Answer
The union is correct, but the sets **overlap**, so they are **not disjoint**.
\[
\boxed{
\{S_a, S_b, S_c, S_\emptyset\} \text{ is not a partition of } \mathcal{P}(S)
}
\]
Work Step by Step
We are given:
- \(S = \{a, b, c\}\)
- \(\mathcal{P}(S)\) is the power set of \(S\), i.e., all \(2^3 = 8\) subsets
Defined sets:
- \(S_a\): all subsets **containing** \(a\)
- \(S_b\): all subsets **containing** \(b\)
- \(S_c\): all subsets **containing** \(c\)
- \(S_\emptyset = \{\emptyset\}\)
---
### ✅ Step 1: List \(\mathcal{P}(S)\)
\[
\mathcal{P}(S) = \{
\emptyset, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}
\}
\]
=
---
### ✅ Step 2: List contents of \(S_a, S_b, S_c, S_\emptyset\)
#### \(S_a =\) subsets that contain \(a\):
\[
S_a = \{ \{a\}, \{a,b\}, \{a,c\}, \{a,b,c\} \}
\]
#### \(S_b =\) subsets that contain \(b\):
\[
S_b = \{ \{b\}, \{a,b\}, \{b,c\}, \{a,b,c\} \}
\]
#### \(S_c =\) subsets that contain \(c\):
\[
S_c = \{ \{c\}, \{a,c\}, \{b,c\}, \{a,b,c\} \}
\]
#### \(S_\emptyset = \{ \emptyset \}\)
---
### 🔍 Step 3: Check if \(\{S_a, S_b, S_c, S_\emptyset\}\) is a partition
To be a **partition** of \(\mathcal{P}(S)\), the sets must:
1. Be **disjoint** (no overlap)
2. Have a union equal to \(\mathcal{P}(S)\)
---
### ❌ Are the sets disjoint?
No!
- \(\{a,b\}\) is in both \(S_a\) and \(S_b\)
- \(\{a,b,c\}\) is in all of \(S_a\), \(S_b\), and \(S_c\)
So the sets **overlap**, meaning they are **not pairwise disjoint**
---
### ✅ Is the union equal to \(\mathcal{P}(S)\)?
Let’s take the union:
\(S_a \cup S_b \cup S_c \cup S_\emptyset\) =
{
\(\emptyset\),
\(\{a\}, \{b\}, \{c\}\),
\(\{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}\)
}
That’s 8 elements → equal to \(\mathcal{P}(S)\)
✅ So the **union** is correct
