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 
is a subset of 
— a collection of subsets of 
— and define 
to be the collection of finite disjoint unions of subsets in . If we impose the following three conditions on :
- The empty set
and the whole space are both in .
- If
and 
are in , then so is their intersection 
.
- If and are in , then their difference
is in 
then is an algebra of sets.
If 
, then 
, and so contains and . We can also find 
, since 
.
Let’s take 
and 
to be two sets in , written as finite disjoint unions of sets in . Then their intersection is
Each of the 
is in , as an intersection of two sets in , and no two of them can intersect. Thus finite intersections of sets in are again in .
If 
, then 
. Since each of the 
are in , their (finite) intersection 
must be as well, and is closed under complements.
And so we can find that if 
and 
are in , then 
and 
are both in , and is thus an algebra of sets.
Like this:
Like Loading...
March 16, 2010 -
Posted by John Armstrong |
Analysis, Measure Theory
[...] here’s where our method of generating an algebra of sets comes in. In fact, let’s generalize the setup a bit. Let’s say we’ve got which [...]
Pingback by Products of Algebras of Sets « The Unapologetic Mathematician | March 17, 2010 |
[...] we know that these properties are exactly what we need to show that the collection of finite disjoint [...]
Pingback by Semiclosed Intervals « The Unapologetic Mathematician | April 14, 2010 |
[...] a countable set of generators, then is separable. Indeed, if is a countable sequence of sets that generate , then we may assume (by -finiteness) that for all . The ring generated by the is itself [...]
Pingback by The Metric Space of a Measure Ring « The Unapologetic Mathematician | August 6, 2010 |