added (here) an example of a regular space which is not Hausdorff

]]>I have re-written the “classical definition” (now here and here) in order to make it read more like a definition in the usual sense. Similarly, I have rewritten the statement and proof that regular $T_0$-spaces are Hausdorff (now here).

Lacking the energy to rewrite the remaining material similarly, I left it untouched but gave it a subsection “Alternative formulations” (now here) with the classical statements now in “Classical formulation” (here).

]]>added mentioning (here) that locally compact Hausdorff spaces are completely regular

]]>Thanks! I think it looks good.

]]>I also added the counterexample from dppes's friend (and some of Mike's related remarks).

]]>I undid some of dppes's edit, because it removed context for the disputed definition. I don't think that I reintroduced any errors, but it would be good for an independent eye to check that.

]]>Thanks!

]]>I broke up the definition via closed neighbourhoods and regular neighbourhoods into less-false parts.

]]>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!

]]>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 $0$.

- This space is not regular since we cannot separate $0$ from $[\frac1 2, 1)\times(0,1)$
- Every point $p = (p_1, p_2) \neq 0$ has the euclidean balls of centre $p$ and radius $\varepsilon \in (0, p_1)$ as regular neighbourhood basis.
- The provided basis for the neighbourhoods of $0$ already is a system of regular open sets.

Am I mistaken in my understanding of the claim? Are we mistaken in our counterexample?

]]>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

Thanks. We could also use a different notation; some people use $\triangleleft$.

]]>I tweaked the notation slightly. The $\subset\subset$ looks bad in the nLab proper without some negative spaces to bring the two symbols together.

]]>I added to regular space a remark that any regular space admits a naturally defined apartness relation.

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