I think this is a dumb question, but I just can't wrap my head around it.
The problem is the following:
Prove that any open set in $\mathbb{R}$ is a countable union of disjoint open intervals.
My problem is at the beginning of the proof. Let $U$ be the open subset of $\mathbb{R}$ in question. Now, for each $x \in U$ let $x$ denote the largest open interval containing $x$ and contained in $U$. Isn't this $U$ itself? Since $U \subset U$, and $x \in U$.
If it helps, I am talking about the proof here:
From http://assets.press.princeton.edu/chapters/s8008.pdf on page 6
Open sets are not the same as open intervals. In general open sets are sets that for each of its points contain some open interval around it, which implies that all open sets are unions of open intervals, e.g. $(1,2) \cup (2,3)$. The point of this theorem is that we can always write it as a disjoint such union, of at most countably many open intervals. (Where a set like $(a,\infty)$ or $\mathbb{R}$ itself must also count as an open interval for the theorem to hold.)