Q:

Is $(\mathbb{Q}\times\{0\}) \cup (\mathbb{R}\times(0,\infty))$ $F_{\sigma},G_{\delta}?$

Is A = $(\mathbb{Q}\times\{0\})\cup (\mathbb{R}\times(0,\infty))$ $F_{\sigma},G_{\delta}?$
I started with saying that it is not $G_{\delta}$ because:
Let's say that it is $G_{\delta}$. Then $(\mathbb{R}\times[0,\infty))\setminus A = \mathbb{R}\setminus \mathbb{Q}$ which is not $F_{\sigma}$. 
But:
$(\mathbb{R}\times[0,\infty))\setminus A = (\mathbb{R}\times[0,\infty))\cap A^{c}$ and $A^{c}$ is $F_{\sigma}$, so 
$(\mathbb{R}\times[0,\infty))\cap A^{c}=(\mathbb{R}\times[0,\infty))\cap\bigcup U_{n}$ where $U_{n}$ are closed. Because $(\mathbb{R}\times[0,\infty))$ is closed we have that $(\mathbb{R}\times[0,\infty))\cap A^{c}=\mathbb{R}\setminus \mathbb{Q}$ is $F_{\sigma}$ - contradiction. 
I thought that $A$ is also not a $F_{\sigma}$ and that the proof would be similar but I can't come up with any. I would be grateful for any help, thanks in advance!

A:

This is an $F_\sigma$ set. To see this note two things:

The union of two $F_\sigma$ sets is an $F_\sigma$ set.
Every open set is an $F_\sigma$ set in $\Bbb R^2$.

And this set is the union of an open set and a countable set.

