Find a bounded infinite subset of Q that does not have an accumulation point in Q.
Favorites|Homepage
Subscriptions | sitemap
HOME > > Find a bounded infinite subset of Q that does not have an accumulation point in Q.

Find a bounded infinite subset of Q that does not have an accumulation point in Q.

[From: ] [author: ] [Date: 12-05-11] [Hit: ]
I also need to ensure that the set does not go to zero as n approaches infinity, since that will make zero an accumulation point if there are infinite points approaching zero, thus the set of rationala should appear to approach irrational numbers. I dont know an example of this set and dont know where to go from here. Thanks for the help-Look for example at the partial sums s(n) = sum(1/k^2)where k = 1,2,......
So, I know I want to make the bounds irrational so they are not in Q.
I also need to ensure that the set does not go to zero as n approaches infinity, since that will make zero an accumulation point if there are infinite points approaching zero, thus the set of rationala should appear to approach irrational numbers. I dont know an example of this set and don't know where to go from here. Thanks for the help

-
Look for example at the partial sums s(n) = sum(1/k^2) where k = 1,2,3...n
Then clearly each s(n) is rational being the finite sum of rationals
lim s(n) = pi^2/6 is clearly irrational and the s(n) are bounded by this value but do not have a rational accumulation point.

There are of course many other examples.

-
How about A = {3.1, 3.14, 3.145, 3.1459, ... } which is the infinite sequence of successive rational approximations to Pi. This a subset of Q which is bounded since each term is smaller than 3.2. Its only accumulation point is Pi which is clearly not in Q.
1
keywords: an,of,that,subset,bounded,point,have,infinite,in,Find,not,accumulation,does,Find a bounded infinite subset of Q that does not have an accumulation point in Q.
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .