The problem involves two sets of real numbers, $F$ and $H$, defined as follows:

• $F = \{x \mid x \geq 2\}$
• $H = \{y \mid y \leq 9\}$

You are asked to:

1. Write $F \cup H$ (union of $F$ and $H$) and $F \cap H$ (intersection of $F$ and $H$) using interval notation.
2. Find the set $F \setminus H$ (set difference of $F$ and $H$).

Step 1: Interval notation for $F$ and $H$

• $F = [2, \infty)$, because $x \geq 2$.
• $H = (-\infty, 9]$, because $y \leq 9$.

Step 2: Union of $F$ and $H$

The union of two sets contains all elements that belong to either set. In interval notation, this is:

• $F \cup H = (-\infty, \infty)$, because $F$ includes all numbers greater than or equal to 2, and $H$ includes all numbers less than or equal to 9. Together, these cover the entire real number line.

Step 3: Intersection of $F$ and $H$

The intersection of two sets contains all elements that belong to both sets. In this case, the intersection is the overlap between $F = [2, \infty)$ and $H = (-\infty, 9]$, which is the set of numbers between 2 and 9, inclusive:

• $F \cap H = [2, 9]$.

Step 4: Set difference $F \setminus H$

The set difference $F \setminus H$ contains all elements in $F$ that are not in $H$. Since $H = (-\infty, 9]$, removing the elements of $H$ from $F = [2, \infty)$ leaves:

• $F \setminus H = (9, \infty)$, because only elements greater than 9 remain from $F$.

Final Answer

1. $F \cup H = (-\infty, \infty)$
2. $F \cap H = [2, 9]$
3. $F \setminus H = (9, \infty)$

Would you like more details or further explanations?

Here are five related questions:

1. What is the union of $F$ and $H$ if $H = (0, 5]$?
2. How would the answer change if $F = (3, \infty)$ and $H = (-\infty, 7)$?
3. Can the intersection of two disjoint sets ever be non-empty? Explain.
4. What is the complement of set $F$ in the real numbers?
5. How would you graph the union and intersection of $F$ and $H$?

Tip: When working with intervals, visualize them on a number line to better understand the union and intersection operations.