Topological Rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:17:14Z http://mathoverflow.net/feeds/question/7103 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7103/topological-rings Topological Rings Ady 2009-11-29T03:01:16Z 2011-02-14T12:13:20Z <p>Is it true that, if S is a subring of a separable topological Noetherian ring R, then S is separable, too ?</p> http://mathoverflow.net/questions/7103/topological-rings/7104#7104 Answer by Simon Rose for Topological Rings Simon Rose 2009-11-29T03:05:01Z 2009-11-29T03:05:01Z <p>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.</p> http://mathoverflow.net/questions/7103/topological-rings/53145#53145 Answer by Michael Blackmon for Topological Rings Michael Blackmon 2011-01-24T22:26:20Z 2011-02-14T12:13:20Z <p>While looking through old questions, I came accross this one, and decided to throw my hat into the "ring."</p> <h2>Partial Answers and Observations:</h2> <ol> <li><p><strong>If</strong> the identity element has a countable neighborhood base, <strong>then</strong> 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.). </p></li> <li><p>Provided, the topology defined on the ring is $T_0$, we have that the topology must also be $T_2$ and completely regular.</p></li> <li><p>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)</p></li> </ol> <p>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) </p> <h2>The interesting Part:</h2> <p>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.) </p> <p><strong>Edit</strong>: While poking around Wiki, I came across something I felt I needed to add</p> <p>However, if you mean that the space is a Noetherian topological space, then we get some gnarlly consequences ( <a href="http://en.wikipedia.org/wiki/Noetherian_topological_space" rel="nofollow">http://en.wikipedia.org/wiki/Noetherian_topological_space</a> ), 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.</p> <h2>The Reality of the Matter:</h2> <p>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 $ZFC$.</p> <h2>Final Conclusion:</h2> <p>Because of this freedom or lack of information, we are left with an answer of <strong>Most Likely No</strong>. (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 <a href="http://www.math.cornell.edu/~justin/Ftp/Lspace.pdf" rel="nofollow">Article</a>)</p>