1
$\begingroup$

How would I prove that every open set in $\mathbb{R}$ is the countable union of open intervals with rational endpoints?

Context: I am trying to see whether the set of rational open subsets of $\mathbb{R}$ is countable.

$\endgroup$
  • 4
    $\begingroup$ How would you write $(e,\pi)$ in that way ? $\endgroup$ – Nitrogen Oct 29 '15 at 0:04
  • $\begingroup$ Delete "disjoint" and try again. $\endgroup$ – B. S. Thomson Oct 29 '15 at 0:12
  • $\begingroup$ I've edited the question. :) $\endgroup$ – Evan Oct 29 '15 at 0:17
  • $\begingroup$ What do you mean by a rational open set? $\endgroup$ – DanielWainfleet May 12 '16 at 2:26
  • $\begingroup$ @Nitrogen you would union subintervals whose endpoints are $\omega$-sequences of rationals converging on the endpoints. For example, decimal expansions that converge down to $e$ and up to $\pi$. $e$ and $\pi$ are actually easy because they are computable. $\endgroup$ – Spencer Jun 1 '16 at 4:37
4
$\begingroup$

Let $U$ be an open set in $\mathbb R$ we have to prove there is a countable family of open intervals $(a_i,b_i)$ with $a_i,b_i \in \mathbb R$ so that $U=\bigcup\limits_{i\in\mathbb N}(a_i,b_i)$.

To do this for each $x\in U$ we have $(a_x,b_x)$ so that $x\in(a_x,b_x) \subseteq U$. This is because $U$ is open. Of course, there is a rational number $a'_x$ between $a_x$ and $x$ and a rational number $b'_x$ between $b_x$ and $x$. This is because the rational numbers are dense in $\mathbb R$.

So we can consider the familiy of open intervals with rational endpoints. $(a'_x,b'_x)$ Why is this family countable? Because the number of intervals with rational coordinates is countable, since its cardinality does not exceed $|\mathbb Q\times \mathbb Q |=|\mathbb N|$.

Moreover $\bigcup\limits_{x\in U}(a'_x,b'_x)$ is clearly a subset of $U$ since every interval in the union is contained in $U$. But this union also contains $U$ since every element $x\in U$ is contained in the interval $(a'_x,b'_x)$ and hence is contained in the union. Hence $U$ is a countable union of rational open intervals.

$\endgroup$
  • $\begingroup$ why we can consider the familiy of open intervals with rational endpoints? $\endgroup$ – DuFong May 12 '16 at 0:54
0
$\begingroup$

The following link shows that opne set can is countable disjoint union of open intervals:

Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]

Let's asume $\mathcal{O}=\bigcup_{i=1}^\infty(a_i,b_i)$,where $a_i,b_i\in \mathbb{R}$, then for each interval $(a_i,b_i)$, since rational number is dense in real number, we can always find rational sequence $\{b_{i,n}\}$ increasingly approaching $b_i$ and $\{a_{i,n}\}$ decreasingly approaching $a_i$, then $(a_i,b_i)=\bigcup_{n=1}^\infty (a_{i,n},b_{i,n})$, then $\mathcal{O}=\bigcup_{i=1}^\infty(a_i,b_i)=\bigcup_{i=1}^\infty\bigcup_{n=1}^\infty(a_{i,n},b_{i,n})$, and countable union of countable union is countable union too.

As for can we find disjoint countable union of rational intervals, I didn't prove it.

$\endgroup$

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.