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I added to regular space a remark that any regular space admits a naturally defined apartness relation.
I tweaked the notation slightly. The $\subset\subset$ looks bad in the nLab proper without some negative spaces to bring the two symbols together.
Thanks. We could also use a different notation; some people use $\triangleleft$.
I gave the entry regular space an Examples-section (here) with the metric space example (previously missing from the entry) and the K-topology counter-example. Similarly at normal space, here
As far as I understand, Definition 2.5 claims that as soon as every point in a topological space $X$ has a neighbourhood basis consisting solely of regular open sets, the space is regular. (It also claims the converse, which is proven immediately before.) There is no proof given for this direction (there seems to have been a plan for further explanation but it never happened) and I had trouble believing this claim. Then, a friend of mine pointed out that the following should provide a counterexample: Let $\bigl((0, 1)\times(0, 1)\bigr)\cup\{0\}$ be equipped with the Euclidean topology on $(0, 1)\times(0,1)$ and have the sets of the form $(0, \frac{1}{2})\times(0, \varepsilon)\cup\{0\}$ (for $\varepsilon \in (0, 1)$) as a basis of open neighbourhoods for the point $0$.
Am I mistaken in my understanding of the claim? Are we mistaken in our counterexample?
I think you’re right. The first sentence of Def 2.5 is correct, as is the restatement of this as “the closed neighborhoods form a base of the neighborhood filter at every point”, but it’s not valid to jump from this to saying that the regular open sets do, since containing a regular open set is (of course) not the same as containing its closure. Please feel free to correct it!
Thanks!
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