Paracompactness of closed subspaces

Question

Suppose that $X$ is a paracompact topological space and let $S \subseteq X$ be a closed subspace.

It is well-known that $S$ itself is paracompact, when equipped with the subspace topology.

But what about the following:

Let $\mathcal{U}$ be a covering of $S$ by sets open in $X$. Does there exist an open locally finite refinement $\mathcal{V}$ of $\mathcal{U}$ that is a covering of $S$?

I have seen authors who say that $S$ is paracompact in $X$ if any such $\mathcal{U}$ has got an open refinement $\mathcal{V}$. But this terminology is not common and I have doubts whether the concept makes sense.

Answer 1

Such concepts have been discussed but they have remained a niche topic.

A paper that might interest you discusses "relative" topological properties: Arhangel'skii, A. V. "Relative topological properties and relative topological spaces." Topology and its Applications 70.2-3 (1996): 87-99.

It seems that your concept of "$S$ paracompact in $X$" is actually a bit stronger than the one by Arhangel'skii: he only requires that the refinement is locally finite over $S$, but you also require that it is locally finite over $X$ as well.

Answer 2

Perhaps I misunderstood you, below I assume $X$ is not necessarily paracompact.

Such subspace $S$ is called strongly paracompact subspace, and for collectionwise normal Hausdorff spaces it can be shown that any closed paracompact subspace is strongly paracompact.

See this paper.

Answer 3

If $X$ is paracompact and $S$ is closed then the property you defined holds. Let $\mathcal{A} =\{U_\alpha\}_{\alpha\in I}$ be an open cover of $S$. Then $\mathcal{A}\cup\{X\setminus S\}$ is an open cover of $X$. Let $\mathcal{B}$ be a locally-finite refinement of $\mathcal{A}\cup\{X\setminus S\}$. Discard from $\mathcal{B}$ any subsets of $X\setminus S$. You're left with a refinement of $\mathcal{A}$ that covers $S$. It is still locally finite for any point in $X$ because you've just discarded sets.