# Is the reciprocal of the even numbers a closed subset of $(0,1)$?

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.
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 ${0}$.