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.

  • $\begingroup$ What is $C(X)$ for you? The set of continuous functions $X \to \mathbb{R}$? $\endgroup$ Aug 12, 2020 at 10:44
  • $\begingroup$ @FrancescoPolizzi Yes, exactly that. $\endgroup$ Aug 12, 2020 at 10:51
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    $\begingroup$ how about existence of weaker second countable topology? $\endgroup$
    – erz
    Aug 12, 2020 at 12:37
  • $\begingroup$ @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. $\endgroup$ Aug 12, 2020 at 14:09
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    $\begingroup$ 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 $\endgroup$
    – erz
    Aug 12, 2020 at 15:11

1 Answer 1


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, $a<f(x)<b<f(y)<c$ 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 $a_i<f_i(x)<b_i$ and we can find $\alpha_i,\beta_i\in\mathbb Q$ such that $a_i<\alpha_i<f_i(x)<\beta_i<b_i$. 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 $\tau^*$.


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