How to prove for every open cover of the real numbers $\mathbb{R}$ there is a countable subcover? Without using more sophisticated results from topology, assuming only a real analysis background.

I've found a proof using second-countable space characterization, but since i never studied general topology before, it's hard to associate a countable base on the real line. My intuition says to transform the open cover into disjoint open subsets, but how to achieve that?

  • 3
    $\begingroup$ Every open cover of what has a countable subcover? $\endgroup$ Nov 11, 2011 at 21:44
  • $\begingroup$ @ChrisEagle: Any $X \subset \mathbb{R}$ $\endgroup$
    – juliohm
    Nov 11, 2011 at 21:48

2 Answers 2


I’m going to assume that you want to prove that $\mathbb{R}$ is Lindelöf. You definitely do not want to try to transform the open cover into disjoint open sets, because it can’t be done: no family of two or more pairwise disjoint non-empty open sets covers $\mathbb{R}$.

Getting a countable base for $\mathbb{R}$ isn’t at all hard, provided that you know that $\mathbb{Q}$, the set of rational numbers, is countable. Just let $\mathscr{B}$ be the set of open intervals with rational endpoints: each pair $\{p,q\}$ of distinct rational numbers determines exactly one such interval, $(p,q)$ if $p<q$, and $(q,p)$ if $p>q$, and there are only countably many pairs of rational numbers, so $\mathscr{B}$ is countable. It only remains to show that $\mathscr{B}$ is a base for the topology of $\mathbb{R}$, which just means showing that every open set in $\mathbb{R}$ is a union of members of $\mathscr{B}$.

Every non-empty open set in $\mathbb{R}$ is a union of open intervals. If we can show that every open interval in $\mathbb{R}$ is a union of members of $\mathscr{B}$, i.e., of open intervals with rational endpoints, it will immediately follow that every non-empty open subset of $\mathbb{R}$ is also such a union. To this end let $(a,b)$ be any non-empty open interval in $\mathbb{R}$. Then there are sequences $\langle p_n:n\in\mathbb{N}\rangle$ and $\langle q_n:n\in\mathbb{N}\rangle$ of rational numbers such that:

  • $p_0>p_1>p_2>\dots\;$;
  • $q_0<q_1<q_2<\dots\;$;
  • $\lim\limits_{n\to\infty}p_n = a\;$;
  • $\lim\limits_{n\to\infty}q_n = b\;$; and
  • $p_0<q_0$.

In other words, $\langle p_n:n\in\mathbb{N}\rangle$ is a decreasing sequence converging to $a$, $\langle q_n:n\in\mathbb{N}\rangle$ is an increasing sequence converging to $b$, and $p_0<q_0$. It easily follows that $$(a,b) = \bigcup_{n\ge 0}(p_n,q_n)\;,$$ and each interval $(p_n,q_n)$ obviously has rational endpoints. Thus, every non-empty open interval in $\mathbb{R}$ is a union of members of $\mathscr{B}$, so every non-empty open set of any kind in $\mathbb{R}$ is such a union, and $\mathscr{B}$ is therefore a countable base for the topology of $\mathbb{R}$.

It’s now trivial to see that $\mathbb{R}$ is Lindelöf: if $\mathscr{U}$ is any open cover of $\mathbb{R}$, just let $$\mathscr{B}_\mathscr{U}=\{B\in\mathscr{B}:\exists U\in\mathscr{U}\big(B\subseteq U\big)\}.$$ Each $U\in\mathscr{U}$ is the union of the members of $\mathscr{B}_\mathscr{U}$ contained in it, so $\mathscr{B}_\mathscr{U}$ covers $\mathbb{R}$. It’s also countable, since it’s a subset of the countable set $\mathscr{B}$. Now for each $B\in\mathscr{B}_\mathscr{U}$ choose some $U(B)\in\mathscr{U}$ such that $B\subseteq U$; the definition of $\mathscr{B}_\mathscr{U}$ guarantees that there is one. Let $\mathscr{U}_0=\{U(B):B\in\mathscr{B}_\mathscr{U}\}$. Then $\mathscr{U}_0$ is countable, since it’s no bigger than $\mathscr{B}_\mathscr{U}$, and $$\bigcup\mathscr{U}_0 = \bigcup_{B\in\mathscr{B}_\mathscr{U}}U(B)\supseteq \bigcup_{B\in\mathscr{B}_\mathscr{U}}B=\mathbb{R},$$ so $\mathscr{U}_0$ is indeed a countable subcover of $\mathscr{U}$.

Added: This idea can be extended to $\mathbb{R}^n$. Instead of open intervals with rational endpoints, you take for your countable base the set of Cartesian products of such intervals. In other words, you take as a base for $\mathbb{R}^n$ the set of open boxes of the form $B_1\times\dots\times B_n$, where $B_1,\dots,B_n\in\mathscr{B}$. Then you show that each open box in $\mathbb{R}^n$ is a union of these ‘rational boxes’. Since there are only countably many rational boxes, and every non-empty open set in $\mathbb{R}^n$ is a union of them, it follows that $\mathbb{R}^n$ is Lindelöf: the remainder of the argument is just like that for $\mathbb{R}$.

