$ such that $k = \sup S$. Since $K \not = \max S$, $K$ is either more or less than $\max S$.
Suppose that $K$ is less than $\max S$. Then $\max S \gt K$, causing a contradiction.
Suppose that $K$ is more that $\max S$. Then there is some number $K^*$ such that $\max S \lt K^* \lt K$. However, since $K^* \gt \max S$, $K^*$ is larger than any element in $S$, which causes a contradiction to the definition of $\sup S$.
Therefore $\sup S = \max S = 0$.