Exercise 1.6.10 (Open subsets of the real line are unions of open intervals)

Consider the following assertion: Every $x\in U$ is contained in a unique maximal open sub-interval of $U$. We verify this assertion.

Let $x\in U$ be arbitrary. Define

We now look at the interval $(a_x,b_x)$ and the nice properties it has.

• $x\in (a_x,b_x)$
  $a_x<x$ since otherwise $a_x=x$ would imply that $x$ is not an interior point of $U$. Similarly, $x<b_x$.
• $(a_x,b_x)\subseteq U$for the sake of contradiction that there is a $z\in (a_x,b_x)$ such that $z\notin U$. Then we have two options: $\mathit{zForall}$a’ < a_x$\mathit{wehave}$(a’,b_x) ⊈U$\mathit{andforall}$b’ > b_x$\mathit{wehave}$(a_x,b’) ⊈U$.\mathit{wewouldhave}$(a’,a_x]$\subseteq U$ - a contradiction to the definition of $a_x$. Similarly with $b^{\prime }$.
• For all $x,y\in U$ the maximal intervals $(a_x,b_x)$ and $(a_y,b_y)$ are either equal or disjoint. Suppose for the sake of contradiction that $(a_x,b_x)\ne (a_y,b_y)$ and $(a_x,b_x)\cap (a_y,b_y)\ne \varnothing$. In other words, we have $z\in (a_x,b_x)\cap (a_y,b_y)$ and $p\in (a_x,b_x)∕(a_y,b_y)$ or $(a_y,b_y)∕(a_x,b_x)$.