Homeomorphic Image of $G_{\delta}$ is $G_{\delta}$

Question

Let $U=\cap_{n \in \mathbb{N}} U_n$ be a non-empty $G_{\delta}$-set; where each $U_n$ is open in a normal topological space $X$. If $\phi:X\rightarrow Y$ is a homeomorphism then is $\phi(U)$ a $G_{\delta}$-subset of $Y$?

Answer 1

Easier: Let $\psi: Y \to X$ be the continuous inverse of $\phi$, using that $\phi$ is a homeomorphism. For each $n$, $\phi[U_n] = \psi^{-1}[U_n]$ is open by continuity of $\psi$ (or use the lemma, which I just reproved, that a homeomorphism is an open map).

And as $\phi$ is 1-1 (so images preserve intersections):

$$\phi[U] = \phi[\bigcap_{n \in \Bbb N} U_n] = \bigcap_{n \in \mathbb{N}} \phi[U_n]$$

which is then a $G_\delta$.

Answer 2

Voila: $$\phi(U)=(\phi^{-1})^{-1}[\cap_{n \in \mathbb{N}} U_n] = \cap_{n \in \mathbb{N}} (\phi^{-1})^{-1}[U_n]= \cap_{n \in \mathbb{N}} \phi[U_n]= \cap_{n \in \mathbb{N}} \phi[U_n];$$ since $\phi$ is a homeomorphism then each $U_n$ is itself open; whence the right-hand side implies that $\phi(U)$ is $G_{\delta}$.