$\sigma$-algebra of sets that are both $G_\delta$ and $F_\sigma$

I'm working on the following problem in my probability theory class:

Let $\Omega$ be a topological space. Show that the system of all sets in $\Omega$ which are $F_\sigma$ as well as $G_\delta$ is a $\sigma$-algebra.

Here, an $F_\sigma$-set is a countable union of closed sets, and a $G_\delta$-set is a countable intersection of open sets. I'm having trouble proving that the system is closed under countable unions. Clearly a countable union of sets which are both $F_\sigma$ and $G_\delta$ is also $F_\sigma$, but how would one show this countable union is again $G_\delta$? Any suggestions?

EDIT: This problem is actually false, as the comments have pointed out.

Solution 1:

Your class of sets cannot be a sigma algebra because an infinite union of $G_{\delta}$ sets may not be $G_{\delta}$

Take for instance $\mathbb{Q}=\{q_1,q_2.....\}$ in the usual topology of the real line.

Then $\mathbb{Q}=\bigcup_{n=1}^{\infty}\bigcap_{i=1}^{\infty}(q_n-\frac{1}{i},q_n+ \frac{1}{i})$ a union of $G$ delta sets.

But it is proved from Baire's category theorem that $\mathbb{Q}$ is not a $G$ delta set.

But your class of sets it is an algebra though.

Lets Denote $\Sigma$ the class of such sets

Let $A \in \Sigma$

We will use the fact that in a topological space the complement of a closed set is open and the complement of an open set is closed.

$A$ is a $G_{\delta}$ set thus from De Morgan Laws $A^c$ is a $F_{\sigma}$ set.

$A$ is a $F_{\sigma}$ thus proceeding as before $A^c$ is a $G_{\delta}$ set.

So $A^c$ belongs in $\Sigma$.

Now let us take a collection of sets $A_1,A_2...A_N \in \Sigma$

$A_n$ is an $F_{\sigma}$ set for all $n \in \{1,2...N\}$ thus it is a union of closed sets i.e $A_n= \bigcup_{i=1}^{\infty}F_{n,i}$

Thus $\bigcup_{n=1}^{N} A_n=\bigcup_{n=1}^{N}\bigcup_{i=1}^{\infty}F_{n,i}$ so it is a union of closed sets.

From this we conclude that the finite union of $A_n$ is an $F_{\sigma}$ set.

.

$A_n$ is an $G_{\delta}$ set for all $n \in \{1,2....N\}$ thus it is an intersection of open sets i.e $A_n= \bigcap_{i=1}^{\infty}G_{n,i}$

Thus $\bigcup_{n=1}^{N} A_n=\bigcup_{n=1}^{N}\bigcap_{i=1}^{\infty}G_{n,i}=\bigcap_{i=1}^{\infty}\bigcup_{n=1}^{N}G_{n,i}=\bigcap_{i=1}^{\infty}B_i$

where $B_i=\bigcup_{n=1}^{N}G_{n,i}$

Note that each $B_i$ is an open set a union of open sets.

In general a finite union of $G_{\delta}$ sets is again $G_{\delta}$

From this we conlude that the union of $A_n$ is a $G_{\delta}$ set,and it is also $F_{\sigma}$

So the finite union of $A_n$ is in $\Sigma$

Now using the fact that in a topological space the whole space and the empty set are both open and closed

you can prove that $\Omega,\emptyset \in \Sigma$