0
$\begingroup$

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:

enter image description here

From http://assets.press.princeton.edu/chapters/s8008.pdf on page 6

$\endgroup$
  • 1
    $\begingroup$ What about $U=(1,3)\cup(4,5)$ and $x=2$? $\endgroup$ – Lord Shark the Unknown Jan 25 at 6:47
  • $\begingroup$ Ohhh! I guess I got confused and thought about $U$ being intervals instead of actually ANY subset $\endgroup$ – The Bosco Jan 25 at 6:48
  • $\begingroup$ So is this like trying to expand one point in every "portion" of the subset $U$ until it covers that specific interval? $\endgroup$ – The Bosco Jan 25 at 6:55
  • 1
    $\begingroup$ The simplest proof from general principles is just to take the connected components of $U$ and show they are open intervals. $\endgroup$ – Henno Brandsma Jan 25 at 6:57
3
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.