4

Let $( M , d )$ be compact. Suppose that $( F_n)$ is a decreasing sequence of nonempty closed sets in $M$, and that $ \bigcap_{n=1}^{\infty} F_n$ is contained in some open set $G$. Show that $F_n \subset G$ for all but finitely many n .

I know how to solve this question. Also, there is a solution here.

But I just don't truly understand this question: It seems that we can just treat the infinite intersection as a limiting process, we don't have to require $M$ to be compact. In any general set $M'$, if we have a nested sequence of subsets $( F_n')$ at hand, and we know $ \bigcap_{n=1}^{\infty} F_n'$ is contained in some set $G'$, then this statement always holds, since, after some $N$, the set $F_n'$ will finally be in $G'$

Hamilton
  • 602

2 Answers2

3

Compactness is necessary and your argument is not valid. For a counter-example consider the real line with the usual metric. Let $F_n=(-\infty, -n]$ and $G=(0,\infty)$. Then $\bigcap F_n=\emptyset \subseteq G$ but no $F_n$ is contained in $G$.

Kavi Rama Murthy
  • 311,013
  • 1
    $\emptyset = (-\infty,1] \cap (-\infty,2]\cap (-\infty,3]\dots,$? – 311411 Dec 07 '21 at 00:18
  • 2
    by replacing n by -n is ok! – stephenkk Dec 07 '21 at 00:42
  • Very very good counter-example. In fact, this example can also be used to show "boundedness" property is necessary for the "nested interval theorem of $R$". However, I'd like to know if it is possible to give a counter example with a non-empty intersection? – Hamilton Dec 07 '21 at 01:39
  • @Beginner Just replace $F_n$ by $(-\infty, -n] \cup {1}$ and keep the same $G$. – Kavi Rama Murthy Dec 07 '21 at 04:55
2

For a counter-example with a non-empty intersection:

Let $F_n=(-\infty,-n]\cup\{7\}$ and $G=(0,\infty)$. Then $\bigcap F_n=\{7\} \subseteq G$, but no $F_n$ is contained in $G$.

311411
  • 3,537
  • 9
  • 19
  • 36