The question is :- Let $K$ be a compact subset of $\mathbb R$ with non empty interior. Show that K is of the form $[a,b]$ or $[a,b] \setminus \bigcup I_n$ , where { $I_n$} is a countale disjoint family of open intervals with end points in K.

Attempt at proof :- By Heine-Borel theorem , we know that a compact set $K$ in $\mathbb R$ is closed and bounded. Since it is bounded, closed sets of the form $[a,\infty]$ and $[-\infty,a]$ are not compact. Also since $K$ has a non empty interior it can not be a finite set.

Since K is a closed set, it can not be of the form of an arbitrary union of closed sets. Since an arbitrary union of closed sets may not necessarily be closed. Thus , it can be a finite union of closed sets.

A finite union of closed sets can be represented in the form $[a,b] \setminus \bigcup I_n$ .

Also K may have some isolated points, for example it is of the form $[c,d] \bigcup${$p$} , but such sets can also be represented using above form. For example if $p > d$ we can express it as $[c,p] \setminus(d,p)$.

Attempt at proof ends here.

This is sort of the general idea which I am having , but I am certain that I have missed something and I am unable to express it mathematically. Please bare with this I am new to writing proofs. I think in my attempt I have not considered countable {${I_n}$} but rather finite, which is one more error in the proof.Thank You.

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    $\begingroup$ just because countable unions of closed sets are not necessarily closed it doesnt mean that your set should not be closed. Just saying. $\endgroup$ Jan 14, 2014 at 6:21
  • $\begingroup$ Ya. I realised it now that you have pointed it out. $\endgroup$
    – ameyask
    Jan 14, 2014 at 6:25
  • $\begingroup$ Actually, “with non-empty interior” is not necessary, “consisting of at least two points” would suffice. Also have a look at the Cantor set to see that compact subsets of $\mathbb R$ can be complicated. Their complements are easier to describe, and that is done is this problem. $\endgroup$
    – Carsten S
    Jan 14, 2014 at 6:51
  • $\begingroup$ Related: math.stackexchange.com/questions/627055/compact-sets-in-r $\endgroup$ Jan 14, 2014 at 12:40

2 Answers 2



  1. Prove that any open subset of $\mathbb{R}$ is an (at most) countable union of disjoint open intervals.

  2. Assume temporarily that the result is true. How would you characterize the numbers $a$ and $b$ in terms of the set $K$? (I am asking this question so that when you return to the actual problem, you can say at the outset "Let $a$ be such or such number," and similarly for $b$.)


Ok , based on the answers and comments received, I am attempting an answer to the question. Please suggest ways on improving it.

As an answer suggests any open subset of $\mathbb R$ is an (at most) countable union of disjoint open intervals. This can be proved using one of the methods outlined here :- Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs] . (For ex:- every point $x$ in open set $O$ is contained in exactly one maximal interval, and using the rational numbers they can be shown to be countable).

Suppose we have a closed set $K$ then $K^C$ is open. Thus $K^C$ can be expressed as $\bigcup {I_n}$ , hence $K$ can be expressed as $(\bigcup I_n)^C$ .

Further $K$ is compact , thus it is closed and bounded by the Heine Borel Theorem. It can be shown that any closed and bounded set contains its infimum and supremum. Let inf$K$ be $a$ and let sup$K$ be $b$. Thus K is a subset of $[a,b]$ .

Thus $K$ can be expressed as $K \bigcap [a,b]$ . But K is $(\bigcup I_n)^C$ . Thus K can be expressed as $(\bigcup I_n)^C \bigcap [a,b]$ . Thus we can express K as $[a,b] \setminus (\bigcup I_n)$ .

Is this proof ok ? I think the last step is a bit suspect. Please help ironing it out. Thanks.


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