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This is just a quick question, as a follow-up to Chris Eagle's answer on this post.

In it, he considered $X=\left \{ \frac{1}{2n} : n \in \mathbb{N} \right\}$ and $Y=\left \{ \frac{1}{2n+1} : n \in \mathbb{N} \right \}.$ And being a requirement to answer that post's question, how does it follow that these are closed subsets of $(0,1)$? It seems that the two would be dense in $(0,1)$ rather than be equal to their closure.

I think I'm missing something obvious or I misinterpreted the OP's question.

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Each of them only have one limit point in $\mathbb{R}$: $0$, but $0\not\in(0,1)$ so neither have any limit point inside the interval. This means that it is nowhere dense inside it. Intuitively, this should make sense: look for any point $x$ inside the interval for which an infinite number of members crowd around, and you can put it in between two consecutive reciprocals like $1/(2n+2)\le x\le 1/(2n)$, which means if you inspect any neigh-borhood of $x$ closer than these two reciprocals you will no longer find any of the two sets' members, contradicting the hypothesis $x$ was infinitely popular among them.

Also, e.g. the first's complement $(1/2,1)\cup(1/4,1/2)\cup(1/6,1/4)\cup\cdots$ is a countable union of open intervals, which means the complement is open, hence the original set is closed.

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Ah, this made the most sense. For some odd reason (perhaps due to my lack of sleep), I was assuming that the limit points for both sets were each other. That is, the limit points for the reciprocal of the evens was the reciprocal of the odds. But this is clearly wrong. Thanks! –  Landyn Oct 18 '11 at 9:02
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Notice that $x \mapsto \frac{1}{x}$ is a homeomorphism $(1,\infty) \to (0,1)$. Since both the even and odd numbers are closed in $(1,\infty)$, the sets $X$ and $Y$ you ask about are closed in $(0,1)$, being images of closed sets under a homeomorphism.

Notice that $X$ and $Y$ are far from dense. Viewed as subsets of $\mathbb{R}$, their only accumulation point is ${0}$.

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