Sigma Algebra on infinite set

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Main Question or Discussion Point

I am trying to better my understanding of [itex]\sigma[/itex]-[itex]algebra[/itex]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 [itex]X[/itex] be an arbitrary set. A collection [itex]\delta\Large[/itex] of subsets of [itex]X[/itex] is an [itex]algebra[/itex] on [itex]X[/itex] if:

(a) [itex]X \in Z[/itex]

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

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

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



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

(a) [itex]X \in Z[/itex]

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

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

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


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

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

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

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

Answers and Replies

  • #2
pwsnafu
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Can you think of an infinite subset of X whose complement is also infinite?
 
  • #3
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Thank you. Let me see if I understand this right,

As before with [itex] X = {1,2,3,...} [/itex] and

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

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

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

Is this logic correct?
 

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