Countable decomposition of a closed subset of real numbers

Question

$\newcommand{\R}{\mathbb{R}}$ I am going to demonstrate how I decomposed a closed set into pairwise disjoint closed intervals. (Here, we also take singletons as intervals).

Let $A$ be a closed subset of $\R$. Then, for all $x \in A$, there is the maximal closed interval $I \subseteq A$ containing $x$. We can form it by taking the union of all closed intervals in $A$ containing $x$. The union is closed: 1) it is trivially closed if it is a singleton; 2) Otherwise if one of its endpoints is assumed to be open, the point is a limit point of the interval, and we can add it to the interval, which contradicts to the claim that it is maximal.

Now, let $C$ be the collection of all such intervals. Then, $C$ is disjoint and $\bigcup C = A$. The countability of $C$ directly comes from the proposition, by which we can pair each element of $C$ with an adjacent open interval in the complement of $A$.

Is this correct?

@user14111 Thank you. I think I made myself obfuscated with a naive concept of adjacency. So, are the others, which seem trivial, all correct except the statement about countability? — Hermis14, Oct 08 '22 at 01:32
@herbsteinberg But if there are any two intervals that share the endpoints, the setup will weld the two intervals then. — Hermis14, Oct 08 '22 at 01:49
@Hermis14 It looks like a trivial question. There can be only a countable number of closed intervals (not discrete points), while discrete points need separation - thus countable. — herb steinberg, Oct 08 '22 at 01:55
I cannot see what the proposition has to do with your question. — Paul Frost, Oct 08 '22 at 13:01

score 2 · Accepted Answer · answered Oct 08 '22 at 13:30

The Cantor set is a counterexample. All its maximal subintervals are singletons, and there are uncountably many of them.

Also note that your subintervals are certainly closed subsets of $\mathbb R$, but they may have the form $\mathbb R$, $(-\infty,a]$ or $[a,\infty)$ with $a \in \mathbb R$. I would not regard as them as closed intervals, but perhaps that is a matter of taste.

All you say about your decomposition is that it can have only countably many non-degenerate maximal subintervals (i.e. non-singletons).

In fact, it impossible that $\mathbb R$ contains uncountably many disjoint intervals.

Let $\{J_\alpha\}$ be an uncountable family of disjoint non-degenerate intervals in $\mathbb R$. Not all of the sets $A_n = \{\alpha \mid J_\alpha \text{ has length } \ge 1/n \}$ can be countable. So let $A_N$ be uncountable. One of the sets $B_n = \{ \alpha \in A_N \mid [-n,n] \cap J_\alpha \ne \emptyset \}$ must be uncountable, thus they are uncountable for $n \ge M$. Take $n = \max(N,M)$. There are at most two $J_\alpha$ with $\alpha \in B_n$ not contained in $[-n,n]$ (one can contain $-n$, another can contain $n$). Thus $[-n,n]$ contains uncountable many disjoint non-degenerate intervals of length $\ge 1/N$. Let their union be $C$. The set $C$ must have infinite Lebesgue measure, which contradicts the fact that it is contained in a set with finite Lebesgue measure.

Countable decomposition of a closed subset of real numbers

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