# The Unapologetic Mathematician

## 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