Show that the set $(\mathbb{I} \cap (-\infty, 0]) \cup (\mathbb{Q} \cap [0, \infty))$ is neither a $F_{\sigma}$ set nor a $G_{\delta}$ set. Note: $\mathbb{I}$ is the set of irrational numbers.
I started by showing $\mathbb{I}$ is a $G_{\delta}$ set and cannot be a $F_{\sigma}$ set, likewise, I showed that $\mathbb{Q}$ is a $F_{\sigma}$ set and cannot be a $G_{\delta}$ set. I also know both $(-\infty, 0]$ and $[0, \infty)$ are both a $G_{\delta}$ and $F_{\sigma}$ set. How do I complete the proof?
Assume $X=(\mathbb{I} \cap (-\infty, 0]) \cup (\mathbb{Q} \cap [0, \infty))$ is a $G_\delta$ set.
Then we have that $Y=(X \cap [0,\infty))=\mathbb{Q}\cap[0,\infty)$ is a $G_\delta$ set as the finite (even the countable) intersection of $G_\delta$ sets is a $G_\delta$ set.
Note that homeomorphisms preserverve open and closed sets, thus $\mathbb{Q}\cap(-\infty,0]$ is also $G_\delta$.
But the finite union of $G_\delta$ sets is a $G_\delta$ set, thus
$Z=(\mathbb{Q}\cap[0,\infty))\cup(\mathbb{Q}\cap(-\infty,0])=\mathbb{Q}$ is a $G_\delta$ set, a contradiction to the fact that $\mathbb{Q}$ is not a $G_\delta$ set.
It is very similar to show that $X$ is not a $F_\sigma$ set, I leave this for you to work out.