Dense
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From: William Elliot on 11 Apr 2010 08:08 It's claimed Q is a dense subset of R. How can that be? It's got more holes in it than it has plugs. Why isn't it called porous? On the other hand, isn't R\Q more dense, less porous than Q? It's got many fewer holes and far many more plugs than Q. In fact R\Q, having more plugs than holes, gives it the appearance of having some substance, of being an irrational materialist, instead of being mostly not there like Q, the rational ghost. Upon magnifying R with the magnifying metric, min{ 1, m|x - y| } where m is the magnifying factor, Q appears as a 1/10 inch mesh and R\Q as a 1/100 inch mesh. Yet neither appear dense. Thus the quandary: what is a real dense set?
From: Tonico on 11 Apr 2010 08:13 On Apr 11, 3:08 pm, William Elliot <ma...(a)rdrop.remove.com> wrote: > It's claimed Q is a dense subset of R. > > How can that be? It's got more holes in it > than it has plugs. Why isn't it called porous? > > On the other hand, isn't R\Q more dense, less porous than Q? > It's got many fewer holes and far many more plugs than Q. > In fact R\Q, having more plugs than holes, gives it the > appearance of having some substance, of being an irrational > materialist, instead of being mostly not there like Q, the > rational ghost. > > Upon magnifying R with the magnifying metric, min{ 1, m|x - y| } > where m is the magnifying factor, Q appears as a 1/10 inch > mesh and R\Q as a 1/100 inch mesh. Yet neither appear dense. > > Thus the quandary: what is a real dense set? In any topological space X, a subset A is called dense if for ANY non- empty subset Y of X, the intersection A /\ Y is non-empty <==> the closure of A is X . You can call Q porous if you like, still it is dense in R with the usual euclidean topology. Tonio
From: George Jefferson on 11 Apr 2010 09:14 "William Elliot" <marsh(a)rdrop.remove.com> wrote in message news:20100411043202.R39951(a)agora.rdrop.com... > It's claimed Q is a dense subset of R. > > How can that be? It's got more holes in it > than it has plugs. Why isn't it called porous? > > On the other hand, isn't R\Q more dense, less porous than Q? > It's got many fewer holes and far many more plugs than Q. > In fact R\Q, having more plugs than holes, gives it the > appearance of having some substance, of being an irrational > materialist, instead of being mostly not there like Q, the > rational ghost. > > Upon magnifying R with the magnifying metric, min{ 1, m|x - y| } > where m is the magnifying factor, Q appears as a 1/10 inch > mesh and R\Q as a 1/100 inch mesh. Yet neither appear dense. > > Thus the quandary: what is a real dense set? > Dense is a relative term. Saying Q is dense is meaningless unless Q happens to be some people that vist sci.math. Y is dense in X because, relatively speaking, X - Y is small. This doesn't work well for infinite sets though and requires a more rigorous definition. Q is dense in R because the points of Q are densely populated with the points of R - Q. Of course, again, one needs a more rigorous definition because R - Q seems to be approximately equal to R. We can see that in the sense that the |X| - |Y| seems to say something about how "dense" Y is in X but the it seems to say too much since we feel that Q is much more dense than |R| - |Q| = |R| seems to suggest. But using the concept of set difference(or really the intersection) doesn't seem to lead to any useful definition of "denseness" since comparing infinite cardinalities does not lead to any useful result. So what does dense really mean? Take a point in Q and form an open neighborhood(or even closed if you want) that contains points in R - Q. The neighborhood will always contain other points of Q and R. In a sense, we cannot separate out Q from R easily. Where we find points in R - Q we find points in Q and vice versa. Hence Q is "dense" in R because Q is not sparse in R even though there are "many more points" in R - Q than in Q. But note the "many more points" is comparing infinite cardinals which doesn't make a whole lot of sense for our concept of denseness. Take Q+. Q+ is not dense in R because there are points in R that have open neighborhoods that do not contain points in Q+. Take N. Same argument. Take the set of points {1/N}. Same argument. For a set to be dense in another set every point must have a dense neighborhood. A dense neighborhood is what it seems. Not so much that there are about equal number of points, again because we can't compare the cardinalities