Prove: every finite cover $\mathcal U$ of $M\subseteq\mathbb R$ by open intervals contains two sets of disjoint intervals whose union covers $M$

Question

It's not hard to show that if three open intervals in $\mathbb R$ have a non-empty intersection, then one of the intervals is contained in the union of the other two. The simplest way to show this is by drawing a picture.

Based on this, I want to show that if $\mathcal U$ is a finite cover of a set $M \subseteq \mathbb R$ by open intervals, then there exist two subfamilies $\mathcal U_1$ and $\mathcal U_2$ of $\mathcal U$ that are each disjointed (i.e. any two open sets in $\mathcal U_1$ are disjoint, and similarly for $\mathcal U_2$) and whose union is a cover of $M$. It's been taking me some time so far.

Specifically, Here is my work: Fix an element $x \in M$. Then the subsets of $\mathcal U$ whose union contains $x$ and has its elements as pairwise disjoint has a maximal element $\mathcal U_1$ by Zorn's lemma. If $M \subseteq \mathcal U_1$ then we may take $U_2 = \emptyset$ and we would be done. If not, there there exists some $y \in M - \bigcup \mathcal U_1$. By similar reasoning from Zorn's lemma on $\mathcal U - \mathcal U_1$, there exists a subset of $\mathcal U - \mathcal U_2$ that is disjointed and is maximal under inclusion in $\mathcal U - \mathcal U_2$. I want to show that $M \subseteq \bigcup \mathcal U_1 \cup \bigcup \mathcal U_2$. So suppose not. Then there exists some element $z \in M - \Big( \bigcup \mathcal U_1 \cup \bigcup \mathcal U_2 \Big)$. Since $M \subseteq \bigcup \mathcal U$, there is some $C \in \mathcal U$ such that $z \in C$. By our construction of $\mathcal U_i$s, there exists $D_1 \in \mathcal U_1$ and $F_2 \in \mathcal U_2$ such that $C \cap D_i \neq \emptyset$. If $D_1 \cap D_2 \cap C$ is nonempty then we would be done.

I am bot sure whether what I have is correct so far and how to proceed from here. I also know that, since $\mathbb R$ is separable, there exists a countable subfamily of sets from $\mathcal U$ that still covers $M$, but I don't know if this would be relevant. Also, I am wondering if there is a method to prove the question without using Zorn's lemma or its equivalent. Can you help?

Answer 1

Take a minimal subcover $\mathcal{V}$ of $\mathcal{U}$ (that is, $\mathcal{V}\subset \mathcal{U}$ covers $M$ but no proper subset of $\mathcal{V}$ does). Enumerate $\mathcal{V}=\{I_1,\ldots, I_n\}$, where $\inf I_1\leqslant \inf I_2\leqslant \ldots \leqslant \inf I_n$.

Note that for each $i\in\{1,\ldots, n\}$, $I_i\neq \varnothing$, otherwise we could omit it and obtain $\mathcal{W}\subsetneq \mathcal{V}\subset \mathcal{U}$ which still covers $M$, contradicting minimality of $\mathcal{V}$.

Fact: For $1\leqslant i<j\leqslant n$, $\inf I_i<\inf_j$ and $\sup I_i<\sup I_j$. Indeed, by our enumeration, $\inf I_i\leqslant \inf I_j$. If we had $\inf I_i=\inf I_j$, then by choosing $k\in\{i,j\}$ such that $\max\{\sup I_i, \sup I_j\}=\sup I_k$ and $l\in \in \{i,j\}\setminus \{k\}$, $I_l\subset I_k$, and we can omit $I_l$ from $\mathcal{V}$ and again contradict minimality. So $\inf I_i<\inf I_j$. Similarly, if $\sup I_i\geqslant \sup I_j$, $I_j\subset I_i$, and we omit it and again contradict minimality.

I claim that $$\mathcal{U}_1:=\{I_k:1\leqslant k\leqslant n,k\text{ odd}\}$$ and $$\mathcal{U}_2:=\{I_k:1\leqslant k\leqslant n,k \text{ even}\}$$ do the trick. By definition, $\cup\mathcal{V}\supset M$.

Now, exactly as in the case of three intervals, if $1\leqslant i<j\leqslant n$ are such that $j-i>1$ and $I_i\cap I_j\neq \varnothing$, then $i<i+1<j$ and $I_{i+1}\subset I_i\cup I_j$, contradicting minimality again. Since $j-i>1$ for any $i,j$ such that $I_i,I_j\in \mathcal{U}_1$ (or such that $I_i,I_j\in \mathcal{U}_2$), the members of each $\mathcal{U}_p$ are pairwise disjoint.