Complement of union of "very disjoint" open sets is perfect?

Question

Let $(I_n)_{n\in\mathbb{N}}$ be a sequence of open intervals in $\mathbb{R}$ such that $\overline{I_j}\cap\overline{I_k}=\emptyset$ if $j\neq k.$ I want to show,

using as elementary means as possible, that $S:=\displaystyle\left(\bigcup_{n=1}^{\infty}I_n\right)^c$ is perfect, that is, $S$ is closed and every point is a limit point of the set. [Recall that by definition an isolated point is not a limit point].

It is clear that $S$ is closed, since $S^c = \displaystyle\bigcup_{n=1}^{\infty}I_n$ is open. But I don't know how to approach the second part, that every point of $S$ is a limit point of $S.$ I believe the condition $\overline{I_j}\cap\overline{I_j}=\emptyset$ if $i\neq j$ is essential for making the question reasonable, as it prevents $S$ from having any isolated points.

I don't know if arguments for showing that the Cantor set is perfect, like the answers to this question, can be generalised to help prove the proposition.

Thanks in advance for your thoughts on the problem.

Answer 1

Another fairly simple proof. The key idea is in "Case 3" below: if a point $x \in S$ is a limit point of the collection of intervals, then a sequence of interval endpoints approaches $x$.

Let's say $I_n = (a_n, b_n)$, always with $a_n < b_n$. Note each endpoint $a_n$ and $b_n$ is an element of $S$: $a_n$ and $b_n$ are obviously not in $I_n$, and if either is an element of another interval $I_k$, this would imply $\overline{I_n} \cap \overline{I_k} = [a_n,b_n] \cap \overline{I_k} \neq \emptyset$, contradicting the initial assumption.

Given a point $x \in S$, we explicitly construct a sequence $(x_n)$ in $S$ converging to $x$. Define set

$$ A := \{a_n : n \in \mathbb{N}, a_n > x\}.$$

Case 1: $A = \emptyset$. Then $(x, \infty) \subset S$. Let $x_n = x + \frac 1 n$.

Case 2: $\inf A > x$. Then $(x, \inf A) \subset S$. Let $x_n = x + \frac 1n (\inf A - x)$.

Case 3: $x \notin A$ and $\inf A = x$. Let $m$ be the smallest index such that $a_m > x$ and define

$$ x_n := \min\left \{a_k : k \leq m+n, a_k > x\right\} .$$

This sequence of numbers in $S$ and larger than $x$ is non-increasing. Since $\inf A = x$, every interval $(x, x+\epsilon)$ contains some $a_k$ and $x_n \leq a_k < x+\epsilon$ for all $n \geq k$, so the sequence converges to $x$.

Case 4: $x \in A$. That is, $x = a_k$ for some $k$.

Then $x$ is not equal to any $b_j$: $b_k>a_k$, and for every other index $j$, $[a_j,b_j] \cap [a_k,b_k] = \emptyset$ implies $x = a_k \neq b_j$.

Consider the sequence of intervals $\tilde{I_n} = (-b_n, -a_n)$. Their closures are disjoint, and $-x$ is not equal to the lower endpoint $-b_n$ of any of the intervals. The corresponding set $\tilde S$ is exactly the negative values of $S$:

$$\tilde S = \left(\bigcup_{n \geq 1} \tilde{I_n} \right)^c = \{ -y : y \in S \} $$

So by one of the three cases above, there is a sequence $(\tilde x_n)_{n\in \mathbb{N}}$ with values in $\tilde S \setminus \{-x\}$ converging to $-x$. Therefore the sequence $(x_n)_{n \in \mathbb{N}} := (-\tilde x_n)_{n \in \mathbb{N}}$ has values in $S \setminus \{x\}$ and converges to $x$.

Answer 2

Let $x\in S$. There are two cases. Firstly, $x$ such that there exists $\varepsilon>0$ such that either $[x,x+\varepsilon)$ or $(x-\varepsilon,x]$ is contained in $S$, which is the trivial case. From now on assume that no such inclusion holds true.

As the closures of the intervals do not intersect, we get that $x$ cannot be the left endpoint of some interval and simultaneously the right endpoint of another interval.

Consider the case where $x$ is not the left endpoint of some interval $I_{n(0)}$. As there exists no $\varepsilon>0$ such that $[x,x+\varepsilon)\subset S$ there exists a sequence $x_m\in S^c$ such that $x_m \rightarrow x$. Wlog we can choose this sequence to be decreasing. Note that every $x_n$ belongs to some unique interval $I_{n(m)}$. We will construct now a suitable subsequence. Namely, we will pick $y_1=x_1$ and remove all $x_j$ which belong to $I_{n(1)}$. As $x$ is not the left endpoint of $I_{n(1)}$ there must still be infinitely many elements left (there is at least one and all to the left of any left over element is also not in $I_{n(1)}$). Now $y_2$ is the largest element which remains. Then we delete all elements contained in the same interval as $y_2$ and keep defining our sequence recursively like that.

