general topology - "Basis" Version of a Local Connectedness Theorem

A space $X$ is locally connected if and only if for every open set $U$ of $X$, each component of $U$ is open in $X$.

The above is a theorem in Munkres' Topology book (theorem 25.3). As usual, I always wonder whether a given theorem is true when open sets are replaced with basis elements. I just proved that $X$ is locally connected if and only if $X$ has a basis entirely comprised of connected sets, thinking that this might help. But I cannot quite connect the pieces. So my question is

Does a "basis" version of the above theorem hold?

Answer

I don't imagine the proof to be that different from the original version: Let $X$ be a topological space with basis $\mathcal{B}$. Suppose for every $B \in \mathcal{B}$ we have that each component of $B$ is open in $X$. Now take any $x \in X$ and any open neighbourhood $U$ of $x$. By definition there is a $B \in \mathcal{B}$ such that $x \in B \subset U$. Now let $C$ be the component of $B$ containing $x$. Then $x \in C \subset U$ and $C$ is open and connected. So $X$ is locally connected.

To prove the converse assume $X$ is locally connected. By the above theorem, we have that every open $U \subset X$, each component of $U$ is open. Since every $B \in \mathcal{B}$ is open, each component of $B \in \mathcal{B}$ must be open too.