# Are there minimal topological conditions on a space 𝑋 for it to have a countable separating set?

Are there minimal topological conditions on a space $$X$$ for it to have a countable separating set?

A separating set here is a set $$D \subset C(X)$$ (where $$C(X)$$ is the space of continuous functions from $$X$$ to $$\mathbb{R}$$) such that for every pair of points $$x \neq y$$ there is a function $$f \in D$$ satisfying $$f(x) \neq f(y)$$. I know that second-countable and normal Hausdorff are sufficient to have a countable separating set, but if one takes $$X$$ to be a reflexive and separable Banach space with the weak topology, there is a countable separating set despite not being even second-countable. So second-countability is not necessary.

• What is $C(X)$ for you? The set of continuous functions $X \to \mathbb{R}$? Aug 12, 2020 at 10:44
• @FrancescoPolizzi Yes, exactly that. Aug 12, 2020 at 10:51
• how about existence of weaker second countable topology?
– erz
Aug 12, 2020 at 12:37
• @erz I guess that if there is a weaker, second-countable topology, there is a countable separating set and the functions there are also continuous with the original topology, hence the original topology is second-countable. Is that what you mean? Is it any easier to prove the existence of a weaker second-countable topology? Thanks. Aug 12, 2020 at 14:09
• Having a countable separating family means that you have a continuous injection into $R^N$. Equivalently, you have a weaker topology that makes your space homeomorphic to a subset of $R^N$. Such subset has to be metrizable and separable. Perhaps Perre PC is talking about something similar
– erz
Aug 12, 2020 at 15:11

With the help of the comments by erz, I will prove the following fact:

$$(X,\tau)$$ admits a countable separating function set if and only if there exists a weaker topology $$\tau^*\subset\tau$$ such that $$(X,\tau^*)$$ is Hausdorff regular (i.e. $$T_3$$) and second countable.

Let me first make a few comments.

• Regular second countable spaces are completely normal, so is it equivalent that $$\tau^*$$ is Hausdorff second countable completely normal (i.e. $$T_5$$).

• In terms of open sets of $$\tau$$, the condition can be rephrased as such: there exists a collection of open sets $$U_i$$, $$i\in I$$ such that

1. (base) for every $$x\in U_i\cap U_j$$, there exists $$k$$ such that $$x\in U_k$$ and $$U_k\subset U_i\cap U_j$$
2. (Hausdorff) it separates points, i.e. for each pair $$x\neq y$$ there are disjoints sets $$U_i$$, $$U_j$$ such that $$x\in U_i$$ and $$y\in U_j$$;
3. (regular) for each $$x\in U_i$$, there exists $$j$$ such that $$x\in U _j$$ and for all $$y\in U_i^\complement$$, $$y\in U_k\subset U_j^\complement$$ for some $$k=k(y)$$ (think $$\overline{U_j}\subset U_i$$, but take the closure with respect to $$\tau^*$$).
4. (second countable) $$I$$ is countable.

Indeed, if such a family exists, then the topology it generates gives a suitable $$\tau^*$$, and if a Hausdorff regular second countable $$\tau^*$$ exists, any of its countable bases gives a suitable $$U_i$$.

• Urysohn's metrisation theorem asserts that a Hausdorff regular second countable space is metrisable. In particular, it means that a Hausdorff space is regular second countable if and only if it is metrisable separable. In other words, a space $$(X,\tau)$$ admits a countable separating function set if and only if there exists a weaker $$\tau^*$$ that is metrisable separable, i.e. it admits a distance $$d$$ such that the associated open balls are open in $$\tau$$ and there exists a countable subset of $$X$$ that intersects every open ball.

### Proof (open sets)

$$(\Rightarrow)$$ For the direct implication, suppose that we are given a countable $$D\subset C(X)$$ that separates points. Then we can define the family $$\mathcal V$$ of open sets of the form $$f^{-1}(a,b)$$, for $$f\in D$$ and $$a,b\in\mathbb Q$$, and the family $$\mathcal U$$ of finite intersections of elements of $$\mathcal V$$. Let us show that the topology $$\tau^*\subset\tau$$ generated by $$\mathcal U$$ is Hausdorff regular second countable. As discussed above, we can reduce the proof to statements about $$\mathcal U$$.

