Does the Everywhere doubled line have a $\sigma$-locally finite $k$-network?

Question

The Everywhere doubled line is an interesting example of a non-Hausdorff manifold that is homogeneous. It is described in section 3 of [BG], similar to the line with two origins but with each point "doubled". The article has some nice pictures.

The space can be described as the union of two real lines $X=A\sqcup B$ with $A = \mathbb R\times\{0\}$ and $B=\mathbb R\times\{1\}$. A set $U\subseteq X$ is open exactly when $U\cap A$ is open in $A$ with the usual Euclidean topology and for each $x=(r,1)\in U\cap B$, the set $U$ contains a deleted neighborhood of $(r,0)$ in $A$. It is a special case of the general construction here.

Question: Does the Everywhere doubled line have a $\sigma$-locally finite $k$-network?

Here, a family $\mathcal N$ of subsets of $X$ is called a $k$-network if for every compact set $K$ and open set $U$ in $X$ with $K\subseteq U$, there exists a finite $\mathcal{N}^* \subseteq \mathcal{N}$ with $K \subseteq \bigcup\mathcal{N}^* \subseteq U$. The family $\mathcal N$ is $\sigma$-locally finite if it is a countable union of locally finite families.

A $k$-network is something intermediate between a base for the topology and a network. A family $\mathcal N$ of subsets of $X$ is called a network if every open set is the union of a subfamily of $\mathcal N$; that is, for every open set $U$ and point $x\in U$ there is some $A\in\mathcal N$ with $x\in A\subseteq U$.

For a general space $X$, we have these implications:

$X$ has a $\sigma$-locally finite base $\implies$ $X$ has a $\sigma$-locally finite $k$-network $\implies$ $X$ has a $\sigma$-locally finite network.

It is already known in pi-base that the Everywhere doubled line does not have a $\sigma$-locally finite base (see here), but does have $\sigma$-locally finite network (see here). But the intermediate property in unknown and it would be very interesting to complete the picture.

Note: See https://math.stackexchange.com/a/4772144. You may find the same shortcut in notation to be useful when answering this.

---

[BG] M. Baillif & A. Gabard, Manifolds: Hausdorffness versus homogeneity, Proc. Amer. Math. Soc. 136 (2008), 1105-1111

Answer 1

Everywhere doubled line does not have a $\sigma$-locally finite $k$-network.

Proof: Suppose $\mathcal N$ is a $\sigma$-locally finite $k$-network for $X$. Define $\mathcal D = \left\{ A \cap (\mathbb R \times \{0\}) \mid A \in \mathcal N \right\}$, then $\mathcal D$ is a $\sigma$-locally finite $k$-network for the subspace $\mathbb R \times \{0\}$, which is homeomorphic to $\mathbb R$.

Since $\mathbb R$ is Lindelöf, and it follows that $\mathcal D$ is countable (for example from Lemma 5.1.24 in Engelking).

Let $x \in (0, 1)$ be arbitrary. Then we can define the compact set $K_x := \{\left< x, 1 \right>\} \cup \bigl( (x, 1] \times \{0\} \bigr)$ and the open set $U_x := \{\left< x, 1 \right>\} \cup \bigl( (\mathbb R \setminus \{x\}) \times \{0\} \bigr)$ such that $K_x \subset U_x$.

By the definition of $k$-network, we know there exists a finite $\mathcal N^* \subseteq \mathcal N$ such that $K_x \subseteq \bigcup \mathcal N^* \subseteq U$. By intersecting with $\mathbb R\times\{0\}$, there exists a finite $\mathcal D^* \subseteq \mathcal D$ such that $(x, 1] \subseteq \bigcup \mathcal D^* \subseteq \mathbb R \setminus \{x\}$.

Since $\mathcal D$ is countable, the collection of finite unions of elements of $\mathcal D$ is still countable. We can assume that $\mathcal D$ is finite-union closed, and consequently, for each $x \in (0, 1)$, there exists $A_x \in \mathcal D$ such that $(x, 1] \subseteq A_x \subseteq \mathbb R \setminus \{x\}$.

Note that $A_x$ satisfies the property $\max \bigl( [0, 1] \setminus A_x \bigr) = x$, this means that all the $A_x$'s are different and there are uncountably many of them. This contradicts that $\mathcal D$ is countable.