An element of a set with a finite cover must be an element of at most two open intervals in a subcover?

Question

Prove:

If a set $A\subseteq\mathbb{R}$ has a cover consisting of a finite number of open intervals, then A has a subcover such that for each $x\in A$, x is an element of at most two of the open intervals in the subcover.

My attempt:

To be honest, I have grappled with this problem for too long; I have no idea how to approach this proof. I only have the definitions of cover, subcover, and compact sets and the Heine-Borel Theorem at my disposal. I am having difficulty connecting this ideas to prove what needs to be proven. Could someone give me an idea on how to begin this proof?

Answer 1

Denote by $\mathcal{C}$ the open cover. Let $I_1\in\mathcal{C}$ denote an interval containing $\inf A$ (or with boundary $\inf A$, if necessary). Next, let $I_2\in\mathcal{C}$ be the interval containing $\inf (A- I_1)$ with maximal upper bound. Then let $I_3\in\mathcal{C}$ be the interval containing $\inf(A- I_2-I_1)$ with maximal upper bound. Since $\sup I_3>\sup I_2$ and $I_2$ has maximal upper bound for intervals containing $\inf(A-I_1)$, it follows that $I_3>\inf(A- I_1)$. Continue inductively until finished.

Answer 2

What if you argued by contradiction? This is a super crude discussion on how I'm thinking one could proceed:

Suppose the statement is false. So assume that for every subcover $T'$, there exists an element $x\in A$ such that $x$ is in at least $3$ of the open intervals of the arbitrary subcover $T'$. Without loss of generality, suppose that there are exactly $3$ intervals containing $x$.

Then, proceed to argue as @bof suggests: refine these three intervals such that one is covered by the other two and discard it from $T'$. Then, note that what remains must be another subcover of $A$ where every element $x\in A$ is in at most two intervals. This contradicts the assumption that every subcover of $A$ contains an $x$ from A that is in more than two open intervals in the subcover.

Answer 3

The following is based on ideas from bof's comment and Black8Mamba23's answer. Since your given cover is finite, consider a minimal subcover $C$. Suppose, toward a contradiction, that some point $x$ is in more than two of the intervals in $C$. Among those (finitely many but more than $2$) intervals from $C$ that contain $x$, let $I$ be one with the smallest left endpoint, and let $J$ be one with the largest right endpoint. There is at least one other interval $K$ from $C$ that contains $x$, because of our "more than $2$" assumption. Any such $K$ is included in $I\cup J$, because (1) our choice of $I$ ensures that the part of $K$ to the left of $x$ is included in $I$ and (2) our choice of $J$ ensures that the part of $K$ to the right of $x$ is included in $J$. So $C-\{K\}$ is still a cover, and that contradicts our choice of $C$ as a minimal subcover.