Is \(\mathbb{R}^\omega\) normal in the product topology? In the uniform topology?
It is not known whether \(\mathbb{R}^\omega\) is normal in the box topology. Mary-Ellen Rudin has shown that the answer is affirmative if one assumes the continuum hypothesis [RM]. In fact, she shows it satisfies a stronger condition called paracompactness.
Solution
In the product topology, \(\mathbb R^\omega\) is metrizable, for example by \[ D(x,y)=\sup_{n\ge1}\frac{\min\{|x_n-y_n|,1\}}{n}. \] Every metric space is normal, so \(\mathbb R^\omega\) is normal in the product topology.
The uniform topology is induced by the uniform metric \[ \bar\rho(x,y)=\sup_n\min\{|x_n-y_n|,1\}. \] Hence it too is metrizable and therefore normal.