Final answer:
To prove 2^A \cup 2^B is a subset of 2^(A \cup B), we show every subset of A or B is also a subset of A \cup B. For proving equality does not hold, an example is given where a subset of A \cup B is not a subset of either A or B. To demonstrate that F is open, we establish that for any point in F, it's possible to find an interval entirely contained in F.
Step-by-step explanation:
To show that 2^A \cup 2^B is a subset of 2^(A \cup B), consider an arbitrary set C in 2^A \cup 2^B. This means that C is a subset of either A or B. Since the union of sets A and B includes all elements of A as well as B, C will also be a subset of the union A \cup B. Thus, every element of 2^A \cup 2^B is in 2^(A \cup B), proving the subset relationship.
However, the equality may not hold, since there can be subsets of A \cup B that are not entirely within A or B. For example, if A = {1, 2} and B = {3, 4}, then {1, 3} is a subset of A \cup B, but it is neither a subset of A nor of B, so it is not in 2^A \cup 2^B. Hence, 2^A \cup 2^B is not equal to 2^(A \cup B).
To prove that F is open, given a set E \subset R and a fixed \(\epsilon > 0\), we need to show that for any y in F, there exists a \(\delta > 0\) such that the interval (y - \delta, y + \delta) is contained in F. For any y in F, there is an x in E such that |x - y| < \epsilon. Choose \delta to be the smaller value between y - (x - \epsilon) and (x + \epsilon) - y. For any point z in this interval, |x - z| < \epsilon is satisfied, ensuring that z is in F, making F open.