When are maximal chains of open sets with nonempty intersection a local basis? Suppose that $C$ is a maximal chain (with respect to $\subseteq$) of nonempty open set in a topological space $X$. Assume that $x \in \bigcap C$.  Is $C$ a local basis for $x$?
I know that this is not generally the case.  For instance, $X = 2^\kappa$ for $\kappa$ an uncountable cardinal does not even have a linearly ordered local basis.  But if I impose some reasonable condition on $X$ or $C$, do I actually have that $C$ is a local basis?
 A: I think you will have to impose some "unreasonable" conditions. Here is a fairly general counterexample.
Let $X$ be a T$_1$ space containing two non-isolated points $x_1\ne x_2.$ Let $C$ be any maximal chain of open supersets of $\{x_1,x_2\}.$ For each $i=1,2$ extend $C$ to $C_i,$ a maximal chain of open sets containing $x_i.$ I claim that at least one of the chains $C_i$ is not a local base at $x_i.$
Assume for a contradiction that $C_1$ is a local base at $x_1$ and $C_2$ is a local base at $x_2.$ Then there are sets $U_1\in C_1$ and $U_2\in C_2$ such that $U_1\subseteq X\setminus\{x_2\}$ and $U_2\subseteq X\setminus\{x_1\}.$ Since $U_i\in C_i$ and $\{x_1,x_2\}\not\subseteq U_i,$ we have that $U_i$ is a subset of every element of $C.$ Hence the set $U=U_1\cup U_2$ is a subset of every element of $C.$ Since $U$ is an open superset of $\{x_1,x_2\},$ if follows from the maximality of $C$ that $U$ is the smallest element of $C,$ and that it is the smallest open superset of $\{x_1,x_2\}.$ Since $X$ is a T$_1$ space, it follows that $U=\{x_1,x_2\}$ and that $x_1,x_2$ are isolated points.
