Equivalence of definitions of star refinements

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Define the star $St(S, \mathcal{U})$ of a set $S$ with respect to the collection of sets $\mathcal{U}$ as $St(S, \mathcal{U}) := \bigcup \{U \in \mathcal{U} \mid S \cap U \neq \varnothing\}$. If $S=\{x\}$, we write $St(x,\mathcal{U})$.

Steen and Seebach define a star refinement of a cover $\mathcal{U}=\{U_\alpha\mid\alpha\in A\}$ of a set $X$ as a cover $\mathcal{V} = \{V_\beta\mid\beta\in B\}$ such that $\{St(x,\mathcal{V})\mid x \in X\}$ is a refinement of $\mathcal{U}$ (basically).

Dugundji defines a star refinement of a cover $\mathcal{U}=\{U_\alpha\mid\alpha\in A\}$ of a set $X$ as a cover $\mathcal{V} = \{V_\beta\mid\beta\in B\}$ such that $\{St(V_\beta,\mathcal{V})\mid \beta \in B\}$ is a refinement of $\mathcal{U}$.

I'm wondering if these are equivalent, and if so, are the collections $\mathcal{U}$ and $\mathcal{V}$ are required to be covers?

A Dugundji star refinement (for lack of a better term) is also a Steen and Seebach star refinement even if we drop the requirement that $\mathcal{U}$ and $\mathcal{V}$ are covers (assuming $\mathcal{U}$ is nonempty): Suppose $x \in X$. Either $x \in \bigcup\mathcal{V}$ or $x \notin \bigcup\mathcal{V}$. If the former, then $x \in V_\beta$ for some $\beta \in B$ and we have $St(x,\mathcal{V}) \subseteq St(V_\beta,\mathcal{V}) \subseteq U_\alpha$ for some $\alpha \in A$ since any $V_\gamma \in \mathcal{V}$ containing $x$ must intersect $V_\beta$. If the latter, then $St(x,\mathcal{V})=\varnothing \subseteq U_\alpha$ for any $\alpha \in A$.

I feel like I must be missing something simple, because I can't seem to prove the converse (with or without the covering assumption).

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Both types of refinements are (implicitly or ecplicitly) only applied to covers.

The type with stars of points is properly called a barycentric refinement.

So $\mathcal{U} \prec_b \mathcal{V}$ means that both are covers of the space $X$ in question (often open, but with paracompactness etc. one also considers closed covers as well) and for every $x \in X$, $\operatorname{St}(x, \mathcal{U}) \subset V$ for some $V \in \mathcal{V}$.

The other type is a real star refinement:

$\mathcal{U} \prec_\ast \mathcal{V}$ iff $\mathcal{U}$ and$\mathcal{V}$ are covers of $X$ and $\forall U \in \mathcal{U}$ there exists some $V \in \mathcal{V}: \operatorname{st}(U,\mathcal{U}) \subseteq V$

See wikipedia for this as well. Note that $\mathcal{U} \prec_\ast \mathcal{V}$ implies $\mathcal{U} \prec_b \mathcal{V}$ if we have covers. So one notion is stronger than the other for covers (if points are uncovered the star of such a point is empty)

A basic fact: if we have 3 covers $\mathcal{U}, \mathcal{V}, \mathcal{W}$ of $X$ then $\mathcal{U} \prec_b \mathcal{V}$ and $\mathcal{V} \prec_b \mathcal{W}$ implies $\mathcal{U} \prec_\ast \mathcal{W}$.

(this is a nice exercise, try it). This means that if we assume that every open cover has a barycentric refinement, we get also that every open cover has a star-refinement: just take two successive barycentric ones. Hence they're both used in some definitions, but he definitions are not equivalent. There are barycentric refinements that are not star-refinements.