Solution: Generated Sigma-Algebra on a Four-Point Space

Exercise: Generated Sigma-Algebra on a Four-Point Space

Part 1

Start from $\mathcal{C} = \{\{1,2\}, \{2,3\}\}$. Apply complementation and union repeatedly:

• Complements: $\{1,2\}^c = \{3,4\}$, $\{2,3\}^c = \{1,4\}$.
• Intersections: $\{1,2\} \cap \{2,3\} = \{2\}$, $\{3,4\} \cap \{1,4\} = \{4\}$.
• Unions: $\{1,2\} \cup \{2,3\} = \{1,2,3\}$, with complement $\{4\}$ (already listed). $\{1,2\} \cup \{1,4\} = \{1,2,4\}$, with complement $\{3\}$. $\{3,4\} \cup \{2,3\} = \{2,3,4\}$, with complement $\{1\}$.

Now every singleton $\{1\}, \{2\}, \{3\}, \{4\}$ is in the collection. Once all singletons are in a $\sigma$-algebra on a finite set, every subset is in it — arbitrary subsets are finite unions of singletons.

Explicitly, all $16$ subsets:

Part 2

$|\sigma(\mathcal{C})| = 16 = 2^4$. The generating collection $\mathcal{C}$ has just $2$ elements, yet the generated $\sigma$-algebra equals the entire power set of a 4-element space — the classic "$\sigma$-algebras blow up fast" phenomenon. Two well-chosen sets can already saturate the full information structure on $\Omega$.

Part 3

Yes, $\sigma(\mathcal{C}) = 2^\Omega$. Every outcome is isolable: $\{1\}, \{2\}, \{3\}, \{4\}$ all sit in $\sigma(\mathcal{C})$ by the construction above. No outcome is lumped with another.

Contrast this with the lesson's Example 1, where $\mathcal{C}' = \{\{1\}, \{2\}\}$ generates only $8$ sets because $\{3\}$ and $\{4\}$ are never separable — the generating sets mention neither one.

Part 4

Takeaways

• Small generating sets can produce large $\sigma$-algebras. Two cleverly chosen sets suffice to generate the full power set on a 4-point space.
• Isolate singletons $\Rightarrow$ full power set. On finite $\Omega$, as soon as every singleton is reachable, the generated $\sigma$-algebra is the entire power set.
• The "information" reading. $\sigma(\mathcal{C})$ captures exactly what you can distinguish by asking the yes/no questions $\{A \in \mathcal{C}\}$ and building up from there. When those questions partition the outcomes individually, no information is hidden — this is the finite analogue of why a natural filtration "sees" every realised path.