Closed $[a,b]⊆\mathbb{R}$ is not a countable union of $≥2$ disjoint closed intervals?

Question

Show that in real line closed intervals cannot be written as a union of countable disjoint closed intervals.

Honestly I cannot see how to show that. If it was given that about a closed sets i know some of the example and Cantor set is also a good example but i dont find any examples to show that a closed interval cannot be written as a union of countable disjoint closed intervals. Although if i consider the trivial representation by singletone sets then i can see that it is uncountable but satisfies given other properties. But i dont know any example. So i want help. Thank you.

Answer 1

Let $I$ be a closed interval of the real line, and assume for a contradiction that $I$ is the union of a collection $\mathcal F=\{F_1,F_2,\dots\}$ of disjoint nonempty closed sets (not necessarily intervals) which has at least two elements and is at most countable.

Lemma. There is a closed interval $I_1\subseteq I$ which is disjoint from $F_1$ and meets at least two elements of $\mathcal F$.

Proof. Choose a point $a\in F_2$, and let $b$ be a point in $F_1$ nearest to $a$. Without loss of generality, assume $a\lt b$; then $(a,b)\subseteq I\setminus F_1$. Since $F_2$ is closed and $b\notin F_2$, $(a,b)\not\subseteq F_2$. Choose a point $c\in(a,b)\setminus F_2$. Then the closed interval $I_1=[a,c]$ is disjoint from $F_1$ and meets at least two elements of $\mathcal F$.

By repeated use of the lemma we can get a sequence $I\supseteq I_1\supseteq I_2\supseteq\cdots$ whose intersection is disjoint from every element $F_n$ of $\mathcal F$. Since the intersection is nonempty by the nested intervals theorem, this is a contradiction.

Corollary. A pathwise connected topological space does not admit a nontrivial partition into countably many closed sets.