$\tau_1\vee\tau_2$ is not normal 
*

*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


*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: 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.
