A space is $F_\sigma$-discrete space if it is the countable union of closed discrete subspaces. Is it true that every subset of an $F_\sigma$-discrete space is of the type $G_\delta$?
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2 Answers
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The answer is yes. Suppose that $X$ is an $F_\sigma$ space as you have defined it, so that $X=\bigcup_n X_n$, where each $X_n$ is a closed discrete space. Suppose that $Y\subset X$ is any subset. For each $n$, note that $X_n-Y$ is closed in the discrete space $X_n$ and hence also closed in $X$, and so the complement $U_n=X-(X_n-Y)$ is open in $X$. Furthermore, $\bigcap_n U_n=Y$, since $Y\subset U_n$ and anything not in $Y$ is in some $X_n$ and hence thrown out of that $U_n$. So $Y$ is a countable intersection of open sets and hence $G_\delta$.
answered Nov 9, 2011 at 13:30
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Yes, since closed discrete sets are hereditarily closed. Let $X$ be the space and $E_n$ the closed discrete sets, so $X = \bigcup\limits_{n=1}^\infty E_n$. Then given $S \subseteq X$, each $E_n \setminus S$ is still closed, so $S = \bigcap\limits_{n=1}^\infty (E_n \setminus S)^\complement$ shows that $S$ is a $G_\delta$.
