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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeDec 15th 2016

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

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeDec 15th 2016
• (edited Dec 15th 2016)

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

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeDec 16th 2016

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

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 30th 2017
• (edited Apr 30th 2017)

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

• CommentRowNumber5.
• CommentAuthordppes
• CommentTimeJul 18th 2019

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?

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeJul 19th 2019
• (edited Jul 19th 2019)

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!

• CommentRowNumber7.
• CommentAuthordppes
• CommentTimeJul 25th 2019

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

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeJul 25th 2019

Thanks!