0
$\begingroup$

I have some doubts in proof.

R is separable, and thus has a countable dense set, namely Q. Let G ⊂ R be any open set.

Then Q∩G is a countable dense set in G by the Archimedean property, and since G is open we can choose an open interval around every rational in G. Then G is the union of that countable collection of intervals. However, we need to find a countable collection of disjoint intervals.

Notice that the union of any intervals which contain the same point is an interval with a lower endpoint equal to the infimum of the lower endpoints of the intervals (possibly −∞) and with an upper endpoint equal to the supremum of the upper endpoints of the intervals (possibly ∞). We create a new countable collection of intervals whose union is G by the following procedure.

Take any point in G ∩Q and take the union of all intervals in G that contain it. Call this interval I1. Now take some point in (G \ I1) ∩Q and take the union of all intervals in G \ I1 that contain it. Repeating this process we get a countable collection of disjoint intervals I1,I2,I3,..., each of which is in G and which together cover G.

Please explain statements in dark / bold letters. Sorry if this topic is repeated but I need to clear this proof

$\endgroup$
1
$\begingroup$

First some quibbles. $\mathbb {R}$ is separable because it has the countable dense subset $\mathbb {Q}$. Not "thus".

The Archimedean principle could be used to prove that $\mathbb {Q}$ is dense, (as it implies there is a rational between any two reals) but having assumed that, $\mathbb {Q} \cap G$ is dense in $G$ follows immediately - as any open set of $G$ is open in $\mathbb {R}$.

Choosing an open interval around each point of $\mathbb {Q} \cap G$ does not necessarily give you a collection which covers $G$. It will if you take the largest such interval within $G$ in each case, which is the point of the statement you have trouble with.

Now, suppose you have a collection of intervals $\{ (a_i, b_i) \}$ each containing a point x. Suppose $U$ is their union. Let $a = inf \{a_i\}$. If $a < y \leq x$ then $a_i < y \leq x < b_i$ for some i, and so $y \in (a_i,b_i) \subset U$. Therefore $(a, x] \subset U$. The same argument applies to $b = sup \{b_i\}$, giving $[x,b) \subset U$, and so $(a,b) \subset U$. (Allowing that a or b may be $\pm \infty$.) The other direction is obvious, so $(a,b) = U$.

Final quibble. Choosing the intervals $I_n$ as you do does not ensure that they cover $G$. You need to do something like enumerate $\mathbb {Q} \cap G$ and then choose the point used at the $(n+1)^{th}$ stage to be the one with the least index that is not yet covered. Alternatively, just note that the maximal intervals within $G$ around any two points of $\mathbb {Q} \cap G$ are either disjoint or equal, so the set of all such intervals is the required collection.

$\endgroup$
1
$\begingroup$

The simplest way to find the promised decomposition of an open set $O$ is to take all connected components of $O$. These are open (as $\mathbb{R}$ is locally connected) and the only connected open subsets of the reals are of the form $(a,b)$ with $a < b$ or $(a,\infty)$ or $(-\infty,b)$. The components are automatically disjoint, and as each of them must contain at least one of countably many rationals there are at most countably many of them.

$\endgroup$
  • $\begingroup$ This doesn't address the question at all. $\endgroup$ – David Hartley Aug 12 '18 at 14:13
  • $\begingroup$ @DavidHartley the OP went via a very roundabout proof idea. I wanted to point out that we need no such sillyness. $\endgroup$ – Henno Brandsma Aug 12 '18 at 14:40

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.