2
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

3
$\begingroup$

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.

Comments

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 $\tau^*$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.