What is the meaning finite minimal cover?

Question

In this lemma, it mentioned finite minimal cover. My question is this:

What is the meaning finite minimal cover?

Thanks.

Answer 1

From the proof, it is pretty clear that they mean a finite cover with no proper subset which also covers. Copying the proof given in Hodel's article:

... suppose Miščenko's lemma is false, and let $\{ \mathscr{A}_\alpha : 0 \leq \alpha < \kappa^+ \}$ be a collection of distinct finite minimal covers of $E$ by elements of $\mathscr{A}$. By the $\Delta$-system lemma, there is a subset $T$ of $\kappa^+$ with $| T | = \kappa^+$, and a subset $\mathscr{F}$ of $\mathscr{A}$ such that $\mathscr{A}_\alpha \cap \mathscr{A}_\beta = \mathscr{F}$ for $\alpha , \beta \in T$, $\alpha \neq \beta$. Let $p \in E$, $p \notin \bigcup \mathscr{F}$.

(To show $\bigcup \mathscr{F} \neq E$, chose $\alpha \in T$ such that $\mathscr{A}_\alpha \neq \mathscr{F}$ and use the fact that $\mathscr{A}$ is a minimal cover of $E$.)

For each $\alpha \in T$ choose $A_\alpha \in \mathscr{A}_\alpha$ such that $p \in A_\alpha$. Let $\alpha , \beta \in T$, $\alpha \neq \beta$. Since $p \notin \bigcup \mathscr{F}$ and $\mathscr{A}_\alpha \cap \mathscr{A}_\beta = \mathscr{F}$, it follows that $A_\alpha \neq A_\beta$. Hence $\mathrm{ord} ( p , \mathscr{A} ) = \kappa^+$, a contradiction.

The separated line, justifying that such a $p \in E$ can be chosen, requires that $\mathscr{F}$ is not a cover of $E$ itself. They do this by noting that $\mathscr{F} \subsetneq \mathscr{A}_\alpha$ for some $\alpha \in T$ and refer to the minimality of $\mathscr{A}_\alpha$, which makes the natural definition that no proper subset of $\mathscr{A}_\alpha$ also covers $E$.