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


3 Answers 3


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.

  • $\begingroup$ This doesn't address the question at all. $\endgroup$ Aug 12, 2018 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. $\endgroup$ Aug 12, 2018 at 14:40

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.


This is just an alternative point of view, not contradicting the already accepted answer by David Hartley. I will make use of an equivalence relation.

So $G$ is a non-empty open set in $\Bbb R$. For every $x\in G$ let $a_x=\inf\{t:(t,x]\subset G\}$ and let $b_x=\sup\{t:[x,t)\subset G\}$. Since $G$ is open we have that $-\infty\le a_x<x<b_x\le\infty$ and if $U_x=(a_x,b_x)$ then $x\in U_x\subseteq G$.

The rest is just a sketch with the proofs left as an exercise.

If $x,y\in G$ and $U_x\cap U_y\not=\emptyset$ then $U_x=U_y$.

Define a relation $\sim$ on $G$ by $x\sim y$ if $U_x\cap U_y\not=\emptyset$. One can prove that $\sim$ is an equivalence relation, and for each $x\in G$ the set $U_x$ is the equivalence class of $x$.

Now, $G$ is the union of all such equivalence classes (all such $U_x$).

The collection of all such intervals (that is, the collection $\{U_x:x\in G\}$) is countable, since any collection of disjoint open intervals is countable. Note that if $U_x=U_y$ we do not count $U_x$ and $U_y$ as different intervals. On the other hand if $U_x\not=U_y$ then $U_x\cap U_y=\emptyset$, that is these intervals are disjoint. Every open interval contains some rational number, and since disjoint intervals contain different rational numbers, it follows that we have at most countably many such disjoint intervals.

Clearly also $G=\cup\{U_x:x\in G\}$ (which was said earlier, $G$ is the union of all equivalence classes).

For that matter, even though I defined the equivalence relation $\sim$ by first defining $U_x=(a_x,b_x)$, where $a_x$ and $b_x$ were defined as certain $\inf$ and $\sup$, one can define the equivalence relation $\sim$ directly, without any reference to $\inf$ and $\sup$.

Namely, given any non-empty open set $G\subseteq\Bbb R$, and any $x,y\in G$ (with $x<y$), define $x\simeq y$ if (and only if) $[x,y]\subset G$. It is easy to verify that $\simeq$ is an equivalence relation on $G$ (and it is the same equivalence relation as $\sim$, defined earlier), and the equivalence classes partition $G$ into disjoint open intervals (and those are at most countably many since different such intervals contain different rational numbers, and the rationals are only countably many).


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