0
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Should $|b|\leq 1$ be $|x|\leq 1$ in your definition of $|x|_b$? $\endgroup$ – kag Oct 25 '16 at 15:57
  • 1
    $\begingroup$ @kag right. I am going to fix my mistake. $\endgroup$ – Hanul Jeon Oct 25 '16 at 17:49

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.