Final answer:
The set S is an open subset of R² because for every point within S, there exists an epsilon neighborhood that is also entirely contained within S, which satisfies the definition of an open set.
Step-by-step explanation:
To prove that the set S = {(x, y) ∈ R² : x > 0 and y > 0} is an open subset of R², one must show that for every point in S, there exists an epsilon neighborhood around that point which is entirely contained within S.
Take any point (a, b) in S, where a > 0 and b > 0. Consider an epsilon neighborhood around the point (a, b) with a radius epsilon smaller than the minimum of a and b. This neighborhood is a circle centered at (a, b) and would not include the boundary because we are taking epsilon to be smaller than both a and b. Hence, the entire neighborhood would stay within the first quadrant where both x and y are positive, meaning it stays entirely within S.
Since the point (a, b) was arbitrary, and we were able to find an epsilon neighborhood that stays within S for it, this means that every point in S has an epsilon neighborhood entirely contained within S. Therefore, S is indeed an open set.