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Thm: Every nonempty open set $G$ of real numbers is the union of a finite or countable family of pairwise disjoint open intervals.

Proof: Let $x$ be a point in a nonempty open set $G$. There is an open interval $(y, z)$ such that $x\in (y, z)\subset G$. Then, $(y, x)\subset G$ and $(x, z)\subset G$. We define (possibly extended) numbers $a_x$ and $b_x$ by $a_x = inf (y : (y, x)\subset G)$ and $b_x = sup (z : (x, z)\subset G)$...

My question. Since $(y,x)$ and $(x,z)$ are intervals in G (hence, connected) wouldn't $a_x = b_x =x$?

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    You seem to have $inf$ and $sup$ confused - if they were the other way round, you'd be right. I suggest you draw a diagram of what's going on.2012-06-18

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No. Suppose we are on the real line, and $G = (0,1)$, and $x = 1/2$ for a moment (I assume this ($\mathbb{R}$) is where the whole question takes place - I don't understand it otherwise). Then $a_x = 0$, as $0$ is the smallest element (usual ordering) such that $(a_x, 1/2)$ is in $(0,1)$. Similarly, $b_x = 1$.