I've been trying to prove the following statement, but the converse has been giving me trouble.

A topological space $X$ is regular if and only if every closed subset $Z\subseteq X$ is the intersection of all open sets $U\subseteq X$ which contain it.

Here, the defintion of regular is that for any point $p\in X$ and any closed subset $Z\subset X$ not containing $p$, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. The space does not necessarily have to be a $T_1$ space.

I believe I showed the forward statement correctly. I suppose $X$ is regular, and clearly $Z\subseteq\cap\mathcal{U}$, where $\mathcal{U}$ is the family of all open sets containing $Z$. Then if I take $p\in\cap\mathcal{U}$, then $p$ is in every open subset containing $Z$. Towards a contradiction, I assume that $p$ is not in $Z$. Then since $X$ is regular, there exist open sets $E,F\subset X$ such that $p\in E$, $Z\subseteq F$, and $E\cap F=\emptyset$. Thus $p\not\in F$, but $F$ is an open set containing $Z$, and thus must contain $p$, a contradiction. Thus $p\in Z$, and so $Z=\cap\mathcal{U}$.

The backwards step has stumped me for a while. I take any point $p$ and a closed set $Z$ which does not contain $p$. So $Z=\cap\mathcal{U}$, and since $p\not\in Z$, there exists an open set $F$ containing $Z$ which does not contain $p$. The only observation I've been able to make is that $F^c$ is a closed set containing $p$, and then $F^c=\cap\mathcal{W}$ where $\mathcal{W}$ is every open set containing $F^c$. Hence $p$ is in every open set containing $F^c$. Taking any open sets $E$ and $F$ such that $p\in E$ and $Z\subseteq F$, and I haven't been able to see a way that there exists two such open sets that are disjoint, to show the space is regular. Showing $F^c$ is an open set would work, but I'm not sure if that's even true. I was hoping someone could point out what I'm missing, thank you.

  • $\begingroup$ $F^C$ certainly need not be open. It is already closed, so that would make it clopen, which isn't going to work in any connected space. $\endgroup$ – Nate Eldredge Nov 17 '10 at 5:30
  • $\begingroup$ @Cromarty: you don't need to prove the forward implication by "contradiction": your argument shows that if $p\notin Z$, then $p\notin\cap\mathcal{U}$. Therefore, the complement of $Z$ is contained in the complement of the intersection, so the intersection contains $Z$; the other inclusion is immediate. $\endgroup$ – Arturo Magidin Nov 17 '10 at 5:55
  • $\begingroup$ Thanks Arturo for pointing that out. $\endgroup$ – yunone Nov 17 '10 at 6:20
  • $\begingroup$ @Cromarty: I misspoke in the next-to-last-clause (and can't edit anymore): it should read "the complement of $Z$ is contained in the complement of the intersection, so $Z$ contains the intersection". $\endgroup$ – Arturo Magidin Nov 17 '10 at 17:25

Ok, let's try another counterexample.

For simplicity, say a space with your property (any closed set is the intersection of its neighborhoods) is "Cromarty".

Let $X$ be an infinite set with the cofinite topology (the closed sets are all the finite sets and $X$ itself). If $Z \subset X$ is closed (hence finite), let $\mathcal{U}$ be the collection of all open sets containing $Z$. For any $x \notin Z$, $\{x\}^c$ is open (because it is cofinite) and contains $Z$, so $\{x\}^c \in \mathcal{U}$, and thus $x \notin \bigcap \mathcal{U}$; hence $Z^C \subset \bigcap \mathcal{U}$, so $\bigcap \mathcal{U}=Z$. So $X$ is Cromarty.

However, $X$ is not regular, since every pair of nonempty open sets has nonempty (in fact, infinite) intersection.

Edit: To amplify, it looks like the Cromarty property is equivalent to being $R_0$, which means: if $x$ has a neighborhood not containing $y$, then $y$ has a neighborhood not containing $x$. This is of course strictly weaker than being regular (as the cofinite topology shows).

