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Math Help - Measure Theory

  1. #1
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    Measure Theory

    Quick question: Does the sigma-algebra generated by a countable number of sets have a countable number of elements?
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  2. #2
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    It can be shown that any infinite sigma algebra has cardinality greater than or equal to that of the continuum.
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  3. #3
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    Thanks, but how? I tried proving that you can't have a bijection from the natural numbers to the sigma-algebra, with no sucess...
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  4. #4
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    Given a countably infinite \sigma-algebra \Sigma over an infinite set X, let \{E_n\}_{n\in\mathbb{N}} be an enumeration of \Sigma. Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in \Sigma. Namely, consider a sequence b=b_1b_2b_3\ldots, and let F_b=\bigcup_{b_i=1}E_i.
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  5. #5
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    Quote Originally Posted by Tikoloshe View Post
    Given a countably infinite \sigma-algebra \Sigma over an infinite set X, let \{E_n\}_{n\in\mathbb{N}} be an enumeration of \Sigma. Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in \Sigma. Namely, consider a sequence b=b_1b_2b_3\ldots, and let F_b=\bigcup_{b_i=1}E_i.
    This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.

    You probably need to use the axiom of choice. Let X be a set and  \mathcal{F} an infinite sigma algebra over X. Then define the equivalence relation ~ as x~y if and only if x \in A \in \mathcal{F} \iff y \in A. Then each equivalence class E(x) is measurable, as it is the countable intersection of sets in the sigma-algebra containing x.

    Let I be the set of representatives from each equivalence class (Axiom of Choice here). Then X= \bigcup_{x \in I} E(x) is a disjoint union. Hence if we suppose that \mathcal{F} is countable then I must either be finite or countable. Thus we have an injection from \{0,1\}^I to  \mathcal{F} as we can pick an element of \mathcal{F} by taking \bigcup_{x\in S \subset I}E(x) (as they are equivalence classes, you do have a guarantee of getting distinct sets).

    This is impossible as |\{0,1\}^I|> |I| if I is countable or finite.
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  6. #6
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    Quote Originally Posted by Focus View Post
    This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.
    I donít understand your objection. If \{E_i\}_{i\in\mathbb{N}} is a countable sequence of disjoint, nonempty sets, then using different index sets, the unions over the E_i will be distinct sets. I.e., if A,B\subset\mathbb{N} where A\neq B, then \bigcup_{i\in A}E_i\neq\bigcup_{i\in B}E_i. Given an arbitrary countable sequence in a \sigma-algebra, you can of course make a new sequence of pairwise disjoint, nonempty sets (without AC).

    Quote Originally Posted by Focus View Post
    You probably need to use the axiom of choice.
    I donít think AC is necessary for the proof (c.f. mine above). Also, here is another proof which takes a different approach.
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