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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

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  • $\begingroup$ the empty set is closed $\endgroup$ – David Holden May 13 '17 at 23:12
  • $\begingroup$ My mistake, I forgot to mention that $A$ should also be non-empty. $\endgroup$ – Ben May 13 '17 at 23:15
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    $\begingroup$ 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. $\endgroup$ – Juanito May 13 '17 at 23:27
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    $\begingroup$ see math.stackexchange.com/questions/356758/… $\endgroup$ – CopyPasteIt May 13 '17 at 23:48
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    $\begingroup$ @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 $x\in A$. $\endgroup$ – Ben May 14 '17 at 0:01
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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.

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  • $\begingroup$ The sets have to be open, as by the question. $\endgroup$ – Juanito May 13 '17 at 23:48
  • $\begingroup$ But, otherwise sharp answer. $\endgroup$ – Juanito May 13 '17 at 23:48
  • $\begingroup$ sorry, have to get in the habit of careful reading of question! $\endgroup$ – CopyPasteIt May 13 '17 at 23:51
  • $\begingroup$ I got it to work @Juanito $\endgroup$ – CopyPasteIt May 14 '17 at 0:20
  • $\begingroup$ haha, nice, upvoting. $\endgroup$ – Juanito May 14 '17 at 1:00

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