decreasing sequence of nonempty closed sets in M

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 $$G'$$

• @311411 Yes, corrected Dec 6 '21 at 22:11
• In your new proposition, are you still assuming the $F'_n$ are closed? Dec 6 '21 at 23:46
• @311411 No... $F_n'$ need not be closed. I don't even consider a topological space. I am assuming a very general situation, in the context of general set theory. Dec 7 '21 at 1:38

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 $$G$$.
• $\emptyset = (-\infty,1] \cap (-\infty,2]\cap (-\infty,3]\dots\,$? Dec 7 '21 at 0:18
• 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? Dec 7 '21 at 1:39
• @Beginner Just replace $F_n$ by $(-\infty, -n] \cup \{1\}$ and keep the same $G$. Dec 7 '21 at 4:55
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 $$G$$.