1. Can someone help me to find examples of two normal topologies $\tau_1$ and $\tau_2$ on a set $X$ such that $(X,\tau_1\vee\tau_2)$ is not a normal space?


  1. Is there a chain of normal topologies $\{\tau_{\alpha}: \alpha\in A\}$ on a set $X$ such that $(X,\vee_{\alpha\in A}\tau_{\alpha})$ is not a normal space?

A convenient way of constructing such examples is from non-normal products of paracompact Hausdorff spaces.

For the first one, let $I = [0,1]$ with the order topology and let $I_S$ be the same set with the Sorgenfrey topology. Then $I \times I_S$ and $I_S \times S$ are paracompact because $I$ is compact and $I_S$ is paracompact. Thus we have two normal topologies on the same set. Since the topology of $I_S$ is finer than that of $I$, the join of these topologies gives $I_S \times I_S$, which is not normal.

For the second question, let $I_d$ be $[0,1]$ with the discrete topology and for $\beta < \omega_1$ put $$\begin{eqnarray} H_\beta = \prod_{\alpha < \beta} I_d \\ T_\beta = \prod_{\beta \le \alpha < \omega_1} I \end{eqnarray}$$ Then $H_\beta$ is metrizable because $\beta$ is countable and as $T_\beta$ is compact Hausdorff, $H_\beta \times T_\beta$ is paracompact Hausdorff.It is easy to verify the following:

  • the products $H_\beta \times T_\beta$ give a chain of topologies on $[0,1]^{\omega_1}$;

  • since $I_d$ has a finer topology than $I$, the join of these topologies is $\prod_{\alpha < \omega_1} I_d$.

Uncountable products of infinite discrete spaces are known not to be normal, which concludes the example.

Useful references:

Sorgenfrey, R. H. On the topological product of paracompact spaces. Bull. Amer. Math. Soc. 53 (1947), no. 6, 631-632.

Stone, A. H. Paracompactness and product spaces. Bull. Amer. Math. Soc. 54 (1948), no. 10, 977-982.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.