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  1. The set $S$ of all rational numbers whose denominators are prime.

  2. $S = \{(-1)^{n}+\frac{1}{n}\mid n\in\mathbb{N}\}$

  3. $S = (0,1) $

I have answers for these from the back of a book but I'm not sure about the intuition behind finding the limit points.

Edit: I understand #3 now, all $x\in[0,1]$ are accumulation points since an open ball around any of those points contains infinitely many points of $S$.

Not sure on how to approach #1 though.

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This community could help you better if you could explain your own thoughts on the problem and where you feel you are stuck. –  Grumpy Parsnip Jan 24 '12 at 4:06

1 Answer 1

The formal definition states that a point $x$ is a limit point of $S$ unlesss there exists a neighborhood of $x$ that doesn't contain any points of $S$ (except, possibly, $x$ itself). Intuitively, if there are elements of $S$ that are really close to $x$, then $x$ is a limit point.

Let's handle your examples one at a time, starting with #3. The point $0$ is a limit point of $S$, because there are points in the open interval $(0,1)$ that are extremely close to $0$. Similarly, $1$ is a limit point, along with everything in $S$ itself. However, $17$ for example is not a limit point, because the neighborhood $(16, 18)$ contains nothing from $S$. This should be intuitively clear.

#2 is more difficult (for some reason, your textbook seems to put the harder problems first!), but you should be able to handle it if you consider the cases where $n$ is even and odd separately. Graph the points if you aren't sure.

#1 is less clear still. Hint: you should be able to show that $S$ is dense in $R$. You might want to use the fact that there are infinitely many primes.

Comment if you want any more help!

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By the way, "unlesss" means "unless and only unless". It's analogous to "iff". I have no idea how standard this notation is, but it's cute. –  Lopsy Jan 24 '12 at 4:13

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