Does Seperable + First Countable + Sigma-Locally Finite Basis Imply Second Countable? A topological space is separable if it has a countable dense subset.  A space is first countable if it has a countable basis at each point.  It is second countable if there is a countable basis for the whole space.  A collection of subsets of a space is locally finite if each point has a neighborhood which intersects only finitely many sets in the collection.  A collection of subsets of a space is sigma-locally finite (AKA countably locally finite) if it is the union of countably many locally finite collections.
My question is, if a space is separable, first countable, and has a sigma-locally finite basis, must it also be second countable?  I think the answer is yes, because I haven't found any counterexample here.
Any help would be greatly appreciated.
Thank You in Advance.
EDIT: I fixed my question.  I meant that the space should have a locally finite basis, not be locally finite itself, which doesn't really mean much.
 A: The Nagata-Smirnov metrization theorem says that a $T_3$ space with a $\sigma$-locally finite base is metrizable (and conversely). A separable metrizable space is second countable. Thus, if your space is $T_3$, the answer to your question is yes.
A: Let $\mathcal{B} = \cup_n \mathcal{B}_n$ be a $\sigma$-locally finite base (all non-empty sets), and let $D = \{d_n: n \in \mathbb{N} \}$ be a dense subset of $X$ (as said in my comment, $X$ is automatically first countable from having such a base, so I won't use that assumption).
For every $n$, $d_n$ is in at most countably many members of $\mathcal{B}$, as in is at most finitely many members of $\mathcal{B}_k$ for every $k$, note that we only need that each of these is point-finite, not locally finite; the same holds for the first countability. Call these members that contain $d_n$: $\mathcal{B}^n$, this is a countable set.
Then, as $D$ is dense, so every member of $\mathcal{B}$ contains some $d_k$, so $\mathcal{B} = \cup_n \mathcal{B}^n$, so $\mathcal{B}$ is already a countable base for $X$.
So in short: yes, this holds, even: every space $X$ with a $\sigma$-point-finite base that is separable has a countable base. Or, a point-countable family of non-empty open subsets in a separable space is (at most) countable...
A: For example:
Helly Space;
Right Half-Open Interval Topology;
Weak Parallel Line Topology.
These space are all separable, first countable and paracompact, but not second countable.
Note that a paracompact is the union of one locall finite collection.
