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I had a question of curiosity. Take the interval $(0,1)$ with the usual metric in $\mathbb{R}$. Is it possible to find closed sets $X$ and $Y$ with $X\cap Y=\varnothing$ such that there is a sequence $\{x_n\}$ in $X$ and $\{y_n\}$ in $Y$ where $\lim_{n\to\infty}\vert x_n-y_n\vert=0$?

I'm having a hard time picturing this, for I see closed sets as closed intervals (or finite unions of them) on the real line. For the limit to approach $0$, I intuitively see that the points eventually all bunch up near some point. With open intervals, I think you could take intervals like $X=(a,b)$ and $Y=(b,c)$ with $0, and essentially let the $x_n$ get arbitrarily close to $b$ from below, and the $y_n$ arbitrarily close from above. But these aren't closed, and when taking the closures, $[a,b]$ and $[b,c]$ are no longer disjoint!

Is there a way to get around this snag?

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    Hint: Look at the sequence $(x_n)$. This sequence has an infinite convergent subsequence $(x_{n_i})$. Look at $(y_{n_i})$. This has an infinite convergent subsequence. Now it should not be difficult to see that we run into problems.2011-10-11

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Yes. For example, take $X=\left \{ \frac{1}{2n} : n \in \mathbb{N} \right\}$ and $Y=\left \{ \frac{1}{2n+1} : n \in \mathbb{N} \right \}$. $X$ and $Y$ are both closed in $(0,1)$. Setting $x_n=\frac{1}{2n}$ and $y_n=\frac{1}{2n+1}$, we have $\lim_{n \to \infty}|x_n-y_n|=\lim_{n \to \infty}\frac{1}{(2n)(2n+1)}=0$.

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    I might be missing something, but how are these $X$ and $Y$'s closed subsets of $(0,1)$? It seems that they both complete $(0,1)$ rather than satisfy $A=\overline{A}$.2011-10-18