I have this exercise;

First part: Let $E$ be a metric space, and $(F_n)$ a decreasing sequence of closed set from $E$ and let $(x_n)$ a convergent sequence such that $x_n\in F_n, $for all $n\geq0$.

Prove that $\lim_{n\rightarrow\infty} x_n\in \bigcap_{n\geq0} F_n$ and give an example such that $\bigcap_{n\geq0} F_n=\emptyset$

----> I know that $\bigcap_{n\geq0} F_n$ is closed but why the limite is exacly in the intersection?

Second part: Let $K_n$ a deceasing sequence of compact nonempty sets from $E$.

Prove that $K=\cap_{n\geq0} K_n$ is nonempty and any open set which contain $K$ then it contain $K_n$ .

Thank you

  • $\begingroup$ The sequence can't be convergent if the intersection is empty, but it could be a Cauchy sequence. $\endgroup$ – Matt Samuel Mar 6 '16 at 19:19
  • $\begingroup$ You only have to prove that the limit is contained in every $K_m$. $\endgroup$ – Friedrich Philipp Mar 6 '16 at 19:20
  • $\begingroup$ Considering you just proved the limit is in the intersect, if the intersection is empty there can be no limit. So an example is impossible. Also when you say any open set contaning $K_n$. Which $n$. Obviously I can find an open set containing K but not containing $K_1$. Do you means some $K_n$. $\endgroup$ – fleablood Mar 6 '16 at 19:34
  • $\begingroup$ Also what does decreasing sequence of sets mean exactly? $\endgroup$ – fleablood Mar 6 '16 at 19:39
  1. For every $n$, and $m>n$, $x_m\in F_n$ since $F_m\subset F_n$, this implies that $x=lim_{m\geq n}x_m=lim_mx_m\in F_n$ since $F_n$ is closed so $x\in \cap_nF_n$.

  2. Let $x_n\in K_n$ we can suppose that $(x_n)$ converges (up to a subsequence) towards $x$, since $x_n\in K_n\subset K_0$ and $K_0$ is compact. The $K_n$ are compact thus closed, so 1. implies that $x\in\cap_n K_n$.

Let $U$ be an open subset which contains $K$ suppose that for every $n$, there exists $x_n\in K_n$ and $x_n$ is not an element of $U$. Again upto a subsequence, we can suppose that $x_n$ converges towards $x$, the first part shows that $x\in K$, thus $x\in U$, since $U$ is open, there exists $N$, such that $n>N$ implies $x_n\in U$. Contradiction.

  • $\begingroup$ is there an example where $\cap F_n=\emptyset$ ? please $\endgroup$ – Vrouvrou Mar 7 '16 at 21:25

Let $\lim x_n = x$. Let $\epsilon > 0$. Then there exists an $x_n$ such that $d(x_n, x) < \epsilon$. $x_n \in F_n \subset \cap_{n \ge 0} F_n$ so $x$ is a limit point of $\cap_{n \ge 0} F_n$ and as $\cap_{n \ge 0} F_n$ is closed $x \in \cap_{n \ge 0} F_n$.

  • $\begingroup$ we have that $\cap F_n\subset F_n$ not the other way! $\endgroup$ – Vrouvrou Mar 7 '16 at 21:22

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.