$\tau_1\vee\tau_2$ is not normal

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?

also

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?

1 Answer

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.