Suppose I have a open set of $\Bbb R$ which is not bounded below but bounded above. Now, I want to show that that open set can be written as atmost countable collection of open intervals.

I have done the proof for the case when it is both bounded above and below. Now, , in this case, I want to first identify the open interval which is not bounded below. How to do that ? Then I can repeat the process for the bounded above and below case.

For the case where the open set $E$ is bounded below and above I do the following : choose any $x \in E$ and consider the sets $\{a|(a,x) \subseteq E\}$ and $\{a|(x,a) \subseteq E\}$ and I know that infimum($y_i$) and supremum ($w_i$) exists for them respectively. So, I can write the open set as union of the open intervals $(y_i, w_i)$. Then I use use the denseness of $\Bbb Q$ in $\Bbb R$ to prove countability. But, the problem is what is the process by which I can write the open set as a union of open intervals when $E$ is not bounded below. If I start with arbitrary $x$ in $E$, then how shall I get these open intervals $y_i, w_i$. How shall I know that $\{a|(a,x) \subseteq E\}$ exists or not ?

  • 3
    $\begingroup$ I believe any open set in ${\mathbb R}$ can be written as a countable union of open intervals. $\endgroup$ Sep 13 '13 at 15:19
  • 2
    $\begingroup$ There need not be an interval contained in the open set that is not bounded below. Consider $U = \bigcup\limits_{n=1}^\infty (-n,-n+1)$. But any open subset of $\mathbb{R}$ is the disjoint union of countably many open intervals nevertheless. $\endgroup$ Sep 13 '13 at 15:24
  • 1
    $\begingroup$ See the answers to this question for a number of proofs; you really don’t have to treat the unbounded interval (if any) separately. $\endgroup$ Sep 13 '13 at 21:53

There need not be a single (semi-)unbounded interval contained in your open set $U$, just think of $\mathbb R\setminus\mathbb Z$. However, your proof for the case of a bounded open set $U$ should work just as well for unbounded $U$. After all $\mathbb R$ is homeomorphic to $(0,1)$, which would take you back to the bounded case.

To write $U$ as a disjoint union of countably many intervals, you can proceed as follows: Enumerate $\mathbb Q$ as $q_1,q_2,\ldots$.Now define open intervals $I_n$ recursively: If $q_n\in U$ and $q_n\notin I_k$ for all $k<n$, then let $I_n=(\sup ((\mathbb R\setminus U)\cap (-\infty,q_n)), \inf ((\mathbb R\setminus U)\cap (q_n,\infty)))$; otherwise let $I_n=\emptyset$. Now show that $I_n\cap I_m=\emptyset$ for $n\ne m$ and that $U=\bigcup_{n\in\mathbb N}I_n$. You may drop the empty intervals from this union if you like.


It's possible that your open set has infinitely many disjoint open intervals extending toward $-\infty$, so you can't necessarily easily reduce to the case of bounded above and below. What you can do is note that for any $a \in A$ in your open set, you can find a "rational" interval $I_a = (q_1,q_2) \subset A$ where $q_1,q_2$ are rational and $a \in I_a$. Then the set of all $I_a$ is countable (because the rationals are countable) and $\cup_{a \in A} I_a = A$. If you want intervals to be disjoint, then just note if you enumerate the intervals you obtained as $I_1,I_2,\ldots$, then you can iterate for $n=1,2,\ldots$, and if $I_n$ intersects any interval you have already created when considering $I_1,\ldots,I_{n-1}$ then merge $I_n$ with that interval to obtain a unioned interval that covers the same points. This process will make all intervals disjoint.


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.