Hello Maria,
Given any two sets A and B, the union of A and B (denoted A U B) is the set of all elements that are in A or B (or both). (Basically, the union operation just combines everything in A or B into one big set.) The intersection of A and B (denoted A ∩ B) is the set of all elements that are in A and B. (That is, the intersection consists of those elements that are common to both A and B.) In your problem, the set F is the set of all real numbers to the left of (and including) 3, while the set H is the set of all real numbers to the right of 6. If you are familiar with interval notation, these sets can also be represented as F = (-∞,3] and
G = (6,∞). You would probably find it helpful to graph these intervals on the number line. Now from the definition of union given above, F U H is the set of real numbers in F or H (or both). If you look at a picture of these two sets on the number line, you will realize there is no shorthand way to represent the set. We can write the result as
F U H = {z|z≤3} U {z|z> 6}
F U H = {z|z≤3 or z> 6}
F U H = (-∞,3] U (6,∞)
Either of the last two forms could be used, depending on whether you want set-builder notation or interval notation. On the other hand, F ∩ H is the set of all real numbers that are in common to the two sets. Geometrically, the intersection represents the overlap of the two intervals (-∞,3] and (6,∞). It is clear that there is nothing in common to these two intervals, so the intersection is the empty set. That is,
F ∩ H = ∅.
Hope that helps!
William