Consider the following relation on subsets of the set $S$ of integers between 1 and 2014. For two distinct subsets $U$ and $V$ of $S$ we say $U < V$ if the minimum element in the symmetric difference of the two sets is in $U$. Consider the following two statements: S1: There is a subset of $S$ that is larger than every other subset. S2: There is a subset of $S$ that is smaller than every other subset. Which one of the following is CORRECT?

Given $S = \{1, 2, \dots, 2014\}$ and the relation $U < V \iff \min(U \Delta V) \in U$, where $U \Delta V = (U \setminus V) \cup (V \setminus U)$.

Let us test the extreme subsets of $S$:

1. Check $U = S$ (the entire set):

For any other subset $V \neq S$, the symmetric difference is:

Since $S \setminus V \subseteq S$, its minimum element must belong to $S$. Therefore, $\min(S \Delta V) \in S$, which means $S < V$ for all $V \neq S$. Hence, $S$ is smaller than every other subset. This makes statement S2 TRUE.

2. Check $V = \emptyset$ (the empty set):

For any non-empty subset $U \neq \emptyset$, the symmetric difference is:

The minimum element $\min(U)$ belongs to $U$. Thus, $U < \emptyset$ for all $U \neq \emptyset$, which means $\emptyset > U$ for all $U \neq \emptyset$. Hence, $\emptyset$ is larger than every other subset. This makes statement S1 TRUE.

Since both S1 and S2 are true, option (A) is correct.