In short, the question is in the title: *is there a Hausdorff space with a $\sigma$-locally finite basis but no $\sigma$-discrete basis?*

**A bit of context:**

Given a topological space $X$, a family $\mathcal{U}\subseteq \mathcal{P}(X)$ is locally finite (resp., discrete) if for every $x\in X$ there is an open neighborhood $U$ of $x$ that intersects only finitely many (resp., at most one) elements of $\mathcal{U}$.

We say $\mathcal{U}$ is $\sigma$-locally finite (resp., $\sigma$-discrete) if it can be partitioned into countably many locally finite (resp., discrete) families.

Nagata-Smirnov-Bing metrization theorem states that the following are equivalent:

- $X$ is metrizable,
- $X$ is $T_3$ and has a $\sigma$-locally finite basis,
- $X$ is $T_3$ and has a $\sigma$-discrete basis.

In particular, the two properties (having a $\sigma$-locally finite basis, and having a $\sigma$-discrete basis) are **equivalent for $T_3$ spaces**.

On the other hand, there are examples of ($T_0$) spaces having a $\sigma$-locally finite basis but no $\sigma$-discrete basis, e.g. the poset $([\omega_1]^{<\omega}, \supseteq)$ with the Alexandrov topology.

So the two properties are **not equivalent in general**.

I tried searching but I could not find any Hausdorff counterexample.

**My question is:** are the two properties still equivalent for $T_2$ spaces?

**Note:**

Say that $\mathcal{U}\subseteq \mathcal{P}(X)$ is star-finite if every $U\in\mathcal{U}$ intersects only finitely many elements of $\mathcal{U}$.

We say $\mathcal{U}$ is $\sigma$-star-finite if it can be partitioned into countably many star-finite families.

Since every star-finite family is also $\sigma$-discrete, we get the following

**Remark:** $\sigma$-discrete $\iff$ $\sigma$-star-finite $\Rightarrow$ $\sigma$-locally finite.

In particular, if a space has a $\sigma$-locally finite basis but not a $\sigma$-discrete basis, the locally finite families should be very far from star-finite.

5more comments