# Dense sets and the empty set

1. Oct 6, 2008

### SW VandeCarr

If A is a dense set, then all subsets of A are dense. However, all sets contain the empty set. Is this a contradiction or is the empty set a compatable subset of a dense set? If it is, why?

2. Oct 6, 2008

### d_leet

Why do you think that all subsets of a dense set are dense?

3. Oct 6, 2008

### HallsofIvy

Yes, that's contradiction! Here's another one: Suppose A is the set of all rational numbers which is dense in the set of real numbers. {1} is a subset of the rational numbers. Is it dense in the set of real numbers? What do those two contradictions tell you about your statement "If A is a dense set, then all subsets of A are dense" (in some given set)?

4. Oct 6, 2008

### SW VandeCarr

Yes, my previous statement did not exclude ordered sets as it should have. {1} and therefore {0} are subsets of the set of rational numbers which are not dense in the set of real numbers.

Now consider a point set as in point set topology. I would argue that here, all subsets of a point set must be dense. How do you isolate a single point from a point set? I'm aware of the Dedekind cut as it applies to ordered sets, but can it be used in a point set? I guess my question boils down to whether there are dense sets which do not contain the null set, or any subset that is not dense.

Last edited: Oct 7, 2008
5. Oct 7, 2008

### CRGreathouse

Of course there are nonempty subsets of dense sets that aren't dense. R is dense in R, but Z (a subset of R) is not dense in R. Q is dense in R, but the rationals less than 0 or greater than 1 are not dense in Q.

6. Oct 7, 2008

### SW VandeCarr

Thank you for your reply. However my question was in regard to topological point sets, not ordered sets. If a point set T is dense is some set A, then are there non-empty in sets in T which are not dense? It seems to me that if an unordered set is dense, every subset must be dense unless there is some way to identify discrete points in T.

Perhaps, another way to state this is: given two (unordered) point sets A and B, can the intersect of A and B be defined such that the intersection contains exactly one point?

7. Oct 7, 2008

### CRGreathouse

I really don't see what you're getting at. Ordered sets are a generalization of sets; just ignore the order and you have a set. My counterexample is just as valid topologically as in the real numbers.

I guess you're just going to have to define for us what sets you're allowed to chose and what your definition of close is:
T is dense in S iff for every s in S there is a sequence of (t_n) with t_i in T so for each positive epsilon there is some N with t_n closer than epsilon to s for all n > N. But it seems like you're rejecting things like real-valued distances...

8. Oct 7, 2008

### SW VandeCarr

I guess I would say there is no general definition of "close" in a dense set. Nevertheless, I see your point that between any two points in a dense set T there is some real valued distance 's' between these points, and that each of these points, or both together, would constitute a non-dense subset of T.

My second question was whether an intersection can be defined for two (dense) point sets T and T' such that the intersection contains exactly one point. Since we are able to select points to define a distance, it seems the answer to this question must be "yes" although I don't see how this would be be defined analytically. The reason I excluded ordered sets such as the real number "line" is that we can define a point in terms of a given sequence such that L < a < U, whereas this type of restriction does not seem to be available for a point set in greater than one integral dimension without the introduction of an external coordinate sytem.

Last edited: Oct 8, 2008
9. Oct 11, 2008

### CRGreathouse

But without a definition for "close", what does it mean to be dense?

I'm sorry if I'm not being very helpful here. Pedagogy is not one of my strengths; I often explain things poorly.

10. Oct 11, 2008

### Hurkyl

Staff Emeritus
I think some clarification of terminology is in order.

First off, I (and CRGreathouse, I believe), are assuming that by the phrase 'point set', you mean to speak of a topological space, and by the word 'dense', you are referring to the topological notion of density.

However, if our assumption is correct, the way you are actually using the adjective 'dense' isn't quite appropriate. "Dense" isn't an adjective that can be applied to sets. Instead, "dense" is an adjective describing a relation between certain sets and topological spaces. (More specifically, between a topological space and certain subsets of its set of points)

11. Oct 11, 2008

### SW VandeCarr

Thanks anyway Greathouse. I'm sure I'm missing something in the way sets are defined. I've read the ZFC axioms and other Set Theory material, but I believe there may be some terminology issues which I'm not understanding. I have training is in mathematical statistics, and while I have interests in other areas of mathematics, I'm certainly no expert.

Jennifer82

I'm going to take a shot at your question and suggest that the set of complex numbers Z(not necessarily hypercomplex though) may satisfy your conditions. This is a dense set for which most pairs would form a partial ordering in that the magnitude of z in Z is the distance (modulus) from the origin in the complex plane. However there are point pairs equidistance from the origin. Therefore, for these pairs the asymmetry condition does not hold in terms of the modulus and these would constitute a subset where neither z1<z2 nor z2<z1 in Z hold. If I'm wrong, I'm sure someone will jump in and set you (and me) straight.

