Every non-empty open set in R is disjoint union of countable collection of open intervals

Question: Prove that every non-empty open set in \mathbb{R} is the disjoint union of a countable collection of open intervals.

The key things to prove are the disjointness and the countability of such open intervals. Otherwise, if disjointness and countability are not required, we may simply take a small open interval centered at each point in the open set, and their union will be the open set.

Elementary Proof: Let U be a non-empty open set in \mathbb{R}.

Let x\in U. There exists an open interval I\subseteq U containing x. Let I_x be the maximal open interval in U containing x, i.e. for any open interval I\subseteq U containing x, I\subseteq I_x. (The existence of I_x is guaranteed, we can take it to be the union of all open intervals I\subseteq U containing x.)

We note that such maximal intervals are equal or disjoint: Suppose I_x\cap I_y\neq\emptyset and I_x\neq I_y then I_x\cup I_y is an open interval in U containing x, contradicting the maximality of I_x.

Each of the maximal open intervals contain a rational number, thus we may write \displaystyle U=\bigcup_{q\in U\cap\mathbb{Q}}I_q. Upon discarding the “repeated” intervals in the union above, we get that U is the disjoint union of a countable collection of open intervals.

There are many other good proofs of this found here (http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv), though some can be quite deep for this simple result.

 

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