This sounds obvious (all balls of all rational radii on rational centres should work, I think) but I can't complete the proof simply from these two properties of $\mathbb R^n$.

  1. $\mathbb R^n$ has a basis of open balls (all balls of all radii on all centres)
  2. $\mathbb R^n$ is second countable and Lindelöf

So from that $\mathbb R^n$ has a countable basis; the cover of $\mathbb R^n$ by open balls has a countable subcover; any open set in $\mathbb R^n$ is the countable union of open balls. But does it easily follow that $\mathbb R^n$ has countable basis of open balls ?

If I take all open sets in $\mathbb R^n$ and express each as a countable union of open balls then I end up with an uncountable set of open balls that cover $\mathbb R^n$, and I can then get a countable subset that covers $\mathbb R^n$, but does this still generate each open set ?

Second thoughts after a hint in a comment. Does this work for a proof ?

  1. $\mathbb R^n$ is second countable so there is a countable basis of open sets {$C_i$}.
  2. $\mathbb R^n$ has a basis of open balls (all balls of all radii on all centres) so each $C_i$ is the union of open balls, and is "covered" by this union.
  3. $\mathbb R^n$ is Lindelöf so the cover of $C_i$ by open balls has a countable subcover.
  4. The set of all open balls in all countable subcovers of {$C_i$} is countable, generates {$C_i$} which generates $\mathbb R^n$ and is therefore a basis for $\mathbb R^n$ .

Thanks for all the feedback. I found a more general answer to the question (more or less in line with my own second thoughts) here: http://www.austinmohr.com/Work_files/hw2-1.pdf referenced in a question here: Bases having countable subfamilies which are bases in second countable space

  • $\begingroup$ No. You might end up with the balls of radius $k\in\Bbb N$ around the origin. A bit more care is required $\endgroup$ Commented Dec 28, 2015 at 13:13
  • $\begingroup$ @HagenvonEitzen Thanks - I think I got it, do you agree ? $\endgroup$ Commented Dec 28, 2015 at 13:37

2 Answers 2


Let $D$ be a dense subset of the metric space $X$ with metric $d.$ Let $B=\{B_d(p,q) :p\in D \land q\in Q^+\}.$ Then $B$ is a base for $X.$ To prove this it suffices to show that whenever $y\in U$ with $U$ open, there exists $ V\in B$ with $ y\in V\subset U.$ Consider that $B_d(y,r)\subset U$ for some $r>0$, and that some $p\in D\cap B_d(y,r/3).$ Take $q\in Q^+$ with $r/3<q<r/2.$ By the triangle inequality, $y\in B_d(p,q)\subset B_d(y,r)\subset U.$...... In particular if $D$ is countable then so is $B$. The set of members of $R^n$ with rational co-ordinates is countable, and dense in $R^n$.


I think you still end up in the same trap that Hagen von Eitzen points out in the comments. It's entirely possible that in step $3$ taking your countable subcover of $C_i$'s could leave you with $C_i = B_{i}(0)$, where each $C_i$ is also an open ball of radius $i$ centered at $0$ from your open-ball basis. If you are comfortable with what you have then ignore what I have to say. If you are looking for some critique and suggestions read on.

I think step $1$ is exactly what you want to do, as is everything you do in step $2$ up until you say "so". At this point I would explicitly show (or claim) that $C_i$ can be written as a union of open balls from your collection of all open balls, $\{\mathcal{B}_\alpha\}_{\alpha \in \lambda}$. More importantly, demonstrate that $C_i$ can be written as a countable union of open balls. That is, $$C_i = \bigcup_{j \in K_i}B_j$$ with $B_j \in \{\mathcal{B}_\alpha\}_{\alpha \in \lambda}$ and $K_i \subseteq \mathbb{N}$ is an index for $C_i$. This can be done by using balls with rational radii, and if need be, placing them carefully inside $C_i$. This will lead to a set that I'll define as $$\mathcal{B}_{C_{i}} = \left\{B_j \in \{\mathcal{B}_\alpha\}_{\alpha \in \lambda}: C_i = \bigcup_{j \in K_i}B_j \right\}$$ which is a subset of countably many open balls whose union gives $C_i$. Since we have countably many $C_i$'s we'll have countably many $\mathcal{B}_{C_{i}}$'s. A countable union of countably many things is still countable, so $\{\mathcal{B}_{C_{i}}\}$ contains countably many open balls. Lastly, any open set $U$ in the topology is such that $$U = \bigcup_{i \in I} C_i = \bigcup_{i \in I}\left(\bigcup_{j \in K_{i}}B_j \right)$$ where $I\subseteq \mathbb{N}$ is an index for $U$. This demonstrates that $\left\{\mathcal{B}_{C_{i}}\right\}$ is a countable basis of open balls for $\mathbb{R}^n$. I do not believe you need the Lindelöf property at any point.


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