Sorry I know this is a stupid question. However I got stuck on this for quite a while. I'm trying to prove that Euclidean spaces have a countable base, which can be constructed by taking all the open balls at all the rational points(points whose coordinates are rational) with all rational radius. However I find it hard to prove that any open set in the Euclidean space can be written as a union of such balls.

Any help?

Kind regards, Evariste

  • $\begingroup$ Don't you think the point that $\mathbb Q^n$ is dense in $\mathbb R^n$ is useful here? $\endgroup$ – mrs Mar 2 '13 at 7:13
  • $\begingroup$ I can't see how they can be useful(although I do agree with you). I can imagine how it works in 1 dimension but not higher dimensions. But thanks for the comment! $\endgroup$ – Evariste Mar 2 '13 at 7:15
  • $\begingroup$ The argument has nothing at all with dimensions... If you know $D$ is dense in a metric space, all rational radius balls around points from $D$ are a base. See my argument. $\endgroup$ – Henno Brandsma Mar 2 '13 at 7:37

Let's call your family of rational-radius balls around all-rational points $\mathcal{B}$. For every open set $O$ and every $x \in O$ we have to find some $B_x \in \mathcal{B}$ such that $x \in B_x$ and $B_x \subset O$. If we can do that we are done, because then $O = \cup_{x \in O} B_x$: all sets $B_x$ lie inside $O$ which gives one inclusion and every $p \in O$ is in "its" $B_p$ so is in the union as well, and gets us the other inclusion.

Now, as all open balls (the complete collection) generates the topology, i.e. all open sets are unions of open balls, we only need to do the above for $O$ of the form $B(p,r)$.

So if $x \in B(p,r)$ you need to find $B \in \mathcal{B}$ (so another ball, from a more limited set) such that $x \in B \subset B(p,r)$. This is a bit more concrete than thinking about arbitrary open sets.

Now, $d(x, p)< r$ ($d$ is the Euclidean distance), so we have some wiggle room $s = r - d(x,p) > 0$. Now, find an all rational point $q$ that has $d(p,q) < \frac{s}{2}$, which can be done (if you know that $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$) and then we can find a rational radius $r' \in \mathbb{Q}$ with $d(p,q) < r' < \frac{s}{2}$. Now if $y \in B(q,r')$ (and that ball lies in your family $\mathcal{B}$) then $d(y,q) < r'$ and we already know $d(x,q) < \frac{s}{2}$, so $$d(y,p) \le d(y,q) + d(q,p) + d(p,r) < r' + \frac{s}{2} + d(p,r) < s + d(p,x) = r$$ so $B(q,r') \subset B(x,p)$ and $x \in B(q,r')$, so we have shown what was needed.

Note that this argument works for any countable dense set in any metric space.


Let $A$ be open, and let $\mathcal{S}$ be the set of all balls $B$ with rational centre and radius ("rational balls") such that $B\subseteq A$. We wish to show that $A$ is the union of all $B\in \mathcal{S}$.

Let $x\in A$. Then there is an $\epsilon\gt 0$ such that the open ball $O$ with centre $x$ and radius $\epsilon$ is a subset of $A$. But $O$ is not necessarily a rational ball.

Let $p$ be a rational point at distance $\lt \epsilon/3$ from $x$, and let $r$ be a rational number strictly between $\epsilon/3$ and $\epsilon/2$. Then the rational ball $B$ with centre $p$ and radius $r$ contains $x$. Moreover, by the Triangle Inequality, $B\subset O$, so $B\subset A$.


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