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I'd like to see some motivation or application of the definition of regular open/closed sets. There are answers already on this site that talk about the "geometric intuition" and the nice properties regular open sets have. However, the "interior of closure is itself" definition does not follow from results in $\mathbb{R}$ as others do(limit points, closure, limit of sequence, etc). I looked up on every single general topology book I know(Willard, Engelking, Kelley, Munkres...) but the only results about regular open/closed sets I can find is just about themselves(Like the "union" of two regular open sets, product of them, etc). If there is no applications of it, and no apparent motivations exist as examples(in $\mathbb{R}$, $\mathbb{R}^n$...), then why do we define such classes of sets in topological spaces?

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  • $\begingroup$ Not sure if this will help, but perhaps you are having some issues with applications of topology outside of metric spaces, i.e., given the open ball definition which is nice and intuitive, why do we need topologies and open/closed sets on their own. To this, I would say it may be a good idea to look at examples in analysis where we have a topology on a set that doesn't have a metric. For example: en.wikipedia.org/wiki/Fr%C3%A9chet_space , en.wikipedia.org/wiki/Weak_topology $\endgroup$ – rubikscube09 Dec 15 '18 at 2:16
  • $\begingroup$ I mistyped in my first sentence. It should be "regular open/closed sets" instead of "open/closed sets". Sorry about that. $\endgroup$ – William Sun Dec 15 '18 at 2:18
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    $\begingroup$ I didn't see that. Sorry! I don't have much experience with general topology, but if it makes you feel any better I don't think the concept of regularity is used much outside general topology in and of itself, i.e. I think you are right to feel it is poorly motivated. $\endgroup$ – rubikscube09 Dec 15 '18 at 2:22
  • $\begingroup$ For some references, see those that I cite in "5. REGULARLY OPEN SETS" in the following 20 September 2006 sci.math post archived at Math Forum. $\endgroup$ – Dave L. Renfro Dec 15 '18 at 9:30
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Regular open sets can be a technical tool: e.g. we can show that in a regular space $X$ we have that $w(X)\le 2^{d(X)}$, where $w(X)$ is the minimal infinite cardinality of a base for $X$ and $d(X)$ is the minimal infinite cardinality of a dense subset of $X$. These are so-called cardinal invariants of topological spaces and their relations have been extensively studied, e.g. in metric spaces $w(X)=d(X)$, but in general $d(X) \le w(X)$ and the aforementioned inequality goes the other way, bounding $w(X)$ in terms of $d(X)$. For that proof we use that the regular open sets form a base for $X$ in a nice way.

But I think the main reason they are studied is the fact that $RO(X)$, the set of regular open sets of $X$, is a natural Boolean algebra, with operations $A \land B = A \cap B$, $A \lor B = \operatorname{int}(\overline{A \cup B)})$ and $\lnot A = X\setminus \overline{A}$ and $\emptyset, X$ as $0$ and $1$.

Another natural Boolean algebra asssociated with a space $X$ is $\operatorname{Clop}(X)$, the set of clopen (closed and open) subsets of $X$. which always contains $\emptyset,X$ (and in connected spaces only these two!) and where we use standard intersection, union and complement as Boolean operations.

One can show that if $B$ is any abstract Boolean algebra there is a compact Hausdorff space $X$ such that $B$ is isomorphic (as a Boolean algebra) to $\operatorname{Clop}(X)$ (and the clopen sets of $X$ form a base for its topology (such spaces are called zero-dimensional). Moreover, $RO(X)$ for that $X$ turns out to be a so-called completion of $B$ (it's clear that clopen sets are regular open so $\operatorname{Clop}(X) \subseteq RO(X)$) in which we can take arbitary $\lor,\land$ operations and all complete Boolean algebras are just regular open algebras of special compact spaces. So regular open sets are used in this connection to set theory/algebra and play an important role there.

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The best motivation for doing anything in general topology is to classify topological spaces up to homeomorphism. In particular, words are defined so that when two spaces are not homeomorphic, we can say why in a sentence, e.g., "one is Hausdorff and the other isn't".

Regular open sets help in this regard. If $X$ is a regular space, then its regular open sets form a base for the topology on $X;$ but the converse statement is not true. There exist spaces which are semiregular, i.e., the regular open sets form a base for the topology, but which are not regular. And in turn, there exist semiregular spaces which are not locally regular.

Now imagine how cumbersome it would be to define "semiregular" without the notion of "regular open set", and then imagine how cumbersome it would be to read/write the previous paragraph if we hadn't defined "semiregular".

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  • $\begingroup$ I agree that it would be much more complicated to define semiregular spaces without regular open sets. However, if regular open sets came from an observation that”A regular space has a base consisting of regular open sets”, then as you stated in the answer, “The best motivation for doing anything in general topology is to classify spaces up to homeomorphism”, how is semiregular property helpful in classifying spaces? Since regular implies semiregular, spaces that cannot be classified by regularity are not distinguishable by semiregularity either. $\endgroup$ – William Sun Dec 15 '18 at 5:58
  • $\begingroup$ So the question becomes”What is the motivation of defining semiregular spaces?” That’s also a question of my concern. $\endgroup$ – William Sun Dec 15 '18 at 5:59
  • $\begingroup$ I stated that wrong. It should be “Semiregular property is good at classifying (some) non-regular spaces(they can still be semiregular).” So what is a nice example of this, that is, spaces that are hard to be distinguished by other common topological properties(compactness, connectedness, countability, separability) but is distinguishable by semiregularity? $\endgroup$ – William Sun Dec 15 '18 at 6:09
  • $\begingroup$ Why does there have to be an application? Why is the reason "because there are semiregular spaces that are not regular" not good enough? One of the goals of topology is to figure out what kinds of topological spaces can and cannot exist, and the definition of "regular open" is useful for this purpose. It feels a bit like you're asking why we define "perfect numbers", even though they have very, very few applictions (if any). The point is just that it's another kind of number, with interesting properties. General topology does not have to be more applicable than number theory. $\endgroup$ – Will R Dec 15 '18 at 15:00

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