# Open set is union of countable disjoint class of open interval

It's been a while since I've touched topology so please bear with me.

Theorem: Every non-empty open set on the real line is the union of countable disjoint class of open intervals.

proof:

Let $$G$$ be a non - empty open set on the real line. Since $$G$$ is non-empty, it's order is at least one. Since $$G$$ is an open set, this implies that for every element $$x$$ in $$G$$, there exists a positive radius $$r$$ such that an open interval $$I$$ may be centred on $$x$$ such that $$I$$ is contained in $$G$$.

But for each of these open intervals $$I_{i}$$, there exists some radius $$\bar{r}$$ such that an open interval $$\bar{I_{i}}$$ is contained in $$I_{i}$$. Hence, $$I_{i}$$ is class of open set. The union of $$\bar{I_{i}}$$ then is an open set. Hence, $$I$$ is the union of open sets.

If y is another point then $$I_{x} = I_{y}$$.

Why is this so?

• what is exactly the question? why the last line holds? or if the proof is correct? if the second then I have to say it is quite cryptically written, in particular since the $I_i$ are never defined. However, if you want i can also write my own proof down, using connected components. Jan 7, 2019 at 13:21

Your solution is a bit unclear, I'm not sure what $$I_x, I_y$$ are so here's another approach:

Let $$G\subseteq\mathbb{R}$$ be a nonempty open subset of $$\mathbb{R}$$. Let $$x\in G$$. Denote by $$C(x)$$ the connected component of $$x$$ in $$G$$, i.e. the maximal connected subset of $$G$$ containing $$x$$.

Lemma 1. Each $$C(x)$$ is an open interval.

Proof. Let $$a,b\in C(x)$$ are such that $$a. Assume that $$a. If $$c\not\in C(x)$$ then $$(-\infty, c)$$ and $$(c,+\infty)$$ would be a nontrivial decomposition of $$C(x)$$ into open subsets. That contradicts $$C(x)$$ being connected. Therefore $$c\in C(x)$$ and so $$C(x)$$ is an interval. We will show it is open:

Since $$G$$ is open then there is an open interval $$I$$ around $$x$$ such that $$I\subseteq G$$. Since $$I$$ is connected and $$x\in I$$ then $$I\subseteq C(x)$$ which completes the proof. $$\Box$$

Lemma 2. Every open subset of $$G\subseteq\mathbb{R}$$ is a countable union of disjoint open intervals.

Proof. Obviously $$G=\bigcup_{x\in G} C(x)$$ and $$C(x)$$ are either disjoint or equal. So the only question is: can we take only countably many of them? The answer is yes. That's because for a given $$C(x)$$ there exists $$q\in\mathbb{Q}$$ such that $$q\in C(x)$$ and $$C(x)=C(q)$$. Indeed, you take an open interval $$I$$ around $$x$$ fully contained in $$C(x)$$ and then you pick any rational from $$I$$ (rationals are dense). Therefore

$$G=\bigcup_{x\in G\cap\mathbb{Q}} C(x)$$

which completes the proof. $$\Box$$