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Can someone please give me an example of a set that lies in a $\sigma$-algebra generated by some set other then the $\sigma$-algebra itself, such that this (the first) set can't be obtained by performing countably many set theoretical operations ? (Obviously with "example of a set" I mean "prove that such a set exist", since this set probably isn't obtainable in a constructive way.)

I'm asking this since I've heard countless times, that generally the elements of a $\sigma$-algebra are not really constructively reachable.

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This is hardly worth an answer, so it'll just have to do as a comment: Have you looked into descriptive set theory? –  kahen Nov 4 '12 at 19:27
    
@kahen outch...what an derogative answer. I looked at the link you provided, but that was way above my level :(. –  user19758 Nov 4 '12 at 19:53
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@user19758: I think kahen was referring to his comment, not to your question! –  Clive Newstead Nov 4 '12 at 19:55
    
@user19758: a link by itself is usually considered a poor answer, so to give you the link, kahen gave it in a comment. I don't think that kahen was implying that your question was "hardly worth an answer". –  robjohn Nov 4 '12 at 22:27

2 Answers 2

I think the answer is negative. You confused two things at here: an element in the $\sigma$-algebra may not be constructively constructed and an element cannot be formed by countable 3 elementary set operations. In general there is no way to expect a $\sigma$-algebra needed to be generated by some set(of subsets). All we know is we have a set in $P(\mathcal{X})$ that satisfies the 3 axioms. Whether there exists such a generating set (of subsets) is not clear. And in particular if you assume you have such a generating set(of subsets), then every element by definition must be generated by countable such set operations.

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Well, every $\sigma$-algebra is generated by itself... –  nonpop Nov 4 '12 at 19:46
    
In this case the answer is clear trivial, though. –  Bombyx mori Nov 4 '12 at 19:50
    
@user32240 So that means, that for any $\sigma$-algebra $\sigma(A)$, where $A$ is some set in the ambient set $X$, all elements of $\sigma(A)$ can be obtained by countable set operations ? If that is so, why do I so often hear, that one would have to do transfinite induction (I don't know the details of that; I only know it is some kind of induction that works when dealing with collections of sets that are more tha countable) to prove that some property hold for $\sigma(A)$ ? –  user19758 Nov 4 '12 at 19:51
    
@nonpop Yes, I should have specified that this case shall be excluded, which I did now –  user19758 Nov 4 '12 at 19:52
    
I do not know why you hear something and did ask for justification, but if $\sigma(A)$ denotes the sigma algebra "generated" by elements in the set of subsets $A$, then I feel this is almost tautology. –  Bombyx mori Nov 4 '12 at 19:54

As has been discussed already, every element of $\sigma(A)$ is reachable by iterating countable unions and complements, starting with $A$, over all countable ordinals-an example of this process is the Borel hierarchy by which the Borel $\sigma$-algebra is constructed from a topology. Note that this is a transfinite process: even once you've gotten $\sigma(A)$ constructed up through $n$ stages for every stage, if you take a union of one set from stage $n$ for every $n$, in general the resulting set won't have already come up, so that you'll have gotten to stage $\omega$. One then goes on through $\omega+1,...,2\omega,...,\omega^2,...,\omega^\omega,...,\omega^{\omega^\omega},...,\varepsilon_0,...$ and on and on up through $\omega_1$ the first uncountable ordinal. We can finally stop here because any countable sequence of countable ordinals has a countable upper bound, so there's no way to "break through" $\omega_1$ via countable unions. One can prove that every stage of this construction is necessary to get, for example, the Borel algebra on the real line, so this might be the sort of non-constructive elements of $\sigma(A)$ you're looking for.

On the other hand, it's much easier to see there are strange elements of the completion of $\sigma(A)$, for instance the Lebesgue $\sigma$-algebra on the real line as the completion of the Borel algebra just discussed. That's because every stage of the construction of the Borel algebra has only $\mathfrak{c}$ members, being a collection of $\mathfrak{c}$ unions of countably many sets; then the whole Borel algebra also has cardinality $\mathfrak{c}$, since we're taking the union over at most $\mathfrak{c}$ stages of sets of cardinality at most $\mathfrak{c}$. (See this question). But the Lebesgue algebra has cardinality $2^{\mathfrak{c}},$ because every subset of any measure-0 set has measure $0$, and in particular is measurable, so that one only has to take some measure-0 set of cardinality continuum, say the Cantor set, to get that the Lebesgue algebra contains the entire powerset of a cardinality-$\mathfrak{c}$ set. So, naturally one cannot construct in any reasonably direct way any but a tiny fraction of the Lebesgue sets: coming up with just one that's not Borel is not that trivial.

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