Every open subset $O \subseteq \mathbb{R}$ is a union of open intervals.

Question

The Theorem:

Every open subset $O \subseteq \mathbb{R}$ is a union of open intervals.

The problem asks for a particular method of proof:

"Complete the proof by showing that $O = \displaystyle\bigcup_{x \in O} (x – r_x, x + r_x)$."

My work:

By the definition of open, we know that, for all $x \in O$, there exists an $r_x > 0$ such that $(x – r_x, x + r_x) \in O$. Since this is true of all $x \in O$, then $O$ precisely is the union of all such intervals. Thus, we have $O = \displaystyle\bigcup_{x \in O} (x – r_x, x + r_x)$.

My question:

Does this work? The result seemed obvious and the proof trivial – enough so to make me nervous and think I must have begged a question somewhere.

Answer 1

Two issues:

1. Possibly a typo, but it should be $(x-r_x,x+r_x) \subseteq O$ where you have written $\in$
2. Formally, your argument as written should only conclude $\bigcup_{x\in O} (x-r_x,x+r_x) \subseteq O.$ A simple argument [that you should include] shows the other inclusion.