I am thinking about the following statement:

"the union of each collection of nontrivial intervals of $\mathbb{R}$ is the union of a countable subset of that collection"

and I came up with the following explanation:

Let $\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$, where $\mathcal{A}$ is uncountable be such a union of nontrivial intervals of $\mathbb{R}$. Remove from this union every interval that is a subset of the union of other intervals: in this way we get a union of nontrivial intervals $\bigcup_{\alpha\in\mathcal{A'}}I_{\alpha},\ \mathcal{A'}\subset\mathcal{A}$ such that $\bigcup_{\alpha\in\mathcal{A'}}I_{\alpha}=\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$, in each interval we can pick a unique rational number (since each $I_{\alpha}$ is not completely inside any union of other intervals) and from this it follows that $\mathcal{A'}$ is countable, as desired.

Now, I know that another question about the same problem has already been asked, and since the proofs posted there by experienced users are much more complicated I guess there must be something wrong with my reasoning above, but since I don't see what that is and I find it to be an appealing argument I would like to see where it fails and/or/how it could be improved, and, if there are any, other proofs of this interesting statement, thanks.

EDIT 09/19: It might be worthwhile to try by contradiction:

Suppose there existed an uncountable collection of nontrivial intervals $(I_{\alpha})_{\alpha\in\mathcal{A}}$ such that for every countable subset $\mathcal{C}\subset \mathcal{A}$ we have that $\bigcup_{\alpha\in\mathcal{C}}I_{\alpha}\subsetneqq\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$. Then it follows that:

  • $(I_{\alpha})_{\alpha\in\mathcal{A}}$ cannot contain $(-\infty,\infty)$;

  • $(I_{\alpha})_{\alpha\in\mathcal{A}}$ cannot contain any countable number of intervals whose union is $(-\infty,\infty)$;

  • the $(I_{\alpha})_{\alpha\in\mathcal{A}}$ cannot be all disjoint, since in that case it is possible to establish an injection to the rational numbers by picking a distinct rational number from each of them, showing that $\mathcal{A}$ is countable, a contradiction.

  • (EDIT 09/20) $\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$ cannot be bounded, because from the fact that every non-trivial interval has positive outer measure and $\bigcup_{\alpha\in\mathcal{C}}I_{\alpha}\subsetneqq\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}$ for every $\mathcal{C}\subset\mathcal{A}$ countable follows that $|\bigcup_{\alpha\in\mathcal{A}}I_{\alpha}|=\infty$.

So, now the task is to get from these two pieces of information (and perhaps others I haven't yet found out) a contradiction.

Comments are welcome.

NOTE: I have never studied topology, only self-studied calculus/linear algebra/real analysis (and currently measure theory) so I am interested in proofs that don't use too much topological machinery, if that is at all possible.

  • 4
    $\begingroup$ "Remove from this union every interval that is a subset of the union of other intervals" -- What are you going to remove from $\bigcup_{x>0}(-x,x)$ and what will you be left with? Similarly what if you start with $\bigcup_{x>0}(-\tanh(x),\tanh(1/x))$? $\endgroup$ Sep 18, 2021 at 18:11
  • 1
    $\begingroup$ If the intervals are open to begin with, this is just a consequence of Lindelof's theorem since $\mathbb{R}$ with the usual topology has a countable basis. $\endgroup$
    – Mittens
    Sep 19, 2021 at 19:35
  • $\begingroup$ @Oliver Diaz thank you for your interest in my question; the intervals can be open, closed, or neither, since by nontrivial interval the book means any interval that contains more than one point. $\endgroup$
    – lorenzo
    Sep 19, 2021 at 19:40
  • $\begingroup$ @OliverDiaz ok thanks, I will look at that theorem (I didn't know about it); however, the thing I find strange is that this exercise (exercise 2D-4a) is taken from Sheldon Axler's real analysis and measure theory book (freely accessible from the author's website) in the chapter about measures (in the section about Lebesgue measure) and the book states as prerequisites only elementary real analysis and not topology so I guess there should be another proof which uses only these tools. $\endgroup$
    – lorenzo
    Sep 19, 2021 at 20:13
  • 1
    $\begingroup$ Lindelof covering theorem for $\mathbb{R} $ is given by Tom Apostol in his Mathematical Analysis. And it is a precursor to the Heine Borel Theorem. However I admit this is not as famous as Heine Borel. $\endgroup$
    – Paramanand Singh
    Sep 20, 2021 at 3:58

1 Answer 1


The main tool to solve the problem is the following result:

Theorem: If $U=\bigcup_{\alpha\in\mathscr{A}}J_\alpha$, where each $J_\alpha$ is an open interval in the real line, then there is countable subset $\mathscr{A}'\subset\mathscr{A}$ such that $U=\bigcup_{\alpha\in\mathscr{A}'}J_\alpha$

This is a particular case of Lindelöf's theorem. The proof is not complicated and I present it here since the OP is not familiar with this result.

Proof: Consider the collection $\mathscr{B}$ of rational intervals ( open bounded intervals with rational endpoints). This is a countable set.

