find the inf/sup of the set A= { n $ \in \mathbb{N} $ | $n^2-3n +1 $}
before finding the inf and sup i checked for the first terms of this set, for n $\in $ {0, . . . 6} we have {1 , -1 , -7 ,5 , 11 ,19} $\subset A$
so i assumed that -7 may be the infimum so based on the defintion
-7 $\le$ a $ \hspace{20mm } $ $ \forall a \in A$
-7 $\le$ $n^2-3n+1 \hspace{15mm}$ $ n^2 -3n +8 \geq 0 \hspace{15mm} \Delta = 9-32 < 0 \implies \forall n \hspace {1cm} -7 \leq a_n \hspace{1cm}\forall a_n \in A $
i don't realize how to find the supremum ( i know it is + $ \infty$ but i dont know how to prove it, i tried with reductio ad absurdum but i failed finding a contradiction)
The least troublesome way to show that your set $A$ doesn't have supremum is to argue that it isn't even bounded from above; recall that the supremum of a set of real numbers is the least upper bound of the set, which cannot exist if there's no upper bound.
Now, to show that there's no upper bound for your set $A$, we can argue by contradiction. Suppose $A$ has an upper bound, say $\alpha > 0$. Then we clearly have $$3n-1 \geq 2 \qquad \text{ for all } n \in \mathbb{N} = \{1,2,3,\ldots\},$$ and so $$n^2 - 2 \leq n^2 - (3n-1) = n^2-3n+1 \leq \alpha \qquad \text{ for all } n \in \mathbb{N},$$ which is equivalent to $$n \leq \sqrt{\alpha +2} \qquad \text{ for all } n \in \mathbb{N}.$$ But this implies that the set $\mathbb{N}$ of natural numbers is bounded from above, which is absurd.