Limit of a sequence in a decreasing family of open sets

Question

Suppose we have a collection of non-empty bounded open sets, $A_{0},A_{1},\dots$ that are subsets of $\mathbb{R}^n$, such that $A_{i}\supseteq A_{i+1}$, and the intersection $$A=\bigcap_{i=0}^{\infty}A_{i}$$ is a non-empty closed set. If we have a convergent sequence $a_{0},a_{1},\dots$ with limit $a$ such that $a_{i}\in A_{i}$, can we conclude that $a\in A$?

This is related to the question I asked yesterday, however that one turned out to have a negative answer, so I am hoping this one turns out to be correct!

Edit: I'm only interested in the standard topology induced by the Euclidean metric.

Suppose $n=1$ and $A_i=(0,1]$ and $a_n=1/n$. Then $A=(0,1]$ and $(a_n)_{n\in N}$ converges to 0 which is not in $A$. Is this a counter-example to your statement? — Meneer-Beer, May 14 '17 at 15:55
Consider the upper limit topology. However, I am not sure if in that case 1/n coverges to 0. — Meneer-Beer, May 14 '17 at 16:04
I'm only using the standard Euclidean topology, not any other spaces. I've added this to the original post. — Ben, May 14 '17 at 16:08
Suppose $A_i=(0,1)$, and $a_n=1/(n+2)$. Then $A=(0,1)$, and $(a_n)_{n \in N}$ converges to 0, which is not in A. Do you think this is a correct counter-example? — Meneer-Beer, May 14 '17 at 16:43
@Meneer-Beer $A$ is not closed, so this is not a valid example. — Henno Brandsma, May 14 '17 at 16:51

score 3 · Answer 1 · answered May 14 '17 at 16:54

3

The same counterexample as to the linked question works:

Let $A_0 = (0, 3)$

Let $A_i = (0, 1/i) \cup (1-1/i, 1+1/i)$ for $i>0$

Then $A = \{1\}$ but if we pick $a_i = \frac{1}{2i} \to 0$, we don't have $0 \in A$.

answered May 14 '17 at 16:54

Henno Brandsma

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Limit of a sequence in a decreasing family of open sets

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