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I am working on a homework problem and am somewhat lost. I know that an answer will not be given on a silver platter and am fine with that - I need to know what I am missing in understanding so that I can solve the problem.

I need an infinite collection of subsets of $\mathbb{R}$ that contains $\mathbb{R}$, is closed under the formation of countable unions and countable intersections, but is not a $\sigma$-algebra.

So I immediately thought that the only requirement not mentioned to make it a $\sigma$-algebra is the closure under complementation. That is why I thought of maybe using $\mathcal{P}(\mathbb{R})-\{\varnothing\}$, the powerset 'minus' the null set. Is this okay? Can you subtract 'nothing' like this? Otherwise I am quite lost and any direction would be greatly appreciated.

Nate

P.S> I could not find suitable suggestions to my question by looking around on the site.

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    $\begingroup$ Your idea of excluding the empty set is sound -- you just have to do a tiny bit more to make it work. Hint: Can you solve the "harder" problem of finding a family that is closed under arbitrary (not just countable) intersections and unions? $\endgroup$ Commented Sep 7, 2011 at 22:36
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    $\begingroup$ For questions like this, I like to start with some "minimal generators" and see what happens. For example, you could see what the minimal such collection is that contains $\mathbb{R}$ and the empty set - it turns out to be just the collection containing $\mathbb{R}$ and the empty set themselves. That doesn't work, since the collection of these two sets is a sigma-algebra and doesn't work since it's finite. But is there some other "small" collection of sets you could start with? $\endgroup$ Commented Sep 7, 2011 at 22:39
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    $\begingroup$ Reg: 'Can you subtract "nothing" like this?'. By the way, if you are wondering if the set $\mathcal P(\mathbb R) - \{ \emptyset \}$ is well-defined, it is. This choice just does not work for the problem. $\endgroup$
    – Srivatsan
    Commented Sep 7, 2011 at 22:50
  • $\begingroup$ Thank you Srivatsan for letting me know that my proposed solution (which doesn't work for this problem) is conceptually okay/ well defined. Makholm: I am intrigued by your suggestion but cannot yet think of such a family. Any more hints? And yes, I do understand what MartianInvader says about the 2 failings in the trivial sigma-algebra. I also know that the "small" collection is still infinite (or countably infinite?) $\endgroup$
    – nate
    Commented Sep 7, 2011 at 22:56
  • $\begingroup$ nate, I hope my answer (on its various revisions) did not scare you. :-) at least the first half... $\endgroup$
    – Asaf Karagila
    Commented Sep 8, 2011 at 0:03

3 Answers 3

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Your proposed answer of $\mathcal{P}(\mathbb{R})-\{\varnothing\}$ is not closed under intersections, because $\{1\}\cap\{0\}=\varnothing$.

Try $\mathcal{P}(\mathbb{N})\cup\{\mathbb{R}\}$.

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  • $\begingroup$ My suggestion does not even contain a $\sigma$-algebra, thus better ;-D $\endgroup$
    – Asaf Karagila
    Commented Sep 7, 2011 at 22:35
  • $\begingroup$ Haha I suppose :) You have my +1 $\endgroup$ Commented Sep 7, 2011 at 22:36
  • $\begingroup$ Thanks. I am in the process of trying to understand how this works! I see that somehow it just doesn't support closure of complementation. $\endgroup$
    – nate
    Commented Sep 7, 2011 at 22:55
  • $\begingroup$ @nate: Note that $\{1\}\in\mathcal{P}(\mathbb{N})\cup\{\mathbb{R}\}$. What is the complement of $\{1\}$? $\endgroup$ Commented Sep 7, 2011 at 23:27
  • $\begingroup$ So on the reals, the complement of $\{1\}$ is: $\{1\}^c = \{\mathbb{R}\}-\{1\}$, which excludes $\{1\}$. But the proposed infinite set (that is uncountably infinite and equinumerous with the powerset of naturals [countably infinite]) is said to contain $\{1\}$ as a member of the powerset, $\mathcal{P}(\mathbb{N})$. So $\{\mathbb{R}\}-\{1\} \notin \mathcal{P}(\mathbb{N}) \cup \{\mathbb{R}\}$. Right? $\endgroup$
    – nate
    Commented Sep 7, 2011 at 23:41
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The freshman solutions:

  1. $\left\{[0,\frac{1}{n}]\mid n\in\mathbb N\setminus\{0\}\right\}\cup\left\{\mathbb R,\{0\}\right\}$
  2. $\{A\subseteq\mathbb R\mid 0\in A\}$. (A suggestion by Henning in the comments)

The second is also known as a principal ultrafilter concentrating on $0$.


The raving set theoretic madness: (Some of the solutions may be fitting for advanced undergrad students, and might be less trivial than the above examples)

My initial approach was to take a ultrafilter which is closed under countable intersections, but clearly not for complements. However the existence of one over the real numbers is equivalent to a certain large cardinal axiom which made me formulate the original solution instead...

  1. $\Sigma^1_1(\mathbb R)$ sets (the analytic sets), which can be obtained as images of Borel sets. These are not closed under complements and contain all the Borel sets of $\mathbb R$.

  2. Take the co-countable filter over the reals, that is $\{A\subseteq\mathbb R\mid |\mathbb R\setminus A|\le\aleph_0\}$.

  3. Assume the Continuum Hypothesis is true. Let $g\colon\omega_1\to\mathbb R$ be some bijection, which is naturally extended to $f\colon\mathcal P(\omega_1)\to\mathcal P(\mathbb R)$. Denote by $\mathcal F$ the club$^+$ filter of $\omega_1$, that is all the subsets of $\omega_1$ which superset a closed and unbounded set.
    The club filter is closed under countable intersections, and any sort of unions. Now consider $\mathcal U=\{f(A)\mid A\in\mathcal F\}$, this would make a filter over the real numbers which is countably closed, closed under any union and since $\omega_1\in\mathcal F$, we have that $\mathbb R\in\mathcal U$. And of course, since $\mathcal U$ is a filter, it cannot be a $\sigma$-algebra.

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  • $\begingroup$ Seems very complex. Since we are now presenting full solutions, what about just $\{ A\subseteq \mathbb R \mid 0\in A \}$? $\endgroup$ Commented Sep 7, 2011 at 22:40
  • $\begingroup$ @Henning: Ah yes. I was too busy thinking about real-measurable cardinals that I forgot atomic measures give this :-) $\endgroup$
    – Asaf Karagila
    Commented Sep 7, 2011 at 22:49
  • $\begingroup$ @Henning: How about the added idea? Does that make my initial one simple? :-) $\endgroup$
    – Asaf Karagila
    Commented Sep 7, 2011 at 23:22
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    $\begingroup$ nate is trying to get union and difference of sets straight and you're coming with filters, the club$^{+}$-filter and the continuum hypothesis? Come on! Cool down... :) $\endgroup$
    – t.b.
    Commented Sep 7, 2011 at 23:28
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    $\begingroup$ By the way, thank you Theo for the apparent sympathy! :) $\endgroup$
    – nate
    Commented Sep 8, 2011 at 0:10
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Almost. But your formulation is not closed under intersections. For instance, the set $\{ 3 \}$ and the set $ \{ \pi \}$ are both there, but their intersection, $\varnothing$, is not.

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    $\begingroup$ Mixedmath: Thank you for pointing that my intersection is not closed! $\endgroup$
    – nate
    Commented Sep 7, 2011 at 22:52

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