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 want some help to show that the Property $K\_1$ is preserved under continuous maps. I tried with the following argument:


Let $X$, $Y$ 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 surjective, $f^{-1}(G)$ 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 works only with a single $G\_\delta$ rather than an arbitrary union (as required for $K\_1$). Any help please?



Property of Mrowka
In the MathSciNet review of Mrowka´s article "*Mazur theorem and m-adic spaces*" it is mentioned that property $K\_1$ is preserved under continuous **closed** maps. I don´t have access to Mrowka´s article but perhaps that is the intended result. To show this, suppose that $X$ has property $K\_1$ and let $f:X \to Y$ be a continuous closed surjective map. If $A$ is a union of $G\_\delta$ subsets of $Y$ and $B$ is the sequential closure of $A$ in $Y$, then $C:=f^{-1}(A)$ is a union of $G\_\delta$ subsets of $X$ and $D:=f^{-1}(B)$ is sequentially closed in $X$. Therefore $\overline{C} \subseteq D$ (using property $K\_1$ in $X$) and hence $A=f(C) \subseteq f(\overline{C}) \subseteq f(D)=B$. Since $f(\overline{C})$ is closed in $Y$ it follows that $\overline{A} \subseteq B$.


