Thick refinements of covers Let $(X,\tau)$ be a topological space, and let ${\cal U}$ be an open cover. We say that ${\cal U}$ is thick if for all $x\in X$ we have $$|\{V\in {\cal U}: x\in V\}| = |X|.$$
Is there a Hausdorff space $(X,\tau)$ with $|X|>1$ and an open cover ${\cal U}$ of $X$ such that every refinement of ${\cal U}$ is thick?
 A: This is impossible for any $T_1$ space $X$. For suppose $X$ is $T_1$ and $\mathcal U$ is an open cover of $X$, and fix $x \in X$. If $U$ is some member of $\mathcal U$ containing $x$, then $\{U\} \cup \{V \setminus \{x\} : V \in \mathcal U,\, V \neq U\}$ is a refinement of $\mathcal U$, but it is not thick because only one set in this refinement contains $x$.
A: This is a partial answer. I would like to note merely that this is not possible in a countable space.
Theorem. In any countable Hausdorff space $X$ and any open
cover $U$, there is a refinement of $U$ to an open cover $U'$ such
that every point $x\in X$ is in only finitely many elements of
$U'$.
Proof. Enumerate the space as $X=\{x_n\mid n\in\omega\}$. At
each stage $n$, consider $x_n$, and let $U_n$ be a set in $U$ with
$x_n\in U_n$. Let $U_n'$ be a refinement of $U_n$ that excludes the
points $x_k$ for $k<n$. This is possible by the separation axiom. Let $U'$ be the set of all $U_n'$. This
refines the original cover, and $x_n$ is an element of at most $n+1$ many
elements of $U'$, since it is excluded after stage $n$. $\Box$
