TITLE: A strange characterization of connected spaces
QUESTION [9 upvotes]: Let $X$ be a topological space. Recall that an open cover $\mathcal U$ of $X$ is point finite if every point of $X$ is contained in finitely many elements of $\mathcal U$.
Let us say that a point finite open cover $\mathcal U$ is fair if any two points of $X$ are contained in the same number of elements of $\mathcal U$. 
Let us call a space unfair, if the only fair covers of $X$ are the trivial covers $\mathcal U = \{X\}$ and $\mathcal U = \{X, \emptyset\}$.

Question: Is it true that $X$ is connected if and only if it is unfair?

If $X$ is not connected, say $X= U \cup V$, with $U, V$ disjoint and nonempty, then $\{U, V\}$ is a nontrivial fair cover of $X$. This establishes one side of the implication. What about the other implication?

REPLY [9 votes]: Suppose that $\mathscr{U}$ is a point finite open cover of $X$ such that each $x\in X$ belongs to exactly $n$ members of $\mathscr{U}$. For each $x\in X$ let $V(x)=\bigcap\{U\in\mathscr{U}:x\in U\}$; clearly $\{V(x):x\in X\}$ is an open cover of $X$. Now suppose that $y\in V(x)$; then if $x\in U\in\mathscr{U}$, we must have $y\in U$, so the $n$ members of $\mathscr{U}$ containing $x$ all contain $y$, and it follows that $V(y)=V(x)$. Thus, $\{V(x):x\in X\}$ is a partition of $X$ into open (and therefore clopen) sets. If this partition has only one element, then $V(x)=X$ for all $x\in X$, $\mathscr{U}=\{X\}$, and $n=1$. Otherwise, the partition is non-trivial, and $X$ is not connected.