# $\mathbb{R^\omega}$ is normal in product topology and uniform topology.

To prove: $\mathbb{R^\omega}$ is normal in product topology and uniform topology.

One theorem says that Every Metrizable Space is Normal. So if we can show that $\mathbb{R^\omega}$ is metrizable with product and uniform topology then we are done.

But I am facing problem in showing it....Help Needed!

Let define $|\cdot |_b :\Bbb{R}\to \Bbb{R}$ as $$|x|_b=\begin{cases} |x| & \text{if }|x|\le 1 \\ 1 & \text{otherwise}\end{cases}$$
and consider the metric $$d(x,y) = \sup_{n\in \Bbb{N}} \frac{|x_n-y_n|_b}{n}$$ and $$\rho(x,y) = \sup_{n\in \Bbb{N}} |x_n-y_n|_b$$ for $x=(x_n)$ and $y=(y_n)$. You can check that $d$ induces product topology and $\rho$ induces uniform topology.
• Should $|b|\leq 1$ be $|x|\leq 1$ in your definition of $|x|_b$? – kag Oct 25 '16 at 15:57