In the given problem, we are provided with two sets of real numbers:

• $E = \{ w \mid w < 3 \}$, which means all real numbers less than 3.
• $F = \{ w \mid w \geq 6 \}$, which means all real numbers greater than or equal to 6.

1. Union of $E$ and $F$ ($E \cup F$):

The union combines all elements from both sets. In this case, $E \cup F$ represents all numbers less than 3 or greater than or equal to 6. Therefore, in interval notation: $E \cup F = (-\infty, 3) \cup [6, \infty)$

2. Intersection of $E$ and $F$ ($E \cap F$):

The intersection includes only those elements common to both sets. However, since there are no numbers that are both less than 3 and greater than or equal to 6, the intersection is empty: $E \cap F = \emptyset$

Final answers:

• $E \cup F = (-\infty, 3) \cup [6, \infty)$
• $E \cap F = \emptyset$

Would you like more details on interval notation or union and intersection of sets?

Here are 5 related questions you might find interesting:

1. How do you represent the complement of a set in interval notation?
2. Can the union of two sets ever be smaller than either of the sets?
3. What happens if the intervals overlap in a union or intersection problem?
4. How do you represent the union of finite sets in interval notation?
5. What is the difference between open and closed intervals in set theory?

Tip: When working with unions and intersections, sketching a number line can help visualize the intervals!