I asked a question from a while back A discrete topological space is a space where all singletons are open $\implies$ all sets are clopen? Closed? but I suddenly find myself unable to wrap my head about this question.

Why is that in a discrete space, every singleton is open $\Rightarrow$ every singleton is closed. But in a $T_1$ space, every singleton is closed $\not\Rightarrow $ every singleton is open?

By discrete space I mean an arbitrary topological space $(X, \mathfrak{F})$ where $\mathfrak{F}$ is the discrete topology

Shouldn't the proof for these be the same?

Suppose $\{x\} \in X$ is open, then $\bigcup_{x \in X \backslash \{y\}} \{x\}$, $x \neq y$, is open as an arbitrary union of open sets , and $X \backslash \bigcup_{x \in X \backslash \{y\}} \{x\} = \{y\}$ is closed. Hence every singleton is closed as well.

For $T_1$ space, we know that every singleton is closed, so we just reverse the above proof and get that every singleton is open. But why doesn't this work?

What is the an assumption we can make on the $T_1$ space (aside from assuming a discrete topology) so we can make all singletons closed as well? Hausdorffness?


If every singleton is open, then each singleton is the complement of an open set and is therefore closed:

$$\{x\}=X\setminus\bigcup_{y\in X\setminus\{x\}}\{y\}\;.$$

This works because the union of any collection of open sets is open.

The same trick does not work for closed sets, because the union of infinitely many closed sets need not be closed. For instance, take $X=\Bbb R$ and $x=0$:

$$\bigcup_{y\in\Bbb R\setminus\{0\}}\{y\}=\Bbb R\setminus\{0\}$$

is definitely not closed, since $0$ is in its closure.

  • $\begingroup$ I see...I didn't expect this loss of symmetry. What minimum assumptions do we need to add to a $T_1$ space to let all singletons be open as well? $\endgroup$ – Olórin Aug 6 '16 at 20:47
  • 2
    $\begingroup$ @John: It’s the fundamental asymmetry between open and closed sets. \\ Saying that all singletons are open is exactly equivalent to saying that the space has the discrete topology. $\endgroup$ – Brian M. Scott Aug 6 '16 at 20:48
  • $\begingroup$ @Milo: Sure do; thanks! $\endgroup$ – Brian M. Scott Aug 6 '16 at 20:49

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