Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
    
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? –  Nagase Apr 26 at 19:19
    
@Nagase: I've rephrased for clarity. (It was a bit more opaque than I had intended.) –  Arthur Fischer Apr 26 at 19:25
    
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? –  Nagase Apr 26 at 20:13
1  
@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? –  Arthur Fischer Apr 26 at 20:26
    
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? –  Nagase Apr 26 at 20:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.