# Topological Rings

Is it true that, if S is a subring of a separable topological Noetherian ring R, then S is separable, too ?

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Is a topological Noetherian ring one in which every ideal is finitely generated or topologically finitely generated? – Qiaochu Yuan Nov 29 '09 at 3:52
I don't quite understand the question, but I don't think it's an "algebraic topology" question. The answer below seems to consider it "general topology", skipping the words "Noetherian ring", so I am retagging it as such, but if there's an algebra or algebraic-topology connection, those tags should be added. – Theo Johnson-Freyd Nov 29 '09 at 4:03
A Noetherian ring is a ring that satisfies the ascending chain condition on ideals. So that the above R simultaneously is a Noetherian ring (in the purely algebraical sense), and a topological ring that is separable as a topological space. – Ady Nov 29 '09 at 4:06
@Jonas Meyer. Well, reading the paper "Density character in topological groups" by W. W. Comfort and G. L. Itzkowitz, (Mathematische Annalen, Volume 226, Number 3 / October, 1977, pp.223-227), replacing the word "closed" in their problem by a richer algebraic structure, and noticing that the modified question has a negative answer for some non-Noetherian rings: that's the motivation. – Ady Nov 29 '09 at 6:40
I also think that the "ring theory" tag should be added. – Jose Brox Nov 30 '09 at 15:09

While looking through old questions, I came accross this one, and decided to throw my hat into the "ring."

1. If the identity element has a countable neighborhood base, then trivially by the continuity of the operation of addition, every point has a countable neighborhood base (as continuous maps take a filter base at some point x, and map this onto a filter base of the image point this is a particular example of homogeneity exhibited by topologically equipped algebraic objects.).

2. Provided, the topology defined on the ring is $T_0$, we have that the topology must also be $T_2$ and completely regular.

3. Moreover, since there is a countable dense subset $D$ by assumption, we have that given the above two assumptions, there exists a countable collection, of countable covers $U_n$, with the following property: Given any closed set $C$ from this space, and point $p\not\in C$ there exists an $n\in\omega$ such that for some $O\in U_n$, $p\in O$ and $O \cap C = \{\}$. (We get such a collection by ordering the neighborhood base of each point in $D$ by reverse inclusion, and taking fixed 'sections of fixed height' from each to form the open covers)

Putting (1), (2), and (3) together produces a topological space which is about as close to a normal Moore space as you can get without actually being one, that is to say, under these assumptions: $R$ is a separable completely regular developable Hausdorff space, (and we haven't even invoked the ACC yet)

## The interesting Part:

Ignoring for the moment the previous assumptions, intuitively, the ACC should somehow produce a covering property for this particular space. However, there is an interesting problem when it comes to the definition of subgroup/subring (which is required to get to the notion of ideal needed to apply the ACC): will they be open, closed, neither? Because of this we cannot really apply the ACC, to produce a nice covering property that might have tied everything together (like Lindelöf.)

Edit: While poking around Wiki, I came across something I felt I needed to add

However, if you mean that the space is a Noetherian topological space, then we get some gnarlly consequences ( http://en.wikipedia.org/wiki/Noetherian_topological_space ), like the fact that the space is compact! Which is exactly the thread we would want to tie everything together, and produces a normal moore space.

## The Reality of the Matter:

The question is ill-posed, in that we do not have enough information to properly deduce a valid and fully general answer. My answer to this question has tried to highlight this point by giving you a case where, you can be about as close as you might ever want to be to something genuinely interesting, and then failing to make it interesting because of the incompatibility of the algebraic assumption with the topological ones. Even if we consider the other possibility we enter one of those strange and beautiful areas in topology where things being to become independent of $ZFC$.

## Final Conclusion:

Because of this freedom or lack of information, we are left with an answer of Most Likely No. (Weak answer I know) But I can make this claim, because we honestly do not completely understand the notion of hereditary separability (in fact it was only in 2006 that J T Moore was able to produce a ZFC example of an L-space Article)

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Boo for that pun. – Harry Gindi Jan 25 '11 at 6:23
Also, "L-space" is another name for locally compact hausdorff totally disconnected spaces, but, for example, profinite groups are L-spaces, so I assume you mean something else? – Harry Gindi Jan 25 '11 at 6:42
an L-Space is a counter example to the implication h-Lindelof => h-Separable. Read the article. – Michael Blackmon Jan 25 '11 at 6:44

Unless I misunderstand something here, this has to be true, but has nothing to do with the ring structure. A subspace of a separable topological space is itself separable. That is, of course, unless separable is in this context an algebraic concept, and not the usual topological one.

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That's not true; see en.wikipedia.org/wiki/Sorgenfrey_plane . – Qiaochu Yuan Nov 29 '09 at 3:08
A subspace of a separable metric space is itself separable. For general topological spaces, separability is not a hereditary property. – Ady Nov 29 '09 at 3:15
Oh, bah. Silly me. That's what I was thinking about. – Simon Rose Nov 29 '09 at 5:27
As another example: the Stone-Cech compactification $\beta X$ of a countable infinite discrete space $X$ will always be separable; but if I recall correctly, $\beta X \setminus X$ is not separable. – Yemon Choi Nov 29 '09 at 9:02
My favorite pathological example is this: take $Y$ to be a space that is not separable. Then $Y$ is a subspace of the space $X=Y\cup\{x_0\}$ with the topology given by inserting $x_0$ into every open set. The countable dense set for $X$ is $\{x_0\}$. Ridiculous. – Elizabeth S. Q. Goodman Dec 13 '09 at 8:09