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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?
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)?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?
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.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.
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.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.
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.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?
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.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...
But without a definition for "close", what does it mean to be dense?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.
I'm sorry if I'm not being very helpful here. Pedagogy is not one of my strengths; I often explain things poorly.That is, there would be NO point pairs such that a<b or b<a and therefore no uniquely identifiable points. But I'm apparently wrong on this. Perhaps the next poster can tell me exactly why.
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.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.
Yes, quite. In post #4 VandeCar wrote "dense in the set of real numbers" which led me to think that topological density was intended.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.
Thank you gentlemen. My Borowski and Borwein Dictionary of Mathematics (Harper Collins 1991 p 150) gives two relevant definitions of 'dense':Yes, quite. In post #4 VandeCar wrote "dense in the set of real numbers" which led me to think that topological density was intended.
I really don't know what you mean here. Are you talking about [itex]\mathbb{Q}\times\mathbb{Q}[/itex] as a (possibly dense) subset of[itex]\mathbb{R}\times\mathbb{R}[/itex] with [itex]d((a, b), (c, d)) = \sqrt{(a-c)^2+(b-d)^2}[/itex]? Because in that case Q^2 does seem to be dense in R^2.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.
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".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.
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?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?
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?
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.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?
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.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".
1. By 'proper' set, I mean a collection which conforms to the ZF or ZFC axioms.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 R^{2} 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.
This appears to be the crucial point. How do you arrive at this? I cannot imagine why you would think it is true.1. By 'proper' set, I mean a collection which conforms to the ZF or ZFC axioms.
2. By '2-space' I mean, in this case, a compact surface such as a the surface of a torus 'T'
which can be smoothly deformed by a topological transformation.
3. This finite surface contains an infinite number of points and any finite area on the surface also contains an infinite number of points (I believe the designation is aleph 1).
4. As I read the ZF(C) axioms, axiom 3 (specification) implies the empty set is a subset of any set. Therefore if I wish to consider a 'space' as a set of points, the formalism of ZF(C) axiomatic set theory appears to require that I state 'T' is dense is some other set.
5. Yes, I agree that, in general, a 'point' is a member of a set, not a subset of a set.
I refer to the following defintions from the previously referenced Borowski and Borwein (1991 p 591) for "topology"This appears to be the crucial point. How do you arrive at this? I cannot imagine why you would think it is true.
The set {a, b} is a perfectly good ZF(C) set and, of course, includes the empty set as subset. If I define its power set, {{}, {a},{b}, {ab}} to be the topology, I have a topological space. Now, for this example, what set is it that you are saying must be dense in some other set? And what does "the empty set is a subset of every set" have to do with density? In this topology, every set is both open and closed. NO set is dense in any other.
This doesn't make much sense -- at least not to me. Are you visualizing T as if it were contained in some other space? Because, for instance, T is dense in itself when we give it its natural topology; in fact, every subset of a topological is dense in itself.I was only arguing that if T is the point set of a surface (ie the surface of a torus), T is dense in some other set. That is not to say that a topological space must contain a dense set.
Morphism,To be honest, I find your questions to be somewhat perplexing. It doesn't seem to me that you actually understand what it means for a set to be dense in another or what a topological space is. For example, in your last post you say
This doesn't make much sense -- at least not to me. Are you visualizing T as if it were contained in some other space? Because, for instance, T is dense in itself when we give it its natural topology; in fact, every subset of a topological is dense in itself.
Are you trying to learn about topology from that dictionary you're referring to? If this is indeed the case, then I've got to warn you, it's not a very good idea!