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Math Help - Sigmaalgebra/Set interpretation question

  1. #1
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    Question 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?
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  2. #2
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    Quote Originally Posted by enumaelish View Post
    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.

    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?
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  3. #3
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    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"?
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