Closure and Compactness in a Metric SpaceDate: 10/08/2002 at 20:24:03 From: Pham Subject: Compactness/ closed/open Please help, Regard Q, the set of all rational numbers, as a metric space, with d(p,q)=|p-q|. Let E be the set of all P in Q such that 2<P^2<3. Show that E is closed and bounded in Q, but that E is not compact. Is E open in Q? Thank you very much. Date: 11/20/2002 at 23:21:58 From: Doctor Nitrogen Subject: Re: Compactness/ closed/open Hi, Pham: Try to see if this reasoning helps. Find convergent sequences of rational numbers P = 1 + a_n, n = 1, 2, 3, ........, with a_n = m/n, m, n integers, and m/n always an element of Q such that sqrt(2) < 1 + a_n < sqrt(3), and, for P = (1 + a_n) and P^2 = (1 + a_n)^2 also both converging: 2 < (1 + a_n)^2 < 3 for each n. Then the set E:= {All P in Q | 2 < P^2 < 3} can be shown to exist. Then E can be shown to contain all its limit points, meaning E must be closed in Q. You will, of course, have to find suitable candidates for a_n and m/n. But also, this set E of elements P, with P = 1 + a_n, would have both P and also P^2 bounded below and above, so E would be bounded in Q. So E is both closed and bounded in Q. But what about E when viewed as a METRIC SPACE in Q? Is it a compact metric space in Q with d(p, q) = |p - q|? Let p and q both be elements satisfying the definition for E, such that p = 1 + a_n, q = 1 + b_n, n = 1, 2, 3, ...., and such that p^2 and q^2 are both convergent sequences with limits inside(2, 3), but for which d(p, q) = |p - q| = sqrt(2), that is, lim n --> oo |(1 + a_n) - (1 + b_n)| = lim n --> oo |a_n - b_n| = sqrt(2). If you can find two such sequences p = 1 + a_n and q = 1 + b_n, maybe so that (1 + a_1) - (1 + b_1) = 1.0, (1 + a_2) - (1 + b_2) = 1.4, (1 + a_3) - (1 + b_3) = 1.41, (1 + a_4) - (1 + b_4) = 1.414, .............................., etc., just to illustrate, this would mean there exists a Cauchy sequence that does not converge either in E nor in Q, but in R (reals), meaning E cannot be a complete metric space inside Q. But this implies E cannot be "sequentially compact" inside Q (you might want to do some research on this term), meaning E cannot be compact in Q. You also ask: Is E open inside Q? To answer this, ask yourself: Is every point inside E interior to E? If every point of E is interior to E, E is open in Q. If the elements of E are not all interior to E, then E is closed in Q. One other alternative to prove E is not compact in Q might be to find an open covering of E which has no finite subcovering for E. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/ |
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