A property between "separable" and "second countable"

Question

Let $(X, \tau)$ be a topological space. It is second countable if it has a countable basis $B \subseteq \tau$. It is separable if there exists a countable $S \subseteq X$ such that $O \cap S \neq \emptyset$ for every nonempty $O \in \tau$. It is well known that second countability is strictly stronger than separability.

I'm working on something hinges on an intermediate property: "there exists a countable subset $C \subseteq \tau$ [edit: with each $C$-member nonempty!] that is dense in $\tau$, in the sense that for all $O \in \tau$, there exists $P \in C$ such that $P \subseteq O$."

Is there a common name for this property? I will call it "property C" for now.

Second countability implies property C (since a countable basis for $\tau$ is dense in $\tau$), which implies separability (choose one member from each $P \in C$ and the set of all the choices serves as the $S$ in the definition of separability). The Moore plane is an example of a topology that has property C but is not second countable.

Are there examples of topological spaces that are separable but do not have property C?

Note that a separable and first countable space will also have your property C. — Henno Brandsma, Sep 08 '19 at 15:19
Minor quibble: in the definition of property C (countable pseudoweight) you need $O$ to be non-empty too, of course. — Henno Brandsma, Sep 08 '19 at 15:53

Ayman Hourieh · Accepted Answer · 2019-09-08T15:27:46.457

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Consider $\mathbb R$ with the finite complement topology. Any infinitely countable subset $A$ of $\mathbb R$ is dense since an open subset of $\mathbb R$ can only miss finitely many points of $A$.

Let $\mathcal B$ be a countable family of non-empty open subsets of $\mathbb R$. Then $\bigcap_{B \in \mathcal B} B$ misses countably many points of $\mathbb R$ at most. It follows that some $x \in \mathbb R$ lies in every element of $\mathcal B$. Thus, the open subset $\mathbb R - \{x\}$ does not contain any element of $\mathcal B$.

edited Sep 08 '19 at 15:27

answered Sep 08 '19 at 15:08

Ayman Hourieh

39,603

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brilliant, thank you. (i think you meant $\bigcap_{B \in \mathcal{B}} B$ rather than $\bigcup_{B \in \mathcal{B}} B$) – Doug McLellan Sep 08 '19 at 15:19
@DougMcLellan happy to help! And yes, you're right. Fixed now. – Ayman Hourieh Sep 08 '19 at 15:22
sorry, i need to take more time to think about this (may be too overcaffeinated to think straight), i think your example produces an open set $O$ such that no $P \in B$ contains $O$, whereas we need an $O$ that contains no $P \in B$. – Doug McLellan Sep 08 '19 at 15:28
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In my example, $O = \mathbb R - {x}$ doesn't contain any element of $\mathcal B$, since every element of $\mathcal B$ contains $x$. Am I missing something? – Ayman Hourieh Sep 08 '19 at 15:31
[you are actually correct & i'm going to log off until my caffeine level exits the red zone!] – Doug McLellan Sep 08 '19 at 15:37
No problem at all. Have a good day. :) – Ayman Hourieh Sep 08 '19 at 15:37

Henno Brandsma · Answer 2 · 2019-09-08T18:08:34.923

What you're looking for is the concept of a $\pi$-base (or pseudobase), i.e. a collection of non-empty (this matters!) open subsets $\mathcal{P}$ such that any non-empty open subset of $X$ contains a member of $\mathcal{P}$. (The collection is downward-dense in the poset $(\mathcal{T}\setminus\{\emptyset\}, \subseteq)$ is another way of putting it)

The minimal size of a $\pi$-base for $X$ is denoted $\pi w(X)$ (rounded up to $\aleph_0$ if necessary, in Juhasz it's $\pi(X)$) , see the cardinal functions sections of this wikipedia page. So property $C$ is countable $\pi$-weight or $\pi w(X)=\aleph_0$ in more conventional terms, and I believe property C is already taken as a name in topology, or at least property (K) is, for sure. (which has the related meaning that every uncountable set of open subsets has an uncountable subset that pairwise intersect; a property implied by but weaker than separability). I prefer countable $\pi$-weight, or having a countable $\pi$-base as a name, being a bit more descriptive.

As to examples: For an $X$ just $T_1$ but not higher, the cofinite topology on an uncountable $X$ is separable and does not have a countable $\pi$-base. A more advanced example (compact Hausdorff): $[0,1]^{\Bbb R}$ in the product topology is separable but has no countable $\pi$-base, as a counting argument involving basic subsets will reveal. That both examples are not first countable is no accident: if $X$ is both separable and first countable, the union of the local bases at the countable dense subset form a countable $\pi$-base, as is easily checked. For metric spaces, having a countable base, being separable and having a countable $\pi$-base are all equivalent.

There also exists the notion of a local $\pi$-base at $x$: a collection of non-empty open subsets of $X$ such that every neighbourhood of $x$ contains a set from it. This is related to notions like tightness at a point etc. We get a similar cardinal invariant of $\pi\chi(x,X)$ for the minimal size of such a collection, etc.

Thanks for this very thorough answer. The topology I'm studying (on an operator algebra's state space, as it happens) is Hausdorff, so it's encouraging to know there are counterexamples that are Hausdorff. — Doug McLellan, Sep 08 '19 at 21:29
@DougMcLellan Maybe a nicer challenge is to find a space $X$ with $\aleph_0=d(X) < \pi w(X) < w(X)$; I couldn't find one off-hand. Maybe a $T_5$ one or a topological group? — Henno Brandsma, Sep 08 '19 at 21:33

A property between "separable" and "second countable"

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