Countable intersection of F-sigma sets

[jbunniii]

My question concerns $F_\sigma$ subsets of $\mathbb{R}$. An $F_\sigma$ set is one which can be expressed as a countable union of closed sets.

I have several books that state that a countable intersection of $F_\sigma$ sets need not be an $F_\sigma$ set (indeed, such sets have their own designation, $F_{\sigma\delta}$), but none of them gives a counterexample. Does anyone know one offhand? This isn't homework/coursework; I'm just curious and haven't been able to come up with one. I would prefer a concrete example if possible, not just an existence proof.

Relevant facts: Clearly a countable UNION of $F_\sigma$ sets is $F_\sigma$. All open and closed sets in $\mathbb{R}$ are $F_\sigma$. $\mathbb{Q}$ is an $F_\sigma$ set as it is a countable union of singletons. The only concrete example of a set I know is not $F_\sigma$ is $\mathbb{R}\setminus\mathbb{Q}$, the set of irrationals.

 

[micromass]

$$\mathbb{R}\setminus \mathbb{Q} = \bigcap_{q\in \mathbb{Q}} \mathbb{R}\setminus \{q\}$$

 

[jbunniii]

$$\mathbb{R}\setminus \mathbb{Q} = \bigcap_{q\in \mathbb{Q}} \mathbb{R}\setminus \{q\}$$

Yes, I just had that realization while thinking about it again. I had been trying some more convoluted way of finding a sequence of open sets that decreased to $\mathbb{R}\setminus\mathbb{Q}$, but missed the most obvious one!

 

[jbunniii]

The above construction also shows that $\mathbb{R}\setminus\mathbb{Q}$ is a $G_\delta$ set (i.e. a countable intersection of open sets), which also follows immediately from the fact that its complement $\mathbb{Q}$ is a countable union of singletons, hence a $F_\sigma$ set.

So $\mathbb{R}\setminus\mathbb{Q}$ is in some sense an atypical member of $F_{\sigma\delta}$ because it is also in $G_\delta$. Similarly, $\mathbb{Q}$ is an atypical member of $F_{\sigma\delta}$ because it is also in $F_\sigma$.

This led me to look for an example of a "pure" $F_{\sigma\delta}$ set, one which is in neither $F_\sigma$ nor $G_\delta$.

I think that $S = (\mathbb{Q} \cap (-\infty,0)) \cup ((\mathbb{R}\setminus\mathbb{Q}) \cap (0,\infty))$ is such a set. (i.e. the union of all negative rationals and all positive irrationals.) Here is my reasoning:

If $S$ were a $F_\sigma$ set, then $S \cap (0,\infty)$ would also be $F_\sigma$, but that is not the case, because $S \cap (0,\infty) = ((\mathbb{R}\setminus\mathbb{Q}) \cap (0,\infty))$. Similarly, $S$ is not a $G_\delta$ set. However, I can form a descending sequence of $F_\sigma$ sets whose intersection is $S$ by starting with $(Q \cap(-\infty,0)) \cup (0,\infty)$ and removing one positive rational at a time.

Then I started thinking about what a "pure" $F_{\sigma\delta\sigma}$ set would look like, and it made me tired.

 

[blue_raver22]

Thank you for your question. You are correct in stating that a countable intersection of F_\sigma sets need not be an F_\sigma set. In fact, this is a well-known result in topology and set theory. One example of an F_{\sigma\delta} set that is not an F_\sigma set is the set of all real numbers that have a decimal expansion consisting only of the digits 0 and 1. This set is not an F_\sigma set, but it can be expressed as a countable intersection of F_\sigma sets.

To understand this example, let's first define what an F_{\sigma\delta} set is. An F_{\sigma\delta} set is a set that can be expressed as a countable intersection of countable unions of closed sets. In other words, it is a countable intersection of F_\sigma sets. Now, let's look at the set of all real numbers that have a decimal expansion consisting only of the digits 0 and 1. This set can be written as the intersection of the sets A_n, where A_n is the set of all real numbers whose decimal expansion contains only 0's and 1's up to the nth digit. Each of these sets is an F_\sigma set, as it is the countable union of the closed sets [0,1], [0.0,0.1], [0.00,0.11], and so on. However, the intersection of these sets is not an F_\sigma set, as it cannot be expressed as a countable union of closed sets.

In general, any set that is not an F_\sigma set can be expressed as a countable intersection of F_\sigma sets. This is because the complement of an F_\sigma set is an F_\sigma set, and the complement of an F_{\sigma\delta} set is an F_{\sigma\delta} set. So, in your example, \mathbb{R}\setminus\mathbb{Q} is an F_{\sigma\delta} set, and its complement \mathbb{Q} is also an F_{\sigma\delta} set.

I hope this explanation helps to clarify the concept of F_\sigma and F_{\sigma\delta} sets. In summary, a countable intersection of F_\sigma sets need not be an F_\sigma set, and the set of all real