# Are there disjoint closed sets which contain sequences that get arbitrarily close to each other?

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 $0<a<b<c<1$, 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?

• 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. – André Nicolas Oct 11 '11 at 22:44

## 1 Answer

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 $\lim_{n \to \infty}|x_n-y_n|=\lim_{n \to \infty}\frac{1}{(2n)(2n+1)}=0$.

• ... which points to an ambiguity in the question: Are $X$ and $Y$ supposed to be closed as subsets of $(0,1)$, or as subsets of $\mathbb R$? – Henning Makholm Oct 11 '11 at 23:00
• @Danielle Intal: The OP has to decide. – André Nicolas Oct 11 '11 at 23:11
• Sorry for the ambiguity! By André's comment I suppose this only makes sense to ask for $X$ and $Y$ to be closed in $(0,1)$, but not all of $\mathbb{R}$. – Danielle Intal Oct 12 '11 at 1:10
• The question makes sense either way, and has a clear answer once once decides whether to use the topology on $\mathbb{R}$ or the relative topology. It is just that the answers are different. – André Nicolas Oct 12 '11 at 2:31
• 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 $A=\overline{A}$. – Dustin Tran Oct 18 '11 at 7:14