and arrive at a meaningful definition, but because of "closeness" between points. Another way to look at it, a more appropriate definition, A set Y is dense in a set X if the union of all open non-singleton neighborhoods for each point in Y is X. Hence we are generating a cover of Y in X that completely covers X. Hence Y is dense in X. We do not use any idea of cardinality between X and Y. Let p \in Q. I_p = (p - e, p + e), I_p \in R, U(I_p, p \in Q) = R The set of I_p's form a complete open cover of R and therefor we can say Q is dense in. Alternatively we can simply look at the closure of Q which gives us R which is the same idea above. http://en.wikipedia.org/wiki/Topological_closure Again, the idea is of "closeness" in a topoligical sense and not of cardinality(which is what you seem to be using). Q is not close to R in cardinality but Q's points densely populate R. Think of a black 2D plane and use your intuitive concept of denseness. If you used a small red dot on each Q then it will cover all of R(Q^2 and R^2 of course) and you would see a red plane. If you did this for N you would just see red dots at the points (n,n) but almost the whole plane would be black. For Q^2, no matter how much you zoomed in you or how small the red dots were you would still get a red map. This is what "dense" means. We can never get some open set in R that just contains points from R - Q even if the cardinality of Q is much smaller than that of R. One more time ;) If you are comparing cardinalities of the sets then your approaching it wrong since it only *works* intuitively for finite sets. When you get to infinite sets it falls apart and no longer makes sense.
From: The Qurqirish Dragon on 11 Apr 2010 09:18 On Apr 11, 8:08 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > It's claimed Q is a dense subset of R. > > How can that be? It's got more holes in it > than it has plugs. Why isn't it called porous? > > On the other hand, isn't R\Q more dense, less porous than Q? > It's got many fewer holes and far many more plugs than Q. > In fact R\Q, having more plugs than holes, gives it the > appearance of having some substance, of being an irrational > materialist, instead of being mostly not there like Q, the > rational ghost. > It appears you are confusing measure with the concept of being dense. Your idea of being "porous" seems to imply the idea that Q (as a subset of R) has measure 0, and R\Q (as a subset of R) has measure 1 - so Q has many more "holes" in this sense. > Upon magnifying R with the magnifying metric, min{ 1, m|x - y| } > where m is the magnifying factor, Q appears as a 1/10 inch > mesh and R\Q as a 1/100 inch mesh. Yet neither appear dense. This analogy doesn't make sense to me. Given any m, your metric looks like (for a given x), a constant for any y more than 1/m away from x, and a copy of R in a segment of length 1/m centered at x. Thus, given a rational m, the non-constant part will be indistinguishable from the pre-magnified version (what was in Q is in the new "Q", what wasn't, isn't). If m is irrational, then any point of the form m*q, q in Q will be your new "Q" and every other point will not. It will still look the same as the original, with respect to being dense.
From: Achava Nakhash, the Loving Snake on 11 Apr 2010 11:57
On Apr 11, 5:08 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > It's claimed Q is a dense subset of R. > > How can that be? It's got more holes in it > than it has plugs. Why isn't it called porous? Mr. Elliot, i am sure you know why Q is dense. You, after all, are pretending to be dense, and yet you aren't. You know tha Q is only claiming to be rational. It's just not real. People think that Q is being honest, but stupid. Hence it is called dense. > On the other hand, isn't R\Q more dense, less porous than Q? > It's got many fewer holes and far many more plugs than Q. > In fact R\Q, having more plugs than holes, gives it the > appearance of having some substance, of being an irrational > materialist, instead of being mostly not there like Q, the > rational ghost. I have seaerched my topology books in vain to find a consistent definition for plugs and holes. You will have to enlighten us further. > Upon magnifying R with the magnifying metric, min{ 1, m|x - y| } > where m is the magnifying factor, Q appears as a 1/10 inch > mesh and R\Q as a 1/100 inch mesh. Yet neither appear dense. > > Thus the quandary: what is a real dense set? Bad grammar. You should have said, "What is a really dense set?" Regards, Achava |