# Sigma Algebra on infinite set

tavrion
I am trying to better my understanding of $\sigma$-$algebra$s, 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?

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?