# Question about a step in a proof (that nonempty open sets are unions of open sets)

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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Open set in what? –  Qiaochu Yuan Jun 18 '12 at 4:40
What is $G$? ${}{}{}$ –  mixedmath Jun 18 '12 at 4:42
Incidentally, the empty (open) set is the union of the empty family of open intervals, so it does not need to be excluded in the theorem. –  Jonas Meyer Jun 18 '12 at 4:43
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. –  Mark Bennet Jun 18 '12 at 5:07

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, $b_x = 1$.