Exercise about closed and compact sets from metric space

Question

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

Answer 1

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 $x \in \cap_{n \ge 0} F_n$.

Answer 2

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.