I'm new to topology (my main interests are in logic), so forgive me if my question is a bit clumsy. Let $\langle X, A(X) \rangle$ be a topology on $X$. Define a regular open set as an open set which is equal to the interior of its own closure. Suppose that, for every $x \in X$, $\{x\} \in F(X)$ ($F(X)$ being the set of all closed sets). Does it follow that every open set is also closed? I've been trying to prove this implication for a while, but to no avail; I suspect it's false (i.e. it does not follow that every open set is clopen). However, I haven't been able to construct a counterexample (probably because of my ignorance about those matters). Any ideas?


Hint. Suppose that $X$ satisies the given conditions, and let $x \in X$. As $\{ x \}$ is closed, then $X \setminus \{ x \}$ is open, and therefore regular open. Clearly $$X \setminus \{ x \} \subseteq \overline{ X \setminus \{ x \} } \subseteq X.$$ That fact that $X \setminus \{ x \}$ is regular open allows us to actually determine what $\overline{ X \setminus \{ x \} }$ is.

This will tell you something new about all singletons in $X$, which will then tell you something about all subsets of the space.

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  • $\begingroup$ I'm still struggling a little bit with it. When you say that "the conditions on the space allow us to actually determine what $\overline{X \setminus \{x\}}$ is", do I have to use both conditions (i.e. that every open set is regular and that every singleton is closed) in order to extract the relevant information? $\endgroup$ – Nagase Apr 26 '14 at 19:19
  • $\begingroup$ @Nagase: I've rephrased for clarity. (It was a bit more opaque than I had intended.) $\endgroup$ – user642796 Apr 26 '14 at 19:25
  • $\begingroup$ Hmm. I've been trying to think contrapositively; suppose there is $y \in X$ such that $y \not \in \text{int}(\overline{X \setminus \{x\}})$. This means that there is an open set (say, $V$) such that $y \in V$ and $V \not \subseteq \overline{X \setminus \{x\}}$. On the other hand, suppose there is $y \in X$ such that $y \not \in \overline{X \setminus \{x\}}$. Then there is an open set (say, $O$) such that $y \in O$ and $O \cap X \setminus \{x\} = \varnothing$. This means that either $O = \varnothing$ or $O = \{x\}$ (and thus $y=x$). But I can't go further than this... am I on the right track? $\endgroup$ – Nagase Apr 26 '14 at 20:13
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    $\begingroup$ @Nagase: You're making it much more difficult than necessary. In particular, we already know what $\mathrm{Int} ( \overline{ X \setminus \{ x \} } )$ is: it's $X \setminus \{ x \}$, since the set is regular open. My hint tells you that $\overline{ X \setminus \{ x \} }$ is either $X \setminus \{ x \}$ or $X$. Can you calculate the interior of either of these sets? $\endgroup$ – user642796 Apr 26 '14 at 20:26
  • $\begingroup$ Ahh, yea, sorry for being dense, I hadn't realized what those set inclusions meant. Well, since both sets are open, they are their own interior. But, if $\overline{X\setminus\{x\}} = X$, then $\mathrm{Int}(\overline{X\setminus\{x\}}) = \mathrm{Int}X = X$, which means that $X \setminus \{x\} = X$, which is absurd. So $\overline{X\setminus\{x\}} = X \setminus\{x\}$, right? $\endgroup$ – Nagase Apr 26 '14 at 20:44

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