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In which topological and metric spaces are the sigletons $G_δ$, and why? One example is $(\mathbb{R},|\cdot|)$, or the Euclidean multidimensional metric space. Thank you.

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  • $\begingroup$ In any metric space, $\{x\} = \bigcap_{n=1}^{\infty}B(x, 1/n)$, where $B(x,1/n)$ is the open ball of radius $1/n$ centered at $x$. So every singleton is a $G_{\delta}$ in any metric space. $\endgroup$ – Bungo Aug 4 '16 at 21:17
  • $\begingroup$ @Bungo And closed, even. $\endgroup$ – Noah Schweber Aug 4 '16 at 21:17
  • $\begingroup$ @NoahSchweber Good point: in any metric space, closed implies $G_{\delta}$. $\endgroup$ – Bungo Aug 4 '16 at 21:20
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Singletons are $G_\delta$'s in a first countable $T_1$ space. That is, for any $x$ there is a sequence $B_n$ of neighbourhoods of $x$ such that every neighbourhood of $x$ contains some $B_n$. For any $y \ne x$, since the complement of $\{y\}$ is a neighbourhood of $x$ there is some $B_n$ that doesn't contain $y$, and so $\cap_n B_n = \{x\}$.

An example of a space that doesn't have this property is the cofinite topology on an uncountable set.

EDIT: A space in which singletons are $G_\delta$'s must be a $T_1$ space, but need not be first countable. A counterexample is the quotient space $\mathbb R/\mathbb N$ (i.e. take the reals and identify all natural numbers as the same point).

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    $\begingroup$ @user359315: As a very limited partial converse, if $X$ is a compact Hausdorff space or a subspace of a linearly ordered space, then singletons are $G_\delta$ sets iff $X$ is first countable. $\endgroup$ – Brian M. Scott Aug 5 '16 at 13:56

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