supremum question

The supremum is smiply the least upper bound of the set A - i.e. it is the smallest real number y such that $x\leq y$ for all $x \in A$



It can be proved that there are an infinite number of primes and so $\frac{1}{p^2}$ will go to zero as we take larger and larger primes - hence sup(P)=1.

Proof:
We need to show:
(1) 1 is an upper bound: $1-\frac{1}{p^2}<1 \forall p\in P$
(2) 1 is the least upper bound: $1-\frac{1}{p^2}\not< 1-\epsilon \forall p\in P$ for any $\epsilon >0$

(1) $1-\frac{1}{p^2}<1$ is true for $\frac{-1}{p^2}<0$ which is for $p^2>0$ and since $p^2$ is always positive (indeed p>1) this is easily assured.
(2) Suppose 1 isn't the least upper bound.

However, this expression is clearly false as
Hence we have a contradiction so our assumption was wrong and so 1 is the least upper bound.