A topological space $X$ satisfies Property $K_1$ (Property of Mrowka) if the closure of the union of arbitrarily many $G_\delta$ sets of $X$ coincides with its sequential closure (the sequential closure of $A \subseteq X $ is the set of all points $x\in X$ for which there is a sequence in $A$ that converges to $x$). I wantwant some helpshelp to show that the Property $K_1$ is preserved under continuous maps. I tried with the following argument:
Let $X,Y$$X$, $Y$ be two topological spaces such that $Y$ is an image of $X$ under the continuous map $f$. Suppose that $X$ has property $K_1$. Let $G$ be a $G_\delta$ set in $Y$ and $y \in \overline{G}$. Since $f$ is continuous and surjectionsurjective, $f^{-1}(G)$ is $G_\delta$ in $X$. Choose $x \in f^{-1}(y),$ so $x \in \overline{f^{-1}(G)}$. As $X$ has property $K_1$, there exist $\langle x_n:~ n<\omega \rangle \subseteq f^{-1}(G)$ which converges to $x$. Thus there is $\langle f(x_n):~ n<\omega \rangle \subseteq G$ which converges to $y$.
This proof just works only with a single $G_\delta$ rather than aritraryan arbitrary union (as required for $K_1$). Any help please?
