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I am trying to better understand mutually dense sets in a topology. I have found very little good information on this and most of that has dealt with ordered sets, such as the mutual density of the rationals and the non-rationals. However I have also seen it stated that C and D are mutually dense sets if they are contained in each other's closures, so that $C \subseteq \overline{D}$ and $D \subseteq \overline{C}$, which is seemingly more general and of more interest to me than definitions based on order. I am particularly interested in the relations of limit points of such sets. Can anyone give me some good explanations, links, salient facts, references or other enlightenment?


In response to the first answer below by Adam: Suppose we have C and D as disjoint sets but each including the other’s boundary. (Thought experiment: Imagine two congruent, two-dimensional sets on two sheets of paper, one above the other so they are disjoint, but infinitely close in the third dimension so that each contains limit points of the other. Then squash the third dimension down so that the points intermingle, but retain their set memberships.) Then it seems to me that $C \cap D = \emptyset$ and since $d(C) \cap D = D$ we have $D = \overline{C}\setminus C$ and hence $D \subset \overline{C}$. By symmetry then also $C \subset \overline{D}$.

@Adam - "congruent" was just being sloppy, meaning they fit on top of one another on the two sheets. I was just trying to give a helpful visualization of the idea, not a mathematical argument.

@Joriki - Thanks for the link. Yes, I'm afraid it is broad. I'm trying to learn more about the necessary and sufficient conditions for sets to be mutually dense, and about the ramifications of their being so. I am having trouble finding much on it, and in particular much that treats with things more generally than a quick example on the real line. For example, it is not obvious to me why sets being subsets of the others' closure would entail anything like a necessity for their points to be such that between any two points of one there was a point of the other and vice versa in more than one dimension.

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This question is rather broad. Is there anything specific you want to know about mutually dense sets? By the way, you may be interested in the order topology. – joriki May 3 '12 at 14:08
Ernie:- What does it mean for two sets two be congruent? – Adam Rubinson May 3 '12 at 19:39

Draw up some examples (from thin air/imagination)... e.g. think about the usual topology over R and consider the subsets

C = [0,1]U[2,3)U(4,5), and D = [0,1)U[2,3]U[4,5].

Or you could change any of the ['s into ('s and any of the ]'s into )'s and vice versa.

Now firstly, as an exercise, write down what $closC$ and $closD$ actually are.

You can see that $closC = closD$, and also both $C \subseteq closD$ and $D \subseteq closC$ are true (think about why). In fact, $C \subseteq closD$ and $D \subseteq closC$ follow directly from the fact that $closC$ = $closD$ (think why). And this fact is true for all topologies (not just metric spaces).

Lemma: It can be proved that closC is the SMALLEST closed subset containing C and the same is true for the set D.

Comment about the above Lemma: There are actually lots of definitions of "clos" (closure). Just look on wiki: The lemma above is kind of obvious (and the formal proof is quite short and easy) if you use the definition: $closA$ = The intersection of all closed subsets that contain A.

So if you want $C \subseteq closD$ and $D \subseteq closC$ to hold, you have lots of options on your sets C and D, BUT you need $closC = closD$, which restricts your freedom of choice for what the sets C and D are. Referring to the lemma, we see that this is because if $closC =/= closD$, there exists an x in D not in closC, leading to a violation of "$D \subseteq closC$ ". We have shown that, for all topological spaces, the following is true:

$C \subseteq closD$ $and$ $D \subseteq closC \Longleftrightarrow closC = closD$

This is the ultimate conclusion. How different can C and D be though? Well I think that depends on the topological space we are working in. For example, in the usual topology we find that we can have infinitely many nint points in D\C. In other topological spaces we might not find may nint points. There are probably spaces where we will find no nint points. Can you think of any (hint: all the open sets will have to be closed)?

Furthermore, if intC is the largest open set contained INSIDE C, then similar results to everything above can be drawn. This is intuitive stuff. Hope you enjoy it.

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Thanks, Adam. Please see my comment appended to the question as I couldn't get it in this field. – ernie May 2 '12 at 16:06
I will try to answer your question at a later date but for now I can leave you with the thought that, if you are interested in limit points and stuff, then you might want to look into Functional Analysis: (which is perhaps better suited to investigating limit points than for example algebraic topology, although I'm not certain this is true). Given that you have some knowledge in Real/Complex analysis, I would recommend the book entitled "Linear Functional Analysis" by Bryan P.Rynne and Martin A Youngson. Edit:I don't know what level you're at – Adam Rubinson May 3 '12 at 21:19
Thank you, Adam. I'm a retired professor/scientist, but NOT a mathematician. My level varies according to what I've had need of. My present interest is in better understanding models of boundaries between disjoint spaces as part of a larger project in causal systems spanning such boundaries. – ernie May 3 '12 at 22:23
To this end, you should probably get advice from someone more qualified than me. Also, what do you mean by "fit on top of one another". Do you mean they are the same set? Otherwise, you can't have two sheets "directly" on top of one another like you think... let me know if you want me to expand on this. – Adam Rubinson May 3 '12 at 23:30
For example, the set R of real numbers has the property that: for any number x in R, there is no number y =/=x which is closest to x. I think that in your reasoning you are assuming there is a closest point, which there isn't. – Adam Rubinson May 3 '12 at 23:34

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