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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?

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The property of Mrowka is not preserved under surjective images without additional hypothesis. If $X$ is any topological space, and $X_{d}$ is the discrete space with the same underlying set as $X$, then the identity map $X_{d}\rightarrow X$ is continuous and $X_{d}$ satisfies the property of Mrowka, but $X$ may not. –  Joseph Van Name Jun 5 '13 at 16:03
    
Thanks for your interesting comments. You are right. But it is strange! Mrowka himself mentioned this result, in his paper entitled Mazur Theorem and m-adic Space, without any additional hypothesis and he pointed it as a trivial without proof unless, I understood it wrongly! –  um Haitham Jun 6 '13 at 8:57
    
The sequential closure of a set $A$ is not just the limits of sequences in $A$ since you may need to take limits of limits, limits of limits of limits, and so on. –  Ramiro de la Vega Jun 6 '13 at 18:47

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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 $\overline{A} \subseteq B$.

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@ramiro,which article? –  R Salimi Jun 7 '13 at 4:37
    
@R salimi: sorry, I thought that the article was mentioned in the OP but it was actually in the comments. I'll edit to include the reference. –  Ramiro de la Vega Jun 7 '13 at 8:30

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