Suppose $X$ is $R_0$, and $Z$ is closed. Suppose $x \notin Z$, $y \in Z$. Then $Z^c$ is a neighborhood of $x$ not containing $y$, so $y$ has a neighborhood $U_y$ that does not contain $x$. Now $U = \bigcup_{y \in Z} U_y$ is a neighborhood of $Z$ that does not contain $x$. As above, it follows that $Z$ is the intersection of its neighborhoods, so $X$ is Cromarty.

Conversely, suppose $X$ is Cromarty. Suppose a point $x$ has a neighborhood $U$ not containing $y$. $U^c$ is closed, hence the intersection of its neighborhoods, so $U^c$ has a neighborhood $V$ not containing $x$. But $V$ is also a neighborhood of $y$, so $X$ is $R_0$.

  • $\begingroup$ Just to clarify, does your edit say the your counterexample is not true, and that the original statement in the question actually is true? $\endgroup$ – yunone Nov 17 '10 at 6:20
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    $\begingroup$ @Cromarty: If I understand correctly, Nate has proved that the following are equivalent: (1) Given any closed subset $Z$ of $X$, and any $x \in X\setminus Z$, there is an open set $U$ containing $Z$ with $x \not\in U$; (2) Any closed set is the intersection of the open sets which contain it; (3) $X$ is $R_0$ in the sense of Nate's answer, i.e. if any point $x$ has a n.h. not containing some point $y$, then also $y$ has a n.h. not containing $x$. (The equivalence of (1) and (2) is more-or-less obvious, and the equivalence of (2) and (3) is what Nate proves in his edit.) It seems that these... $\endgroup$ – Matt E Nov 17 '10 at 6:40
  • $\begingroup$ ... conditions are weaker than $X$ being regular (as the example of an infinite set with the cofinite topology shows). $\endgroup$ – Matt E Nov 17 '10 at 6:42
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    $\begingroup$ @Matt E: That's right. The counterexample is a counterexample. I pointed out the equivalence of Cromarty's property with the $R_0$ property just because the latter is a (somewhat) standard entry in the usual hierarchy of separation axioms, where it falls on the weak side of regularity. $\endgroup$ – Nate Eldredge Nov 17 '10 at 14:00
  • $\begingroup$ Thanks for the thought you put into this question Nate, very helpful. Thanks too Matt for clarifying. $\endgroup$ – yunone Nov 17 '10 at 22:32

Nate's example is a good one and actually solves the question as it is. But suppose you think that the Cromarty property, as Nate called it, is not good enough since 'almost all' spaces satisfy it. Then you try to substitute it for a stronger condition. You can require that every closed subset $Z \subseteq X$ is a countable intersection of open sets. That's exactly what a $G_{\delta}$ space is. In this situation the example Nate gave still works, you just need to require that the space $X$ is countable. In fact, if $F$ is a closed subset of $X$ with the cofinite topology then $X-F$ is the countable union of the closed sets $\{x\}$, with $x \in X - F$. So

$$F = \bigcap_{x \in X-F} (X - \{x\}).$$

Thus, a countable set $X$ with the cofinite topology is a $G_{\delta}$ space. But then you conclude that your condition isn't strong enough. Now you try to impose some kind of separation. The question becomes: is there a Hausdorff $G_{\delta}$ space that isn't regular?

The answer is yes. Let $K = \{ 1/n : n \in \mathbb{N} \}$ and consider the collection $\mathcal{B}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. It's easy to show that $\mathcal{B}$ is a basis for a topology on $\mathbb{R}$, which is called the K-topology. When $\mathbb{R}$ is endowed with this topology it's denoted by $\mathbb{R}_K$.

The K-topology is finer than the standard topology on the real line. So it's a Hausdorff space (actually a completely Hausdorff space). Also, $\mathbb{R}_K$ is second countable, since the colletion $\mathcal{B'}$ of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$ and with $a,b \in \mathbb{Q}$ is a basis for the K-topology. $\mathbb{R}_K$ isn't regular, since the closed set and the point $0$ cannot be separated by disjoint open sets.

Now let's see why $\mathbb{R}_K$ is a $G_{\delta}$ space. It's easier to prove that it's an $F_{\sigma}$ space, that is, a space such that every open set can be written as a countable union of closed sets, which is equivalent to prove that it's a $G_{\delta}$ space. Since $\mathbb{R}_K$ is second countable, every open set can be written as a countable union of basis elements. If we prove that every basis element is an $F_{\sigma}$ set, then will follow that $\mathbb{R}_K$ is an $F_{\sigma}$ space.

