Example of not being a sigma algebra as complement property does not hold

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.

Solution 1:

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

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

Solution 2:

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.

Solution 3:

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.