Counterexample for P being a closed set with isolated points in real analysis

Valerie
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For my homework, I have to find a counterexample for this: (with S being a subset of the reals.)
If P is the set of all isolated points of S, then P is a closed set.

I don't quite understand the concept of isolated points, which might be why I can't figure out a counterexample.
 
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If it > 0 is any real number and x < y, show that there exists a rational number r such that
x < ru < y. (Hence the set (ru: r e Q) is dense in R.)
 
Valerie said:
For my homework, I have to find a counterexample for this: (with S being a subset of the reals.)
If P is the set of all isolated points of S, then P is a closed set.

I don't quite understand the concept of isolated points, which might be why I can't figure out a counterexample.

An "isolated" is just a point that is "away" from other points in the set. More technically, there is some neighborhood (open set) containing the point that does not contain any other point of the set. For example, [0, 1]\cup {2}, the set of all numbers between 0 and 1 (inclusive) and the number 2, has "2" as an isolated point. If a set is a sequence, say, {1, 2, 3, 4, 5}, then every point is an isolated point.

Think about a sequence of numbers that converge to some limit. What are its isolated points?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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