1
$\begingroup$

I am asked to find the smallest field that contains the sets {$1$} and {$2,3$}, if $S=${$1,2,3,4$}.

I understand that with the field, I want to find the unions, intersections, complements, and then the empty set and S.

The answer is as follows: $F=${{}, {$1$}, {$4$}, {$1,4$}, {$2,3$}, {$1,2,3$}, {$2,3,4$}, {$1,2,3,4$}}

My question is: Where did the {$4$} element come from? Is it from the Union of {$1,4$}?

  • 2
    You must be using the term "field" in a way I haven't yet heard, connected to set theory. Define this, please: what's for you "a field"? – 2017-01-19
  • 0
    I think the field here refers to $\sigma$-field. – 2017-01-19
  • 0
    @DonAntonio Since the OP put the tag `probability-theory` under the question, would it be clear that it means "field" in $\sigma$-field as "algebra" in $\sigma$-algebra? – 2017-01-19
  • 0
    Sorry, you are all correct. I meant that it is a sigma field! – 2017-01-19

2 Answers 2

2

It is the complement of $\{1\}\cup\{2,3\}$ in $S$.

  • 0
    Okay, so I am supposed to also take not only the complement of {1}, {2,3}, but also the complement of the Union of {1} and {2,3} when finding the sigma field? – 2017-01-19
  • 0
    By definition, yes. You might want to be careful: 'field' and $\sigma$-field are two different concepts in general. – 2017-01-19
2

To calculate a field generated by a family of sets $\mathcal M$ containing $S$, you need to:

1) add the complements of the sets to $\mathcal M$

2) Take any two sets and add their intersection to $\mathcal M$ until all the intersections are in $\mathcal M$

2) Take any two sets and add their union to $\mathcal M$ until all the unions are in $\mathcal M$

In other words the field generated of $\mathcal M$ is:

$$(\mathcal M\cup\mathcal M^c)_{ds}$$

Where the elements of $\mathcal M^c$ are the complement of each set of $\mathcal M$, and $d$ and $s$ means the closure to finite intersection and finite union respectively.

If by field you understand $\sigma$-field, then it is more complicated than this, although for finite $S$ (which is your case), the field and $\sigma$-field concepts are identical.