# If $X$ is a topological separable space then every open set is a union of a countable number of open balls?

I am studying some real analysis (I am in the part before defining Borel's $\sigma$-algebra) and there is a reminder in my lecture notes that states that if $X$ is a separable metric space then every open set is a union of a countable number of open balls (where separable is defined as having a countable dense subset).

Can we remove the requirement that $X$ is metric ?

To be honest, I don't really know the proof even with requiring that $X$ is metric, I guess that we move from any union to a countable union by using somehow the fact that every open ball have an element in the dense set, this is what also led me to believe that the claim doesn't have much to do with $X$ being metric.

So there are two questions here, is it true for any topological space and how can we prove this claim ?

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How would you define 'open balls'? – Berci Nov 20 '12 at 19:44
The problem can arise when you define arbitrary topology on X . So to be on the safer side and the statement to hold always true you need metric on $X$ . This is what i can see . – Theorem Nov 20 '12 at 19:48

## 3 Answers

For the proof, consider a point $x$ in an open subset $U$, then there is a ball $B_\epsilon$ around it $\subseteq U$, then look at a smaller ball around $x$, $B_{\epsilon/2}$, choose a point $d$ in $B_{\epsilon/2}\cap D$ where $D$ is the countable dense set, then consider $B_{\epsilon/2}(d)$, it contains $x$.

And one more trick, choose $\epsilon\in\Bbb Q$.

It seems this proof could be generalized to a kind of generalized metric space in the sense that the space of distances, $\Bbb R$ is replaced to something else (I guess something like topological cancellable commutative semigroup), but still having a countable dense subset.

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Thanks, I didn't even notice that we need the space to be metric to talk about open balls, been a while since I studied topology. This proof is really nice and easy to understand, thanks again! – Belgi Nov 20 '12 at 19:59

The point is that is $X$ is metric, then the family of open balls forms a base for the topology. If $X$ is in additional separable, then by taking open balls of radius $\frac{1}{n}$ ($n \in \mathbb{N}$) centred at the elements of a countable dense subset, this family is also a base for the topology. In the end, the proof relies on the fact that separable metric spaces are second-countable.

In a certain sense, the condition "separable metric" may be replaced by any property ensuring second-countability. This comes from the following

Fact: If $X$ is any second-countable topological space, and $\mathcal{B}$ is any base for the topology on $X$, then there is a countable subfamily $\mathcal{B}_0 \subseteq \mathcal{B}$ which is also a base for the topology.

So by replacing "separable metric" by "second-countable" and "open balls" by "open sets in some pre-determined base" you get a similar result.

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Thanks for the answer, I will take a look at this property of being second countable! – Belgi Nov 20 '12 at 20:00

Berci has pointed out in a comment that, in order to talk about open balls, one needs a metric; there is no notion of ball in a general topological space. It might also be worth pointing out that, in separable metric spaces, one has the stronger property that there is a countable family of open balls (namely those with rational radii and centers in a given countable dense set) such that every open set is a union of a subfamily of this particular family. One could try to generalize this last observation by asking whether, in a separable topological space, there must be a countable family $\mathcal B$ of open sets such that every open set is the union of a subfamily of $\mathcal B$. This property of a space is called the second axiom of countability (and such spaces are called second-countable), and it is known not to follow from separability. Perhaps the nicest counterexample is the product (with the usual product topology) of $\kappa$ copies of a 2-element discrete space. This is separable for all cardinal numbers $\kappa$ up to and including the cardinality of the continuum (and not for any larger $\kappa$), but it is second-countable only for countable $\kappa$.

Some tangential comments: Instead of generalizing the observation that I began with, one could similarly try to generalize the original, weaker statement: Must there be a family $\mathcal B$ of open sets such that every open set is a union of a subfamily? Unfortunately, this generalization is silly; the answer is trivially affirmative (whether or not the space is separable) as we can just take $\mathcal B$ to contain all the open sets.

Although I described what I consider the "nicest" counterexample above, I must admit that my personal favorite is another counterexample: the Stone-Cech compactification of a countably infinite discrete space.

Finally, I emphasize that, although I'm now a senior citizen, terrible terminology like "first-" and "second-countable", "first" and "second category", was around long before I learned it. I officially disclaim any responsibility.

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Responsibility is easy: first and second category was used by Baire himself (item 59). After discussing the axioms $(A)$-$(C)$ for neighborhood systems and the separation axiom $(D)$, Hausdorff introduced the two axioms $(E)$ and $(F)$ for topological spaces which he called the first and second axiom of countability, respectively. Finally, Fréchet is responsible for separable, which I consider to be similarly awful... – commenter Nov 21 '12 at 4:47