TITLE: Countable topological basis
QUESTION [6 upvotes]: Let $(E,\mathcal{T})$ be a topological space with a countable basis (a second countable space). If $\mathcal{U}$ is a topology on $E$ with $\mathcal{U} \subset \mathcal{T}\,$ ($\mathcal{U}$ is coarser than $\mathcal{T}$), does $(E,\mathcal{U})$ have a countable basis?

REPLY [2 votes]: I think that William Chan's example doesn't work, for the reason francis-jamet points out in his comment.
But here is something that works, along the same lines, giving a countable space.
Let $U$ be a non-principal ultrafilter on $\mathbb N$.
Consider the space $\mathbb N\cup\{U\}$ with the following basis for a topology:
for all $n\in\mathbb N$, $\{n\}$ is in the basis.  For all $A\in U$, 
$A\cup\{U\}$ is in the basis.
Let $V$ be a countable collection of sets from $U$.
There is $B\subseteq$ such that for each $A\in V$, $B\setminus A$ is finite.
Either $B$ or its complement is in $U$.  The complement clearly includes no element of $V$.
If $B\in U$, split $B$ into two infinite pieces.  One of the pieces is in $U$.  Both pieces contain no element of $V$.  In either case, $U$ has an element that contains no member of $V$.  
This information can be used to show that the space $\mathbb N\cup U$ has no countable basis for its topology.  And clearly the topology is contained in the discrete topology on the space.  Also note that these spaces are obtained by adding a single new point to a countable discrete space.