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Let $X$ be a set and let ${\frak T}$ be a collection of paracompact topologies on $X$ such that for any $\tau, \tau'\in {\frak T}$ we have $\tau\subseteq \tau'$ or $\tau'\subseteq \tau$. Let $\sigma$ be the topology having $\bigcup {\frak T}$ as a base.

Is $(X,\sigma)$ necessarily paracompact?

EDIT. Thanks to Tomek Kania for observing that $\bigcup {\frak T}$ need not be a topology.

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    $\begingroup$ Hold on, the union of a chain of topologies need not be a topology. $\endgroup$ – Tomasz Kania Oct 24 '18 at 8:07
  • $\begingroup$ That's right...! Never occurred to me. Will reformulate the question. Thanks for making me aware of it, @TomekKania $\endgroup$ – Dominic van der Zypen Oct 24 '18 at 8:10
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Here is an easy counterexample under the Continuum Hypothesis, where the union of the topologies generates a non-normal topology. Take the upper half plane $\mathbb{H}$ (including $x$-axis) and enumerate $\mathbb{R}$ as $\langle x_\alpha:\alpha<\omega_1\rangle$. Now let topology $\mathcal{T}_\alpha$ be generated by the normal topology plus the tangent disk neighbourhoods at all points $(x_\beta,0)$ for $\beta\le\alpha$. Each topology $\mathcal{T}_\alpha$ is still separable and metrizable (regular with a countable base) but the union generates the Tangent Disk topology, which is not (even) normal.

This example can be made `real' as follows: take an injective sequence $\langle x_\alpha:\alpha<\omega_1\rangle$ of real numbers and consider the subset $X$ of $\mathbb{H}$ consisting of all points above the $x$-axis and the points $(x_\alpha,0)$ on the x-axis. Perform exactly the same construction as above; the intermediate topologies are again separable and metrizable and their union generates the Tangent Disk topology on $X$. In this topology the set $\{(x_\alpha,0):\alpha<\omega_1\}$ is closed and discrete but there is no pairwise disjoint family $\{U_\alpha:\alpha<\omega_1\}$ of open sets such that $(x_\alpha,0)\in U_\alpha$ for all $\alpha$. So the space is not collectionwise Hausdorff and hence not paracompact.

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    $\begingroup$ Thanks for your answer @kphart! I'll wait a short time before accepting it - maybe an example that doesn't use (CH) can be found $\endgroup$ – Dominic van der Zypen Oct 24 '18 at 8:34
  • $\begingroup$ That's quite likely; but none of the `standard' examples seem to offer an easy solution. $\endgroup$ – KP Hart Oct 24 '18 at 8:40
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    $\begingroup$ I was wrong: the Niemytzki plane does offer an easy solution. $\endgroup$ – KP Hart Oct 24 '18 at 9:40

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