The sets $(a,b)$ are $F_{\sigma}$, since $(a,b) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, b - \frac{1}{n}]$. Now consider $(a,b) - K$; we do not need to worry with the end point $b$, so suppose $b > 1$. If $a>0$, let $n_0$ be the largest natural such that $a < 1/n_0$. Then

$$ (a,b)-K = (a, \frac{1}{n_0}) \cup \bigcup_{n=1}^{n_0-1} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b).$$

Each of the sets in the decomposition above are $F_{\sigma}$, thus $ (a,b)-K$ is $F_{\sigma}$. If $a=0$, then

$$ (0,b)-K = \bigcup_{n=1}^{\infty} (\frac{1}{n+1}, \frac{1}{n}) \cup (1,b),$$

so $(0,b)-K$ is $F_{\sigma}$. If $a<0$, then

$$ (a,b)-K = (a,0] \cup ((0,b)-K) = \bigcup_{n=1}^{\infty} [a + \frac{1}{n}, 0] \cup ((0,b)-K),$$

so $(a,b)-K$ is an $F_{\sigma}$ and we conclude that $\mathbb{R}_K$ is an $F_{\sigma}$ space.

  • $\begingroup$ Thanks for the response Nuno. I'm not familiar with many of the ideas you just mentioned, but I'd be curious to see such a proof, if it would in fact provide a counter example to the statement. The statement is actually found in a set of notes I am reading, so I assumed it to be true, but perhaps it is incorrect, and thus giving me so much difficulty. $\endgroup$ – yunone Nov 17 '10 at 2:40
  • $\begingroup$ @Cromarty: I added some links. I hope it helps. Perhaps wikipedia is incorrect. But I'll think about it tomorrow. $\endgroup$ – Nuno Nov 17 '10 at 2:52
  • $\begingroup$ Thanks also Nuno, I'd like to see what you proved, but only if it's not too much trouble to post. $\endgroup$ – yunone Nov 17 '10 at 22:34
  • $\begingroup$ @Cromarty: Posted it. Let me know if there is something unclear. $\endgroup$ – Nuno Nov 18 '10 at 0:08

Isn't the trivial (indiscrete) topology a counterexample? Perhaps there are more assumptions somewhere.

Edit: I had a different definition of "regular" in mind from the asker. The question has been clarified, and the trivial topology is not a counterexample.

  • $\begingroup$ Indiscrete topologies have points separated from closed sets (trivially), and to some that is the definition of regular. $\endgroup$ – Jonas Meyer Nov 17 '10 at 3:55
  • $\begingroup$ So the only closed sets are $\emptyset$ and $X$, correct? Then $\emptyset=\emptyset\cap X$ and $X=\cap X$, so the hypothesis is satisfied. Then for any $p$, the only closed set that doesn't contain $p$ is the empty set. But couldn't I take $X$ to be an open set containing $p$, $\emptyset$ to be the open set containing $\emptyset$, and then $X\cap\emptyset=\emptyset$?, so there's no problem? Or does the open set containing $p$ have to be a proper subset of $X$? $\endgroup$ – yunone Nov 17 '10 at 3:59
  • $\begingroup$ @Cromarty: According to some conventions, regular spaces are assumed to be Hausdorff. Some say that T_3 means points and closed sets can be separated, and regular is T_3 + Hausdorff (e.g., Counterexamples in Topology). Some say the exact opposite (e.g., Wikipedia and presumably some of its references). Others say that T_3 and regular are synonyms, and both are Hausdorff. So it never hurts to specify precisely which definition you're using. $\endgroup$ – Jonas Meyer Nov 17 '10 at 4:12
  • $\begingroup$ @Jonas, I apologize, I didn't know that there were different conventions for the definition. I edited in the one that I am using to hopefully make my question more clear. $\endgroup$ – yunone Nov 17 '10 at 4:18
  • $\begingroup$ @Cromarty: Thank you, it is very clear. $\endgroup$ – Jonas Meyer Nov 17 '10 at 4:24

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