Intersection of closure of decreasing family of sets

Question

Suppose we have a countably infinite family of non-empty, bounded open sets, $A_{0},A_{1},\dots$ in $\mathbb{R}^n$ such that $A_{i}\supseteq A_{i+1}$ for all $i$. If the intersection $$A=\bigcap_{i=0}^{\infty}A_{i}$$ is non-empty and closed, is it true that we have $$\bar A=\bigcap_{i=0}^{\infty}\bar{A_{i}}$$ where $\bar X$ denotes the closure of $X$? I think it is true intuitively, but I can not seem to prove it.

Edit: $A$ should be non-empty

My mistake, I forgot to mention that $A$ should also be non-empty. — Ben, May 13 '17 at 23:15
Here's my intuition: For the equality to not hold, there must a limit point of all $A_i$'s, which is not a limit point of A. This should be ruled out by the boundedness of the sets and the limiting condition on sets. — Juanito, May 13 '17 at 23:27
see https://math.stackexchange.com/questions/356758/when-is-the-closure-of-an-intersection-equal-to-the-intersection-of-closures — CopyPasteIt, May 13 '17 at 23:48
@Juanito That's actually what I was trying to do when I ended up with this problem - to show that if we have $(x_0,x_1,\dots)$ converging to $x$, where $x_{i}\in A_{i}$, then $x\in A$. — Ben, May 14 '17 at 00:01

CopyPasteIt · Answer 1 · 2017-05-16T14:58:47.913

3

It is not true.

Counterexample:

Looking at the union of two open real intervals (except for $A_0$):

Let $A_0 = (0, 3)$

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

The intersection is $\{1\}$ but the intersection of the closures is $\{0, 1\}$.

But this is true:

Let a sequence of subsets

$A_{0},A_{1}, A_2\dots$ in $\mathbb{R}$ be bounded, open and nested (i.e. $A_{i}\supseteq A_{i+1}$ for all $i$) intervals. Assume also than each $A_{i+1}$ has different endpoints than $A_{i}$.

THEN

The intersection of the $A_i$ is either a singleton or a closed interval $[a,b]$. Also, you get the same result by intersecting the closures of the $A_i$ (i.e. adding the interval endpoints).

Proof: On reflection, you can simply apply the Nested Interval Theorem to this situation.

Note that with these assumptions, the open intervals are squeezing down from both the left and right sides.

edited May 16 '17 at 14:58

answered May 13 '17 at 23:41

CopyPasteIt

11,366

The sets have to be open, as by the question. – Juanito May 13 '17 at 23:48
But, otherwise sharp answer. – Juanito May 13 '17 at 23:48
sorry, have to get in the habit of careful reading of question! – CopyPasteIt May 13 '17 at 23:51
I got it to work @Juanito – CopyPasteIt May 14 '17 at 00:20
haha, nice, upvoting. – Juanito May 14 '17 at 01:00

Intersection of closure of decreasing family of sets

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