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Let $(X,\tau)$ be a Hausdorff paracompact space. Let $\tau'$ be the smallest $P$-topology refining $(X,\tau)$, i.e. the topology which has for base the $G_\delta$-subsets of $(X,\tau)$.

Is it true that $(X,\tau')$ is again paracompact?

If not, does the result hold under extra assumptions (e.g. hereditary paracompactness or even more)?

EDIT: Given the first answer received, I'll add a question:

Is there a Hausdorff paracompact space $(X,\tau)$ where every point has a local base of size $\leq \omega_1$ such that $(X,\tau')$ is not paracompact?

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    $\begingroup$ The space $(X,\tau')$ is commonly known as the $P$-space coreflection of $(X,\tau)$. $\endgroup$ Oct 2, 2022 at 15:51

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A counterexample to this claim is stated in the paper Normality and paracompactness in box products by Eric K.van Douwen and Kenneth Kunen (1990).

If $\kappa$ is an infinite cardinal, then the $<\kappa$-box topology on a product $\prod_{i\in I}X_i$ of topological spaces is the topology where the basic open sets are of the form $\prod_{i\in I}U_i$ where each $U_i$ is an open subset of $X_i$ and where $|\{i\in I\mid U_i\neq X_i\}|<\kappa$. The $P$-space coreflection of a space $(X,\tau)$ is the space $(X,\tau')$ where $\tau'$ is generated by the $G_\delta$ subsets of $(X,\tau)$.

Theorem: If $\kappa$ is an infinite cardinal, $\kappa$ is given the discrete topology, and $\kappa^{(\kappa^+)}$ is given the $<\kappa$-box topology, then $\kappa^{(\kappa^+)}$ is not normal.

In particular, if we assume the continuum hypothesis, then $\mathfrak{c}^{(\mathfrak{c}^+)}$ is not normal, but $\mathfrak{c}^{(\mathfrak{c}^+)}$ is the $P$-space coreflection of the compact space $[0,1]^{\mathfrak{c}^+}$ with the product topology.

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  • $\begingroup$ Thank you! That's more than I was hoping for. By the way, do you know by chance whether there is also a counterexample where every point has a local base of size $\leq\omega_1$? $\endgroup$
    – Cla
    Oct 2, 2022 at 16:54
  • $\begingroup$ I am not sure about how to get a space with a local basis of size at most $\aleph_1$. $\endgroup$ Oct 5, 2022 at 12:55

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