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There was a similar thread on the neighbour forum StackExchange on sufficient conditions for a topological space to be completely regular $T_{3^1/_2}$.

Please, let me know any known condition(s) that a topological space is regular $T_3$. Any known approach or a standard strategy for showing that a topological space is regular would be very welcome and helpful answer too.

Thank you in advance!

EDIT: The topology I have in mind is a sequential ($\implies$ $k$-space, in the sense of Engelking) Lusin ($\implies$ Souslin, in the sense of Fernique) space topology defined via the Kantorovich-Vulih-Pinsker-Kisyński recipe; see e.g. Engelking's book Subsection 1.7.18-21 and references therein for the aforementioned K-V-P-K recipe.

UPDATE: Sufficient conditions for a topological space to be regular $T_3$ include:

-Compact Hausdorff spaces (admit a unique compatible uniformity $\implies T_{4}$),

-Uniform (e.g. metrizable $\implies T_6$) spaces,

-Topological groups (e.g topological vector spaces $\implies T_{3^1/_2}$),

-Arbitrary subspaces of a regular space,

-Arbitrary products of regular spaces.

I will keep updating the list for similar answers (as Todd's answer below). I thank everybody for their time and valuable help.

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    $\begingroup$ I actually think it would be more sensible posting at Math.SE than here. It's not very clear what conditions would be useful to you. The first condition that comes to mind is uniform space structure; these include metric spaces, compact Hausdorff spaces, and topological groups. Arbitrary products of regular spaces are regular. Arbitrary subspaces of regular spaces are regular. In practice this may give all conditions you'll ever need to know. For an example of a regular but not completely regular space, see Munkres's book. There are more abstruse categorical theorems that could be stated. $\endgroup$
    – Todd Trimble ♦
    Sep 22 '18 at 23:56
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    $\begingroup$ I can't tell what the $S$-topology is at a quick glance, but I'm glad you mentioned that this is part of actual research. (Not doing so is like talking about a medical condition with a doctor and not mentioning you have medical training.) You might edit in the fact about closure under products. It can be shown that the category of $T_3$ spaces is a reflective subcategory of $\text{Top}$ and therefore admits colimits as well (although they are not necessarily computed as in $\text{Top}$). $\endgroup$
    – Todd Trimble ♦
    Sep 23 '18 at 17:47
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    $\begingroup$ By the way, compact Hausdorff spaces admit a unique compatible uniformity and are therefore $T_{3\frac1{2}}$. (Edit: oh, silly me. Of course compact Hausdorff implies $T_4$ as well.) $\endgroup$
    – Todd Trimble ♦
    Sep 23 '18 at 17:48
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    $\begingroup$ Thanks, Todd. Those are now added. Thank you also your comments on posting this forum. I'm still quite new and learning... :) $\endgroup$
    – Matti Kiiski
    Sep 23 '18 at 18:42
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    $\begingroup$ The topology S is a topology on the Skorokhod space introduced by Adam Jakubowski in 1997. The Skorokhod space is a canonical space for stochastic processes that consists of real-valued functions on the real axis that have only discontinuities of the first kind and that are normalized in a certain way, meaning that, that the function admits a finite limit from the left and the right hand-side at every point and the function is continuous from the right at every point, respectively. Roughly speaking, the topology S is obtained by turning those characteristics as a (pre-)compactness criteria. $\endgroup$
    – Matti Kiiski
    Sep 23 '18 at 19:01
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The list of sufficient conditions in the question neatly avoids addressing the real issue I think. From the comments I see that it is regularity of a very specific topology that you are after: Jakubowski's $S$-topology on Skorokhod space. None of the conditions that you have will help you very much in that case. The reason is that proving normality or (complete) regularity requires fairly detailed knowledge of the nature of the open sets. The Kisynski construction yields a topology but with very little information on what the closed sets look like, other than that they are closed under taking limits. Hausdorffness of the $S$-topology is forced by having a point-separating family of continuous functions. For regularity (to prove it or disprove it) you need to know what arbitrary open sets look like; that would require a very detailed study of the convergence used (I must admit that I did not have the time for that).

To see how difficult it may be to deduce properties of the topology from properties of the convergence have a gander at this example: a $T_1$-space that is Frechet-Urysohn, with unique sequential limits but that is very far from Hausdorff.

In short: if you want to (dis)prove regularity of the $S$-topology you'll have to get your hands dirty.

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  • $\begingroup$ Thank you very much for your answer! This is indeed the case. As you write: a separation axiom is rather a statement about "the nature of an arbitrary open set". Even "an arbitrary S-convergent sequence" is a very difficult object to work with, not only due to K-V-P-K, but also due to the fact that there is an existence statement in the definition of the S-convergence (those BV functions). $\endgroup$
    – Matti Kiiski
    Sep 27 '18 at 20:30
  • $\begingroup$ Disproving the regularity: An obvious (easy) approach would be show that the topology violates some non-trivial property that can be deduced from the regularity in conjunction with some other property that the topology has... but, for instance, as Adam Jakubowski already in his original 1997 article writes: "In the absence of regularity almost nothing of the theory for Lusin spaces can be used." $\endgroup$
    – Matti Kiiski
    Sep 27 '18 at 20:45
  • $\begingroup$ Proving the regularity: The most promising approach is to show that the Skorokhod space endowed with the $S$-topology is a topological vector space. This approach was pursued just recently by Adam Jakubowski himself by considering a weaker topology $\Sigma$ generated by a family of seminorms. It is an open conjecture if $S$ and $\Sigma$ are equivalent; cf. Question 3.7 in arxiv.org/abs/1609.00215 The topology $\Sigma$ provides "an educated guess" of "the nature of an arbitrary open set", but the topology $\Sigma$ is non-sequential. Thus, the difficulty as you described it persists. $\endgroup$
    – Matti Kiiski
    Sep 27 '18 at 21:12

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