The Unapologetic Mathematician

Mathematics for the interested outsider

Generating Algebras of Sets

We might not always want to lay out an entire algebra of sets in one go. Sometimes we can get away with a smaller collection that tells us everything we need to know.

Suppose that \mathcal{R} is a subset of P(X) — a collection of subsets of X — and define \mathcal{E}\subseteq P(X) to be the collection of finite disjoint unions of subsets in \mathcal{R}. If we impose the following three conditions on \mathcal{R}:

  • The empty set \emptyset and the whole space X are both in \mathcal{R}.
  • If A and B are in \mathcal{R}, then so is their intersection A\cap B.
  • If A and B are in \mathcal{R}, then their difference A\setminus B is in \mathcal{E}

then \mathcal{E} is an algebra of sets.

If A\in\mathcal{R}, then A\in\mathcal{E}, and so \mathcal{E} contains \emptyset and X. We can also find A^c\in\mathcal{E}, since A^c=X\setminus A.

Let’s take E_1=\bigcup_{i=1}^m R_i and E_2=\bigcup_{j=1}^nS_j to be two sets in \mathcal{E}, written as finite disjoint unions of sets in \mathcal{R}. Then their intersection is

\displaystyle E_1\cap E_2=\bigcup\limits_{i=1}^m\bigcup\limits_{j=1}^n R_i\cap S_j

Each of the R_i\cap S_j is in \mathcal{R}, as an intersection of two sets in \mathcal{R}, and no two of them can intersect. Thus finite intersections of sets in \mathcal{E} are again in \mathcal{E}.

If E=\bigcup_{i=1}^n R_i, then E^c=\bigcap_{i=1}^n R_i^c. Since each of the R_i^c are in \mathcal{E}, their (finite) intersection E^c must be as well, and \mathcal{E} is closed under complements.

And so we can find that if E_1 and E_2 are in \mathcal{E}, then E_1\setminus E_2=E_1\cap E_2^c and E_1\cup E_2=(E_1\setminus E_2)\cup E_2 are both in \mathcal{E}, and \mathcal{E} is thus an algebra of sets.

March 16, 2010 - Posted by | Analysis, Measure Theory


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