When Is Refinement of Open Covers a Partial Ordering on the Set of Open Covers?

Question

Rick Miranda's Algebraic Curves and Riemann Surfaces has a problem in the $\check{C}ech$ $Cohomology$ section (Problem IX.3.C) which says: Show that refinement gives a partial ordering on the set of all open coverings of a space.

I proved the transitivity and reflexivity properties, but I don't know what to do for a proof of antisymmetry. Letting $\mathcal{U},\mathcal{V}$ be two covers of a topological space $X$ such that $\mathcal{U}\prec\mathcal{V}$ and $\mathcal{V}\prec\mathcal{U}$ (i.e. $\mathcal{U}$ is a refinement of $\mathcal{V}$ and $\mathcal{V}$ is a refinement of $\mathcal{U}$), I suppose that I have a set/element $U\in\mathcal{U}$ and would like to show that $U\in\mathcal{V}$ as well to show set containment $\mathcal{U}\subset\mathcal{V}$ (and then to show the other direction and thus the set equality of the open covers by a similar argument). But... I'm stuck. By the refinement conditions I have that there exists $V\in\mathcal{V}$ such that $U\subset V$ and there exists similarly a $U'\in\mathcal{U}$ such that $V\subset U'$. But a) I don't know how to prove that $V = U$ or anything like that from here, and b) the Wiki for "Open Cover" explicitly states that refinement is a preorder (rather than a partial ordering) so that antisymmetry is probably the condition which fails.

I'm thinking there is just a slight typo in the text but wanted to ask the community. Do note that I am working in a Riemann surfaces book, so perhaps some of the conditions on Riemann surfaces like Hausdorff, second countable, etc. could come into play here to make refinement into a partial ordering. (However, the problem doesn't say we have a Riemann surface $X$ but rather just a space $X$, so I'm thinking he meant just a topological space in general considering we're in the chapter on sheaves which just require a topological space as their domain.)

Answer 1

It's just a mistake. The family of open covers is pre-ordered, but except in trivial cases not partially ordered by the refinement relation.

Any two open covers containing the whole space are refinements of each other.

For a less trivial example, consider the two covers of $\mathbb{R}$ by the families of bounded open intervals with rational endpoints, where in the first family we require the denominator of both endpoints (in the reduced form) to be odd, and in the second family the denominator of both end points shall be even. These are mutual refinements, but they don't have a set in common.