Sigmaalgebra/Set interpretation question

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August 18th 2008, 04:49 PM
enumaelish

Sigmaalgebra/Set interpretation question

Hi there,

I'm faced with two questions concerning sigma algebras but i believe my problem is with the interpretation of the sets themselves!

1. F = {A subset Omega : A is finite} . Is F always a sigma algebra?
The way I'm reading this is that F contains only the set A so that would mean the complement of A (i.e. A') isn't inside F and hence won't fulfil the condition of "closed under complements" for a sigma algebra right away, right?

2. Omega = R (real line)
F = {A subset R : A is finite or A' is finite}. Is F a sigma algebra?
This is where my suspicion about my interpretation grew. I mean, there wouldn't be a need to mention A' if it isn't even in F right? So, the question is, what exactly is contained in F in both cases?
August 18th 2008, 09:03 PM
NonCommAlg

Quote:

Originally Posted by enumaelish

Hi there,

I'm faced with two questions concerning sigma algebras but i believe my problem is with the interpretation of the sets themselves!

1. F = {A subset Omega : A is finite} . Is F always a sigma algebra?
The way I'm reading this is that F contains only the set A so that would mean the complement of A (i.e. A') isn't inside F and hence won't fulfil the condition of "closed under complements" for a sigma algebra right away, right?

the answer, as you said, is no. suppose $\Omega$ is an infinite set. then $\Omega \notin F,$ and thus $F$ is not a sigma algebra.

Quote:

2. Omega = R (real line)
F = {A subset R : A is finite or A' is finite}. Is F a sigma algebra?
This is where my suspicion about my interpretation grew. I mean, there wouldn't be a need to mention A' if it isn't even in F right? So, the question is, what exactly is contained in F in both cases?

let $A_j=\{j\}, \ j \in \mathbb{Z}.$ then $\forall j: A_j \in F,$ because each $A_j$ is a finite set. but $\cup A_j = \mathbb{Z} \notin F,$ because neither $\mathbb{Z}$ nor the complement of $\mathbb{Z}$ in $\mathbb{R}$ is finite. so $F$ is not a sigma algebra.

now try this exercise: let $\Omega$ be any set and $F=\{A \subseteq \Omega: \ A \ \text{is at most countable or} \ A' \ \text{is at most countable} \}.$ is $F$ a sigma algebra?
August 18th 2008, 10:29 PM
enumaelish

Hmm i get what you are saying, but my question is a little more fundamental/basic.

What i really want to know is what exactly is inside F - whether it's just A alone (my initial interpretation), or contains Omega, Null, A and A' or perhaps something else. To put it bluntly, it seems a little odd that the question does not explicitly say Omega, Null and A' are inside. And from your answers, you included the check on the union of Ajs and its complement to see whether it fulfils the condition of F or not - even though there is no mention that F contains these elements - that's why i suspect even more now that i'm missing something from reading the question with my own interpretation.

Also, i'm guessing "at most countable" (in your exercise) means "finite or countable or both"?