Answer
\(S_0 = \{ \emptyset \}\)
\(S_1 = \{ \{a\}, \{b\}, \{c\} \}\)
\(S_2 = \{ \{a,b\}, \{a,c\}, \{b,c\} \}\)
\(S_3 = \{ \{a,b,c\} \}\)
\[
\boxed{
\{S_0, S_1, S_2, S_3\} \text{ is a partition of } \mathcal{P}(S)
}
\]
Work Step by Step
We are given the set:
\[
S = \{a, b, c\}
\]
And for each \(i = 0, 1, 2, 3\), define \(S_i\) as the set of all subsets of \(S\) that have exactly \(i\) elements.
---
### ✅ List all elements of \(S_0, S_1, S_2, S_3\)
---
#### 🔹 \(S_0\): subsets with 0 elements
Only the empty set:
\[
S_0 = \{ \emptyset \}
\]
---
#### 🔹 \(S_1\): subsets with 1 element
\[
S_1 = \{ \{a\}, \{b\}, \{c\} \}
\]
---
#### 🔹 \(S_2\): subsets with 2 elements
\[
S_2 = \{ \{a, b\}, \{a, c\}, \{b, c\} \}
\]
---
#### 🔹 \(S_3\): subsets with 3 elements
That’s just the full set:
\[
S_3 = \{ \{a, b, c\} \}
\]
---
### ✅ Is \(\{S_0, S_1, S_2, S_3\}\) a partition of \(\mathcal{P}(S)\)?
A **partition** of a set must satisfy:
1. Each part is non-overlapping (pairwise disjoint)
2. The union of all parts equals the full set
Let’s check:
- All subsets of \(S\) are:
\[
\mathcal{P}(S) = \{
\emptyset, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}
\}
\]
That’s \(2^3 = 8\) subsets.
Now total from \(S_0 \cup S_1 \cup S_2 \cup S_3\):
\(S_0\): 1 subset
\(S_1\): 3 subsets
\(S_2\): 3 subsets
\(S_3\): 1 subset
Total: 1 + 3 + 3 + 1 = 8 → ✅ matches \(\mathcal{P}(S)\)
Also, no subset appears in more than one group → ✅ disjoint
