Skip to main content
MathOverflow

Return to Answer

one format correction
Source Link
Nathan
Nathan
  • 269
  • 3
  • 8

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$S$ is a $G_\delta$.

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 $G_\delta$.

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 $G_\delta$.

Source Link
Nathan
Nathan
  • 269
  • 3
  • 8

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 $G_\delta$.