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What is name for those topological spaces that satisfy the following property?

  • For every point $x$, $\emptyset$ is the only closed set properly contained in the closure of $\{x\}$ (the $\subseteq$-minimal closed set containing $x$).

Any space with all singleton sets closed satisfies this property, but the converse need not hold.

Are there any interesting characterization of these spaces?

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    $\begingroup$ By the "least" closed set containing $x$ you just mean the closure of $\{x\}$, right? $\endgroup$ – Randall Apr 4 at 18:03
  • $\begingroup$ Yes. The closure of the singleton. I updated the OP for clarity $\endgroup$ – fundagain Apr 4 at 18:04
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Suppose that $X$ has this property, and let $x\in X$. If $y\in\operatorname{cl}\{x\}$, then of course $\operatorname{cl}\{y\}\subseteq\operatorname{cl}\{x\}$, so we must actually have $\operatorname{cl}\{y\}=\operatorname{cl}\{x\}$. Thus, the specialization order on $X$ is symmetric, and it’s clear that the converse also holds: $X$ has the property in question if and only if its specialization order is symmetric.

The specialization order is actually a pre-order: it is reflexive and transitive but not necessarily antisymmetric. Points that are equivalent with respect to this order are topologically indistinguishable: they have the same open nbhds. The order is symmetric precisely when topologically distinguishable points are separated, meaning that each of them has an open nbhd that does not contain the other. Such a space is said to be $R_0$ (or symmetric), and its Kolmogorov quotient is $T_1$.

In other words, it’s basically a $T_1$ space in which some of the points may have been ‘fattened up’ into sets of topologically indistinguishable points.

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  • $\begingroup$ Thanks, I appreciate it. Clearly explained. $\endgroup$ – fundagain Apr 4 at 19:43
  • $\begingroup$ @fundagain: You’re welcome. $\endgroup$ – Brian M. Scott Apr 4 at 19:46
  • $\begingroup$ Just to be clear, these are the 𝑅0 (or symmetric) spaces? $\endgroup$ – fundagain Apr 4 at 19:47
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    $\begingroup$ @fundagain: Yes, that’s right. $\endgroup$ – Brian M. Scott Apr 4 at 19:47

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