Answer
**(a)** \(S_1 = \mathcal{P}(\{t, u, v\})\)
**(b)** \(S_2 = \{ X \cup \{w\} \mid X \subseteq \{t,u,v\} \}\)
**(c)** Yes, \(S_1\) and \(S_2\) are disjoint.
**(d)** \(|S_1| = |S_2| = 8\)
**(e)** \(|S_1 \cup S_2| = 16\)
**(f)** \(S_1 \cup S_2 = \mathcal{P}(A)\)
Work Step by Step
We are given a set:
\[
A = \{t, u, v, w\}
\]
Define:
- \(S_1\): all subsets of \(A\) **that do not contain \(w\)**
- \(S_2\): all subsets of \(A\) **that contain \(w\)**
Let’s answer each part:
---
### **a. Find \(S_1\)**
Subsets of \(A\) **not containing** \(w\) are just all subsets of \(\{t, u, v\}\). The power set of a 3-element set has \(2^3 = 8\) elements.
So:
\[
S_1 = \mathcal{P}(\{t, u, v\}) =
\left\{
\emptyset, \{t\}, \{u\}, \{v\}, \{t,u\}, \{t,v\}, \{u,v\}, \{t,u,v\}
\right\}
\]
---
### **b. Find \(S_2\)**
Subsets **that contain \(w\)** can be described by taking each subset of \(\{t, u, v\}\), and adding \(w\) to it.
So:
\[
S_2 = \{X \cup \{w\} \mid X \subseteq \{t, u, v\}\}
\]
This gives:
\[
S_2 =
\left\{
\{w\}, \{t, w\}, \{u, w\}, \{v, w\}, \{t, u, w\}, \{t, v, w\}, \{u, v, w\}, \{t, u, v, w\}
\right\}
\]
---
### **c. Are \(S_1\) and \(S_2\) disjoint?**
Yes.
- Every element in \(S_1\) **does not contain** \(w\)
- Every element in \(S_2\) **does contain** \(w\)
⟹ So no element appears in both.
\[
\boxed{\text{Yes, } S_1 \cap S_2 = \emptyset}
\]
---
### **d. Compare the sizes of \(S_1\) and \(S_2\)**
Each is the set of all subsets of a 3-element set, so:
\[
|S_1| = |S_2| = 2^3 = 8
\]
\[
\boxed{|S_1| = |S_2| = 8}
\]
---
### **e. How many elements are in \(S_1 \cup S_2\)?**
Since \(S_1\) and \(S_2\) are disjoint:
\[
|S_1 \cup S_2| = |S_1| + |S_2| = 8 + 8 = 16
\]
\[
\boxed{|S_1 \cup S_2| = 16}
\]
---
### **f. What is the relation between \(S_1 \cup S_2\) and \(\mathcal{P}(A)\)?**
- \(\mathcal{P}(A)\): the power set of \(A\), i.e., all \(2^4 = 16\) subsets of \(A\)
- We've just shown that \(S_1 \cup S_2\) contains all subsets of \(A\): those that contain \(w\), and those that don’t.
So:
\[
\boxed{S_1 \cup S_2 = \mathcal{P}(A)}
\]