Denote by $a_m$ the left endpoint of the interval containing $y_m$. Then we have $a_m\in S$ (as the closures of the intervals don't overlap) and we have $$x<a_m<y_m.$$ Thus, by the squeeze theorem we get $a_m\rightarrow x$.

By the same argument we deal with the case where $x$ is the not the right endpoint of an interval.

Answer 3

Suppose $I_n = (a_n, b_n)$ with $a_n < b_n$ and let $U = \bigcup_{n=1}^\infty I_n$.

First we show that if $I$ is an open interval contained in $U$ then there exists an $n$ such that $I \subseteq I_n$. Suppose to the contrary that there exist $c < d$ in $I$ and $j \neq k$ with $c \in I_j$ and $d \in I_k$. Then because $\overline{I_j} \cap \overline{I_k} = \emptyset$ it must be that $c < b_j < a_k < d$ and it follows that $a_k \in (c, d) \subseteq I \subseteq U$. So there exists an $m$ with $a_k \in I_m$. If $m = k$ this says that an open interval contains one of its endpoints and if $m \neq k$ then the condition $\overline{I_m} \cap \overline{I_k} = \emptyset$ is violated. Both are unacceptable.

Now suppose $x \in S$ is isolated, so there is a $\delta > 0$ with $(x - \delta, x + \delta) \cap S = \{x\}$. Then, by the previous paragraph, there exist $j \neq k$ with $(x - \delta, x) \subseteq I_j$ and $(x, x + \delta) \subseteq I_k$. It follows that $x \in \overline{I_j} \cap \overline{I_k}$, contradicting $\overline{I_j} \cap \overline{I_k} = \emptyset$.

Answer 4

• Fix $x \in S$. We can write the intervals in terms of their endpoints as $I_k \equiv (a_k, b_k)$.

• Because $x$ is not in any $I_k$, each interval $I_k$ must lie entirely to the "left" of $x$ (meaning $a_k < b_k \leq x$) or entirely to the "right" of $x$ (meaning $x \leq a_k < b_k$).

• Let $b_{max}$ be the largest of the $b_k \leq x$. Symmetrically, let $a_{min}$ be the smallest of the $a_k \geq x$.

• By design, we must have $b_{max} \leq x \leq a_{min}$.

• So that we have a concrete representative, pick any $m$ such that $b_m = b_{max}$ and pick any $n$ such that $a_n=a_{min}$.

• We know that $b_{max}\leq a_{min}$. Is it possible that equality holds so that $b_{max} = a_{min}$? No, because then the closures of $(a_m, b_m)$ and $(a_n, b_n)$ would intersect at that point. So instead, $b_{max} \lneqq a_{min}$.

• And so the interval $J^* \equiv (b_{min}, a_{min})$ of points between them is nonempty.

• The interval $J^*$ also does not intersect any of the $I_k$—the intervals to the left of $x$ are also to the left of $J^*$. The intervals to the right of $x$ are also to the right of $J^*$. This means that $J^* \subset S$.

• The point $x$ therefore must be perfect in $S$— that is, every interval around $x$ must contain some other point of $S$. Why?

  We know that $b_{max} \leq x \leq a_{min}$. Whether $x=b_{max}$ or $x=a_{min}$ or $b_{max}\lneqq x \lneqq a_{min}$, we know that each open ball around $x$ will overlap $J^*=(b_{min},a_{max})$ at an open interval, and we know $J^*\subset S$. So each open ball around $x$ will in fact contain a whole interval of other points from $S$.

  To rephrase in gritty detail, we know that $b_{max} \leq x \leq a_{min}$. There are three cases. If $x=b_{max}$, then for sufficiently small $\epsilon>0$, the point $(x+\epsilon)\in J^* \subset S$. Similarly, if $x=a_{min}$, then for sufficiently small $\epsilon >0$, the point $(x-\epsilon)\in J^* \subset S$. And certainly if $x$ is in the interior of $J^*$, for sufficiently small $\epsilon>0$, both $(x-\epsilon)$ and $(x+\epsilon)$ are in $J^*\subset S$.

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Pedantic aside: by "let $b_{max}$ be the largest of the $b_k \leq x$", I mean let $b_{max} = \sup \{b_k : k\in \mathbb{N}, b_k \leq x\}$, which has a value of $-\infty$ if no such $b_k$ exist, and similarly for $a_{min}$.