4
$\begingroup$

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 $G'$

$\endgroup$
3
  • 1
    $\begingroup$ @311411 Yes, corrected $\endgroup$
    – Beginner
    Dec 6 '21 at 22:11
  • $\begingroup$ In your new proposition, are you still assuming the $F'_n$ are closed? $\endgroup$
    – 311411
    Dec 6 '21 at 23:46
  • 1
    $\begingroup$ @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. $\endgroup$
    – Beginner
    Dec 7 '21 at 1:38
3
$\begingroup$

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 $G$.

$\endgroup$
4
  • 1
    $\begingroup$ $\emptyset = (-\infty,1] \cap (-\infty,2]\cap (-\infty,3]\dots\,$? $\endgroup$
    – 311411
    Dec 7 '21 at 0:18
  • 2
    $\begingroup$ by replacing n by -n is ok! $\endgroup$
    – stephenkk
    Dec 7 '21 at 0:42
  • $\begingroup$ 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? $\endgroup$
    – Beginner
    Dec 7 '21 at 1:39
  • $\begingroup$ @Beginner Just replace $F_n$ by $(-\infty, -n] \cup \{1\}$ and keep the same $G$. $\endgroup$ Dec 7 '21 at 4:55
2
$\begingroup$

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 $G$.

$\endgroup$

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.