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?

  • $\begingroup$ 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. $\endgroup$ – André Nicolas Oct 11 '11 at 22:44

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

  • 3
    $\begingroup$ ... 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 $\mathbb R$? $\endgroup$ – Henning Makholm Oct 11 '11 at 23:00
  • $\begingroup$ @Danielle Intal: The OP has to decide. $\endgroup$ – André Nicolas Oct 11 '11 at 23:11
  • $\begingroup$ 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 $\mathbb{R}$. $\endgroup$ – Danielle Intal Oct 12 '11 at 1:10
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Oct 12 '11 at 2:31
  • $\begingroup$ 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 $A=\overline{A}$. $\endgroup$ – Dustin Tran Oct 18 '11 at 7:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.