A result using compactness and LUB axiom

Question

Let $a,b \in \mathbb{R}, a<b$ and $\mathcal{A}$ be a collection of open sets in $\mathbb{R}$ such that $[a,b] \subseteq \cup_{A \in \mathcal{A} }A, C=\{x \in [a,b] : [a,x]$ is covered by finitely many members of $\mathcal{A} \}$. Now $C$ is a non empty bounded subset of $\mathbb{R}$ implies by LUB Axiom $z=sup C=lub C$ exists. Prove that $z=b$ and $b \in C.$

Can someone help me with this?

Answer 1

Suppose $z < b$. Then $z$ is covered by some $A_z \in \mathcal{A}$. So there is some open interval $(L,R) \subseteq A_z$ that contains $z$.

For some $c \in C$, we know that $L < c \le z < R$ (why?). Pick $ R'$ in the open interval $(z,R)$ as well.

$[a,c]$ is covered by finitely many members of $\mathcal{A}$. Now show that $[a,R']$ also is (why?). This is a contradiction (why?).

So $z = b$.

Now apply a similar argument to see that $b \in C$ as well.