# Cantor intersection theorem where sets aren't necessarily closed

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 $$A_n \subseteq B_x(r)$$?

• The intersection of balls centered at $x$ is $\{x\}$, so how can one $A_n$ be contained in all of them? Commented Jul 9, 2023 at 16:48
• 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 $\operatorname{diam}(A_n)\to 0$. Commented Jul 9, 2023 at 17:36
• @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 $z\in A_n$. Commented Jul 9, 2023 at 17:54
• @Mr.GandalfSauron see en.wikipedia.org/wiki/Cantor%27s_intersection_theorem variant in complete metric spaces Commented Jul 9, 2023 at 18:08
• @SaimFaigol: $x\in\overline{A_n}$, so $x$ is a limit point of $A_n$, so every neighborhood of $x$ intersects $A_n$. Commented Jul 9, 2023 at 18:10

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 $$A_n \subseteq B_x(r)$$.