There is another way to proceed, if you know that closed, bounded subsets of $\mathbb{R}^n$ are compact, meaning that every open cover of such a set has a finite subcover. I’ll do it for $\mathbb{R}$; the generalization to $\mathbb{R}^n$ is pretty straightforward. $\mathbb{R}$ is the union of the closed intervals $[n,n+1]$ for $n\in\mathbb{Z}$. There are only countably many such intervals, and each of them is compact. Now let $\mathscr{U}$ be any open cover of $\mathbb{R}$. For each $n\in\mathbb{Z}$ let $\mathscr{U}_n = \{U\in\mathscr{U}:U\cap [n,n+1]\ne\varnothing\}$. Then $\mathscr{U}_n$ is an open cover of $[n,n+1]$, so it has a finite subcover, $\mathscr{V}_n$. Finally, let $$\mathscr{V}=\bigcup_{n\in\mathbb{Z}}\mathscr{V}_n\;;$$ $\mathscr{V}$ is the union of countably many finite sets, so it’s a countable subset of $\mathscr{U}$, and it clearly covers $\mathbb{R}$.

  • $\begingroup$ Thank you for your answer, it's really helpful. I'll take a better look to investigate all the details and then back to probably mark the topic as solved. $\endgroup$
    – juliohm
    Nov 11, 2011 at 22:21
  • $\begingroup$ Can't you just, given an uncountably infinite cover of the real numbers, just choose one open set containing each rational? $\endgroup$
    – Carl
    Nov 12, 2011 at 0:26
  • $\begingroup$ That is, unless you have an objection to the axiom of choice I suppose. $\endgroup$
    – Carl
    Nov 12, 2011 at 0:32
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    $\begingroup$ @Carl: No, you can’t: the union of the resulting sets will be a dense open set, but it won’t necessarily be all of $\mathbb{R}$. For a very simple example, the union might be $\mathbb{R}\setminus\{n\sqrt{2}:n\in\mathbb{Z}\}$. In that case you could just throw in one open set for each $n\sqrt{2}$, but the uncovered bit could be uncountable, e.g., a Cantor set contained in the irrationals. (They do exist.) $\endgroup$ Nov 12, 2011 at 4:58
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    $\begingroup$ @MickG: I just checked: $\mathsf{CC}(\Bbb R)$ is equivalent to every second countable space being Lindelöf. $\endgroup$ Apr 23, 2015 at 21:33

NOTE: I don't know if this is what the user would call "sophisticated results". It uses some ostensibly highfalutin language, but the ideas are very simple. I hope it suffices.

I think what the user is asking is to prove that every second countable space is Lindelof (in more common notation).

Really, the user is asking to prove that "If $X$ is second countable and $A$ is a subset of $X$, the any open cover of $A$ admits a countable subcover." To see how $\text{Second Countable}\implies \text{Lindelof}$ gives us this merely note that if $\Omega$ is a $X$-open cover of $A$ then $\Omega$ induces a $A$-open cover of $A$ and since $A$ is second countable (since second countability is hereditary) our result $\text{Second Countable}\implies\text{Lindelof}$ gives us what we want.

So, let's prove $\text{Second Countable}\implies\text{Lindelof}$. So, let $X$ be second countable with countable basis $\mathscr{B}$, and let $\Omega=\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be an open cover for $X$. By assumption, for each $\alpha\in\mathcal{A}$ we can cover $U_\alpha$ with some collection $B_\alpha$ of elements of $\mathscr{B}$. Note then that $\displaystyle \Sigma=\bigcup_{\alpha\in\mathcal{A}}B_\alpha$ is a countable open cover for $X$. So, for each element $O$ of $\Sigma$ choose an element $U$ of $\Omega$ containing it. Then, this subset, call it $\Gamma$, of $\Omega$ is an open cover of $X$ (since its union contains the union over all the elements of $\Sigma$ which is $X$) and is countable since there is a surjection $\Sigma\to\Gamma$ and $\Sigma$ is countable. Thus, $\Gamma$ is our desired countable subcover of $\Omega$.

  • $\begingroup$ Thank you for your answer, it's very elucidative. Unfortunely, i'm asking a more simple question. $\endgroup$
    – juliohm
    Nov 11, 2011 at 22:06
  • $\begingroup$ Ah, I see. I'm sorry that it was overly complicated! $\endgroup$ Nov 11, 2011 at 22:08
  • $\begingroup$ @AlexYoucis I suppose by $\bigcup_{\alpha\in\frak A}B_{\alpha}$ you mean the union of the sets in that collection. However, I would read that as a new collection of sets containing all the open sets in the various $B_\alpha$. I suggest you add an extra $\bigcup$ before the one already present in that $\Sigma$. $\endgroup$
    – MickG
    Apr 23, 2015 at 19:38
  • $\begingroup$ @juliohm in fact this answer looks like just a generalization to a generic topological space of the above answer, which was specific to $\mathbb{R}$ first and $\mathbb{R}^n$ then. The arguments are much the same. $\endgroup$
    – MickG
    Apr 23, 2015 at 19:39
  • $\begingroup$ And the answer's start is the generalisation of this comment :). $\endgroup$
    – MickG
    Apr 23, 2015 at 19:43

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