What is the issue with the example of \sigma-algebras on infinite sets?

[tavrion]

I am trying to better my understanding of \sigma-algebras, and I have a bit of an issue with one of the examples. This is from Cohn Measure Theory, and before I give the problem, here are two definitions:

Let X be an arbitrary set. A collection \delta\Large of subsets of X is an algebra on X if:

(a) X \in Z

(b) for each set A that belongs to \delta\Large, the set A^c belongs to \delta\Large

(c) for each finite sequence A_{1}, ... , A_{n} of sets that belong to \delta\Large the set \bigcup_{i=1}^{n}A_{i} belongs to \delta\Large and

(d) for each finite sequence A_{1}, ... , A_{n} of sets that belong to \delta\Large the set \bigcap_{i=1}^{n}A_{i} belongs to \delta\Large



Let X be an arbitrary set. A collection \delta\Large of subsets of X is a \sigma-algebra on X if:

(a) X \in Z

(b) for each set A that belongs to \delta\Large, the set A^c belongs to \delta\Large

(c) for each infinite sequence \{A_{i}\} of sets that belong to \delta\Large the set \bigcup_{i=1}^{\infty}A_{i} belongs to \delta\Large and

(d) for each infinite sequence \{A_{1}\} of sets that belong to \delta\Large the set \bigcap_{i=1}^{\infty}A_{i} belongs to \delta\Large


Okay, with all that. Here is what I am having issues with.

If X is an infinite set, and \delta is the collection of all subsets A such that either A or A^c is finite. Then \delta is an algebra on X but not closed under the formation of countable unions, and so not a \sigma-algebra.

So, if I take, for example X to be the set of all positive integers, that is X = {1,2,3,...} and define A_{i} = i.

Then, I have \bigcup_{i=1}^{n}A_{i} = {1,2,3,...,n} which belongs to \delta but \bigcup_{i=1}^{\infty}A_{i} = {1,2,3,...} belongs to \delta as well, so why does this fail to be a \sigma-algebra? Where have I gone wrong?

 

[pwsnafu]

Can you think of an infinite subset of X whose complement is also infinite?

 

[tavrion]

Thank you. Let me see if I understand this right,

As before with X = {1,2,3,...} and

if A_{i} = 2i and A_{j} = 2j-1, then A = \bigcup_{i}^{\infty}A_{i} = 2,4,6,... and A_{\infty}^c = \bigcup_{j}^{\infty}A_{j} = 1,3,5,...

If A^c is defined to be finite, we have A_{n}^c = \bigcup_{j}^{n}A_{j} = 1,3,5,...,2n-1.

So, we have X \notin A_{n}^c\cup{A} which fails to be a \sigma-algebra but X \in A_{\infty}^c\cup{A} is a \sigma-algebra?

Is this logic correct?