• (base) $$\mathcal V$$ is stable by finite intersection.
• (Hausdorff) For a given pair $$x\neq y$$, because $$D$$ separates points, we have $$f(x)\neq f(y)$$ for some $$f\in D$$; without loss of generality, $$a for some $$a,b,c\in\mathbb Q$$, and $$f^{-1}(a,b)$$, $$f^{-1}(b,c)$$ are disjoint sets in $$\mathcal U$$ containing respectively $$x$$ and $$y$$.
• (regular) Let $$U_1,\ldots,U_n$$ be elements of $$\mathcal V$$, i.e. $$U_i=f_i^{-1}(a_i,b_i)$$, $$f_i\in D$$, $$a_i,b_i\in\mathbb Q$$. If $$x$$ belongs to the intersection $$U$$ of the $$U_i$$, then $$a_i and we can find $$\alpha_i,\beta_i\in\mathbb Q$$ such that $$a_i<\alpha_i. Then the intersection $$U'$$ of the sets $$U'_i:=f_i^{-1}(\alpha_i,\beta_i)$$ contains $$x$$. Suppose $$y$$ is not in $$U$$, for instance $$f_1(y)\geq b_1$$. Then $$y\in f_1^{-1}(\beta_1,M)\subset (U')^\complement$$ for some $$M\in\mathbb Q$$ large enough. Other possibilities for $$y$$ are treated similarly.
• (second countable) Elements of $$\mathcal V$$ are described by finite sequences of elements of $$\mathcal U$$, which in turn are described by elements of $$D\times\mathbb Q\times\mathbb Q$$.

$$(\Leftarrow)$$ In the other direction, let $$\tau^*\subset\tau$$ be a Hausdorff regular second countable topology on $$X$$, and $$(U_n)_{n\geq0}$$ a countable basis of $$\tau^*$$. For each $$(n,m)$$, choose if possible a continuous $$f_{nm}:(X,\tau)\to\mathbb R$$ such that $$(f_{nm})_{|U_n}\equiv 0$$, $$(f_{nm})_{|U_m}\equiv 1$$. If there is no such function, have $$f_{nm}\equiv 1/2$$. The set $$D:=\lbrace f_{nm},n,m\in\mathbb N\rbrace$$ is obviously countable; let us show that it separates points.

We work in $$\tau^*$$ in this paragraph. Choose any $$x\neq y$$ in $$X$$. Because $$X$$ is Hausdorff, there exist $$U,V$$ disjoint open sets such that $$x\in U$$ and $$y\in V$$. Because it is regular, we have $$x\in U'\subset\overline{U'}\subset U$$ for some open set $$U'$$, and similarly for $$y$$. Since $$(U_n)_{n\geq0}$$ is a basis, we find $$n,m$$ such that $$x\in U_n\subset U'$$ and $$y\in U_m\subset V'$$. It follows that the closures $$\overline {U_n}$$ and $$\overline {U_m}$$ are disjoint (they belong to $$\overline{U'}\subset U$$ and $$\overline{V'}\subset V$$ respectively). Since $$X$$ is normal (regular second countable spaces are completely normal hence normal), Urysohn's lemma shows that there exists some continuous function $$f:(X,\tau^*)\to\mathbb R$$ such that $$f_{|\overline{U_n}}\equiv 0$$ and $$f_{|\overline{U_m}}\equiv 1$$. But then $$f:(X,\tau)\to\mathbb R$$ is continuous, so $$f_{nm}$$ is not 1/2 but a function that is 0 (resp. 1) when restricted to $$U_n$$ (resp. $$U_m$$). In particular, $$f_{nm}(x)=0\neq1=f_{nm}(y)$$ for some $$f_{nm}\in D$$.

### Proof (metric spaces)

As discussed, the condition on $$(X,\tau)$$ is equivalent to the existence of some separable metrisable $$\tau^*\subset\tau$$.

$$(\Rightarrow)$$ This elegant proof is due to erz. Let $$D$$ be a countable separating function set. There is an obvious continuous function $$(X,\tau)\to\mathbb R^D$$ that sends $$x$$ to the collection of $$f(x)$$ for $$f\in D$$. Let $$\tau^*$$ be the pulls back of the topology of $$\mathbb R^D$$. Because $$D$$ separates points, this map is injective, so $$(X,\tau^*)$$ has the topology of a subset of $$\mathbb R^D$$ (its image). Since second countability and metrisability are hereditary properties (a subset of a metric/second countable space is metric/second countable) and a separable metric space is second countable, it suffices to show that $$\mathbb R^D$$ is metrisable separable. This is well known: $$d(x,y):=\sum_{k\geq0}\min(|y(f_k)-x(f_k)|,2^{-k})$$, for $$D=\lbrace f_k\rbrace_{k\geq0}$$, is a metric generating the topology, and the set $$\mathbb Q^{(D)}$$ of rational sequences with finite support is countable dense.

$$(\Leftarrow)$$ Take $$D=\lbrace y\mapsto d(x_n,y) \rbrace$$, for $$d$$ a metic generating $$\tau^*$$ and $$x_n$$ a dense sequence with respect to $$\tau^*$$.

### For fun

There is no explicit use of Urysohn's metrisation theorem in the proof above, but one can suspect it is lurking in the shadows. Indeed, the proof I know of this result goes as follows. Suppose $$(X,\tau^*)$$ is Hausdorff regular second countable. Construct a countable family $$(f_n)_{n\geq0}$$ of functions that separates points, by following the proof given above. Then $$d(x,y):=\sum_{n\geq0}\min(|f(y)-f(x)|,2^{-k})$$ is a distance inducing $$\tau^*$$.