I was motivated by this question: Is a covering space of a manifold second countable?
I still think that a covering space of a manifold is not necessarily second countable. I define base $\mathcal B$ of $\mathbb R^2$ as:
$$\mathcal B := \{U \times y \mid \text{$U$ is open in $\mathbb R$ (Euclidean), $y \in \mathbb R$}\}.$$
$\mathcal B$ satisfies both axioms of a base. Let $\tau$ be a topology formed by $\mathcal B$.
A few observations that I made:
- Every singleton is a closed set.
- Any non-horizontal line segment is a closed set (regardless of endpoint included in the line segment or not).
Now, consider a projection $p: \mathbb R^2 \rightarrow \mathbb R^1$. $p$ is a covering map since every $U \subset \mathbb R^1$ is evenly covered by $U \times y$ for every $y \in \mathbb R$, and $(U \times y) \cap (U \times y') = \varnothing$ if $y \neq y'$. Clearly, $\mathbb R^2$ with our new topology is not second countable. Am I missing anything?
