# Is the topology generated by the union of a chain of paracompact topologies paracompact?

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.

• Hold on, the union of a chain of topologies need not be a topology. – Tomasz Kania Oct 24 '18 at 8:07
• That's right...! Never occurred to me. Will reformulate the question. Thanks for making me aware of it, @TomekKania – Dominic van der Zypen Oct 24 '18 at 8:10

## 1 Answer

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.

• 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 – Dominic van der Zypen Oct 24 '18 at 8:34
• That's quite likely; but none of the standard' examples seem to offer an easy solution. – KP Hart Oct 24 '18 at 8:40
• I was wrong: the Niemytzki plane does offer an easy solution. – KP Hart Oct 24 '18 at 9:40