Claim I: Any open interval $I$ in the real line is the union of intervals in $\mathscr{B}$: if $x\in I$ then there is $\delta>0$ such that $(x-2\delta,x+2\delta)\subset I$. Let $q_x\in (x-\delta,x+\delta)\cap\big(\mathbb{Q}\setminus\{x\}\big)$. Choose a rational number $r_x>0$ such that $|x-q_x|<r_x<\delta$. Then $x\in(q_x-r_x,q_x+r_x)\subset (x-2\delta,x+2\delta)\subset I$. Then $I=\bigcup_{x\in I}(q_x-r_x,q_x+r_x)$. Notice that this last union is a countable union since there are $\mathscr{B}$ is countable. $\Box$.

Let $\mathscr{B}'$ be the collection of all rational intervals that are contained in some interval in $\{J_\alpha:\alpha\in\mathscr{A}\}$.

Claim II: $U=\bigcup\{B:B\in\mathscr{B}'\}$. If $x\in U$, then there is $\alpha\in \mathscr{A}$ such that $x\in J_\alpha$. Since $J_\alpha$ is an open interval, Claim I shows there is a rational interval $B_x$ such that $x\in B_x$ and $B_x\subset J_\alpha$. Then $U=\bigcup_{x\in U}B_x$ and so, $U=\bigcup\{B:B\in\mathscr{B}'\}$. $\Box$

For each $B\in\mathscr{B}'$ choose $J_{\alpha_B}\in\{J_\alpha:\alpha\in\mathscr{A}\}$ such that $B\subset J_{\alpha_B}$. The collection of sets $\{J_{\alpha_B}:B\in\mathscr{B}'\}$ is countable since $\mathscr{B}'$ is countable, and $U=\bigcup_{B\in\mathscr{B}'}J_{\alpha_B}$. This concludes the proof of the Theorem. $\Box$.

As in the OP, suppose $\{I_\alpha:\alpha\in\mathcal{A}\}$ is an arbitrary collection of non degenerate intervals, and $$U=\bigcup_{\alpha\in \mathcal{A}} I_\alpha$$ The $I_\alpha$'s may be the form $(a_\alpha,b_\alpha)$,$[a_\alpha, b_\alpha)$, $(a_\alpha,b_\alpha]$, $[a_\alpha,b_\alpha]$ with $-\infty<a_\alpha<b_\alpha<\infty$, or of the form $(a_\alpha,\infty)$, $[a_\alpha,\infty)$, $(-\infty,b_\alpha)$, $(-\infty,b_\alpha]$ with $a_\alpha,b_\alpha\in\mathbb{R}$.

Let $J_\alpha$ be the interior of $I_\alpha$, i.e., the largest open interval contained in $I_\alpha$. For example, if $I_\alpha=[a_\alpha,b_\alpha)$, $J_\alpha=(a_\alpha,b_\alpha)$, etc. Let $V=\bigcup_{\alpha\in\mathcal{A}}J_\alpha$. By the theorem above, there is a countable collection $\{\alpha_n:n\in\mathbb{N}\}\subset\mathcal{A}$ such that $$V=\bigcup_n J_{\alpha_n}$$ If $x\in U\setminus V$ and $x\in I_\alpha$, then $x$ the endpoint of some $I_\alpha$. Let $L$ be the collection of points in $U\setminus V$ that are left endpoints of intervals $I_\alpha$ which are bounded below. Similarly, let $R$ be the collection of all points in $U\setminus V$ that right endpoints of intervals $I_\alpha$ which are bounded above.

If $x\in L$, there is $I_{\alpha_x}$ such that $x\in I_{\alpha_x}$ and $J_{\alpha_x}=(x,b_x)$ where $x<b_x\leq\infty$. There may be several $I_\alpha$'s having $x$ as a left endpoint, but it suffices to choose one for each $x\in L$

Claim: The sets in $\{J_x: x\in L\}$ are pairwise disjoint. Suppose there are $x,x'\in L$, $x<x'$, such that $J_x\cap J_{x'}=(x,b_x)\cap(x',b_{x'})\neq\emptyset$. This implies that $x'\in (x,b_x)=J_x$ and so, $x'\in V$, in contradiction to $x'\in L\subset U\setminus V$. $\Box$

A similar conclusion holds for the points in $R$. For $x\in R$, choose $I_x:=I_{\alpha_x}$ with $x\in I_x$. The sets in $\{J_x: x\in R\}$ are pairwise disjoint.

Claim: The collection $\{J_x: x\in L\}$ is countable. For each $x\in L$, choose a rational number $q_x\in J_x$. Since the $J_x$'s are pairwise disjoint, all the $q_x$'s are distinct. Since the set od rational numbers is countable, it follows that $\{q_x:x\in L\}$ and thus $\{J_x:x\in L\}$ is countable.

A similar result holds for $\{J_x:x\in R\}$.

Let $\{J_n:n\in\mathbb{N}\}$ be an enumeration of the sets that make up $V$; let $\{J'_n: n\in\mathbb{N}\}$ be an enumeration of the sets $\{J_x: x\in L\}$; let $\{J''_n:n\in\mathbb{N}\}$ be an enumeration of the sets $\{J_x:x\in R\}$. Then $\{I_n,I'_n, I''_n:n\in\mathbb{N}\}$ is a countable sub collection of $\{I_\alpha:\alpha\in\mathcal{A}\}$ whose union is also $U$.

  • $\begingroup$ Thank you so much :-). $\endgroup$
    – Koro
    May 21 at 6:58

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