Given $G_{\delta}$ set $G$ and a point $x$, $G \cup \{x\}$ is also a $G_{\delta}$ set

Question

I thought of this in many different ways. By definition of $G_{\delta}$, $G=\bigcap_{i\geq 1}U_i$ where $U_i\in G$ are open sets. Then $G \cup \{x\}=\bigcup_{i\geq 1}U_i\cup\{x\}=\bigcup_{i\geq 1}(U_i\cup\{x\})$. So I think will be sufficient to just show that for any $i$, $U_i\cup\{x\}$ is open. But I don't see how this is valid.

Any help is appreciated.

Answer 1

Let $B_i$ be any collection of decreasing open sets such that $\{x\}=\cap_i B_i$, (for examples balls of radius $\epsilon_i=1/i$). Then $G\cup\{x\}=(\cap_i U_i) (\cap_j B_j)=\cap_i(U_i\cap B_i)$