Final answer:
The set S is proved to be closed in the Euclidean topology on \(\mathbb{R}\) by showing that its complement is open, which means for every point not in S, there is an interval around it that does not intersect with S.
Step-by-step explanation:
To prove that the set S is closed in the Euclidean topology on \(\mathbb{R}\), we need to show that its complement, \(\mathbb{R} \setminus S\), is open. In this context, a set is open if for every point in the set, there is an epsilon neighborhood around it that is wholly contained within the set. Considering any point in \(\mathbb{R} \setminus S\), we can find a small enough epsilon such that the interval (x-\epsilon, x+\epsilon) does not intersect with S. The points in S get arbitrarily close to zero but do not include numbers less than zero. Hence, any interval around a negative number or any positive number greater than 1 would not intersect S. Additionally, for the numbers in \(\mathbb{R}\) that are not of the form 1/n, we can choose an epsilon that is smaller than the difference between x and the nearest fractions of the form 1/n. This shows that \(\mathbb{R} \setminus S\) contains an epsilon neighborhood for every one of its points, hence S is closed in the Euclidean topology.