# A stronger version of paracompactness

Given a topological space $$(X,\tau)$$, recall that a cover $$\mathcal{U}$$ of $$X$$ is locally finite if for every point $$x\in \mathcal{U}$$ has a neighborhood $$U$$ that intersects finitely many elements of $$\mathcal{U}$$. Instead, we call a cover $$\mathcal{U}$$ finitely intersecting if every member of $$\mathcal{U}$$ intersect finitely many elements of $$\mathcal{U}$$.

Recall that $$X$$ is paracompact if every open cover $$\mathcal{U}$$ has a locally finite open refinement. Just for the purpose of this question, let us call $$X$$ strongly paracompact if every open cover $$\mathcal{U}$$ has a finitely intersecting open refinement.

My question is: does this notion coincide with paracompactness?

If the answer is no, I would be very courious to know more about this property. For example:

• Has strong paracompactness been studied before, and what is the right name for it?

• Is it true that all metrizable second countable spaces are strongly paracompact?

• The metric Hedgehog with uncountably many spines is a famous example of a completely metrisable (hence paracompact) space which is not strongly paracompact (this answers your first two questions). On the other hand, every regular Lindelof space is strongly paracompact (so your last question has an affirmative answer). The usual terminology for a 'finitely intersecting' cover is 'star-finite'. Commented Nov 4, 2022 at 13:58
• Thank you. Searching for star-finite refinement I see now that a similar question has been asked already on the other forum: math.stackexchange.com/questions/4305301/… Maybe we can mark this question as a duplicate, I will close it.
– Cla
Commented Nov 4, 2022 at 16:15
• @Tyrone: I don't see how a Hedgehog with countably many spines is strongly paracompact. Take such a Hedgehog, and cover it with an $\epsilon$-ball around the joint and countably many open sets, one for each spine. What is the star-finite refinement? Commented Jul 26, 2023 at 2:26
• @Tyrone: here, hedgehog means metric hedgehog. The (metric) hedgehog with countably many spines is supposed to be strongly paracompact, since it is separable. Commented Jul 26, 2023 at 9:57

Lemma: Let $$(U_{\alpha})_{\alpha\in A}$$ be a finitely intersecting open cover of a space $$X$$. Then there is some partition $$P$$ of $$A$$ where $$(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$$ is a partition of $$X$$ into clopen sets and where each $$R\in P$$ is finite or countable.
Proof: Let $$E$$ be the smallest equivalence relation on $$A$$ where if $$U_\alpha\cap U_\beta\neq\emptyset$$, then $$(\alpha,\beta)\in E$$. Let $$P$$ be the partition associated with $$E$$. Then since $$(U_\alpha)_{\alpha\in A}$$ is finitely intersecting, each $$R\in P$$ is finite or countable. Clearly, $$(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$$ covers $$X$$. Furthermore, if $$R,S\in P,R\neq S$$, then $$(\bigcup_{\alpha\in R}U_\alpha)\cap(\bigcap_{\beta\in S}U_\beta) =\bigcup_{\alpha\in R,\beta\in S}(U_\alpha\cap U_\beta)=\emptyset.$$ Therefore, $$(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$$ is a partition into open (and hence clopen) sets. $$\square$$
Proof: Let $$X$$ be a connected strongly paracompact space. Let $$\mathcal{U}$$ be an open cover of $$X$$. Let $$\mathcal{V}$$ be a finitely intersecting open refinement of $$\mathcal{U}$$. By the above lemma, $$\mathcal{V}$$ must be countable or finite, so $$\mathcal{U}$$ has a countable subcover. $$\square$$