# Limit of a sequence in a decreasing family of open sets

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
• $A_i$ are open sets and their intersection is closed. – Ben May 14 '17 at 16:02
• 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
• @Meneer-Beer $A$ is not closed, so this is not a valid example. – Henno Brandsma May 14 '17 at 16:51

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 $0 \in A$.