1
$\begingroup$

I came across a problem that states that given $(X,d)$ a complete metric space and a decreasing sequence of non-empty bounded sets $A_n \subseteq X$ such that $\lim_{n\to\infty } diam(A_n) = 0$ then there is some $x \in X: A_n \subseteq B_x(r)$ for some set $A_n$ in the sequence and for every ball centered at $x$.

I know that Cantor's intersection theorem states that if the sets $A_n$ are closed then $\bigcap_{n \in \mathbb{N}} A_n = \{x\}$. But here the sets aren't closed, so why is it that $A_n \subseteq B_x(r)$?

$\endgroup$
13
  • $\begingroup$ The intersection of balls centered at $x$ is $\{x\}$, so how can one $A_n$ be contained in all of them? $\endgroup$
    – Chad K
    Commented Jul 9, 2023 at 16:48
  • 2
    $\begingroup$ Show that the closures $\overline{A_n}$ meet the conditions of Cantor's intersection theorem and take $x\in\bigcap_n \overline{A_n}$. Then $x$ is a limit point of $A_n$ for all $n$. So given $r>0$ the neighborhood $B(x,r)$ intersects $A_n$ for all $n$. Now find a sufficiently large $n$ so that $A_n$ is completely contained in $B(x,r)$. This can be done since $\operatorname{diam}(A_n)\to 0$. $\endgroup$
    – Chad K
    Commented Jul 9, 2023 at 17:36
  • 1
    $\begingroup$ @SaimFaigol: I got the order a little bit wrong. Start by choosing large enough $n$ such that $\operatorname{diam}(A_n)<r/2$. Then take $y\in A_n\cap B(x,r/2)$ (which is non-empty). Then by the triangle inquality $d(z,x)<r$ for all $z\in A_n$. $\endgroup$
    – Chad K
    Commented Jul 9, 2023 at 17:54
  • 1
    $\begingroup$ @Mr.GandalfSauron see en.wikipedia.org/wiki/Cantor%27s_intersection_theorem variant in complete metric spaces $\endgroup$ Commented Jul 9, 2023 at 18:08
  • 1
    $\begingroup$ @SaimFaigol: $x\in\overline{A_n}$, so $x$ is a limit point of $A_n$, so every neighborhood of $x$ intersects $A_n$. $\endgroup$
    – Chad K
    Commented Jul 9, 2023 at 18:10

1 Answer 1

1
$\begingroup$

No need even to bring in the big guns of the Cantor intersection theorem.

Just choose an $x_n\in A_n$. When $m > n, x_m \in A_m \subseteq A_n$, and therefore $d(x_m, x_n) \le \text{diam}(A_n) \to 0$ as $n \to \infty$, so the sequence is Cauchy and has limit $x$.

Since $x \in \overline{A_n}$ for all $n$, for any $r > 0$, when $n$ is large enough that $\text{diam}(A_n) < r$, it must be that $A_n \subseteq B_x(r)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .