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
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
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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.
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.
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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.