12. Oct 11, 2008

### CRGreathouse

Yes, quite. In post #4 VandeCar wrote "dense in the set of real numbers" which led me to think that topological density was intended.

13. Oct 12, 2008

### SW VandeCarr

Thank you gentlemen. My Borowski and Borwein Dictionary of Mathematics (Harper Collins 1991 p 150) gives two relevant definitions of 'dense':

1) (of a set in an ordered space) having the property that between any two comparable elements a third can be interposed.

2) (of a set in topology) having a closure that contains a given set. More simply, one set is dense in another if the second is contained in the closure of the first.

Topological density was intended, but I did not make the distinction between the first and second definition and I must confess I still don't fully understand it. The first indicates (to me) density on the number line which is (to me) the ordered set of the reals in a one dimensional space such that a>b and b<a always hold. However, in the complex plane, this does not always hold if we use the modulus as the basis for ordering. I called the set Z a partially ordered "dense" set ( in the topological sense) in my response to Jennifer82 which is probably the wrong terminology. However it seems clear that among complex numbers there are those for which we can say z(i) is greater or less than z(j) and those where we cannot say z(i)' is greater or less than z(j)' in terms of their moduli. We may not be able to say they are "equal" in the sense that there is a distance between z(i)' and z(j)', but they have the same magnitude.

Last edited: Oct 12, 2008
14. Oct 12, 2008

### CRGreathouse

I really don't know what you mean here. Are you talking about $\mathbb{Q}\times\mathbb{Q}$ as a (possibly dense) subset of$\mathbb{R}\times\mathbb{R}$ with $d((a, b), (c, d)) = \sqrt{(a-c)^2+(b-d)^2}$? Because in that case Q^2 does seem to be dense in R^2.

15. Oct 12, 2008

### HallsofIvy

Thank you for stating the definitions. The second was the one I had in mind- but it still is NOT true that "If A is dense in B and C is a subset of A, then C is dense in B".

16. Oct 12, 2008

### jennifer82

Sorry about the question which I wrote down here. I'm new here so I just wrote down here. Thanks for your help.

17. Oct 12, 2008

### Bob3141592

I don't think all sets contain the empty set. The empty set is a subset of every set, but it does not have to be contained in the set as an element, and quite often it isn't. Does that make a difference to your question?

18. Oct 12, 2008

### HallsofIvy

Because of the reference to "all subsets of A", it is clear that "all sets contain the empty set" should have been "all sets have the empty set as a subset". However, since it is the case that the poster's statement "If A is a dense set, then all subsets of A are dense" is false the whole question becomes moot.

Last edited by a moderator: Oct 16, 2008
19. Oct 15, 2008

### SW VandeCarr

Thank HallsofIvy et al. From what has been said, I believe my error was in confusing a topological space from a proper set. It seems a topological 'point set' is not a proper set. My understanding is that in a 2-space topological point 'set', given a radius 'r' originating from any arbitrary point, the circle so defined will contain an area 'dense' in points for any value of 'r' greater than zero. The question remains (for me) whether any such arbitrary point, not identified in in terms of a number (real or complex), could be considered a subset of a point set so defined. My understanding, is that if such a point set is not a proper set, any arbitrary point in the topological space is not a subset of any set.

20. Oct 16, 2008

### HallsofIvy

Now I am wondering what you mean by a "proper set"! I don't believe I have ever seen such a term (except perhaps in distinguishing "sets" from "classes" which surely doesn't apply here!). A "topological space" is simply a set with a given topology (a certain collection of its subsets). Also by a "2-space topological point set", do you mean a subset of R2 with the usual metric?

"the circle so defined will contain an area 'dense' in points for any value of 'r' greater than zero"
Once again, saying an area is "dense in points" makes no sense. One set may or may not be "dense" in another set. No set can be called dense without saying dense in what set.

"The question remains (for me) whether any such arbitrary point, not identified in in terms of a number (real or complex), could be considered a subset of a point set so defined."
I don't understand this at all. If you have some given point set, each point in it is a MEMBER, not a subset. Yes, if you have a point set A and p is a point in it, then the set {p} is a subset of A.

"My understanding, is that if such a point set is not a proper set, any arbitrary point in the topological space is not a subset of any set. "
I still don't understand what you mean by a "proper set" but in general a point is NOT a set! A point is IN a set of points, not a subset of it.